### RE: Observation selection effects

```Brent Meeker and Jesse Mazer and others wrote:
Well, lots and lots of complex mathematical argument on the two envelope
problem...

But no-one has yet pointed out a flaw in my rather simplistic analysis:
(1) One envelope contains x currency units, so the other contains 2x
currency units;

(2) If you stop at the first envelope you choose, expected gain is: 0.5*x +
0.5*2x = 1.5x;

(3) If you open the first envelope then switch to the second, your expected
gain is: 0.5*2x + 0.5*x = 1.5x - as above, just in a different order,
obviously;

(4) If, in a variation, the millionaire flips a coin to give you double or
half the amount in the first envelope if you switch envelopes, expected gain
is: 0.25*2x + 0.25*0.5x + 0.25*x + 0.25*4x = 1.875x.

In the latter situation you are obviously better off switching, but it is a
mistake to assume that (4) applies in the original problem, (3) - hence, no

Is the above wrong, or is it just so obvious that it isn't worth discussing?
(I'm willing to accept either answer).

Stathis Papaioannou
_
Searching for that dream home? Try   http://ninemsn.realestate.com.au  for

```

### RE: Observation selection effects

```

-Original Message-
From: Stathis Papaioannou [mailto:[EMAIL PROTECTED]
Sent: Thursday, October 14, 2004 7:36 AM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED];
[EMAIL PROTECTED]
Subject: RE: Observation selection effects

Brent Meeker and Jesse Mazer and others wrote:

Well, lots and lots of complex mathematical argument on
the two envelope
problem...

But no-one has yet pointed out a flaw in my rather
simplistic analysis:

(1) One envelope contains x currency units, so the other
contains 2x
currency units;

(2) If you stop at the first envelope you choose,
expected gain is: 0.5*x +
0.5*2x = 1.5x;

(3) If you open the first envelope then switch to the
gain is: 0.5*2x + 0.5*x = 1.5x - as above, just in a
different order,
obviously;

(4) If, in a variation, the millionaire flips a coin to
give you double or
half the amount in the first envelope if you switch
envelopes, expected gain
is: 0.25*2x + 0.25*0.5x + 0.25*x + 0.25*4x = 1.875x.

In the latter situation you are obviously better off
switching, but it is a
mistake to assume that (4) applies in the original
problem, (3) - hence, no

Is the above wrong, or is it just so obvious that it
isn't worth discussing?
(I'm willing to accept either answer).

Stathis Papaioannou

It's not wrong - I just don't think it addresses the paradox.  To
resolve the paradox you must explain why it is wrong to reason:

I've opened one envelope and I see amount m.  If I keep it my gain
is m.  If I switch my expected gain is 0.5*m/2 + 0.5*2m = 1.25m,
therefore I should switch.

To say that in another, similiar game (4) this reasoning is
correct, doesn't explain why it is wrong in the given case.

Your (2) and (3) aren't to the point because they don't recognize
that after opening one envelope you have some information that
seems to change the expected value.

In my analysis, it is apparent that the trick of showing the
expected value doesn't change depends on the feature of the
problem statement that the distribution of the amount of money is
scale free - i.e. all amounts are equally likely.  If you accept
this, then a Bayesian analysis of your rational belief shows that
the expected value doesn't change when you open the envelope and
see amount m.  Intuitively, observing a value from a distribution
that is flat from zero to infinity *doesn't* give you any
information.  Solving the paradox is to show explicitly why this
is so.

As Jesse and others have pointed out this scale-free (all amounts
are equally likely) aspect of the problem as stated is unrealistic
and in any real situation your prior estimate of the scale of the
amounts will cause you to modify your expected value after you see
the amount in first envelope.  This modification may prompt you to
switch or not - but it's a different problem.

Brent Meeker

```

### RE: Observation selection effects

```

-Original Message-
From: Jesse Mazer [mailto:[EMAIL PROTECTED]
Sent: Tuesday, October 05, 2004 11:01 PM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: RE: Observation selection effects

-Original Message-
From: Jesse Mazer [mailto:[EMAIL PROTECTED]
Sent: Tuesday, October 05, 2004 8:45 PM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: RE: Observation selection effects

If the range of the smaller amount is infinite,
as in my P(x)=1/e^x
example, then it would no longer make sense to say that
the range of the
larger amount is r times larger.

Sure it does; r*inf=inf.  P(s)=exp(-x) - P(l)=exp(-x/r)

But it would make just as much sense to say that the
second range is 3r
times wider, since by the same logic 3r*inf=inf. In
other words, this step
in your proof doesn't make sense:

In other words, the range of possible
amounts is such that the larger and smaller amount do
not overlap.
Then, for any interval of the range (x,x+dx) for the smaller
amount with probability p, there is a corresponding
interval (r*x,
r*x+r*dx) with probability p for the larger amount.  Since the
latter interval is longer by a factor of r

P(l|m)/P(s|m) = r ,

In other words, no matter what m is, it is r-times more
likely to
fall in a large-amount interval than in a small-amount interval.

As for your statement that P(s)=exp(-x) -
P(l)=exp(-x/r), that can't be
true. It doesn't make sense that the value of the second
probability
distribution at x would be exp(-x/r), since the range of
possible values for
the amount in that envelope is 0 to infinity, but the
integral of exp(-x/r)
from 0 to infinity is not equal to 1, so that's not a
valid probability
distribution.

Also, now that I think more about it I'm not even sure
the step in your
proof I quoted above actually makes sense even in the
case of a probability
distribution with finite range. What exactly does the equation
P(l|m)/P(s|m) = r mean, anyway?

For any give amount of money, m, found in the first envelope, it
is more probable by a factor of r that it came from the Larger
envelope - where probable means degree of rational belief, not
fraction in a statistical ensemble.

It can't mean that if
I choose an envelope
at random, before I even open it I can say that the
amount m inside is r
times more likely to have been picked from the larger
distribution, since I
know there is a 50% chance I will pick the envelope
whose amount was picked
from the larger distribution. Is it supposed to mean
that if we let the
number of trials go to infinity and then look at the
subset of trials where
the envelope I opened contained m dollars, it is r times
more likely that
the envelope was picked from the larger distribution on
any given trial?
This can't be true for every specific m--for example, if
the smaller
distribution had a range of 0 to 100 and the larger had
a range of 0 to 200,

But whole point is that there is no specific m from which you
can reason.

if I set m=150, then in every single trial where I found
150 dollars in the
envelope it must have been selected from the larger
distribution. You could
do a weighted average over all possible values of m,
like integral over all
possible values of m of P('I found m dollars in the envelope I
selected')*P('the envelope I selected had an amount
taken from the smaller
distribution' | 'I found m dollars in the envelope I
selected'), which you
could write as integral over m of P(m)*P(s|m), but I
don't think it would
be true that the ratio integral over m of
P(m)*P(l|m)/integral over m of
P(m)*P(s|m) would be equal to r, in fact I think both
integrals would
always come out to 1/2 so the ratio would always be
1...and even if I'm
wrong, replacing P(l|m)/P(s|m) with this ratio of
integrals would mess up

Jesse

No, it doesn't depend on assuming a flat distribution for the
money, only for our knowledge (or on our acceptance of problem as
stated).  Here's the more explicit (but less intuitive) proof - I
hope the formatting doesn't get chopped up too much by your mail

Without loss of generality, we can describe our prior density
functions for the amounts in the two envelopes in terms of a
density function, fo(x), the ratio r of the larger amount to the
smaller, and a scale factor, k.  Let L be the event that the
evelope with the larger amount is picked and S the event that the
envelope with the smaller amount if picked.  Then our prior
density functions for the amount m in the envelope is:

For the smaller amount our prior is:f(m|S k) = k fo(km)
and for the larger amount:  f(m|L k) = (k/r) fo(km/r)

Our uncertainity about the scale factor, k, is described by a
density g(k).  So

f(m|S) = INT k fo(km) g(k) dk

,where INT is integral zero-to-infinity

f(m|L) = INT (k/r) fo(km/r) g(k) dk

Now in the first equation make a change of variable in the
integral by y=km

f(m|S) = INT (y/m) fo(y) g(y/m) dy/m = (1/m^2) INT y fo(y) g(y/m)
dy```

### Re: observation selection effects

```Thanks, Kory, that takes care of my confusion.
The same to Jesse's post.
John Mikes
- Original Message -
From: Kory Heath [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Sunday, October 10, 2004 7:17 PM
Subject: Re: observation selection effects

At 02:57 PM 10/10/2004, John M wrote:
Then it occurred to me that you made the same
assumption as in my post shortly prior to yours:
a priviledge of ME to switch, barring the others.

I think this pinpoints one of the confusions that's muddying up this
discussion. Under the Flip-Flop rules as they were presented, the Winning
Flip is determined before people switch, and the Winning Flip doesn't
change based on how people switch. In that scenario, my table is correct,

We can also consider the variant in which the Winning Flip is determined
after people decide whether or not to switch. But that game is
functionally
identical to the game where there is no coin-toss at all - everyone just
freely chooses Heads or Tails, then the Winning Flip is determined and the
winners are paid. Flipping a coin, looking at it, and then deciding
whether
or not to switch it is identical to simply picking heads or tails! The
coin-flips only matter in the first variant, where they determine the
Winning Flip *before* people make their choices.

In this variant, it doesn't matter whether you switch or not (i.e. whether
you choose heads or tails) - you are more likely to lose than win. We can
use the same 3-player table we've been discussing to see that there are
eight possible outcomes, and you only win in two of them. Once again,
there's no paradox, although you might *feel* like there is one. You might
reason that the Winning Flip is equally likely to be heads or tails, so no
matter which one you pick, your odds of winning will be 50/50. What's
missing from this logic is the recognition that no matter what you pick,
your choice will automatically decrease the chances of that side being in
the minority.

-- Kory

```

### Re: observation selection effects

```At 04:47 PM 10/10/2004, Jesse Mazer wrote:
If I get heads, I know the only possible way for the winning flip to be
heads would be if both the other players got tails, whereas the winning
flip will be tails if the other two got heads *or* if one got heads and
the other got tails.
I agree with this, but I want to add a subtle point: it's correct to switch
*even if I haven't looked at my own coin*. That's because, despite the fact
that I don't know whether my own coin is heads or tails, I know that,
whichever it is, it's more likely to be in the majority than the minority.

That's not to say that *nobody* needs to look at my coin. In order to
determine whether or not my choice to switch puts me in the heads or the
tails group, *someone's* going to have to look at my coin. But the rules of
the game allow me to pass off this act of looking to someone else - they
essentially allow me to tell the casino worker hey, take a look at my
coin, will ya, and assign me to the opposite. The important point is that
I can safely pass off this instruction to switch without even knowing the
result of my coin-flip, because I know that, whatever my coin is, it's more
likely to be in the majority group.

If we change the rules of the game slightly, and say that, instead of
choosing whether or not to switch, you have to actually choose heads or
tails, then, of course, you yourself do need to see the result of your
own coin-flip.

-- Kory

```

### re: observation selection effects

```You're right, as was discussed last week. It seems I clicked on the wrong
thing in my email program and have re-sent an old post. My apologies for
taking up the bandwidth!

--Stathis
From: Kory Heath [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: re: observation selection effects
Date: Sat, 09 Oct 2004 18:17:50 -0400
At 10:35 AM 10/9/2004, Stathis Papaioannou wrote:
From the point of view of typical player, it would seem that there is not:
the Winning Flip is as likely to be heads as tails, and if he played the
game repeatedly over time, he should expect to break even, whether he
switches in the final step or not.
That's not correct. While it's true that the Winning Flip is as likely to
be heads as tails, it's not true that I'm as likely to be in the winning
group as the loosing group. Look at the case when there are only three
players. There are eight possible outcomes:

Me: H  Player 1: H  Player 2: H - WF: T
Me: H  Player 1: H  Player 2: T - WF: T
Me: H  Player 1: T  Player 2: H - WF: T
Me: H  Player 1: T  Player 2: T - WF: H
Me: T  Player 1: H  Player 2: H - WF: T
Me: T  Player 1: H  Player 2: T - WF: H
Me: T  Player 1: T  Player 2: H - WF: H
Me: T  Player 1: T  Player 2: T - WF: H
I am in the winning group in only two out of these eight cases. So my
chances of winning if I don't switch are 1/4, and my chances of winning if
I do switch are 3/4. There's no paradox here.

-- Kory
_
Enter our Mobile Babe Search and win big!  http://ninemsn.com.au/babesearch

```

### Re: observation selection effects

```At 07:17 PM 10/10/2004, Kory Heath wrote:
We can also consider the variant in which the Winning Flip is determined
after people decide whether or not to switch.
In a follow-up to my own post, I should point out that your winning chances
in this game depend on how your opponents are playing. If all of your
opponents are playing randomly, then you have a negative expectation no
matter what you do. If your opponents are not playing randomly, then you
may be able to exploit patterns in their play to generate a positive
expectation.

-- Kory

```

### re: observation selection effects

```

Here is a similar paradox to the traffic lane example:

In the new casino game called Flip-Flop, an odd number of
players pay \$1 each to gather in individual cubicles and flip a coin (so no
player can see what another player is doing). The game organisers tally up the results, and the result which is in the minority is said to
be the Winning Flip, while the minority result is said to be the Losing Flip.
For example, if there are 101 players and of these 53 flip heads while 48 flip
tails, tails is the Winning Flip and heads is the Losing Flip. Before the
result of the tally is announced, each player must commit to either keep the
result of their original coin flip, whether heads or tails, or switch to the
opposite result. The casino then announces what the Winning Flip was, and
players whose final result (however it was obtained) corresponds with this are
paid \$2, while the rest get nothing.

The question now: is there anything to be gained by
switching at the last step of this game? From the point of view of typical
player, it would seem that there is not: the Winning Flip is as likely to be
heads as tails, and if he played the game repeatedly over time, he should
expect to break even, whether he switches in the final step or not. On the
other hand, it seems clear that if nobody switches, the casino is ahead, while
if everbody switches, the players are ahead; so switching would seem to be a
winning strategy for the players. This latter result is not due to any
cooperation effect, as only those players who switch get the improved (on
average) outcome.

Stathis Papaioannou

```

### Re: observation selection effects

```

Stathis, in this new FLip-Flop I see some slight
merit beyond the symmetry of switching from one unknown to another
unknown:
If I got heads,I can THINK of the majority
getting heads. (No justifiacation, however, but a slight idea that I am the
'average guy',
not the exceptional minority).
In that case it is allowable to switch, to get
into the less likely group, the paying minority.
I would not include any calculations: the
conditions are not quantizable in my opinion. In the moment you start
quantizing, the situation reverts into a perfect symmetry with the
opposite.UNLESS there are hidden parameters pointing to a NON-50-50 situation.
The 'insufficient number of sampling' is unsatisfactory it works both
ways.

John Mikes

- Original Message -
From:
Stathis Papaioannou
To: [EMAIL PROTECTED]
Sent: Saturday, October 09, 2004 10:35
AM
Subject: re: observation selection
effects

Here is a similar paradox to the
traffic lane example:

In the new casino game called
Flip-Flop, an odd number of players pay \$1 each to gather in individual
cubicles and flip a coin (so no player can see what another player is doing).
The game organisers tally up the results, and the
result which is in the minority is said to be the Winning Flip, while the
minority result is said to be the Losing Flip. For example, if there are 101
players and of these 53 flip heads while 48 flip tails, tails is the Winning
Flip and heads is the Losing Flip. Before the result of the tally is
announced, each player must commit to either keep the result of their original
coin flip, whether heads or tails, or switch to the opposite result. The
casino then announces what the Winning Flip was, and players whose final
result (however it was obtained) corresponds with this are paid \$2, while the
rest get nothing.

The question now: is there
anything to be gained by switching at the last step of this game? From the
point of view of typical player, it would seem that there is not: the Winning
Flip is as likely to be heads as tails, and if he played the game repeatedly
over time, he should expect to break even, whether he switches in the final
step or not. On the other hand, it seems clear that if nobody switches, the
casino is ahead, while if everbody switches, the players are ahead; so
switching would seem to be a winning strategy for the players. This latter
result is not due to any cooperation effect, as only those players who switch
get the improved (on average) outcome.

Stathis
Papaioannou

```

### re: observation selection effects

```At 10:35 AM 10/9/2004, Stathis Papaioannou wrote:
From the point of view of typical player, it would seem that there is
not: the Winning Flip is as likely to be heads as tails, and if he played
the game repeatedly over time, he should expect to break even, whether he
switches in the final step or not.
That's not correct. While it's true that the Winning Flip is as likely to
be heads as tails, it's not true that I'm as likely to be in the winning
group as the loosing group. Look at the case when there are only three
players. There are eight possible outcomes:

Me: H  Player 1: H  Player 2: H - WF: T
Me: H  Player 1: H  Player 2: T - WF: T
Me: H  Player 1: T  Player 2: H - WF: T
Me: H  Player 1: T  Player 2: T - WF: H
Me: T  Player 1: H  Player 2: H - WF: T
Me: T  Player 1: H  Player 2: T - WF: H
Me: T  Player 1: T  Player 2: H - WF: H
Me: T  Player 1: T  Player 2: T - WF: H
I am in the winning group in only two out of these eight cases. So my
chances of winning if I don't switch are 1/4, and my chances of winning if
I do switch are 3/4. There's no paradox here.

-- Kory

```

### Re: Observation selection effects

```Hal Finney writes:
Not to detract from your main point, but I want to point out that
sometimes there is ambiguity about how to count worlds, for example in
the many worlds interpretation of QM.  There are many examples of QM
based world-counting which seem to show that in most worlds, probability
theory should fail.
I'm not sure what examples you have in mind here, but this is actually the
general point I was trying to make: probability theory doesn't seem to work
the same way in a many worlds cosmology, due to complications such as
observers multiplying and then not being able to access the entire
probability space after the event of interest.

Consider these three examples:
(A) In a single world cosmology, I claim that using my magic powers, I have
bestowed on you, and you alone, the ability to pick the winning numbers in
this week's lottery. If you then buy a lottery ticket and win the first
prize, I think it would be reasonable to concede that there was probably
some substance to my claim (if not magic powers, then at least an effective
way of cheating).

(B) In a single world cosmology, I announce that using my magic powers, I
have bestowed on some lucky gambler the ability to pick the winning numbers
in this week's lottery. Now, someone does in fact win the first prize this
week, but that is not surprising, because there is almost always at least
one winner each week. I cannot reasonably claim to have helped the winner
unless I had somehow tagged him or otherwise uniquely identified him before
the lottery was drawn, as in (A).

(C) In a many worlds cosmology, I seek you out as in (A) and make the same
claim about bestowing on you the ability to pick the winning numbers in this
week's lottery. You buy a ticket, and win first prize. Should you thank me
for helping you win, as in (A)? In general, no; this situation is actually
more closely analogous to (B) than to (A). For it is certain that at least
one future version of you will win, just as it is very likely that at least
one person will win in the single world example. I can only claim that I
helped you win if I somehow identified which version in which world is going
to win before the lottery is drawn, and that is impossible.

Stathis Papaioannou
_
Check out Election 2004 for up-to-date election news, plus voter tools and
more! http://special.msn.com/msn/election2004.armx

```

### Re: Observation selection effects

```Stathis Papaioannou writes:
Hal Finney writes:
Not to detract from your main point, but I want to point out that
sometimes there is ambiguity about how to count worlds, for example in
the many worlds interpretation of QM.  There are many examples of QM
based world-counting which seem to show that in most worlds, probability
theory should fail.

I'm not sure what examples you have in mind here,

The specific kind of example goes like this.  Suppose you take a
vertically polarized photon and pass it through a polarizer that is
tilted slightly from the vertical.  Quantum mechanics predicts that
there is a high chance, say 99%, that the photon will pass through,
and a low chance, 1%, that it will not make it and be absorbed.

Now, the many worlds interpretation can be read to say that the universe
splits into two when this experiment occurs.  There are two possible
outcomes: either it passes through or it is absorbed.  So there are two
universes corresponding to the two results.

However, the universes are not of equal probability, according to QM.
One should be observed 99% of the time and the other only 1% of the
time.

The discrepancy gets worse if we imagine repeating the experiment multiple
times.  Each time the multiverse splits again in two.  If we did it, say,
20 times, there would be 2^20 or about 1 million universes.  In only
one of those universes did the photon take the 99% chance each time,
yet that is the expected result.  By a counting argument, the chance
of getting that result is only one in a million since only one world
out of a million sees it.  This is the apparent contradiction between
the probability predictions of orthodox quantum mechanics and the MWI,
assuming that we count worlds in this way.

but this is actually the
general point I was trying to make: probability theory doesn't seem to work
the same way in a many worlds cosmology, due to complications such as
observers multiplying and then not being able to access the entire
probability space after the event of interest.

Consider these three examples:

(A) In a single world cosmology, I claim that using my magic powers, I have
bestowed on you, and you alone, the ability to pick the winning numbers in
this week's lottery. If you then buy a lottery ticket and win the first
prize, I think it would be reasonable to concede that there was probably
some substance to my claim (if not magic powers, then at least an effective
way of cheating).

(B) In a single world cosmology, I announce that using my magic powers, I
have bestowed on some lucky gambler the ability to pick the winning numbers
in this week's lottery. Now, someone does in fact win the first prize this
week, but that is not surprising, because there is almost always at least
one winner each week. I cannot reasonably claim to have helped the winner
unless I had somehow tagged him or otherwise uniquely identified him before
the lottery was drawn, as in (A).

(C) In a many worlds cosmology, I seek you out as in (A) and make the same
claim about bestowing on you the ability to pick the winning numbers in this
week's lottery. You buy a ticket, and win first prize. Should you thank me
for helping you win, as in (A)? In general, no; this situation is actually
more closely analogous to (B) than to (A). For it is certain that at least
one future version of you will win, just as it is very likely that at least
one person will win in the single world example. I can only claim that I
helped you win if I somehow identified which version in which world is going
to win before the lottery is drawn, and that is impossible.

I'm afraid I don't agree with the conclusion in (C).  I definitely should
thank you.  To see this, let's make my thanks a little more sincere,
in the form of a payment.  Suppose I agree in advance to pay you \$1000
if you succeed in helping me win the lottery.  I say that is a wise
decision on my part.  It doesn't cost me anything if you don't help,
and if you do have some way of rigging the lottery then I can easily
afford to pay you this modest sum out of my winnings.

But I think your reasoning suggests that it is unwise, since I will win
anyway, so why should I pay anything to you?  I don't need to thank you
in this way.  Do you agree that this follows from your reasoning?

Hal Finney

```

### Re: Observation selection effects

```Jesse Mazer wrote:
I don't think that's a good counterargument, because the whole concept of
probability is based on ignorance...

No, I don't agree! Probability is based in a sense on ignorance, but you
must make full use of such information as you do have. If you toss a fair
anything between zero and one, because it is possible I am lying about it
being a fair coin!

Here is another two envelope example:
Two envelopes, A and B, contain two doses of the drug Lifesavium, the
Correct Dose and the Half Dose. If you give the patient more than 1.5 times
the Correct Dose you will certainly kill him, while if you give him the Half
Dose you will save his life, although he won't make an immediate recovery as
he would if you gave him the Correct Dose. If you don't give him any
medication at all, again, he will surely die. Once you open an envelope, the
medication in in such a form that you must give the full dose, or nothing.

You are faced with the two envelopes, the above information and the sick
patient, with no other help, on a desert island. There is one further
complication: if you open the first envelope, and then decide to open the
second envelope, you must destroy the contents of the first envelope in
order to get to the second envelope.

OK: so you open envelope A and find that it contains 10mg of Lifesavium. You
don't know whether this is The Correct Dose or the Half Dose; so envelope B
may have either 5mg or 20mg, right? And if 10mg is the Correct Dose, then if
you discard envelope A in favour of envelope B, there is a 50% chance that
envelope B will have double the Correct Dose and you will kill the patient -
so you had better stick with envelope A, right?

I think you can see the error in the above argument. You already know that
the amount in each envelope is fixed, so even though you have no idea of the
actual dosages involved, or which envelope contains which dose, even after
opening the first envelope, there is NO WAY you can give the patient an
overdose. There is no way envelope B can contain 20mg of Lifesavium, but
even though you cannot know this, you can use the above reasoning to deduce
that there is no expected benefit from choosing a strategy of switching or
not switching - as you can also see intuitively from the symmetry of the
situation, whether you choose envelope A or B first.

In the game with the envelopes and the money, the analogous error is to
think that there is a possibility of doubling your money when you have
actually picked the envelope containing the larger sum first. As I said in
my previous post, if this assumption is valid, then you are playing a
different game in which our eccentric millionaire flips a coin to decide
(without telling you which) if he will put double or half the sum you find
on opening envelope A into envelope B. You would then certainly be better
off, on average, if you switched envelopes.

Stathis Papaioannou
_
Check out Election 2004 for up-to-date election news, plus voter tools and
more! http://special.msn.com/msn/election2004.armx

```

### Re: Observation selection effects

```Stathis Papaioannou wrote:
Jesse Mazer wrote:
I don't think that's a good counterargument, because the whole concept of
probability is based on ignorance...

No, I don't agree! Probability is based in a sense on ignorance, but you
must make full use of such information as you do have.
Of course--I didn't mean it was based *only* on ignorance, you must
incorporate whatever information you have into your estimate of the
probability, but no more. Your argument violates the but no more rule,
since it incorporates the knowledge of an observer who has seen how much
money both envelopes contain, while I only know how much money one envelope
contains.

If you toss a fair coin, is Pr(heads)=0.5? According to your argument, it
could actually be anything between zero and one, because it is possible I
am lying about it being a fair coin!
My argument implies nothing of the sort. But your argument would seem to
imply that if I am watching a videotape of a fair coin toss, then if someone
else has already watched the tape, it would be permissible to incorporate
their knowledge of the outcome of the toss into a probability calculation,
even if I myself don't have this knowledge.

Here is another two envelope example:
Two envelopes, A and B, contain two doses of the drug Lifesavium, the
Correct Dose and the Half Dose. If you give the patient more than 1.5 times
the Correct Dose you will certainly kill him, while if you give him the
Half Dose you will save his life, although he won't make an immediate
recovery as he would if you gave him the Correct Dose. If you don't give
him any medication at all, again, he will surely die. Once you open an
envelope, the medication in in such a form that you must give the full
dose, or nothing.

You are faced with the two envelopes, the above information and the sick
patient, with no other help, on a desert island. There is one further
complication: if you open the first envelope, and then decide to open the
second envelope, you must destroy the contents of the first envelope in
order to get to the second envelope.

OK: so you open envelope A and find that it contains 10mg of Lifesavium.
You don't know whether this is The Correct Dose or the Half Dose; so
envelope B may have either 5mg or 20mg, right? And if 10mg is the Correct
Dose, then if you discard envelope A in favour of envelope B, there is a
50% chance that envelope B will have double the Correct Dose and you will
kill the patient - so you had better stick with envelope A, right?

I think you can see the error in the above argument. You already know that
the amount in each envelope is fixed, so even though you have no idea of
the actual dosages involved, or which envelope contains which dose, even
after opening the first envelope, there is NO WAY you can give the patient
an overdose. There is no way envelope B can contain 20mg of Lifesavium, but
even though you cannot know this, you can use the above reasoning to deduce
that there is no expected benefit from choosing a strategy of switching or
not switching - as you can also see intuitively from the symmetry of the
situation, whether you choose envelope A or B first.
This case is not analogous to the two-envelope problem, because in this case
it is part of *my* knowledge that one envelope contains the Correct dose and
the other contains the Half dose, and neither contains a Double dose. In
contrast, your analysis of the two-envelope problem relied on information I
don't have, namely whether the two envelopes contained \$50 and \$100 or \$25
and \$50.

Suppose I know that the envelope-stuffer flipped a fair coin to decide
whether to put \$25 or \$50 in one envelope, then put double that amount in
the other. I randomly choose one envelope and open it, and find \$50. Do you
disagree that my average expected return from switching would now be
(0.5)(25) + (0.5)(100) = 62.5? If this experiment was repeated many times
and we looked only at the subset of cases where the first envelope I opened
contained \$50 and I chose to switch, wouldn't my average winnings in this
subset of cases be \$62.50?

In the game with the envelopes and the money, the analogous error is to
think that there is a possibility of doubling your money when you have
actually picked the envelope containing the larger sum first.
But *you* don't know that the envelope you picked was the one with the
larger sum. This is akin to arguing that it is an error to think there is
a possibility of winning if you bet \$100 on heads in a videotaped coin toss,
since someone who's already watched the tape knows it comes up tails, even
though you don't know that. Would you indeed say it's an error to believe my
average expected return is \$50 in this case?

Jesse

```

### Re: Observation selection effects

```
Jesse Mazer wrote:
I don't think that's a good counterargument, because the whole concept of
probability is based on ignorance...

No, I don't agree! Probability is based in a sense on ignorance, but you
must make full use of such information as you do have.
Of course--I didn't mean it was based *only* on ignorance, you must
incorporate whatever information you have into your estimate of the
probability, but no more. Your argument violates the but no more rule,
since it incorporates the knowledge of an observer who has seen how much
money both envelopes contain, while I only know how much money one envelope
contains.
Sorry Jesse, I can see in retrospect that I was insulting your intelligence
as a rhetorical ploy, and we shouldn't stoop to that level of debate on this
list.

You say that you must incorporate whatever information you have, but no
more in the envelopes/money example. The point I was trying to make with my
envelope/drug example is that you need to take into account the fact that
the amount in each envelope is fixed, but again you are right, it was not
exactly analogous. But you have passed over the final point in my last post,
which I now restate:

(1) The original game: envelope A and B, you know one has double the amount
of the other, but you don't know which. You open A and find \$100. Should you
switch to B, which may have either \$50 or \$200?

(2) A variation: everything is the same, up to the point where you are
pondering whether to switch to envelope B, when the millionaire walks in,
and hidden from view, flips a coin to decide whether to replace whatever was
originally in envelope B with either double or half the sum in envelope A,
i.e. either \$50 or \$200.

Now, which game would you prefer to play, (1) or (2)? They are not the same.
In game (1), if the \$100 in A is actually the higher amount, if you switch
you will get \$50 for sure; but in game (2) if the \$100 is actually the
higher amount you have a 50% chance of getting \$200 if you switch. It works
in the reverse way if the \$100 is the lower amount - you could lose \$50
rather than gain \$100 - but the possible gain outweighs the possible loss.

Look at it another way: game (2) is actually asymmetrical. The amount you
win if you play it many times will be different if you switch, because you
really do have more to gain than to lose by switching (and the millionaire
will have to pay out more on average). On the other hand, intuitively, you
can see that your expected gains in game (1) should be the same whether you
switch or not. The paradox comes from reasoning as if you are playing game
(2) when you are really playing game (1).

Stathis Papaioannou
_
Discover how everyone  everything in our world's connected:
http://www.onebigvillage.com.au?obv1=hotmail

```

### Re: Observation selection effects

```Addition to my last post:
(1) The original game: envelope A and B, you know one has double the amount
of the other, but you don't know which. You open A and find \$100. Should
you switch to B, which may have either \$50 or \$200?

(2) A variation: everything is the same, up to the point where you are
pondering whether to switch to envelope B, when the millionaire walks in,
and hidden from view, flips a coin to decide whether to replace whatever
was originally in envelope B with either double or half the sum in envelope
A, i.e. either \$50 or \$200.
Say one envelope contains \$x and the other \$2x. If you keep the first
envelope in game (2), and if you keep the first one OR switch in game (1),
you should expect to win \$1.5x. If you switch in game (2), you should expect
to win 0.25*(\$0.5x + \$2x + \$x +\$4x) = \$1.875x.

Stathis Papaioannou
_
Enter our Mobile Babe Search and win big!  http://ninemsn.com.au/babesearch

```

### Re: Observation selection effects

```This has been an interesting thread so far, but let me bring it back to
topic for the Everything List. It has been assumed in most posts to this
list over the years that our current state must be a typical state in some
sense. For example, our world has followed consistent laws of physics for as
long as anyone has been able to determine - the old no white rabbit worlds
observation. In the face of ensemble type theories such as the MWI of QM,
this is seen as presenting a problem: if anything that can happen, does
happen, why does our experience of the world include only a very limited,
orderly subset of this anything?

There have been many attempts to answer the above question, eg. see Russell
Standish' paper Why Ockham's Razor? But does our current orderly world
imply that most possible worlds are orderly? It seems to me that there is an
asymmetry between (a) what we can expect for the future, and (b) what we can
deduce about the probability implicit in (a) from what actually does happen.

Suppose that according to X-Theory, in the next minute the world will split
into one million different versions, of which one version will be the same
sort of orderly world we are used to, while the rest will be worlds in which
it will be immediately obvious to us that very strange things are happening,
eg. dragons materialising out of thin air, furniture levitating, the planet
Jupiter hurling itself into the sun, etc. I think it is reasonable to expect
that if X-Theory is correct, we will very likely see these bizarre and
frightening things happen in the next minute.

Now, here we are, a minute later, and nothing bizarre has happened after
all. Does this mean that X-Theory is probably wrong? Perhaps not. After all,
the theory did predict with 100% certainty that one version of the world
will continue as before. The objection to this will no doubt be, yes, but
how likely is it that WE end up in that particular version? And the answer
to this objection is, it is 100% certain that WE end up in that particular
version; just as it is 100% certain that 999,999 copies of us end up in the
bizarre versions. Those 999,999 copies are not continuing to type away as I
am, because they are running around in a panic.

The above is simply a version of the Anthropic Principle as applied to
intelligent life in the universe. A particular ensemble theory may predict
that it is overwhelmingly unlikely that a particular universe will allow the
development of intelligent life. Does the fact that we are here, appparently
intelligent and alive, count as evidence against that theory? No, because
the theory predicts that although unlikely, it is certain to happen in at
least ONE universe, and obviously that universe will be the one we find
ourselves in.

Stathis Papaioannou
_
In the market for a car? Buy, sell or browse at CarPoint:

```

### Re: Observation selection effects

```Stathis Papaioannou wrote:
Sorry Jesse, I can see in retrospect that I was insulting your intelligence
as a rhetorical ploy, and we shouldn't stoop to that level of debate on
this list.
No problem, I wasn't insulted...
You say that you must incorporate whatever information you have, but no
more in the envelopes/money example. The point I was trying to make with
my envelope/drug example is that you need to take into account the fact
that the amount in each envelope is fixed
Well, I think that's like saying that in the videotaped coinflip example,
you need to take into account the fact that the outcome of the flip is
already fixed. I don't think it matters whether it's really fixed or not,
(they're epistemological, not ontological), and since you are equally
ignorant of the outcome regardless of whether the flip happens in realtime
or on video, your probabilistic reasoning should be the same.

But you have passed over the final point in my last post, which I now
restate:

(1) The original game: envelope A and B, you know one has double the amount
of the other, but you don't know which. You open A and find \$100. Should
you switch to B, which may have either \$50 or \$200?

(2) A variation: everything is the same, up to the point where you are
pondering whether to switch to envelope B, when the millionaire walks in,
and hidden from view, flips a coin to decide whether to replace whatever
was originally in envelope B with either double or half the sum in envelope
A, i.e. either \$50 or \$200.
Well, my argument about the two-envelope paradox all along has been that you
need to think about the probability distribution the millionaire uses to
stuff the two envelopes, and that once you do that the apparent paradox
disappears. So we need to consider what probability distributions the
millionaire used in these examples. Let's say, for example, that the
millionaire flips a coin to decide whether to put \$50 or \$100 in one
envelope, and then puts double that amount in the other. In that case, if I
pick an envelope randomly, there is a 1/4 chance I'll find \$50 inside, a 1/2
chance I'll find \$100 inside, and a 1/4 chance I'll find \$200 inside. If I
find either \$50 or \$200, I know with complete certainty how much the other
envelope contains; but if I find \$100, then from my point of view there's a
1/2 chance the other envelope contains \$50 and a 1/2 chance the other
envelope contains \$200.

Now, if you assume that in game (2) the millionaire *only* replaces the
amount in the second envelope with a new amount based on a coinflip *if* I
found \$100 in the first envelope, but doesn't mess with the second envelope
if I found \$50 or \$200 in the first one, then both games are exactly equal,
from my point of view. In both cases, whenever I find \$100 in the first
envelope, my average expected return from switching would be (0.5)(\$50) +
(0.5)(\$200) = \$125, so it's better to switch.

On the other hand, if in game (2) the millionaire replaced the amount in the
second envelope with a new amount based on a coinflip *regardless* of how
much I found in the first envelope, this would change my strategy if I found
either \$50 or \$200 in the first envelope. In this case, it will be to my
advantage to switch no matter how much I find in the first envelope, since
my average expected return from switching will always be 1.25 times however
much I found in that envelope.

In contrast, in game (1) my average expected return from switching would be
\$100 if I found \$50 in the first envelope and \$100 if I found \$200 in the
first envelope, while my average expected return from switching if I found
\$100 would still be \$125, so my total average expected return from switching
regardless of what I find in the first envelope is (1/4)(\$100) + (1/2)(\$125)
+ (1/4)(\$100) = \$112.50, while my average expected return from sticking with
my first choice regardless of how much I find is (1/4)(\$50) + (1/2)(\$100) +
(1/4)(\$200) = \$112.50 as well. Again, in game (1) the resolution of the
apparent paradox must be that for any possible probability distribution
the millionaire uses to pick the amounts in the envelopes, your average
expected return from sticking with your first choice if you don't open it to
see how much is inside must always be equal to your average expected
winnings if you decide to switch without first checking how much was inside

Now, which game would you prefer to play, (1) or (2)? They are not the
same.
With the conditions I mentioned--the millionaire flips a coin to decide
whether to put \$50 or \$100 in one envelope, then puts double in the other,
and in game (2) he only replaces the amount in the second envelope if you
find \$100 in the first envelope you choose--then the two games actually are
exactly the same, in terms of probabilities and average expected returns. If
you want to suggest some ```

### Re: Observation selection effects

```Stathis Papaioannou writes:
Suppose that according to X-Theory, in the next minute the world will split
into one million different versions, of which one version will be the same
sort of orderly world we are used to, while the rest will be worlds in which
it will be immediately obvious to us that very strange things are happening,
eg. dragons materialising out of thin air, furniture levitating, the planet
Jupiter hurling itself into the sun, etc. I think it is reasonable to expect
that if X-Theory is correct, we will very likely see these bizarre and
frightening things happen in the next minute.

Not to detract from your main point, but I want to point out that
sometimes there is ambiguity about how to count worlds, for example in
the many worlds interpretation of QM.  There are many examples of QM
based world-counting which seem to show that in most worlds, probability
theory should fail.

Now, here we are, a minute later, and nothing bizarre has happened after
all. Does this mean that X-Theory is probably wrong? Perhaps not. After all,
the theory did predict with 100% certainty that one version of the world
will continue as before. The objection to this will no doubt be, yes, but
how likely is it that WE end up in that particular version? And the answer
to this objection is, it is 100% certain that WE end up in that particular
version; just as it is 100% certain that 999,999 copies of us end up in the
bizarre versions. Those 999,999 copies are not continuing to type away as I
am, because they are running around in a panic.

Would you agree that those who assume that such an outcome (no bizarre
events) disproves X-theory will be right more often than they are wrong?
Hence adopting such a policy will generally be successful, and beings
who base their decisions on such a rule will become more numerous and
influential in the multiverse.

Even though there are universes where this rule (that is, the rule a
theory which predicts something we don't see is probably an incorrect
theory) does not work, still it is a good rule to follow.

Hal Finney

```

### Re: Observation selection effects

```Norman Samish writes:
QUOTE-
Assume an eccentric millionaire offers you your choice of either of two
sealed envelopes, A or B, both containing money.  One envelope contains
twice as much as the other.  After you choose an envelope you will have the
option of trading it for the other envelope.
Suppose you pick envelope A.  You open it and see that it contains \$100.
Now you have to decide if you will keep the \$100, or will you trade it for
whatever is in envelope B?
You might reason as follows: since one envelope has twice what the other one
has, envelope B either has 200 dollars or 50 dollars, with equal
probability.  If you switch, you stand to either win \$100 or to lose \$50.
Since you stand to win more than you stand to lose, you should switch.
-ENDQUOTE
The problem is that you are reasoning as if the amount in each envelope can
vary during the game, whereas in fact it is fixed. Suppose envelope A
contains \$100 and envelope B contains \$50. You open A, see the \$100, and
then reason that B may contain either \$50 or \$200, each being equally
likely. In fact, B cannot contain \$200, even though you don't know this yet.
It is easy enough for an external observer (who does know the contents of
each envelope) to calculate the probabilities: if you keep the first
envelope, your expected gain is 0.5*\$100 + 0.5*\$50 = \$75. If you switch,
your expected gain is 0.5*\$100 (if you open B first)  + 0.5*\$50 (if you open
A first) = \$75, as before.

Ignorance of the actual amounts may lead you to speculate that one of the
envelopes may contain \$200, but it won't make the money magically
materialise! And even if you don't know the actual amounts, the above
analysis should convince you that nothing is to be gained by switching
envelopes.

If the game changes so that, once you have opened the first envelope, the
millionaire decides by flipping a coin whether he will put half or double
that amount in the second envelope, then you are actually better off
switching.

Stathis Papaioannou
_
Discover how everyone  everything in our world's connected:
http://www.onebigvillage.com.au?obv1=hotmail

```

### RE: Observation selection effects

```-Original Message-
Norman Samish:

The Flip-Flop game described by Stathis Papaioannou
strikes me as a
version of the old Two-Envelope Paradox.

Assume an eccentric millionaire offers you your choice
of either of two
sealed envelopes, A or B, both containing money.  One
envelope contains
twice as much as the other.  After you choose an
envelope you will have the
option of trading it for the other envelope.

Suppose you pick envelope A.  You open it and see that
it contains \$100.
Now you have to decide if you will keep the \$100, or
whatever is in envelope B?

You might reason as follows: since one envelope has
twice what the other
one
has, envelope B either has 200 dollars or 50 dollars, with equal
probability.  If you switch, you stand to either win
\$100 or to lose \$50.
Since you stand to win more than you stand to lose, you
should switch.

But just before you tell the eccentric millionaire that
you would like to
switch, another thought might occur to you.  If you had
picked envelope B,
you would have come to exactly the same conclusion.  So
if the above
argument is valid, you should switch no matter which
envelope you choose.

Therefore the argument for always switching is NOT
valid - but I am unable,
at the moment, to tell you why!

Of course in the real world you have some idea about how much
money is in play so if you see a very large amount you infer it's
probably the larger amount.  But even without this assumption of
realism it's an interesting problem and taken as stated there's
still no paradox.  I saw this problem several years ago and here's
my solution.  It takes the problem as stated, but I do make one

Let:  s = envelope with smaller amount is selected.
l = envelope with larger amount is selected.
m = the amount in the selected envelope.

Since any valid resolution of the paradox would have to work for
ratios of money other than two, also define:

r = the ratio of the larger amount to the smaller.

Now here comes the restrictive assumption, which can be thought of
as a restrictive rule about how the amounts are chosen which I
hope to generalize away later.  Expressed as a rule, it is this:

The person putting in the money selects, at random (not
necessarily uniformly), the smaller amount from a range (x1, x2)
such that x2  r*x1.  In other words, the range of possible
amounts is such that the larger and smaller amount do not overlap.
Then, for any interval of the range (x,x+dx) for the smaller
amount with probability p, there is a corresponding interval (r*x,
r*x+r*dx) with probability p for the larger amount.  Since the
latter interval is longer by a factor of r

P(l|m)/P(s|m) = r ,

In other words, no matter what m is, it is r-times more likely to
fall in a large-amount interval than in a small-amount interval.

But since l and s are the only possibilities (and here's where I
need the non-overlap),

P(1|m) + P(s|m) = 1

which implies,

P(s|m) = 1/(1+r)  and P(1|m) = r/(1+r) .

Then the rest is straightforward algebra. The expected values are:

E(don't switch) = m

E(switch) = P(s|m)rm + P(l|m)m/r
= [1/(1+r)]rm + [r/(1+r)]m/r
= m

Brent Meeker

```

### [Fwd: RE: Observation selection effects]

```I always forget to reply-to-all in this list.
So below goes my reply which went only to Hal Finney.

-Forwarded Message-
From: Eric Cavalcanti [EMAIL PROTECTED]
To: Hal Finney [EMAIL PROTECTED]
Subject: RE: Observation selection effects
Date: Tue, 05 Oct 2004 12:57:14 +1000

On Tue, 2004-10-05 at 10:20, Hal Finney wrote:
Stathis Papaioannou writes:
In the new casino game Flip-Flop, an odd number of players pays \$1 each to
individually flip a coin, so that no player can see what another player is
doing. The game organisers then tally up the results, and the result in the
minority is called the Winning Flip, while the majority result is called the
Losing Flip. Before the Winning Flip is announced, each player has the
opportunity to either keep their initial result, or to Switch; this is then
called the player's Final Flip. When the Winning Flip is announced, players
whose Final Flip corresponds with this are paid \$2 by the casino, while the
rest are paid nothing.

Think about if the odd number of players was exactly one.  You're guaranteed
to have the Winning Flip before you switch.

Then think about what would happen if the odd number of players was three.
Then you have a 3/4 chance of having the Winning Flip before you switch.
Only if the other two players' flips both disagree with yours will you not
have the Winnning Flip, and there is only a 1/4 chance of that happening.

Exactly.

It is interesting to note that, even though you are
more likely to be in the Winning Flip, there is no
disadvantage in Switching. To understand that, we can
look at the N=3 case, and see that if I am in the
Winning Flip with someone else, then if I change I
will still be in the Winnig Flip with the other person.

As opposed to Stathis initial thought, even though the
Winning Flip is indeed as likely to be Heads as Tails,
each individual is more likely to be in the
Winning Flip as in the Losing Flip in any given run.

So that this would never make it into a Casino game,
because the house would lose money in the long run.

Eric.

```

### RE: Observation selection effects

```Brent Meeker wrote:
Of course in the real world you have some idea about how much
money is in play so if you see a very large amount you infer it's
probably the larger amount.  But even without this assumption of
realism it's an interesting problem and taken as stated there's
still no paradox.  I saw this problem several years ago and here's
my solution.  It takes the problem as stated, but I do make one
Let:  s = envelope with smaller amount is selected.
l = envelope with larger amount is selected.
m = the amount in the selected envelope.
Since any valid resolution of the paradox would have to work for
ratios of money other than two, also define:
r = the ratio of the larger amount to the smaller.
Now here comes the restrictive assumption, which can be thought of
as a restrictive rule about how the amounts are chosen which I
hope to generalize away later.  Expressed as a rule, it is this:
The person putting in the money selects, at random (not
necessarily uniformly), the smaller amount from a range (x1, x2)
such that x2 lt; r*x1.  In other words, the range of possible
amounts is such that the larger and smaller amount do not overlap.
Then, for any interval of the range (x,x+dx) for the smaller
amount with probability p, there is a corresponding interval (r*x,
r*x+r*dx) with probability p for the larger amount.  Since the
latter interval is longer by a factor of r
P(l|m)/P(s|m) = r ,
In other words, no matter what m is, it is r-times more likely to
fall in a large-amount interval than in a small-amount interval.
But since l and s are the only possibilities (and here's where I
need the non-overlap),
P(1|m) + P(s|m) = 1
which implies,
P(s|m) = 1/(1+r)  and P(1|m) = r/(1+r) .
Then the rest is straightforward algebra. The expected values are:
E(don't switch) = m
E(switch) = P(s|m)rm + P(l|m)m/r
= [1/(1+r)]rm + [r/(1+r)]m/r
= m
This is right, but it's a pretty special case--there are an infinite number
of possible probability distributions the millionaire could use when
deciding how much money to put in one envelope, even if we assume he always
puts double in the other. For example, he could use a distribution that
gives him a 1/2 probability of putting between 0 and 1 dollars in one
envelope (assume the dollar amounts can take any positive real value, and he
uses a flat probability distribution to pick a number between 0 and 1), a
1/4 probability of putting in between 1 and 2 dollars, a 1/8 probability of
putting in between 2 and 3 dollars, and in general a 1/2^n probability of
putting in between n-1 and n dollars. This would insure there was some
finite probability that *any* positive real number could be found in either
envelope.

The basic paradox is that the argument tries to show that the average
expected payoff from picking the second envelope is higher than the average
expected payoff from sticking with the first one, *regardless of what amount
you found in the first envelope*--in other words, even without opening the
first envelope you'd be better off switching to the second, which doesn't
make sense since the envelopes are identical and your first pick was random.
But it's not actually possible that, regardless of what you found in the
first envelope, there would always be a 50% chance the other envelope
contained half that and a 50% chance it contained double that...for that to
be true, the amount in the first envelope would have to be picked using a
flat probability distribution which is equally likely to give any number
from 0 to infinity, and as I said that's impossible. But my argument was not
really sufficiently general either, because it doesn't rule out other
possibilities like a 55% chance the other envelope contained half what was
found in the first envelope and a 45% chance it contained double, in which
case your average expected payoff would still be higher if you switched.

A truly general argument would have to show that, for any logically possible
probability distribution the millionaire uses to pick the amounts in the
envelopes, the average expected payoff from switching will always be exactly
equal to the average expected winnings from sticking with your first choice.
There are two different ways this can be true:

Possibility #1: it may be that you know enough about the probability
distribution that opening the envelope and seeing how much is inside allows
you to refine your evaluation of the average expected payoff from switching.
I gave an example of this in my post, where the millionaire picks an amount
from 1 to a million to put in one envelope and puts double that in the
other; in that case, if you open your first pick and find an amount greater
than a million, the average expected payoff from switching is 0. But even if
the average expected payoff may vary depending on what you find in the ```

### Re: Observation selection effects

```Dear Brent,

I enjoyed your description which was quite a show for my totally qualitative
(and IN-formally intuitive) logic - how you, quanti people substitute common
sense for those letters and numbers.
Have a good day

John Mikes
PS: to excuse my lingo: my 1st Ph.D. was Chemistry-Physics-Math. J
- Original Message -
From: Brent Meeker [EMAIL PROTECTED]
To: Everything-List [EMAIL PROTECTED]
Sent: Monday, October 04, 2004 6:19 PM
Subject: RE: Observation selection effects

-Original Message-
Norman Samish:

The Flip-Flop game described by Stathis Papaioannou
strikes me as a
version of the old Two-Envelope Paradox.

Assume an eccentric millionaire offers you your choice
of either of two
sealed envelopes, A or B, both containing money.  One
envelope contains
twice as much as the other.  After you choose an
envelope you will have the
option of trading it for the other envelope.

Suppose you pick envelope A.  You open it and see that
it contains \$100.
Now you have to decide if you will keep the \$100, or
whatever is in envelope B?

You might reason as follows: since one envelope has
twice what the other
one
has, envelope B either has 200 dollars or 50 dollars, with equal
probability.  If you switch, you stand to either win
\$100 or to lose \$50.
Since you stand to win more than you stand to lose, you
should switch.

But just before you tell the eccentric millionaire that
you would like to
switch, another thought might occur to you.  If you had
picked envelope B,
you would have come to exactly the same conclusion.  So
if the above
argument is valid, you should switch no matter which
envelope you choose.

Therefore the argument for always switching is NOT
valid - but I am unable,
at the moment, to tell you why!

Of course in the real world you have some idea about how much
money is in play so if you see a very large amount you infer it's
probably the larger amount.  But even without this assumption of
realism it's an interesting problem and taken as stated there's
still no paradox.  I saw this problem several years ago and here's
my solution.  It takes the problem as stated, but I do make one

Let:  s = envelope with smaller amount is selected.
l = envelope with larger amount is selected.
m = the amount in the selected envelope.

Since any valid resolution of the paradox would have to work for
ratios of money other than two, also define:

r = the ratio of the larger amount to the smaller.

Now here comes the restrictive assumption, which can be thought of
as a restrictive rule about how the amounts are chosen which I
hope to generalize away later.  Expressed as a rule, it is this:

The person putting in the money selects, at random (not
necessarily uniformly), the smaller amount from a range (x1, x2)
such that x2  r*x1.  In other words, the range of possible
amounts is such that the larger and smaller amount do not overlap.
Then, for any interval of the range (x,x+dx) for the smaller
amount with probability p, there is a corresponding interval (r*x,
r*x+r*dx) with probability p for the larger amount.  Since the
latter interval is longer by a factor of r

P(l|m)/P(s|m) = r ,

In other words, no matter what m is, it is r-times more likely to
fall in a large-amount interval than in a small-amount interval.

But since l and s are the only possibilities (and here's where I
need the non-overlap),

P(1|m) + P(s|m) = 1

which implies,

P(s|m) = 1/(1+r)  and P(1|m) = r/(1+r) .

Then the rest is straightforward algebra. The expected values are:

E(don't switch) = m

E(switch) = P(s|m)rm + P(l|m)m/r
= [1/(1+r)]rm + [r/(1+r)]m/r
= m

Brent Meeker

```

### [Fwd: Re: Observation selection effects]

```

Original Message

Subject:
Re: Observation selection effects

Date:
Sat, 04 Sep 2004 02:29:54 -0400

From:
Danny Mayes [EMAIL PROTECTED]

To:
[EMAIL PROTECTED]

References:
[EMAIL PROTECTED]

These problems remind me of the infamous Monty Hall problem that got
Marilyn vos Savant in some controversy.  Someone wrote and asked the
following question:

You are on "lets make a deal", and are chosen to select a door among 3
doors, one of which has a car behind it.  You randomly select door 1.
Monty, knowing where the car is, opens door 2 revealing an empty room,
and asks if you want to stay with door one.  The question was:  Is there
any benefit in switching from door 1 to door 3.  Common sense would
suggest Monty simply eliminated one choice, and you have a 50-50 chance
either way.  Marylin argued that by switching, the contestant actually
increases his odds from 1/3 to 2/3.  The difference coming about through
the added information of the car not behind door 2.  This example is
discussed in the book "Information:  The New Language of Science" by
Hans Christian von Baeyer, which I am trying to read, but only getting
through bits and pieces as usual due to my work schedule.  According to
the book, vos Savant still gets mail arguing her position on this
matter.  It seems to me it would be very easy to resolve with a friend,
letting one person play Monty and then keeping a tally of your success
in switching vs. not switching (though I haven't tried this- my wife
didn't find it intriguing enough, unfortunately).

I think these games provide good examples of how our common sense often
works against a deep understanding of what is really going on around
here.  I also think they point to a very fundamental level of importance
of the role of information in understanding the way our world  (or
multiverse) works

Jesse Mazer wrote:

Norman Samish:

The "Flip-Flop" game described by Stathis Papaioannou strikes me as a
version of the old Two-Envelope Paradox.

Assume an eccentric millionaire offers you your choice of either of two
sealed envelopes, A or B, both containing money.  One envelope contains
twice as much as the other.  After you choose an envelope you will
have the
option of trading it for the other envelope.

Suppose you pick envelope A.  You open it and see that it contains \$100.
Now you have to decide if you will keep the \$100, or will you trade
it for
whatever is in envelope B?

You might reason as follows: since one envelope has twice what the
other one
has, envelope B either has 200 dollars or 50 dollars, with equal
probability.  If you switch, you stand to either win \$100 or to lose
\$50.
Since you stand to win more than you stand to lose, you should switch.

But just before you tell the eccentric millionaire that you would
like to
switch, another thought might occur to you.  If you had picked
envelope B,
you would have come to exactly the same conclusion.  So if the above
argument is valid, you should switch no matter which envelope you
choose.

Therefore the argument for always switching is NOT valid - but I am
unable,
at the moment, to tell you why!

Basically, I think the resolution of this paradox is that it's
impossible to pick a number randomly from 0 to infinity in such a way
that every number is equally likely to come up. Such an infinite flat
example, if you picked two positive integers randomly from a flat
probability distribution, and then looked at the first integer, then
there would be a 100% chance the second integer would be larger, since
there are only a finite number of integers smaller than or equal to
the first one and an infinite number that are larger.

For any logically possible probability distribution the millionaire
uses, it will be true that depending on what amount of money you find
in the first envelope, there won't always be an equal chance of
finding double the amount or half the amount in the other envelope.
For example, if the millionaire simply picks a random amount from 0 to
one million to put in the first envelope, and then flips a coin to
decide whether to put half or double that in the other envelope, then
if the first envelope contains more than one million there is a 100%
chance the other envelope contains less than that.

For a more detailed discussion of the two-envelope paradox, see this
page:

http://jamaica.u.arizona.edu/~chalmers/papers/envelope.html

I don't think the solution to this paradox has any relation to the
solution to the flip-flop game, though. In the case of the flip-flop
game, it may help to assume that the players are all robots, and that
each player can assume that whatever decision it m```

### RE: Observation selection effects

```Brent Meeker wrote:
-Original Message-
From: Jesse Mazer [mailto:[EMAIL PROTECTED]
Sent: Tuesday, October 05, 2004 6:33 PM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: RE: Observation selection effects
Brent Meeker wrote:
years), I think it works without the restrictive assumption that
the range of distirbutions not overlap.  It's still
necessary that
P(l|m)+P(s|m)=1 and P(l|m)/P(s|m)=r, which is all that
is required
for the proof to go thru.
gotten the idea that you
were assuming a uniform probability of the
envelope-stuffer picking any
number between x1 and x2 for the first envelope, but I
see this isn't
necessary, so your proof is a lot more general than I
thought. Still not
completely general though, because the envelope-stuffer
can also use a
distribution which has no upper bound x2 on possible
amounts to put in the
first envelope, like the one I mentioned in my last post:
For example, he could use a
distribution that
gives him a 1/2 probability of putting between 0 and 1
dollars in one
envelope (assume the dollar amounts can take any
positive real value, and he
uses a flat probability distribution to pick a number
between 0 and 1), a
1/4 probability of putting in between 1 and 2 dollars, a
1/8 probability of
putting in between 2 and 3 dollars, and in general a
1/2^n probability of
putting in between n-1 and n dollars. This would insure
there was some
finite probability that *any* positive real number could
be found in either
envelope.
Likewise, he could also use the continuous probability
distribution P(x) =
1/e^x (whose integral from 0 to infinity is 1). And if
you want to restrict
the amounts in the envelope to positive integers, he could use a
distribution which gives a 1/2^n probability of putting
exactly n dollars in
the first envelope.
Jesse
That doesn't matter if I can do without the no-overlap
assumption - which I think I can.  Do you see a flaw?
When I first did it I was drawing pictures of distributions and I
thought I needed non-overlap to assert that P(l|m)/P(s|m)=r and
P(l|m)+P(s|m)=1.  But now it seems that the last follows just from
the fact that the amount was either from the larger or the
smaller.  The ratio doesn't depend on the ranges not overlapping,
it just depends on the fact that the larger amount's distribution
must be a copy of the smaller amount's distribution stretched by a
factor of r.
But in order for the range of the larger amount to be double that of the
smaller amount, you need to assume the range of the smaller amount is
finite. If the range of the smaller amount is infinite, as in my P(x)=1/e^x
example, then it would no longer make sense to say that the range of the
larger amount is r times larger.

Also, what if the envelope-stuffer is only picking from a finite set of
numbers rather than a continuous range? For example, he might have a 1/3
chance of putting 100, 125 or 150 dollars in the first envelope, and then he
would double that amount for the second envelope. In this case, your
assumption no matter what m is, it is r-times more likely to fall in a
large-amount interval than in a small-amount interval wouldn't seem to be
valid, since there are only six possible values of m here (100, 125, 150,
200, 250, or 300) and three of them are in the smaller range.

Jesse

```

### RE: Observation selection effects

```

-Original Message-
From: Jesse Mazer [mailto:[EMAIL PROTECTED]
Sent: Tuesday, October 05, 2004 8:45 PM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: RE: Observation selection effects

Brent Meeker wrote:

-Original Message-
From: Jesse Mazer [mailto:[EMAIL PROTECTED]
Sent: Tuesday, October 05, 2004 6:33 PM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: RE: Observation selection effects

Brent Meeker wrote:

years), I think it works without the restrictive
assumption that
the range of distirbutions not overlap.  It's still
necessary that
P(l|m)+P(s|m)=1 and P(l|m)/P(s|m)=r, which is all that
is required
for the proof to go thru.

gotten the idea that you
were assuming a uniform probability of the
envelope-stuffer picking any
number between x1 and x2 for the first envelope, but I
see this isn't
necessary, so your proof is a lot more general than I
thought. Still not
completely general though, because the envelope-stuffer
can also use a
distribution which has no upper bound x2 on possible
amounts to put in the
first envelope, like the one I mentioned in my last post:

For example, he could use a
distribution that
gives him a 1/2 probability of putting between 0 and 1
dollars in one
envelope (assume the dollar amounts can take any
positive real value, and he
uses a flat probability distribution to pick a number
between 0 and 1), a
1/4 probability of putting in between 1 and 2 dollars, a
1/8 probability of
putting in between 2 and 3 dollars, and in general a
1/2^n probability of
putting in between n-1 and n dollars. This would insure
there was some
finite probability that *any* positive real number could
be found in either
envelope.

Likewise, he could also use the continuous probability
distribution P(x) =
1/e^x (whose integral from 0 to infinity is 1). And if
you want to restrict
the amounts in the envelope to positive integers, he
could use a
distribution which gives a 1/2^n probability of putting
exactly n dollars in
the first envelope.

Jesse

That doesn't matter if I can do without the no-overlap
assumption - which I think I can.  Do you see a flaw?

When I first did it I was drawing pictures of
distributions and I
thought I needed non-overlap to assert that P(l|m)/P(s|m)=r and
P(l|m)+P(s|m)=1.  But now it seems that the last
follows just from
the fact that the amount was either from the larger or the
smaller.  The ratio doesn't depend on the ranges not
overlapping,
it just depends on the fact that the larger amount's
distribution
must be a copy of the smaller amount's distribution
stretched by a
factor of r.

But in order for the range of the larger amount to be
double that of the
smaller amount, you need to assume the range of the
smaller amount is
finite.

I only had to assume it was finite to avoid overlap in the ranges.
Dropping that assumption the range of the lower amount can be
infinite.

If the range of the smaller amount is infinite,
as in my P(x)=1/e^x
example, then it would no longer make sense to say that
the range of the
larger amount is r times larger.

Sure it does; r*inf=inf.  P(s)=exp(-x) - P(l)=exp(-x/r)

Also, what if the envelope-stuffer is only picking from
a finite set of
numbers rather than a continuous range? For example, he
might have a 1/3
chance of putting 100, 125 or 150 dollars in the first
envelope, and then he
would double that amount for the second envelope. In
this case, your
assumption no matter what m is, it is r-times more
likely to fall in a
large-amount interval than in a small-amount interval
wouldn't seem to be
valid, since there are only six possible values of m
here (100, 125, 150,
200, 250, or 300) and three of them are in the smaller range.

Yes, that's true.  The proof depends on smooth, integrable
distributions.

Brent

```

### RE: Observation selection effects

```-Original Message-
From: Jesse Mazer [mailto:[EMAIL PROTECTED]
Sent: Tuesday, October 05, 2004 8:45 PM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: RE: Observation selection effects
If the range of the smaller amount is infinite,
as in my P(x)=1/e^x
example, then it would no longer make sense to say that
the range of the
larger amount is r times larger.
Sure it does; r*inf=inf.  P(s)=exp(-x) - P(l)=exp(-x/r)
But it would make just as much sense to say that the second range is 3r
times wider, since by the same logic 3r*inf=inf. In other words, this step
in your proof doesn't make sense:

In other words, the range of possible
amounts is such that the larger and smaller amount do not overlap.
Then, for any interval of the range (x,x+dx) for the smaller
amount with probability p, there is a corresponding interval (r*x,
r*x+r*dx) with probability p for the larger amount.  Since the
latter interval is longer by a factor of r
P(l|m)/P(s|m) = r ,
In other words, no matter what m is, it is r-times more likely to
fall in a large-amount interval than in a small-amount interval.
As for your statement that P(s)=exp(-x) - P(l)=exp(-x/r), that can't be
true. It doesn't make sense that the value of the second probability
distribution at x would be exp(-x/r), since the range of possible values for
the amount in that envelope is 0 to infinity, but the integral of exp(-x/r)
from 0 to infinity is not equal to 1, so that's not a valid probability
distribution.

Also, now that I think more about it I'm not even sure the step in your
proof I quoted above actually makes sense even in the case of a probability
distribution with finite range. What exactly does the equation
P(l|m)/P(s|m) = r mean, anyway? It can't mean that if I choose an envelope
at random, before I even open it I can say that the amount m inside is r
times more likely to have been picked from the larger distribution, since I
know there is a 50% chance I will pick the envelope whose amount was picked
from the larger distribution. Is it supposed to mean that if we let the
number of trials go to infinity and then look at the subset of trials where
the envelope I opened contained m dollars, it is r times more likely that
the envelope was picked from the larger distribution on any given trial?
This can't be true for every specific m--for example, if the smaller
distribution had a range of 0 to 100 and the larger had a range of 0 to 200,
if I set m=150, then in every single trial where I found 150 dollars in the
envelope it must have been selected from the larger  distribution. You could
do a weighted average over all possible values of m, like integral over all
possible values of m of P('I found m dollars in the envelope I
selected')*P('the envelope I selected had an amount taken from the smaller
distribution' | 'I found m dollars in the envelope I selected'), which you
could write as integral over m of P(m)*P(s|m), but I don't think it would
be true that the ratio integral over m of P(m)*P(l|m)/integral over m of
P(m)*P(s|m) would be equal to r, in fact I think both integrals would
always come out to 1/2 so the ratio would always be 1...and even if I'm
wrong, replacing P(l|m)/P(s|m) with this ratio of integrals would mess up

Jesse

```

### RE: [Fwd: RE: Observation selection effects]

```On Tue, 2004-10-05 at 19:31, Brent Meeker wrote:

I always forget to reply-to-all in this list.
So below goes my reply which went only to Hal Finney.

-Forwarded Message-
From: Eric Cavalcanti [EMAIL PROTECTED]

Think about if the odd number of players was exactly
one.  You're guaranteed
to have the Winning Flip before you switch.

No, you're guranteed NOT to be in the winning flip.

Then think about what would happen if the odd number
of players was three.
Then you have a 3/4 chance of having the Winning
Flip before you switch.
Only if the other two players' flips both disagree
with yours will you not
have the Winnning Flip, and there is only a 1/4
chance of that happening.

Exactly.

It is interesting to note that, even though you are
more likely to be in the Winning Flip, there is no
disadvantage in Switching. To understand that, we can
look at the N=3 case, and see that if I am in the
Winning Flip with someone else, then if I change I
will still be in the Winnig Flip with the other person.

As opposed to Stathis initial thought, even though the
Winning Flip is indeed as likely to be Heads as Tails,
each individual is more likely to be in the
Winning Flip as in the Losing Flip in any given run.

So that this would never make it into a Casino game,
because the house would lose money in the long run.

I think you've confused the definitions of winning flip and
losing flip.  The winning flip is the *minority at the time of
the flip*  For N=3 you can't be in the winning flip with someone
else at the time of the flip - but you can switch to it.

Yes, you're right.
Hal and I have confused the definitions. It is still
not a paradox, though. You are more likely to be
in the Losing Flip.

So that this could indeed be a Casino game.

Eric.

```

### Re: Observation selection effects

```Norman Samish:
The Flip-Flop game described by Stathis Papaioannou strikes me as a
version of the old Two-Envelope Paradox.
Assume an eccentric millionaire offers you your choice of either of two
sealed envelopes, A or B, both containing money.  One envelope contains
twice as much as the other.  After you choose an envelope you will have the
option of trading it for the other envelope.
Suppose you pick envelope A.  You open it and see that it contains \$100.
Now you have to decide if you will keep the \$100, or will you trade it for
whatever is in envelope B?
You might reason as follows: since one envelope has twice what the other
one
has, envelope B either has 200 dollars or 50 dollars, with equal
probability.  If you switch, you stand to either win \$100 or to lose \$50.
Since you stand to win more than you stand to lose, you should switch.

But just before you tell the eccentric millionaire that you would like to
switch, another thought might occur to you.  If you had picked envelope B,
you would have come to exactly the same conclusion.  So if the above
argument is valid, you should switch no matter which envelope you choose.
Therefore the argument for always switching is NOT valid - but I am unable,
at the moment, to tell you why!
Basically, I think the resolution of this paradox is that it's impossible to
pick a number randomly from 0 to infinity in such a way that every number is
equally likely to come up. Such an infinite flat probability distribution
positive integers randomly from a flat probability distribution, and then
looked at the first integer, then there would be a 100% chance the second
integer would be larger, since there are only a finite number of integers
smaller than or equal to the first one and an infinite number that are
larger.

For any logically possible probability distribution the millionaire uses, it
will be true that depending on what amount of money you find in the first
envelope, there won't always be an equal chance of finding double the amount
or half the amount in the other envelope. For example, if the millionaire
simply picks a random amount from 0 to one million to put in the first
envelope, and then flips a coin to decide whether to put half or double that
in the other envelope, then if the first envelope contains more than one
million there is a 100% chance the other envelope contains less than that.

http://jamaica.u.arizona.edu/~chalmers/papers/envelope.html
I don't think the solution to this paradox has any relation to the solution
to the flip-flop game, though. In the case of the flip-flop game, it may
help to assume that the players are all robots, and that each player can
assume that whatever decision it makes about whether to switch or not, there
is a 100% chance that all the other players will follow the same line of
reasoning and come to an identical decision. In this case, since the money
is awarded to the minority flip, it's clear that it's better to switch,
since if everyone switches more of them will win. This problem actually
reminds me more of Newcomb's paradox, described at
http://slate.msn.com/?id=2061419 , because it depends on whether you assume
your choice is absolutely independent of choices made by other minds or if
you should act as though the choice you make can cause another mind to
make a certain choice even if there is no actual interaction between you.

Jesse

```

### RE: Observation selection effects

```Hal Finney writes:
Stathis Papaioannou writes:
Here is another example which makes this point. You arrive before two
adjacent closed doors, A and B. You know that behind one door is a room
containing 1000 people, while behind the other door is a room containing
only 10 people, but you don't know which door is which. You toss a coin
to
decide which door you will open (heads=A, tails=B), and then enter into
the
corresponding room. The room is dark, so you don't know which room you
are
now in until you turn on the light. At the point just before the light
goes
on, do you have any reason to think you are more likely to be in one
room
rather than the other? By analogy with the Bostrom traffic lane example
you
could argue that, in the absence of any empirical data, you are much
more
likely to now be a member of the large population than the small
population.
However, this cannot be right, because you tossed a coin, and you are
thus
equally likely to find yourself in either room when the light goes on.

Again the problem is that you are not a typical member of the room unless
the mechanism you used to choose a room was the same as what everyone
else did.  And your description is not consistent with that.

This illustrates another problem with the lane-changing example, which
is that the described mechanism for choosing lanes (choose at random)
is not typical.  Most people don't flip a coin to choose the lane they
will drive in.
Yes, this is correct. The typical observer must be typical in the way he
makes the choice of room or lane. With the traffic example, given that there
are slower and faster lanes on most roads, even in the absence of road works
or accidents, this may mean that for whatever reason the typical driver on
that day is more likely to choose the slower lane on entering the road. If
this is so, then a winning strategy for getting to your destination faster
could be to pick the lane with the most immediate appeal, then reflect on
this (having participated in the present discussion) and choose a
_different_ lane. This is analogous to counter-cyclical investing in the
stock market, where you deliberately try to do the opposite of what the
typical investor does.

But there may be a problem with the above argument. Suppose everyone really
did flip a perfectly fair coin to decide which lane of traffic to enter. It
is then still very most that one lane would be more crowded than the other
at any given time, purely through chance. Now, every driver might reason,
everyone including me has flipped a coin to decide which lane to enter, so
there is nothing to be gained by changing lanes. However, most of the
drivers reasoning thus would, by chance, be in the more crowded lane, and
therefore most would in fact be better off changing lanes.

--Stathis Papaioannou
_
Protect yourself from junk e-mail:
http://microsoft.ninemsn.com.au/protectfromspam.aspx

```

### RE: Observation selection effects

```Eric Cavalcanti writes:
QUOTE-
And this is the case where this problem is most paradoxical.
We are very likely to have one of the lanes more crowded than
the other; most of the drivers reasoning would thus, by chance,
be in the more crowded lane, such that they would benefit from
changing lanes; even though, NO PARTICULAR DRIVER would benefit
from changing lanes, on average. No particular driver has basis
for infering in which lane he is. In this case you cannot reason
as a random sample from the population.
-ENDQUOTE
I find this paradox a little disturbing, on further reflection. You enter
the traffic by tossing a coin, so you are no more likely to end up in one
lane than the other, and you would not, on average, benefit from changing
lanes. Given that you are in every respect a typical driver, what applies to
you should apply to everyone else as well. This SHOULD be equivalent to
saying that if every driver decided to change lanes, on average no
particular driver would benefit - as Eric states. However, this is not so:
the majority of drivers WOULD benefit from changing. (The fact that nobody
would benefit if everyone changed does not resolve the paradox. We can
restrict the problem to the case where each driver individually changes, and
the paradox remains.) It seems that this problem is an assault on the
foundations of probability and statistics, and I would really like to see it
resolved.

Stathis Papaioannou
_
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```

### RE: Observation selection effects

```On Mon, 2004-10-04 at 10:42, Stathis Papaioannou wrote:
Eric Cavalcanti writes:

QUOTE-
And this is the case where this problem is most paradoxical.
We are very likely to have one of the lanes more crowded than
the other; most of the drivers reasoning would thus, by chance,
be in the more crowded lane, such that they would benefit from
changing lanes; even though, NO PARTICULAR DRIVER would benefit
from changing lanes, on average. No particular driver has basis
for infering in which lane he is. In this case you cannot reason
as a random sample from the population.
-ENDQUOTE

I find this paradox a little disturbing, on further reflection. You enter
the traffic by tossing a coin, so you are no more likely to end up in one
lane than the other, and you would not, on average, benefit from changing
lanes. Given that you are in every respect a typical driver, what applies to
you should apply to everyone else as well. This SHOULD be equivalent to
saying that if every driver decided to change lanes, on average no
particular driver would benefit - as Eric states. However, this is not so:
the majority of drivers WOULD benefit from changing. (The fact that nobody
would benefit if everyone changed does not resolve the paradox. We can
restrict the problem to the case where each driver individually changes, and
the paradox remains.) It seems that this problem is an assault on the
foundations of probability and statistics, and I would really like to see it
resolved.

I found the answer of why you should be more likely
to enter in the crowded lane in this case. The answer
came after I tried to think about an example for few
people (which turned out not to work as I thought it
would)

Suppose a coin is toss for N people, which enter one
of two rooms according to the result. Suppose first
N=3. Then it is more likely that I will be in
the crowded room, even though there was no particular
bias in each coin toss. But still, if I am given the
option to change, and if I am in the crowded room,
I'll probably still be in the crowded room after I
change!

Now as N grows large, it is still more likely that I
will be in the crowded room, only it is less so. I was
neglecting the effect that you make yourself when you
enter the room/lane.

When N is large and even, it is equally likely that the
lane I enter is slower/faster. But it may be that both
lanes have same numbers, so my entering will make that
lane be the slower, and that's where the effect comes
from.

If it is odd, and I enter the fast lane, it is possible
that they become equal. If I enter the slower lane, it
will become even slower.

A minute of thought shows that my changing lanes does not
affect the result, though, as much as changing rooms does
not make me more likely to be in the less crowded when
N=3.

Therefore it is not a good advice for people to change
lanes in this case, even though it is more likely that
they are in the slower lane!

Eric.

```

### RE: Observation selection effects

```
Eric Cavalcanti writes:
From another perspective, I have just arrived at the
road and there was no particular reason for me to
initially choose lane A or lane B, so that I could just
as well have started on the faster lane, and changing
would be undesirable. From this perspective, there
is no gain in changing lanes, on average.
Here is another example which makes this point. You arrive before two
adjacent closed doors, A and B. You know that behind one door is a room
containing 1000 people, while behind the other door is a room containing
only 10 people, but you don't know which door is which. You toss a coin to
decide which door you will open (heads=A, tails=B), and then enter into the
corresponding room. The room is dark, so you don't know which room you are
now in until you turn on the light. At the point just before the light goes
on, do you have any reason to think you are more likely to be in one room
rather than the other? By analogy with the Bostrom traffic lane example you
could argue that, in the absence of any empirical data, you are much more
likely to now be a member of the large population than the small population.
However, this cannot be right, because you tossed a coin, and you are thus
equally likely to find yourself in either room when the light goes on.

--Stathis Papaioannou
_
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```

### RE: Observation selection effects

```Stathis Papaioannou writes:
Here is another example which makes this point. You arrive before two
adjacent closed doors, A and B. You know that behind one door is a room
containing 1000 people, while behind the other door is a room containing
only 10 people, but you don't know which door is which. You toss a coin to
decide which door you will open (heads=A, tails=B), and then enter into the
corresponding room. The room is dark, so you don't know which room you are
now in until you turn on the light. At the point just before the light goes
on, do you have any reason to think you are more likely to be in one room
rather than the other? By analogy with the Bostrom traffic lane example you
could argue that, in the absence of any empirical data, you are much more
likely to now be a member of the large population than the small population.
However, this cannot be right, because you tossed a coin, and you are thus
equally likely to find yourself in either room when the light goes on.

Again the problem is that you are not a typical member of the room unless
the mechanism you used to choose a room was the same as what everyone
else did.  And your description is not consistent with that.

Suppose we modify it so that you are handed a biased coin, a coin which
will come up heads or tails with 99% vs 1% probability.  You know about
the bias but you don't know which way the bias is.  You flip the coin
and walk into the room.  Now, I think you will agree that you have a
good reason to expect that when you turn on the light, you will be in
the more crowded room.  You are now a typical member of the room so the
same considerations that make one room more crowded make it more likely
that you are in that room.

This illustrates another problem with the lane-changing example, which
is that the described mechanism for choosing lanes (choose at random)
is not typical.  Most people don't flip a coin to choose the lane they
will drive in.  Instead, they have an expectation of which lane they will
start in based on their long experience of driving in various conditions.
It's pretty hard to think of yourself as a typical driver given the wide
range of personality, age and experience among drivers on the road.

Hal Finney

```

### Re: Observation selection effects

```Eric Cavalcanti writes regarding
http://plus.maths.org/issue17/features/traffic/index.html:

I agree with the general conclusion:
when we randomly select a driver and ask her
whether she thinks the next lane is faster, more
often than not we will have selected a driver from
the lane which is in fact slower and more densely
packed.
...
From another perspective, I have just arrived at the
road and there was no particular reason for me to
initially choose lane A or lane B, so that I could just
as well have started on the faster lane, and changing
would be undesirable. From this perspective, there
is no gain in changing lanes, on average.

That's a good question.  One thing I would note is that if everyone
entering the road chose between the two lanes with equal probability,
and stayed in their lane, then neither lane would be more crowded
If everyone behaved like this, one lane wouldn't be faster than the other.

Extending the argument, suppose I drive for a couple
of miles, and get to another point where I want to decide
if I should change lanes. Since I had no reason to
change lanes a couple of miles ago, I still have no reason
to do so now. Unless, of course, I can clearly see that
the next lane is faster, but adding that assumption changes
the problem completely.

I think this is true as well, assuming you have not changed lanes yet.

Let's go on and suppose that you drive for a while and change lanes
occasionally based on how the traffic seems to be moving at that moment,
and that you are a typical driver in this regard.  Then your alarm clock
rings and you ask yourself, am I more likely to be in the more crowded
lane.  I think you will agree that in that case, the answer is yes.