# Re: Why you should do the unexpected bet in front of a QS experiment ?

```On 1/9/2013 11:52 AM, Quentin Anciaux wrote:
```

```

2013/1/9 meekerdb <meeke...@verizon.net <mailto:meeke...@verizon.net>>

On 1/9/2013 3:10 AM, Quentin Anciaux wrote:

Hi,

let us start with the proposed QS experiment by Tegmark, a QS machine
with a
99/100 chance of a *perfect* kill (so let's put aside HP failure or
whatever so
to have either the experimenter is killed with the given probabilities
or it is
not, no in between, so in 1/100 he is not killed and perfectly well,
99/100 he
is killed).

You are a witness of such experiment, and you're asked to make a bet on
the
experimenter surviving (or not).

So you bet 100\$, if you bet on the experimenter surviving, if he
survive, you'll
get 200\$, if he does not you'll lose your bet, likewise if you bet on
him die.

What you should do contrary to what seems reasonable, is to bet on the
experimenter will survive for the following reason:

If MWI is true:

1st Test: in 99/100 worlds you lose 100\$ (and the bet ends here, there
is no
experimenter left for a second round), in 1/100 worlds you win 200\$
2nd Test: well... you cannot play again in the 99/100 worlds where you
did lose
100\$, so you start already with 200\$ in your pocket for this 2nd test,
so you
should do the same, no here in 99/100 worlds, you did make a draw (you
put 100\$
in 1st test + 100\$ win on the 1st test - 100\$ you did lose now because
the
experimenter is dead), in 1/100 you win again 200\$, that make 300\$ in

>From the 3rd test on, you can only get richer, weither the
experimenter lives

In QM+collapse, if the guy luckily survive two tests, you win money...
you'll
only lose money if he is killed at the first test.

So contrary to what you may think, you should bet the experimenter
should live,
because in MWI, it is garanteed that you'll win money in a lot branches
after
only two succeeded test, and as in QM+collapse, only the 99/100 of the
first
test lose money, all the others either make no loss or win money.

Did you bother to calculate the expected value of playing this game?  It's
\$98/0.99
whether you bet on survival or death.  And since \$98/0.99<\$100 you had to
start
with, it's better not to play at all.

??

```
you only lose on first bet if the experimenter die, which in MWI happens in 99% of the worlds... so discounting that *first* and only bet where you lose, you win 100\$ every time till the experimenter die.
```
```
On 2nd bet, you win nothing if the experimenter die (100\$ (from first bet) +100\$ (from winning first bet)-100\$(from losing second bet).
```
```
At the third bet, you win 100\$ if the experimenter die... and 100\$ more every time you see the experimenter survive. Only on the first bet when the experimenter die you lose 100\$ (and in that case, there is no more bet possible as there is no more experimenter).
```

Let E=expected value of playing the game by always betting on survival

E = 0.99(-\$100) + 0.01(\$100 + E)

```
Solve for E ==> E=98/0.99 Let F=expected value of playing the game and always betting on death
```
F = 0.99(\$100) + 0.01(-\$100 + F)

solution is left as an exercise to the reader.

Brent

```
```
```
But after the second bet, all worlds following that 2nd bet if MWI is true, contains *only* winner witness.
```
Quentin

Brent

```
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