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 would lead to paradoxical conclusions--for 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 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




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