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If I offered you the chance to play
this lottery game, just one time, would you take it?
For $1,000, you may have a 1/1,000
chance to win $10,000,000.
Your expected winnings are +$9,000! What
have you got to lose?
Oh, right: $1000. Indeed, you have a
99.9% probability of losing exactly that.
An answer often given is that the
marginal utility of wealth is not constant, that $1,000,000 isn't
really worth 1,000 times as much as $1,000. I fail to see how,
really, but even if that's true, it doesn't solve the problem.
For, on the other hand, suppose I
offered you the chance to play that game 10,000 times? If you really
had the $10,000,000 to play the game 10,000 times (or say I offered
you credit on which to do this), you'd be crazy not to! You really
would make $90,000,000 this way, with very high probability. Your
expected winnings would be your actual winnings.
If we assign a marginal utility
function so that you won't play the first game, this means that we
have U(x) such that (1/1000)*U($10,000,000) < (1)*U($1,000). But
since this is the same utility value at each play of the second game,
then you shouldn't play the second game either!
The problem is clearly expected
utility itself. For games that you only play once, you can't use
expectation values! Expectation values only make sense when you can
play many times.
Instead, I propose the principle of
most probable outcome of strategy.
If you play a long
sequence of games, this works out the same: On a large number of
plays, the most probable outcome of your strategy will be in fact the
expectation value of that strategy. (This is why you should play the
second game.)
But on fewer
plays, it is often quite different: In the first game, for instance,
the most probable outcome is clearly that you'll lose $1,000.
Moreover, this
also provides a continuous progression of intermediate states, and
there is a point at which you should just barely play: When your
probability of winning more than you lose goes above 50%.
In the games above,
your probability of losing it all on n plays in a row is
(999/1000)^n; thus, your probability of winning at least once
in n plays is 1-(999/1000)^n. Your probability of
winning more than you lose is the same as the probability that
(wins/10,000) > (losses); this is a little sticky to calculate
exactly, but as long as we play fewer than about 5,000 times, it
works out very close to the probability of winning at least
once—since one win more than covers 5,000 losses, and the
probability of winning twice in only 5,000 times is negligible.
Thus, we can say,
to a good approximation:
1-(999/1000)^n >
0.50
0.999^n > 0.50
n*ln(0.999) >
ln(0.50)
n >
ln(0.50)/ln(0.999)
n >= 693
Thus, if you can
play at least 693 times, you should play. If you can't, you
shouldn't. The expected utility is always the same—but the most
probable outcome is what you should actually be using.
Hence, I propose
that we abandon expected-utility calculations in favor of the most
probable outcome, since clearly the latter much better fits how
rational people really behave.
(This
is also why you should never play a Martingale strategy in real life:
If your win probability is above 50% on each round—e.g: you are
counting cards—you may as well bet normally. If not, a Martingale
won't help you: No matter how much money you have to lose, your most
probable outcome is still losing more than you win.)
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