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scott wrote:
> Here's an interesting puzzle.
>
> A game is played by repeatedly tossing a coin until it lands heads. If
> it lands heads on the first try, you win $1 and the game is over. If it
> lands heads on the 2nd try, you win $2 and the game is over. On the 3rd
> try $4, and in general if you get the head on the nth try, you win
> $2^(n-1).
>
> I simulated this game in C++ and after 1e6 goes the average win-per-go
> settles down quite nicely to $8.09 and stays there up to 1e8 goes. Is
> that correct? How much should you be willing to pay for each go? Does
> it depend on how many goes you are going to have?
The expected payoff is infinite of course. Personally, I'd pay about
$10 to play the game, which indicates that I'm not attempting to
maximize my expected payoff in making this decision. The best I can
formalize my reasoning is that the `utility' to me of winning a sum of
money levels off asymptotically as that sum of money decreases. For
example, as far as I'm concerned there's no difference between winning
$10^20 and $10^30, despite the fact that the latter is a vastly larger sum.
Furthermore, since I have only a very limited amount of money, losing $x
has a higher negative utility than gaining $x -- as indicated by the
fact that I wouldn't play a game with zero expected payoff (unless the
game itself was fun of course).
To properly estimate how much I should pay to play under this view I
should make a guess at my payoff vs. utility curve and then calculate an
expected utility, but that's a bit more effort than I's probably care to
spend were I presented with such a situation, so I'll take the
satisficing approach and just say $10.
Also, I assume that you're familiar with the name of this puzzle, but on
the off chance that you're not:
http://en.wikipedia.org/wiki/St._Petersburg_paradox
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