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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