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?

By "average" you mean "arithmetic mean"?

The probability of N tosses is 2^(-N). So the probability of winning 
2^(N-1) dollas is 2^(-N).

[This simple analysis ignores the fact that if, say, you tossed a coin 
100 times without getting any heads at all, almost any *real* human 
would question whether the coin is loaded...]

Just as the probability of individual scores have a probability 
distribution, so the average scores should have one. And, by the central 
limit theorum, I would suggest that these converge on being normally 
distributed for larger and larger numbers of turns.

But what would the mean and standard deviation be? I have no idea.

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