"scott" <sco### [at] scottcom> wrote in message news:4965e43e@news.povray.org...

> 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?

No. More properly, it's useless. You won't be able to solve this empirically
as such. Most influential outcomes on payoff are least likely, which you
will not ever come across in finite time.

> How much should you be willing to pay for each go?

Infinite. Sum(n,0,inf)( (1/2)^n * 2^(n-1) ) = sum(n,0,inf)(1/2) = inf.

Of course it only makes sense to play if both you and the casino have
infinite time and infinite money.

> Does it depend on how many goes you are going to have?

No, each round of game is independent, is it not?

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