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 probability to win at n:th round is (1/2)^n. If you win, you get
2^(n-1). So, the theoretical predicted win-per-go is
(1/2)^1*2^(0)+(1/2)^2*2^1+(1/2)^3*2^2.... witch is the same as 1/2 + 1/2
+ 1/2 + 1/2... => infinite.
So, if you are able to have an unlimited number of goes and the other
part of the game is able to pay you what ever it has to, it doesn't
matter how much you pay for the each round: you will eventually make profit.

At least that's how our maths book explained it to me like two months ago :)

-- Arttu Voutilainen

Post a reply to this message