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I'm trying to solve a problem involving the payout from a probability game.
Here it is:
The game goes something like this:
pay a dollar
Flip a coin
If it's heads, you win 2$, otherwise you win 0$
If (some event which has chance P)
you are forced to play again
You must stop if your initial dollar is lost. (Essentially, you either move up, down, stop, or are forced to stop if you hit 0).
At first glance this looks like a fair game, and maybe it is, but what is the EXACT payout?
All I've tried to calculate so far is the payout if you always win (but stop) (not including cost of playing).
<br /> payout = \sum_{n=1}^\infty (\frac{1}{2})^n*p^{n-1}*(n+1)<br />
<br /> = \sum_{n=0}^\infty (\frac{1}{2})^{n+1}*p^n*(n+2)<br />
<br /> = \frac{1}{2} \sum_{n=0}^\infty (\frac{1}{2})^n*p^n*(n+2)<br />
<br /> = \frac{1}{p} \sum_{n=0}^\infty (\frac{p}{2})^{n+1}*(n+2)<br />
<br /> = \frac{1}{p} \frac{d}{dp}{\int \sum_{n=0}^\infty (\frac{p}{2})^{n+1}*(n+2) dp}<br />
<br /> = \frac{1}{2*p} \frac{d}{dp} \sum_{n=0}^\infty (\frac{p}{2})^{n+2}<br />
<br /> = \frac{1}{2*p} \frac{d}{dp} ( \frac{1}{1 - p/2} - 1 - \frac{p}{2} )<br />
<br /> = \frac{1}{2*p} \frac{1}{(1 - p/2)^2} * \frac{-1}{2} - \frac{1}{2}<br />
<br /> = \frac{1}{2*p} *( \frac{-1}{(2 - p)^2} - \frac{1}{2} )<br />
<br /> = \frac{-1}{2*p} * ( \frac{1}{(p - 2)^2} - \frac{1}{2} )<br />
I know this is wrong because it's negative for all p from 0 to 1 (and the game should be at least fair in this case)
Here it is:
The game goes something like this:
pay a dollar
Flip a coin
If it's heads, you win 2$, otherwise you win 0$
If (some event which has chance P)
you are forced to play again
You must stop if your initial dollar is lost. (Essentially, you either move up, down, stop, or are forced to stop if you hit 0).
At first glance this looks like a fair game, and maybe it is, but what is the EXACT payout?
All I've tried to calculate so far is the payout if you always win (but stop) (not including cost of playing).
<br /> payout = \sum_{n=1}^\infty (\frac{1}{2})^n*p^{n-1}*(n+1)<br />
<br /> = \sum_{n=0}^\infty (\frac{1}{2})^{n+1}*p^n*(n+2)<br />
<br /> = \frac{1}{2} \sum_{n=0}^\infty (\frac{1}{2})^n*p^n*(n+2)<br />
<br /> = \frac{1}{p} \sum_{n=0}^\infty (\frac{p}{2})^{n+1}*(n+2)<br />
<br /> = \frac{1}{p} \frac{d}{dp}{\int \sum_{n=0}^\infty (\frac{p}{2})^{n+1}*(n+2) dp}<br />
<br /> = \frac{1}{2*p} \frac{d}{dp} \sum_{n=0}^\infty (\frac{p}{2})^{n+2}<br />
<br /> = \frac{1}{2*p} \frac{d}{dp} ( \frac{1}{1 - p/2} - 1 - \frac{p}{2} )<br />
<br /> = \frac{1}{2*p} \frac{1}{(1 - p/2)^2} * \frac{-1}{2} - \frac{1}{2}<br />
<br /> = \frac{1}{2*p} *( \frac{-1}{(2 - p)^2} - \frac{1}{2} )<br />
<br /> = \frac{-1}{2*p} * ( \frac{1}{(p - 2)^2} - \frac{1}{2} )<br />
I know this is wrong because it's negative for all p from 0 to 1 (and the game should be at least fair in this case)
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