Interesting probability problem

[Alkatran]

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).

$$payout = \sum_{n=1}^\infty (\frac{1}{2})^n*p^{n-1}*(n+1)$$
$$= \sum_{n=0}^\infty (\frac{1}{2})^{n+1}*p^n*(n+2)$$
$$= \frac{1}{2} \sum_{n=0}^\infty (\frac{1}{2})^n*p^n*(n+2)$$
$$= \frac{1}{p} \sum_{n=0}^\infty (\frac{p}{2})^{n+1}*(n+2)$$
$$= \frac{1}{p} \frac{d}{dp}{\int \sum_{n=0}^\infty (\frac{p}{2})^{n+1}*(n+2) dp}$$
$$= \frac{1}{2*p} \frac{d}{dp} \sum_{n=0}^\infty (\frac{p}{2})^{n+2}$$
$$= \frac{1}{2*p} \frac{d}{dp} ( \frac{1}{1 - p/2} - 1 - \frac{p}{2} )$$
$$= \frac{1}{2*p} \frac{1}{(1 - p/2)^2} * \frac{-1}{2} - \frac{1}{2}$$
$$= \frac{1}{2*p} *( \frac{-1}{(2 - p)^2} - \frac{1}{2} )$$
$$= \frac{-1}{2*p} * ( \frac{1}{(p - 2)^2} - \frac{1}{2} )$$

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)

 

[CRGreathouse]

Payoff for an n-round game:

$$P_1=0/2+2/2-1=0$$
$$P_2=0/2+0/4+4/4-1=0$$
$$P_3=0/2+0/4+2/8+6/8-1=0$$
$$P_3=0/2+0/4+0/16+4/16+4/16+8/16-1=0$$
. . .

You can then sum these

$$\sum_{i=1}^\infty(1-p)^{n-1}pP_n$$

 

[Alkatran]

Payoff for an n-round game:

$$P_1=0/2+2/2-1=0$$
$$P_2=0/2+0/4+4/4-1=0$$
$$P_3=0/2+0/4+2/8+6/8-1=0$$
$$P_3=0/2+0/4+0/16+4/16+4/16+8/16-1=0$$
. . .

You can then sum these

$$\sum_{i=1}^\infty(1-p)^{n-1}pP_n$$

I think you have the wrong game in mind. You can only win or lose a dollar in this game, not double or nothing. I already tried the double-or-nothing and found you broke even.

 

[davidseed]

are you sure you can differentiate w.r.t. p

this equation looks like the expectation of a geometric series which I posted at
https://www.physicsforums.com/showthread.php?t=70705

 

[Alkatran]

I'm not differentiating with respect to p: I'm integrating AND differentiating with respect to p, effectively doing nothing. We did this in calculus to solve series with factors of n+x in them.

 

[Hurkyl]

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)

It's not negative for all p from 0 to 1. (For example, p = 1/2)

Did you notice your initial sum only converges for -1/2 <= p < 1/2? Frankly, I'd be surprised if your answer wasn't negative for p = 1.

(It disturbs me that it's positive for p=1/2 and undefined for p = 0, though)

I'm not yet convinced your initial sum is right though...

 

[Hurkyl]

I see two mistakes in your derivation:

(1) You got the derivative of 1/(1-p/2) wrong.
(2) When you factored out the minus sign, you only pulled it from one of the terms.

(I'm not promising there aren't more!)

 

[CRGreathouse]

I think you have the wrong game in mind. You can only win or lose a dollar in this game, not double or nothing. I already tried the double-or-nothing and found you broke even.

I'm not calculating for double or nothing -- look at my numbers again. The possible payoffs for the game I'm calculating are -1, 1, 3, 5, ...

Can you give an example of play? You said "Essentially, you either move up, down, stop, or are forced to stop if you hit 0", but this was a little confusing.

 

[Hurkyl]

Oh, I know why I wasn't convinced your sum was right -- I didn't notice you were only modelling the case of always winning!