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Consider a game where you flip a biased coin. You win a dollar if it lands on tails and you lose a dollar if it lands on heads. The probability that it lands on tails is $p$ where $0<p<\frac{1}{2}$.

You enter the game with $A$ dollars and leave when you have either reached $2A$ dollars, or if you are broke.

If we leave the game after $k$ turns, then it is equally likely that we are either leaving broke or with $2A$ dollars.

I am trying to prove this result formally. However, I am not sure where to begin with equations.

CURRENT PROGRESS: If $P(A)$ is the probability that we leave broke after entering with $A$ dollars, then

$$P(A)=pP(A+1)+(1-p)P(A-1)$$

And this recurrence is constructed by partitioning the event space based on our first toss of the coin. With a little more work, I have managed to prove that

$$P(A)=\frac{w^{A}}{w^{A}+1}$$

where $w=\frac{1-p}{p}$.

But from here I am not certain how I can proceed with the proof.

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  • $\begingroup$ If we leave the game after $k$ turns, then the probabilities to leave broke or with $2A$ dollars are in the proportions $$(1-p)^A:p^A$$ hence very far from being equally likely. $\endgroup$ – Did Aug 23 '16 at 8:41
  • $\begingroup$ Okay so it is then clear that this can only work if the coin is fair. The textbook had a previous part about computing probabilities if the coin was fair, but it was not made clear that this condition carried on to the next part too. $\endgroup$ – Trogdor Aug 23 '16 at 8:47
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This statement is not true for all $k$. Let $X_k$ denote the amount of money after $k$ games. For $k<A$ we have $P(X_k = 0) = P(X_k = 2A) = 0$. However, for example, for $k=A$ we have $P(X_A = 0) = p^A \neq (1-p)^A = P(X_A = 2A)$, since $p<1/2$.

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