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Suppose A has \$2 and B has \$3. They play a game, each game gives the winner \$1 from other. A has a probability $\frac{3}{5}$ to win each game. They play this game until one of them is bankrupt. What's the probability that A wins finally?

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2 Answers 2

up vote 2 down vote accepted

Hint: denote $X_t$ as the cash amount of A, then $\left(\frac{1-3/5}{3/5}\right)^{X_t}$ is a martingale.

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Which use is the hint (first line)? –  Did Apr 2 '13 at 8:14
    
No use to this question I think, but it can be used to calculate the expected winning time. –  Chen Yang Apr 2 '13 at 10:17
    
Then the "Hint" caption is misleading. –  Did Apr 2 '13 at 15:15
    
OK, I will edit it. Thanks for the comment. –  Chen Yang Apr 3 '13 at 0:09
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How to do it depends on the tools you have. We use only basic techniques.

Let $p_1$ be the probability that A (ultimately) wins the game if she currently has $1$ dollar.

Let $p_2$ be the probability she ultimately wins if she has $2$ dollars. Define $p_3$ and $p_4$ analogously. We only want to know $p_2$, but the others will be helpful.

To introduce fancier language, $p_i$ is the probability she ultimately wins given that she is in State $i$.

We have $p_1=\frac{3}{5}p_2$. For with probability $\frac{2}{5}$ she will lose the game, and be broke. With probability $\frac{3}{5}$ she will win the game, will have $2$ dollars, and then her probability of ultimately winning is by definition $p_2$.

Similarly, $p_2=\frac{2}{5}p_1+\frac{3}{5}p_3$.

Write down analogous equations for $p_3$ and $p_4$, carefully.

You will end up with $4$ (easy) linear equations in $4$ unknowns. Solve for $p_2$.

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