There are two players. Player 1 has 2 dollars and Player 2 has 3 dollars.
The Game:
They toss a coin - If it is heads Player 1 gives 1 dollar to Player 2. - Otherwise Player 2 gives 1 dollar to Player 1.
H : Heads $$P\left ( H \right ) = \frac{1}{3}$$
I couldnt find a pattern for this question. Do you have any ideas? What is the probability that Player 1 is (ultimately) ruined?
Thanks in advance.
We assume that the question will be about the probability that Player 1 (or perhaps Player 2) is (ultimately) ruined. Player 1 and Player 2 are not nice names, so let us call them Xavier and Yolande, or more briefly $X$ and $Y$.
Let $a$ be the probability that $X$ will ultimately be ruined, given that he has $1$ dollars and $Y$ has $4$.
Let $b$ be the probability that $X$ will ultimately be ruined, given that he has $2$ dollars and $Y$ has $3$.
Let $c$ be the probability that $X$ will ultimately be ruined, given that he has $3$ dollars and $Y$ has $2$.
Let $d$ be the probability that $X$ will ultimately be ruined, given that he has $4$ dollars and $Y$ has $1$.
It is at least intuitively clear that the numbers $a,b,c,d$ exist, since the probability that the game continues forever is $0$.
We happen to want $b$, but it is useful to bring in $a$, $c$, and $d$.
If $X$ has $1$ dollar, then with probability $\frac{1}{3}$ she has to give $1$ dollar to $Y$ on the next toss, and he is ruined. With probability $\frac{2}{3}$ he wins a dollar, and then the probability he is ultimately ruined is $b$. Thus $$a=\frac{1}{3}+\frac{2}{3}b$$
Similarly, $$b=\frac{1}{3}a+\frac{2}{3}c$$ $$c=\frac{1}{3}b+\frac{2}{3}d$$ $$d=\frac{1}{3}c$$ We have four linear equations in four unknowns. Solve for $b$.
