Probability of going broke I have a biased coin that comes heads with probability $2/3$ and tails with probability $1/3$.
I start with $\$1$, and you start with $\$2$. If it comes up heads, you lose $\$1$ and I gain $\$1$. If it comes up tails, I lose $\$1$ and you gain $\$1$. 
What's the probability that you lose all your money? 
I tried to make a recurrence relation between $p$ but I got nowhere. Can someone please help me?
 A: Hint
If $H$ is head and $T$ is the event tail, I denote $T_1=HH$, $T_2=HTHH$, $T_3=HTHTHH$, $T_4=HTHTHTHH$...
Then the event "you loose first" is $\bigcup_{n=1}^\infty T_n$.
A: Let $p_k$ be the probability that player $2$ goes broke, if he currently has $k$ dollars, for $k=0,1,2,3$.  Then $p_0=1$ and $p_3=0$.  We have $$\begin{align}
p_2&=pp_1+(1-p)p_3=pp_1\\
p_1&=pp_0+(1-p)p_2=p+(1-p)pp_1\end{align}$$
Solving gives $$\begin{align}
p_1&={p\over 1-p(1-p)}\\
p_2&={p^2\over 1-p(1-p)}\end{align}$$
Since player $2$ starts with $2$ dollars, we want the value of $p_2$ when $p=\frac23$, which is $${4/9\over7/9}=\boxed{\frac47}$$
A: Your problem looks not well defined: does the game end when one of the two players goes broke? In that case, you are asking for the probability that player 1 will win.
The probability of the event "Player 2 will lose all of his money before Player 1 does" is the sum of the probabilities of the 'paths' (sorry for the sloppy notation!):


*

*$a_0$ = (1,2) --> (2,1) --> (3,0)

*$a_1$ = (1,2) --> (2,1) --> (1,2) --> (2,1) --> (3,0)

*...

*$a_k$ = $[$(1,2) --> (2,1) --> $]^k$ --> (1,2) --> (2,1) --> (3,0)


Now, this translates to:
$$ P(p1\; wins) = \sum_{k=0}^\infty\bigg[\bigg(\frac{2}{3}\cdot\frac{1}{3}\bigg)^k \bigg(\frac{2}{3}\bigg)^2\bigg] = \frac{9}{7}\cdot \frac{4}{9} = \frac{4}{7} $$ 
A: Assume that player 1 starts with \$1, player 2 starts with \$2, and the game ends if either player reaches 0.  
Let $p_1$ be the probability that player 1 will win given that he has \$1.
Let $p_2$ be the probability that player 1 will win given that he has \$2.
Then 
$$
p_1 = 2/3\cdot p_2 + 1/3\cdot 0 =  2/3 p_2,\quad\mathrm{and}
$$
$$
p_2 = 2/3\cdot1 + 1/3\cdot p_1 =  (2+p_1)/3.
$$ 
By substitution,
$$ p_1 = 2/3\cdot (2+p_1)/3 $$
$$ p_1 =  (4+ 2p_1)/9 $$
$$ 9 p_1 = 4+ 2 p_1 $$
$$ 7 p_1 = 4  $$
$$  p_1 = 4/7.  $$
