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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?

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 $$beginalign
p_2&=pp_1+(1-p)p_3=pp_1\
p_1&=pp_0+(1-p)p_2=p+(1-p)pp_1endalign$$
Solving gives $$beginalign
p_1&=pover 1-p(1-p)\
p_2&=p^2over 1-p(1-p)endalign$$
Since player $2$ starts with $2$ dollars, we want the value of $p_2$ when $p=frac23$, which is $$4/9over7/9=boxedfrac47$$

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:
$$ P(p1; wins) = sum_k=0^inftybigg[bigg(frac23cdotfrac13bigg)^k bigg(frac23bigg)^2bigg] = frac97cdot frac49 = frac47 $$

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 $bigcup_n=1^infty T_n$.

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
$$
p_1 = 2/3cdot p_2 + 1/3cdot 0 = 2/3 p_2,quadmathrmand
$$
$$
p_2 = 2/3cdot1 + 1/3cdot p_1 = (2+p_1)/3.
$$
By substitution,
$$ p_1 = 2/3cdot (2+p_1)/3 $$
$$ p_1 = (4+ 2p_1)/9 $$
$$ 9 p_1 = 4+ 2 p_1 $$
$$ 7 p_1 = 4 $$
$$ p_1 = 4/7. $$