A game wins you \$1 with probability 0.6 and takes away \$1 with probability 0.4. The game stops when you have won \$4 or lost all your money.
Let $p_n$ be the probability that you acquire \$4 if you start the game with \$n, where $n = 1,2,3$. Using the conditional probability given the result of the first game, express $p_2$ in terms of $p_1$ and $p_3$.
I know that the conditional probability is equal to $$\mathsf P(A\mid B)=\frac{\mathsf P(A\bigcap B)}{\mathsf P(B)}$$ and the first time we will stop our game is equal to 0.4 and 0.6 to continue the game but I don't know how to get involve with this to $p_2$. And how to derive three equations with three unknowns $p_1,p_2,p_3$ and calculate the probabilities?
Well, $p_2$ is the probability that you ultimately win if you have $\$2$.
Let $S_2$ be the event of starting with $\$2$. Let $R_4$ be the event of reaching $\$4$. Then $p_2=\mathsf P(R_4\mid S_2)$
Condition on the next game you play.
Let $W_n$ be the event of winning the next game. $\mathsf P(W_n)=0.6$
Then:
$$\begin{align}\mathsf P(R_4\mid S_2)~=~&\mathsf P(W_n)\,\mathsf P(R_4\mid W_n,S_2)+\mathsf P(W_n^\complement)\,\mathsf P(R_4\mid W_n^\complement,S_2)\\[1ex] ~=~&\mathsf P(W_n)\,\bbox[cornsilk, 1pt,border:solid 1pt lemonchiffon]{\color{cornsilk}{\mathsf P(R_4\mid S_3)}?}+\mathsf P(W_n^\complement)\;\bbox[cornsilk, 1pt,border:solid 1pt lemonchiffon]{\color{cornsilk}{\mathsf P(R_4\mid S_1)}?}\\[1ex] p_2 ~=~&0.6\;\bbox[cornsilk, 1pt,border:solid 1pt lemonchiffon]{\color{cornsilk}{p_3\quad}?}+0.4\;\bbox[cornsilk, 1pt,border: lemonchiffon solid 1pt]{\color{cornsilk}{p_1\quad}?}\end{align}$$
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