Gambler's ruin problem but with many extra conditions

Question

$A$ and $B$ play a series of games. They stop playing the game when the total number of wins of player $A$ is four greater than that of the player $B$ or when the total number of wins of player $B$ is six greater than that of the player $A$. When the game is over, player with greater number of total wins is declared the winner of the series. Find the probability that $A$ is the winner of the series if

(i) $A$ and $B$ are likely to win each game.

Since $A$ needs less games to win than $B$, the probability that $A$ is the winner will not be $0.5$ right? unlike in problems where both players need the same amount to win. How do I find this probability then?

(ii) Each game is independently won by $A$ with probability $0.47$ and by $B$ with probability $0.53$

Normally, the formula would be $\frac {(1-q/p)^i}{ (1- q/p)^N}$, where $q$ = probability that $B$ wins ($0.53$), $p$ = probability that $A$ wins ($0.47$), $i$ = number of games $A$ takes to win, and $N$ = total number of games. The part I'm stuck is deciding what is $N$. Since the $4$ more wins $A$ gets can be for any number of total games starting from $4$ games.

(iii) In each independent game, $A$ wins with probability $0.54$, $B$ wins with probability $0.36$, and there is no winner with probability $0.1$

I'm stuck completely for part 3.

Answer 1

First notice that we can describe the state in terms of just the difference between A's wins and B's wins. If A has 3 wins and B has 2 wins, this is just as far from A winning 4 more than B or B winning 6 more than A as a scenario where A has 1 win and B has no wins. So I'll say $D = (\mathrm{times\ A\ won} - \mathrm{times\ B\ won})$. Next define $P(d)$ to be the probability $A$ will win the series if $D=d$. Then we know $P(4)=1$ and $P(-6)=0$, and we want to find $P(0)$.

For problem (i), when $-6 < d < 4$, the next game has probability $\frac{1}{2}$ of transitioning from $D=d$ to $D=d-1$, or probability $\frac{1}{2}$ of transitioning from $D=d$ to $D+1$. Therefore

$$ P(d) = \frac{1}{2} P(d-1) + \frac{1}{2} P(d+1) $$

and solving for $P(d+1)$,

$$ P(d+1) = 2 P(d) - P(d-1) $$

Knowing $P(-6) = 0$, we can work our way up the system of equations to write each $P$ value in terms of $P(-5)$:

$$ \begin{align*} P(-4) &= 2 P(-5) \\ P(-3) &= 3 P(-5) \\ P(-2) &= 4 P(-5) \\ P(-1) &= 5 P(-5) \\ P(0) &= 6 P(-5) \\ & ... \\ 1 = P(4) &= 10 P(-5) \end{align*} $$

So $P(-5) = \frac{1}{10}$, and the answer to problem (i) is $P(0) = \frac{6}{10} = \frac{3}{5}$.

In problem (i) it might be easy to notice $P$ must be linear, but this step-by-step solving is more general and will work the same for problem (ii) where $P$ will not be linear.

For problem (iii), there's a chance $D$ doesn't change at all after one game, so the transition equation can be written giving $P(d)$ in terms of $P(d-1)$, $P(d)$, and $P(d+1)$. Or, you could start out by an argument to ignore the games without a winner: at each point, the probability A will win the next game that has a winner is $0.54 / (0.54+0.36) = 3/5$, and the probability B will win the next game that has a winner is $0.36 / (0.54+0.36) = 2/5$.

Answer 2

For problem (i), you already know that by linearity, P(A) wins $= \frac6{6+4} = \frac35$

For problem (ii), treating it as a random straight line walk where $A$ loses is at $0,\;A$ is currently at $6$, and needs to take $4$ steps to the right to reach $10$ to win, let us derive a formula

$L = $ left bound, $R=$ right bound, $n =$ steps away from loss for $A$,
$p = Pr$ (step to right), $q = Pr$ (step to left), $k = q/p,$

Then $P(n) = a + bk^n$ is the general solution,
with $P(L) = 0, P(R) = 1$ as the boundary conditions.

$\displaylines{a+bk^L = 0, a + bk^R = 1\\bk^R - bk^L = 1,\;\;b = \dfrac{1}{k^R-k^L}\\a +bk^n = bk^n -bk^L = \boxed{\dfrac{k^n-k^L}{k^R-k^L}}}$

Just plug in $k = \frac{0.47}{0.53}, L =0, n = 6, R = 10$ into the boxed formula to get $P(A \;wins) \approx 0.7428$

And for part (iii), just change $k$ to $\frac{0.36}{0.54}$