2 players, A and B, play a series of games

Question

Question:-

2 players, A and B, play a series of games where player A has a 60% chance of winning each time. The winner of the series is the first player to win 3 games in a row. What is the probability of player A winning the series? Further question: 2 players, A and B, play a series of games where player A has a 60% chance of winning each time. The winner of the series is the first player to win 2 more games than their opponent. What is the probability of player A winning the series?

MyApproach:-

I've tried so many ways to solve this problem but it does not seem to resemble any other "probability game" I've encountered. So here are my thoughts:

Suppose there are n games. The result of last 4 games is determined: BAAA (representing winner). Starting from there, we fill in A or B back without placing three in a row.

I am stuck here and tried solving the problem for when n=4,5,6..... But it does not seem to work out, where I find no pattern.

For the second part, I got: Suppose there are n games, so A wins n/2+1 and B wins n/2-1. Therefore the probability is 0.6^(n/2+1) x 0.4^(n/2-1).

Answer 1

For the contest where winning the series requires winning $3$ consecutive games . . .

For $-2 \le n \le 2$, define state $n$ as follows . . .

• If $n > 0$, player $A$ has accumulated $n$ consecutive (immediately prior) winning games.$\\[4pt]$
• If $n < 0$, player $B$ has accumulated $|n|$ consecutive (immediately prior) winning games.$\\[4pt]$
• If $n = 0$, no games have yet been played.

Let $p[n]$ be the probability that player $A$ wins the series, assuming the current state is $n$.

Then we have the following equations \begin{align*} p[-2]&=\bigl({\small{\frac{3}{5}}}\bigr)p[1]\\[4pt] p[-1]&=\bigl({\small{\frac{3}{5}}}\bigr)p[1]+\bigl({\small{\frac{2}{5}}}\bigr)p[-2]\\[4pt] p[0]&=\bigl({\small{\frac{3}{5}}}\bigr)p[1]+\bigl({\small{\frac{2}{5}}}\bigr)p[-1]\\[4pt] p[1]&=\bigl({\small{\frac{3}{5}}}\bigr)p[2]+\bigl({\small{\frac{2}{5}}}\bigr)p[-1]\\[4pt] p[2]&=\bigl({\small{\frac{3}{5}}}\bigr)+\bigl({\small{\frac{2}{5}}}\bigr)p[-1]\\[4pt] \end{align*} So $5$ linear equations in $5$ unknowns.

Solving the system yields $p[0]={\large{\frac{1053}{1445}}}$.

For the contest where winning the series requires having won $2$ games more than the opponent . . .

Define states $1,0,-1$ as follows . . .

• If $n =1$, player $A$ has won $1$ more game than player $B$.$\\[4pt]$
• If $n = 0$, the players have won the same number of games.
• If $n = -1$, player $B$ has won $1$ more game than player $A$.$\\[4pt]$

Let $p[n]$ be the probability that player $A$ wins the series, assuming the current state is $n$.

Then we have the following equations \begin{align*} p[-1]&=\bigl({\small{\frac{3}{5}}}\bigr)p[0]\\[4pt] p[0]&=\bigl({\small{\frac{3}{5}}}\bigr)p[1]+\bigl({\small{\frac{2}{5}}}\bigr)p[-1]\\[4pt] p[1]&=\bigl({\small{\frac{3}{5}}}\bigr)+\bigl({\small{\frac{2}{5}}}\bigr)p[0]]\\[4pt] \end{align*} So $3$ linear equations in $3$ unknowns.

Solving the system yields $p[0]={\large{\frac{9}{13}}}$.