Probability of winning a game that ends after one player gets 3 wins in a row

Question

Question:

In a game with two people, there is a draw half of the time. The other half of the time, player A wins with 2/3 and player B wins with 1/3. The game ends once a player wins three games in a row. What is the probability A wins?

Approach:

I am thinking that this is somewhat similar to asking what is the probability of getting HHH before TTT. However, you need to compute the probability of each player winning in a round and not having a draw.

Let $P(A)$ probability A wins, let $P(B)$ be probability $B$ wins:

$$P(A) = \frac{1}{2}\frac{2}{3} = \frac{2}{6} \quad P(B) = \frac{1}{2}\frac{1}{3} = \frac{1}{6}$$

Let $E$ be the event that we get $AAA$ before $BBB$. However, I am not sure if we can use the below approach because I think we may violate the total law of probability. I am not sure how I can include a third draws? Do I add $P(\text{draw})P(E)$ to each of these? because if it is a draw, we restart...

$$P(E) = P(A)P(E|A_1) + P(B)P(E|B_1)$$

$$P(E|A_1) = P(A)P(E|A_1A_2) + P(B)P(E|A_1B_2) = P(A)P(E|A_1A_2) + P(B)P(E|B_1)$$

$$P(E|A_1A_2)= P(A)P(E|A_1A_2A_3) + P(B)P(E|A_1A_2B_3) = P(A) + P(B)P(E|B_1)$$

$$P(E|B_1) = P(A)P(E|B_1A_2) + P(B)P(E|B_1B_2) = P(A)P(E|A_1) + P(B)P(E|B_1B_2)$$

$$P(E|B_1B_2)= P(A)P(E|B_1B_2A_3) + P(B)P(E|B_1B_2B_3) = P(A)P(E|A_1) + 0$$

Substituting:

$$P(E|B_1) = P(A)P(E|A_1) + P(B)P(A)P(E|A_1)$$

$$P(E|A_1A_2)= P(A) + P(B)P(A)P(E|A_1) + P(B)P(B)P(A)P(E|A_1)$$

$$P(E|A_1) = P(A)P(A) + P(A)P(B)P(A)P(E|A_1) + P(A)P(B)P(B)P(A)P(E|A_1) + P(B)P(A)P(E|A_1) + P(B)P(B)P(A)P(E|A_1)$$ etc...

But this seems very messy ...

Answer 1

Note: I am assuming that draws kill the current winning streak. That is to say, if $A$ wins the first two games, draws the third, and then $A$ wins the fourth, $A$ is now just on a one game streak.

Proceed by states. Denote the current state of a game by $(X,n)$ Where $X\in \{A,B\}$ denotes the player currently on a winning streak and $n$ denotes the length of the winning streak. We let $\emptyset$ denote the starting state as well as any state in which the last round ended in a draw.

For a state $(X,n)$ we let $P_{X,n}$ denote the probability that $A$ will eventually win given that we are currently in that state.

We remark that \begin{align} P_{A,2} &= \frac 26\times 1+\frac 16\times P_{B,1}+\frac 12\times P_{\emptyset} \\ P_{A,1} &= \frac 26\times P_{A,2}+\frac 16\times P_{B,1}+\frac 12\times P_{\emptyset} \\ P_{B,2} &= \frac 26\times P_{A,1}+\frac 16\times 0+\frac 12\times P_{\emptyset} \\ P_{B,1} &= \frac 26\times P_{A,1}+\frac 16\times P_{B,2}+\frac 12\times P_{\emptyset} \\ P_{\emptyset} &= \frac 26\times P_{A,1}+\frac 16\times P_{B,1}+\frac 12\times P_{\emptyset} \end{align}

That gives us a solvable linear system. Barring arithmetic error, I get that $$\boxed{P_{\emptyset}=\frac {86}{99}}$$

here is the computation. There, $X=P_{A,2}, Y=P_{A,1}, Z=P_{B,2}, W=P_{B,1}, V=P_{\emptyset}$

Answer 2

Here’s another way to do this, using generating functions.

A match that $A$ wins has the form

$$ \left(B_0^2\left(A_1^2B_1^2\right)_0^\infty A_0^2D\right)_0^\infty B_0^2\left(A_1^2B_1^2\right)_0^\infty A_3^3\;, $$

where $A$, $B$ and $D$ denote a win for $A$, a win for $B$ and a draw, respectively, and $X_a^b$ means anything from $a$ to $b$ repetitions of $X$. That is, we optionally start with up to two wins by $B$; then we have any number of repetitions of first $A$ then $B$ winning one or two games; then $A$ optionally wins up to two games; then a draw; this entire sequence can be repeated any number of times (possibly none) and then the same once more except now ending in three wins by $A$.

With $x$, $y$, $z$ denoting the probabilities for $A$ winning, $B$ winning and a draw, respectively, and with $\sum_{k=0}^\infty q^k=(1-q)^{-1}$, this corresponds to the generating function

$$ \left(1-\left(1+y+y^2\right)\left(1-\left(x+x^2\right)\left(y+y^2\right)\right)^{-1}\left(1+x+x^2\right)z\right)^{-1}\\\cdot\left(1+y+y^2\right)\left(1-\left(x+x^2\right)\left(y+y^2\right)\right)^{-1}\cdot x^3 \\ =\frac{x^3}{\frac{1-\left(x+x^2\right)\left(y+y^2\right)}{1+y+y^2}-z\left(1+x+x^2\right)}\;. $$

Substituting the given probabilities $x=\frac13$, $y=\frac16$, $z=\frac12$ yields the probability

$$ \frac{\left(\frac13\right)^3}{\frac{1-\left(\frac13+\left(\frac13\right)^2\right)\left(\frac16+\left(\frac16\right)^2\right)}{1+\frac16+\left(\frac16\right)^2}-\frac12\left(1+\frac13+\left(\frac13\right)^2\right)}=\frac{86}{99}\approx87\% $$

for $A$ to win the match.