I would like to know how to find the answer to this probability problem.

Two players, $A$ and $B$, are playing an arbitrary game (no draw is possible). The winner is the player who wins two consecutive games. Player $A$ has $2/3$ chances of winning a single game and player $B$ $1/3$.

Example: Player $A$ loses the first game, but wins the two next games, so he wins the overall game.

What is the probability that Player $A$ wins the overall game?

  • $\begingroup$ Thank you TNM. My english isn't at his best so far :) $\endgroup$ – Provost Oct 21 '12 at 20:11

Let $E$ be the probability that $A$ wins the game finishes in an even number of games, and $O$ the probability that $A$ wins the game in a odd number of games. We have $\mathbb{P}(A \text{ wins})=E+O.$

Let us have a look at $E$. Denots $a$ a victory for $A$ and $b$ a victory for $B$ on each round.

If the game finishes in $2$ games, then it is $aa$.

If the game finishes in $4$ games, then it is $abaa$.

If the game finishes in $6$ games, then it is $ababaa$.

Now we begin to see a pattern. The probability of winning in $2k$ matches, is $(2/3)^{k+1}(1/3)^{k-1}$ This gives $$ E=\sum_{k=1}^{\infty} (2/3)^{k+1}(1/3)^{k-1}=\frac{4}{7} $$ I leave the odd part to you, but you should find $O=\frac{4}{21}$ for a total of $\frac{16}{21}$

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Let $S_A$ denote the probability of $A$ winning, given he has won the previous game (but not the game before that; he has not yet won). Let $S_B$ be the probability of $A$ winning, given $B$ won the previous game (but also $B$ has not yet been declared winner). Then we get a set of recurrence relations as

$$S_A = \frac{2}{3} \cdot 1 + \frac{1}{3} \cdot S_B,$$


$$S_B = \frac{1}{3} \cdot 0 + \frac{2}{3} \cdot S_A.$$

(Can you explain the above relations?)

These are two linear equations in two unknowns, so we can solve them for $S_A$ and $S_B$. Finally, the answer to the question is given by considering the possible outcomes of the first single game, and adding up the probabilities:

$$S = \frac{2}{3} \cdot S_A + \frac{1}{3} \cdot S_B.$$

Solving the first set of equations in $S_A$ and $S_B$ results in $S_A = \frac{18}{21}$ and $S_B = \frac{12}{21}$. The last equation thus gives us the solution $\boxed{S = \frac{16}{21}}$.

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  • 1
    $\begingroup$ I found $S_A = 18/21$ and $S_B = 12/21$ which gives me $S=16/21$. $\endgroup$ – Provost Oct 21 '12 at 20:36

Let $N$ denote the number of games needed to find an overall winner. Now the probability of $A$ winning can be written as an infinite sum, as

$$P(A \text{ wins}) = \sum_{N = 2}^{\infty} P(A \text{ wins after exactly } N \text{ games}).$$

Can you calculate these separate probabilities, and compute the sum?

(Hint: What is the sequence of outcomes of the single games if player $A$ wins after exactly $3$ games? Or $4$ games? Or $N$ games?)

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  • $\begingroup$ For 3 games, the probability that I found was (2/3)^3, so on for N games it would be (2/3)^N ? $\endgroup$ – Provost Oct 21 '12 at 20:21
  • $\begingroup$ What is the sequence of outcomes of the 3 games if player $A$ wins after exactly 3 games? ($A$ can't have won three times, because then he would have won after 2 games already!) $\endgroup$ – TMM Oct 21 '12 at 20:22
  • $\begingroup$ It would be 1 L + 2 W $\endgroup$ – Provost Oct 21 '12 at 20:23
  • $\begingroup$ But what is the probability of $A$ losing the first game and then winning the next two games? $\endgroup$ – TMM Oct 21 '12 at 20:30
  • $\begingroup$ (1/3)*(2/3)*(2/3) + (1/3)*(2/3)*(2/3) = (2/3)^3 $\endgroup$ – Provost Oct 21 '12 at 20:32

This problem is actually a special case of a more general version. Consider $p$ as the chance for $A$ to win a single game. We can compute the probability that $A$ first wins $n$ games. Suppose that they keep playing even if one player wins $n$ points first, until the total number of games achieves $2n-1$. Then

{$A$ first wins $n$ points} = {$A$ wins $k$ points in the first $2n-1$ games, for $k = n,\cdots,2n-1$}.

The probability of the right hand side can be easily computed since $k$ has a binomial distribution, thus

$$P(LHS) = P(RHS) = \sum_{k = n}^{2n-1}C_{2n-1}^k p^k(1-p)^{2n-1-k},$$ where $C_b^a$ is the binomial coefficient, and is sometimes denoted by $_b C_a$ or $(^b_a)$.

You can see a more general problem and my answer here: Probability of winning first n games.

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  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$ – Saad Feb 11 '18 at 3:34
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    $\begingroup$ Included now. @Alex $\endgroup$ – Wanshan Feb 11 '18 at 5:10

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