Play until you win (or lose)

Probability Level 2

A A and B B agree to play a match that consists of a series of games you either win or lose. They go on playing until one of the players wins two games in a row. Suppose player A A has a probability p p of winning each game.

What are the chances of player A A to win the match?

p p p 2 1 − p 2 \dfrac{p^2}{1-p^2} p 2 1 − p + p 2 \dfrac{p^2}{1-p+p^2} p 2 ( 2 − p ) 1 − p + p 2 \dfrac{p^2(2-p)}{1-p+p^2} ( 1 − p ) 2 ( 1 + p ) 1 − p + p 2 \dfrac{(1-p)^2(1+p)}{1-p+p^2} p ( 1 − p ) 1 − p ( 1 − p ) \dfrac{p(1-p)}{1-p(1-p)}

Gabriel Chacón by Gabriel Chacón , 55, Spain

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2 solutions

Chris Lewis
Apr 3, 2019

An alternative approach without summing the series is as follows.

Let X Y X_Y denote the probability that player X X wins overall given that player Y Y won the first game.

We have A A = P ( A A ) + P ( A B A A ) + P ( A B A B A A ) + ⋯ A_A=P(AA)+P(ABAA)+P(ABABAA)+\cdots

and A B = P ( B A A ) + P ( B A B A A ) + P ( B A B A B A A ) + ⋯ = ( 1 − p ) A A A_B=P(BAA)+P(BABAA)+P(BABABAA)+\cdots = (1-p)A_A

Similarly B A = p B B B_A=pB_B

Also, A A + B A = p A_A+B_A=p and A B + B B = 1 − p A_B+B_B=1-p . We now have four equations in four unknowns; solving gives

A A = p 2 1 − p + p 2 A_A=\frac{p^2}{1-p+p^2}

A B = ( 1 − p ) p 2 1 − p + p 2 A_B=\frac{(1-p)p^2}{1-p+p^2}

and so the probability that A A wins overall is A A + A B = p 2 ( 2 − p ) 1 − p + p 2 A_A+A_B=\boxed{\frac{p^2 (2-p)}{1-p+p^2}} .

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Gabriel Chacón
Apr 2, 2019

Consider the sequences of games that yield A A as the winner of the match (each letter denotes the winner of each game):

  • A A , A B A A , A B A B A A , A B A B A B A A , … AA, ABAA, ABABAA, ABABABAA, \ldots , with probability P 1 = p 2 [ 1 + p ( 1 − p ) + p 2 ( 1 − p ) 2 + … ] = p 2 1 − p ( 1 − p ) P_1=p^2\left[1+p(1-p)+p^2(1-p)^2+ \ldots \right]=\dfrac{p^2}{1-p(1-p)}
  • B A A , B A B A A , B A B A B A A , B A B A B A B A A , … BAA, BABAA, BABABAA, BABABABAA, \ldots , with probability P 2 = ( 1 − p ) p 2 [ 1 + ( 1 − p ) p + ( 1 − p ) 2 p 2 + … ] = ( 1 − p ) p 2 1 − p ( 1 − p ) P_2=(1-p)p^2 \left[1+(1-p)p+(1-p)^2p^2+ \ldots \right]=\dfrac{(1-p)p^2}{1-p(1-p)}

Therefore, P A = P 1 + P 2 = p 2 1 − p ( 1 − p ) + ( 1 − p ) p 2 1 − p ( 1 − p ) = p 2 ( 2 − p ) 1 − p + p 2 P_A=P_1+P_2=\dfrac{p^2}{1-p(1-p)}+\dfrac{(1-p)p^2}{1-p(1-p)}=\boxed{\dfrac{p^2(2-p)}{1-p+p^2}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Perhaps this is another problem, but any thoughts on how to extend this to the case where a player needs to win n n games in a row? Even the case n = 3 n=3 is significantly more complicated than n = 2 n=2 .

Chris Lewis - 2 years, 2 months ago

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I was suspecting the general case was solved using some kind of recurrence but I was not able to work out a solution myself, so I looked it up. I found this answer to an equivalent problem: https://math.stackexchange.com/questions/201531/whats-the-probability-of-tossing-r-heads-in-a-row-before-s-tails-in-a-row

Gabriel Chacón - 2 years, 2 months ago

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