# Probability of winning a game

$A$ and $B$ play a game.

• The probability of $A$ winning is $0.55$.
• The probability of $B$ winning is $0.35$.
• The probability of a tie is $0.10$.

The winner of the game is the person who first wins two rounds. What is the probability that $A$ wins?

The answer is $0.66$. I don't know how it's coming $0.66$.

EDIT :

The right combinations according to me are

{null,T,TT,TTT....}A{null,T,TT,TTT....}A

{null,T,TT,TTT....}A{null,T,TT,TTT....}B{null,T,TT,TTT....}A

{null,T,TT,TTT....}B{null,T,TT,TTT....}A{null,T,TT,TTT....}A

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I edited it to show you – Rajarshi Sarkar Apr 30 '13 at 15:19

Ties don't count, don't record them. So in effect we are playing a game in which A has probability $p=\frac{0.55}{0.90}$ of winning a game, and B has probability $1-p$ of winning a game.

Now there are several ways to finish. The least thinking one is that A wins with the pattern AA, or the patterns ABA, or BAA.

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They will not play exactly $3$ games in which there is a winner. If A wins the first two non-tie games, the series is over. If A loses one of the first two non-tie games, she wins the series precisely if she wins the third non-tie game. Hence the $3$ cases. Alternately, we can take the approach taken by Ross Millikan, unfortunately at this time deleted, in which we continue the series until there is a total of $3$ wins. Then A wins the original series precisely if she wins $2$ or $3$ games in the modified series. – André Nicolas Apr 30 '13 at 15:29
SO, it should be (p)^2 + 2[(1-p)*(p)^2] = .6645 – Rajarshi Sarkar Apr 30 '13 at 15:36
Got the concept , Thanks :) – Rajarshi Sarkar Apr 30 '13 at 15:37

During the match, we swicth between three states: $t=$"tie", $a=$"$A$ leads", $b=$"$B$ leads". Let $p_t, p_a, p_b$ be the probabilizty that $A$ wins if the starting position is as indicated.

If we start in $t$, then we continue with $a$, $b$, or $t$ according to the given probabilities afor a single game, that is $$p_t = .55p_a+.35p_b+.10 p_t$$ If we start in $a$, we have essentially the same, except that a won game decides the match. Thus $$p_a = .55\cdot 1+.35p_t+.10 p_a$$ Finally, $$p_b = .55p_t+.35\cdot 0+.10 p_b$$ and this gives us three equations in three unknowns. The number we are looking for is $p_t$ (as we actually start in a tie situation).

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