# Probability of winning a 7 game series in 6 games

Team A and B are playing a best of 7 series, with the first team to win in 4 games winning the series. Team A has the probability $\dfrac{1}{2}$ of winning a game. If the series lasts 6 games, what is the probability that Team A wins?

I am confused with my solution, which doesn't feel right with my intuition which suggests Team A or Team B can win the series with a probability of $\dfrac{1}{2}$.

My solution,

Game 6 is the decider, implying that Game 6 went to Team A. Hence of the earlier 5 games 3 won by Team A and 2 by Team B.

This can be done in $\binom{5}{3} = 10$ ways

Probability of any 1 such sequence is, $(\dfrac{1}{2})^6 = \dfrac{1}{64}$

Hence Probability of winning in all 10 ways = $10(\dfrac{1}{64}) = \dfrac{5}{32}$

I think I am probably doing something really stupid here, but can't put my finger on it.

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You are computing the probability that "A wins at the 6th game"

This is not the same as the probability that "A wins at the 6th game, given that the series ends at the 6th game"

In fact:

$$P(A\mbox{ wins at 6th g. | series ends at 6th g.})=\frac{P(A\mbox{ wins at 6th g. and series ends at 6th g.})}{P(\mbox{series ends at 6th g.})}$$ $$=\frac{P(A\mbox{ wins at 6th g.})}{P(\mbox{series ends at 6th g.})}$$ $$=\frac{P(A\mbox{ wins at 6th g.})}{P(A\mbox{ wins at 6th})+P(B\mbox{ wins at 6th)}}$$ $$=\frac{5/32}{5/32+5/32}=1/2$$

Of course your "intuitive" argument would suffice and is rigorous enough, since exchanging "$A$" with "$B$" does not change the probabilities.

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Very well explained, Thank you. – mathguy80 Jan 6 '12 at 7:37

Your computation is correct, only it does not compute exactly what you think it does. You correctly state that $\color{red}{\frac5{32}}$ is the probability that the series lasts 6 games and that team $\color{red}{\text A}$ wins it. By symmetry, $\color{blue}{\frac5{32}}$ is also the probability that the series lasts 6 games and that team $\color{blue}{\text B}$ wins it.

Now, you know that the series lasted 6 games (an event of probability $\color{red}{\frac5{32}}+\color{blue}{\frac5{32}}$) and you wonder what are the chances that team $\color{red}{\text A}$ won it (an event of probability $\color{red}{\frac5{32}}$ included in the former event). Bayes rule tells you this is the ratio $\frac{\color{red}{\tfrac5{32}}}{\color{red}{\tfrac5{32}}+\color{blue}{\tfrac5{32}}}=\frac12$.

Note One can ask the same question if team A wins each game with probability $p$. You should be able to show that the answer becomes $\frac{p^2}{p^2+(1-p)^2}$.

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Thanks, it makes sense now, I'll try to work out the exercise. – mathguy80 Jan 6 '12 at 7:38

Actually, your intuition is right here (although intuition in probability is often wrong !)

since both have equal chances of winning, the indicated probability will remain half by symmetry regardless of the number of games played to find the winner.

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Thanks. Yes, I have learnt that lesson about intuition and probability! – mathguy80 Jan 6 '12 at 7:39