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Suppose that two sports teams were to play each other in a series, and the winning team is the one that wins two games. One of the teams, Team A, has a probability of 80% chance of winning any one game. What is the probability of Team A winning the series?

The answer that I reached is 64%, and my reasoning is that for Team A to win, there must be two conditions:

  1. Team A must win its first game
  2. Team A must win its second game

With that in mind, I multiplied 80% (the chance of Team A winning its first game) with 80% again (because it has 80% chance of winning its second game). Is this correct? I'm getting a bit tripped up because the probability of Team A winning its second game is not dependent on its probability of winning the first game, so that's why I'm sort of leaning on the total probability being 80% again (because if we can assume Team A wins its first game then everything else is irrelevant; the chances of them winning again is 80%).

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  • $\begingroup$ A doesn't have to win the first game in order to win the series. The winners could be, in sequence, BAA. $\endgroup$ – symplectomorphic Jul 5 '16 at 8:08
  • $\begingroup$ Yes I understand that. I apologize for not explicitly saying that. However, which game they win does not influence the probability, correct? $\endgroup$ – Tyler Jul 5 '16 at 8:08
  • $\begingroup$ Just consider all the possible outcomes that would result in A winning the series: AA, ABA, BAA. $\endgroup$ – symplectomorphic Jul 5 '16 at 8:16
  • $\begingroup$ There are three ways team A could win against B : AA --> $80\% \cdot 80\%$ ABA --> $80\% \cdot 20\% \cdot 80\%$ BAA --> $20\% \cdot 80\% \cdot 80\%$ Total = ? $\endgroup$ – Zubzub Jul 5 '16 at 8:16
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i will solve a general case , say probability of $A$ winning is $p$ and for a team to win it needs to win $n$ matches. now say $k$ matches have been played , and $A$ won the series , it means $A$ won the last match , thus out of the rest $k-1$ matches $A$ won $n-1$ games. thus for $A$ to win the series in $k$ matches probability is $$\binom{k-1}{n-1}p^{n}(1-p)^{k-n}$$

now $k>=n$ and $k<=2n-1$ ($A$ needs to win before $B$) thus our total probability is $$\sum_{k=n}^{2n-1} \binom{k-1}{n-1}p^{n}(1-p)^{k-n}$$

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If $2$ games wins a series, the series has to be of $3$ games

and $A$ can win with $AA, ABA, BAA$

Actually, we can simply compute $P(X\ge 2)$ for bin$(3,0.8)$ to get the result.

[ because $P(AA)$ has the same probability as $P(AAA) + P(AAB)$ ]

ADDED

In general, if a series has a maximum of $2n-1$ matches to decide the winner,
and $A$ has a probability $p$ of winning a single match,

P(A wins) $= P(X\ge n), X\sim bin(2n-1,p)$

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I stumbled upon this when I hit the same dilemma - probability of winning game 1 and game 2 are independent , its like picking a red ball from a jar that has 3 red and 2 blue with replacement - a general example problem.

However, on thinking more, even though the individual game result is independent, the overall series win is dependent of the result of game 1 and game 2.

Thus to get the probability of winning the series, the P(winning game 1) has to be multiplied with P(winning game 2), which is:

0.8 * 0.8 = 0.64.

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