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Suppose two teams play a series of games, each producing a winner and a loser, until one team has won two more games than the other. Let $G$ be the total number of games played. Assuming each team has chance $0.5$ to win each game, independent of results of the previous games. Find the expected value of $G$.

I think $G$ takes even values and I need to use negative binomial.But I am unable to find probability distribution and expectation. Please help.

A.D
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3 Answers3

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Let the required expectation be $a$. We assume without proof that $a$ is finite. (The proof is not hard.)

If the same team wins the first two games (probability $1/2)$, then the number of games, and hence the expected number, is $2$. Otherwise, $2$ games have been "wasted" and the expected number is $2+a$. It follows that $$a=\frac{1}{2}(2)+\frac{1}{2}(2+a).$$ Solve for $a$.

André Nicolas
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  • Thank you. Now I understand how can we find expectation avoiding the distribution. I will be thankful to you if you help me to find probability distribution of $G$. – A.D May 10 '13 at 04:32
  • The distribution is close to geometric. Let $X$ be the number of games. If $k$ is odd, $\Pr(X=k)=0$. If $k\ge 2$ is even, $\Pr(X=k)=\frac{1}{2^{k/2}}$. Try to see why this is true. If by tomorrow you haven't seen it, send a message and I will add to the answer. – André Nicolas May 10 '13 at 05:45
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    Why is it that the probability of one team winning first two games is 1/2?why not 1/4? – kris91 May 11 '13 at 02:31
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    It can be Team A or Team B. Each has probability $\frac{1}{4}$. I have edited the answer, replacing "one team" by "the same team." Thanks for pointing out the possible ambiguity. – André Nicolas May 11 '13 at 02:33
  • ok.I got it now. – kris91 May 11 '13 at 10:40
  • but say team A wins 3 games and team B wins 1 or vice versa then shouldnt the probability be 1/2^3? Or should only the no of wins of the winning team be considered? that is A has 3 wins or B has 3 wins.Then the probability is 1/2^2 which is in accordance with the above formula. – kris91 May 11 '13 at 10:48
  • The series ends in $4$ if it didn't end in $2$ (probability $1/2$) and some team the last two games (probability $1/2$). So probability series lasts exactly $4$ is $(1/2)(1/2)$. – André Nicolas May 11 '13 at 15:17
  • @AndréNicolas Why's that the probability distribution? Could you explain it once? – Tapi May 11 '23 at 17:55
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The current net position (wins of team 1 minus wins of team 2) could be modeled as a simple 1-dimensional random walk. You are looking for the expected number of steps the walk takes before reaching $2$ or $-2$ (the absorbing states). In this case $G$ is referred to as the stopping time.

A result (maybe not well known since I had to look it up again) gives the expected stopping time as $a*b$ where $a$ is the upper boundary and $-b$ the lower. Here, the expected stopping time is $4$. This conforms to @AndréNicolas answer.

CommonerG
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Use the linearity of expectation. $G_n = X _1+X_2+X_3+\dots$ where $X_n = 2$ if the $n$-th 2-game set has to be played (i.e. each of the first $n-1$ 2-game sets is tied), $X_n = 0$ otherwise. Clearly $E(X_n) = 2(0.5)^{n-1}$, whence $E(G) = E(X_1)+E(X_2)+\dots=2+1+1/2+\dots=4$.