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In the final of the World Series Baseball, two teams play a series consisting of at most seven games until one of the two teams has won four games. Two unevenly matched teams are pitted against each other and the probability that the weaker team will win any given game is equal to 0.45. Assuming that the results of the various games are independent of each other, calculate the probability of the weaker team winning the final. What are the expected value and the standard deviation of the number of games the final will take?

I know how to find the probability that the weaker team wins. This is a negative binomial distribution so I can just apply the formula for p = 0.45 and r = 4. But I am not sure how to calculate the expected value.

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  • $\begingroup$ If all else fails, you could find the probability distribution and apply the definition of expected value directly. The following question and answer should help as well: math.stackexchange.com/questions/139181/… $\endgroup$
    – JMoravitz
    Commented Mar 17, 2016 at 2:49

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Let $P_n$ be the probability that the final takes $n$ games. $$ P_4 = (0.45)^4+(0.55)^4 $$

$$ P_5=\binom 41 \left[(0.45)^4(0.55)+(0.45)(0.55)^4 \right]$$

$$ P_6=\binom 52 \left[(0.45)^4(0.55)^2+(0.45)^2(0.55)^4 \right]$$

$$ P_7=\binom 63 \left[(0.45)^4(0.55)^3+(0.45)^3(0.55)^4 \right]$$

The expected value is then

$$\mu = 4P_4+ 5P_5+ 6P_6+ 7P_7 $$

and the variance

$$ \sigma^2 = (4-\mu)^2P_4+ (5-\mu)^2P_5+ (6-\mu)^2P_6+ (7-\mu)^2P_7 $$

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  • $\begingroup$ Thanks! This is really helpful! $\endgroup$
    – Ben Thompson
    Commented Mar 17, 2016 at 4:10
  • $\begingroup$ How would you get the standard error for the number of games played $\endgroup$
    – pino231
    Commented Jul 7, 2018 at 20:34

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