You are the team manager for the CU intramural Quidditch team, the Boulder Beate

Question

You are the team manager for the CU intramural Quidditch team, the Boulder Beaters, and you've just made it to the national championship series against the Stanford Snitching Snitches. The typical Quidditch championship is a best of 11 series, so the first team to win 6 matches is declared the winner. Furthermore, based on a rigorous statistical analysis of regular season play, the probability that the Boulder Beaters win any one match against the Snitching Snitches is given by p = 0.55 Part A What is the probability that you win the 11-game series on the 8th match played? Part B: What is the probability that you win the 11-game series in any valid number of matches? Part C: The winner of the champsionship will take home 1$10,000 to use for equipment and victory parties in the following season. Because the stakes are so high, you've decided to hire a famous Quidditch consultant, Bludger Bob, to help you prepare for each match. With Bob's help your probability of winning any given match increases to p 0.7. Bludger Bob charges a fee of 1$100 per game, but as a testament to his confidence in his abilities, only charges you the 1$100 for games that you win and only if you win the series. So, if you play a 11-game series and win you will take home I$10,000 less I$600 for Bludger Bob's fees, for a total payout of $9,400. What is the expected payout for an 11-game series? Part D: This year the National Intramural Quidditch Association (NIQA) is implementing an experimental rule. As the team with the best record in the regular season you will get to choose the length of the championship series. Specifically, you can choose the series length to be anywhere from N-3, 5, 7, ..., 99 games long. Assuming that you plan to use Bludger Bob's services and considering the fact that you have a better than even chance of winning each game, explain the trade-offs between choosing a short series vs. a long series in terms of Expected Payout.

Explanation / Answer

This is a Binomial distribution with p = 0.55

Let X denote the RV for the number of games won.

(a)

Probability of winning the series on 8th game played = P(X=5)*0.55, given that n = 7 and p = 0.55

So,

Reqd probability = 7C5*0.555*(1-0.55)2*0.55 = 0.118

(b)

Probability of winning the series in any possible way = 1 - Probability of losing

Probability of losing = P(X<=5) = P(X=0) + P(X=1) + P(X=2) + P(X=4) + P(X=5)

Now,

P(X=0) = 0.45^11 = 0.00015

P(X=1) = 11C1*0.551*0.4510 = 0.00206

P(X=2) = 11C2*0.552*0.459 = 0.0125

P(X=3) = 11C3*0.553*0.458 = 0.046

P(X=4) = 11C4*0.554*0.457 = 0.112

P(X=5) = 11C5*0.555*0.456 = 0.193

So,

Probability of losing = 0.193+0.112+0.046+0.0125+0.00206+0.00015 = 0.366

So,

Probability of winning the series = 1-0.366 = 0.634

Hope this helps !

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