NBA finals are played between 2 teams, say team A and team B. The finals is 7 ga

Question

NBA finals are played between 2 teams, say team A and team B. The finals is 7 games maximum and terminates as soon as one team wins 4 games out of these 7 games. Assume X is random variable that represents number of games played, so X is 4 5, 6, 7. Assume each team equally matched and each game can go either way and each game result is independent of each other. (a) Please find the entropy of X, (H(X))? b) How much memory do you need to store number of games played in NBA finals in last 50 years? (c) If one of the teams has slight advantage over the other one and winning chance of team A is 60 %, please calculate the entropy, and if it is different from Part A, why and how much?

Explanation / Answer

a) Probability of winning A and B is same i.e. 1/2.

For number of matches to be 4, all four games should be won by either A or B.

P(A) = P(B) = (1/2)(1/2)(1/2)(1/2) = 1/16

P(X=4) = Total number of games equal to be 4 = P(A) + P(B) = 1/8

For number of matches to be 5, four games should be distributed among A and B as 3wins and 1wins while the 5th match should be won by the player who has won the 3 matches.

P(A) = P(B) = (4C3(1/2)3(1/2))(1/2) {3 wins out of 4 matches and then win in 5th game.

P(X=5) = P(A)+P(B) = 1/4

Similarly for X=6 matches, intial five matches should be won as 3 and 2.

P(A) = P(B) = (5C3(1/2)3(1/2)2)(1/2)

P(X=6) = 5/16

For total number of matches to be 7, 3-3 matches should be won and the winner of last game will be the ultimate winner.

P(A) = P(B) = (6C3(1/2)3(1/2)3)(1/2)

P(X=7) = P(A)+P(B) = 5/16

entropy = sum(-plog2(p)) = -[(1/8)log(1/8)+(1/4)log(1/4)+(5/16)log(5/16)+(5/16)log(5/16)] = 1.9237

b) Expected number of match = sum(X.P(X))

= 4(1/8) + 5(1/4) + 6(5/16)+7(5/16) = 5.8125

Since match cannot be in fraction. So lets round off to nearest integer i.e. 6. Total bits required to represent 6 is 3. for last 50 years, total bits required are 3*50 = 150 bits or 1580/8 Bytes or 18.75B.

c) This part is similar to part a expect P(A) is not equal to P(B). P(A) = 0.6, P(B) = 0.4

For X=4, either A can win all four or B can win all four.

P(A) = (0.6)4, P(A) = (0.4)4

P(X=4) = Total number of games equal to be 4 = P(A) + P(B) = 0.1552

For X=5, A can win 3 out of four and win last one (same for B)

P(A) = (4C3(.6)3(0.4))(.6)

P(B) = (4C3(.4)3(0.6))(.4)

P(X=5) = Total number of games equal to be 5= P(A) + P(B) = 0.2688

Similarly for X=6,

P(A) = (5C3(.6)3(0.4)2)(.6)

P(B) = (5C3(.4)3(0.6)2)(.4)

P(X=5) = Total number of games equal to be 6= P(A) + P(B) = 0.2995

Similarly for X=7,

P(A) = (6C3(.6)3(0.4)3)(.6)

P(B) = (6C3(.4)3(0.6)3)(.4)

P(X=5) = Total number of games equal to be 7= P(A) + P(B) = 0.2764

entropy = sum(-plog2(p)) = -[(.1552)log(.1552)+(.268)log(.268)+(.2965)log(.2965)+(0.2768)log(0.2768)] =1.959

This entropy is different because the probabilities have been changed

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