Question

the NBA final is a seven game series, and the team that wins four games

Chap. 2 Review of Probability 2.70 The NBA final is a seven-game series, and the team that wins four games first wins the series. Denote by p the probability that the Eastern Conference team wins a game and by q = 1- p the probability that the Western Conference team wins a game. Assume that these probabilities remain constant from game to game and that the outcomes of the games are mutually independent (a) Find the probability that the Eastern Conference team wins the series. (b) Show that the probability that the series ends in j games is (i-1) [pq;-4 + q*p/--), j = 4, 5, 6, 7. 1 3 /

Answer 1

There is a seven game series and the team that first win four games win the series.

• Eastern Conference wins game with probability p
• Western Conference wins game with probability q = 1 - p

Since , probabilities are constant and outcomes are mutually independent , we could model the game using a Binomial Distribution .

P ( X = x ) = ⁿC_x p ^x q ^n-x ; x = 0 , 1 , ..... , n , where n is the total no. of observations

( a ) Probability that Eastern Conference wins the series

Eastern conference need to win four games to win the series .

$\because$ Probability of winning each game is p , Therefore Possibilities are as follows :

• Win 4 games out of first 4 games

$\Rightarrow$  ⁴C₄ p ⁴ q ^4-4 =  p ⁴

• Win 4 games out of first 5 games

$\Rightarrow$  ⁵C₄ p ⁴ q ^5-4 = 5 p ⁴ q

• Win 4 games out of first 6 games

$\Rightarrow$  ⁶C₄ p ⁴ q ^6-4 = 15 p ⁴ q ²

• Win 4 games out of first 7 games

$\Rightarrow$  ⁷C₄ p ⁴ q ^7-4 = 35 p ⁴ q ³

Therefore ,

P ( Eastern Conference Wins Series ) =   p ⁴ + 5 p ⁴ q + 15 p ⁴ q ² + 35 p ⁴ q ³

=   p ⁴ ( 1 + 5 q + 15 q ² + 35 q ³ )

( b ) Probability that series end in j games

The series ends if either of the teams wins the four games first .

Series end in j ^th game if a particular team has already won 3 games in the first ( j - 1 ) ^th games .

• Probability of winning 3 games in the first ( j - 1 ) ^th games by Eastern Conference

=  ^{j - 1} C ₃ p ³ q ^{j - 4}   ; ( j - 1 ) = 3 , 4 , 5 , 6 , 7

• Probability of winning 3 games in the first ( j - 1 ) ^th games by Western Conference

=  ^{​​​​​​​j - 1} C ₃ p ^{j - 4} q ³ ; ( j - 1 ) = 3 , 4 , 5 , 6 , 7

P (Eastern Conference wins series in j games ) =   ^{​​​​​​​j - 1} C ₃ p ³ q ^{j - 4} * p =    ^{​​​​​​​j - 1} C ₃ p ⁴ q ^{j - 4}

P (Western Conference wins series in j games ) =   ^{​​​​​​​j - 1} C ₃ p ^{j - 4} q ³ * q =    ^{​​​​​​​j - 1} C ₃ p ^{j - 4} q ⁴

^{$\Rightarrow$}  P ( Series end in j games ) =  ^{​​​​​​​j - 1} C ₃ p ⁴ q ^{j - 4}  +     ^{​​​​​​​j - 1} C ₃ p ^{j - 4} q ⁴  

=  ^{​​​​​​​j - 1} C ₃ { p ⁴ q ^{j - 4}  +  p ^{j - 4} q ⁴  } ;  j = 4 , 5 , 6 , 7 ..Since ,at least 4 games must be played to win series

Hence Proved

This is a case of Negative Binomial Distribution .