# Bayes Probability - Yankees and World Series

• Jan 30th 2010, 04:30 PM
lindah
Bayes Probability - Yankees and World Series
The Yankees are playing the Dodgers in a world series. The Yankees win each
game with probability .6. What is the probability that the Yankees win the
series? (The series is won by the first team to win four games.)

May anyone start me off on this question? I'm feeling very dumb and do not know where to start :(
My apologies that I don't even have suggestions to begin with

TIA
• Jan 30th 2010, 06:22 PM
mr fantastic
Quote:

Originally Posted by lindah
The Yankees are playing the Dodgers in a world series. The Yankees win each
game with probability .6. What is the probability that the Yankees win the
series? (The series is won by the first team to win four games.)

May anyone start me off on this question? I'm feeling very dumb and do not know where to start :(
My apologies that I don't even have suggestions to begin with

TIA

Read this: Negative binomial distribution - Wikipedia, the free encyclopedia
• Jan 30th 2010, 07:21 PM
Soroban
Hello, lindah!

Quote:

The Yankees are playing the Dodgers in the World Series.
The Yankees win each game with probability 0.6.
What is the probability that the Yankees win the series?
(The series is won by the first team to win four games.)

Let: .$\displaystyle \begin{Bmatrix}W &=& \text{win} \\ L &=& \text{loss} \end{Bmatrix}$

There are 4 possible scenarios . . .

(1) They win the first 4 games: .$\displaystyle WWWW$
. . . $\displaystyle P(4W) \:=\:(0.6)^4 \:=\:0.1296$

(2) They win in the 5th game: .$\displaystyle LWWWW$
. . . The one Loss can occur in any of the first 4 games.
. . . $\displaystyle P(\text{4W, 1L}) \:=\:4(0.6)^4(0.4)^1 \:=\:0.20736$

(3) They win in the 6th game: .$\displaystyle LLWWWW$
. . The 2 Losses can occur in any of the first 5 games: .$\displaystyle _5C_2 \,=\,10$ ways..
. . . $\displaystyle P(4W, 2L) \:=\:10(0.6)^4(0.4)^2 \:=\:0.20736$

(4) They win in the 7th game: .LLLWWWW[/tex]
. . . The 3 Losses can occur in any of the first 6 games: .$\displaystyle _6C_3 \,=\,20$ ways.
. . . $\displaystyle P(\text{4W, 3L}) \:=\:20(0.6)^4(0.4)^3 \:=\:0.165888$

Therefore: .$\displaystyle P(\text{Yankees win}) \;=\;0.1296 + 0.20736 + 0.20736 + 0.165888 \;=\;\boxed{0.710208}$