## Thursday, December 5, 2013

### Converting odds to probabilities for the 1996 Boston Red Sox.

Question:  The odds of the Boston Red Sox winning a game at home in 1996 were 1.382 while the odds of winning on the road were 0.884.  The Red Sox played 81 games at home and 81 games on the road.  What is the probability that the Red Sox would have won a game at home and the probability that the Red Sox would have won a game on the road?  How many games did the Red Sox win at home, on the road, and in total?

Answer:  The odds an event occurs is defined as

odds=p/(1-p)

where p is the probability that an event occurs.

The problem for home games is solved by plugging in odds = 1.382 and then solving for p.

1.382=p/(1-p)

1.382*(1-p)=p

1.382 = (1+1.382) x p

p = 1.382/2.382 = 0.5802

The problem for away games is solved by plugging in odds=0.884 and rearranging

0.884=p/(1-p)

Based on the above math i am pretty sure this will rearrange to

p = 0.884/(1+0.884) =  0.4692

To get the number of victories at home and away multiply the respective victory probabilities by the number of home and away games, which is 81.   I get 47 home victories, 38 away victories and  a total of 85 victories.

The Boston Red Sox did win 85 games in1996 coming third in their division.

Author's Note:

This problem first appeared in my book Statistical Applications of Baseball, published in 1996.  It is available at a very low price on kindle.

https://www.amazon.com/Statistical-Applications-Baseball-Statistics-Sports-ebook/dp/B006M3PQWQ

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