# Bayes Theorem & Fantasy Football

Recommended: Listen to the Podcast Episode that accompanies this analysis

Bayes Theorem allows us to predict the probability of a particular event (A) occurring given that we know that another condition (B) has already occurred. There are two basic formulations of the theorem that can be used

Example 1: What is the probability that the Green Bay Packers will win a game (A) given that it is snowing during the game?

Known quantities:
p(A), probability Green Bay wins any game based on their win percentage for the season = 0.6
p(B), probability of it snowing during a Packers game = 0.1
p(B|A), probability of their being snow in a game the Packers happen to win = 0.12

What we want to find?
p(A|B), probability Green Bay wins game given that it is snowing=?

Formula:
p(A|B) = p(A)*p(B|A)/p(B)
=0.6*0.12/0.1
=0.72

In other words, there is a 72% chance Green Bay will win the game if it is snowing

Example 2: In deciding which players to draft in our fantasy league we want to know which running backs are most likely to gain 1000 yards. To help we our going to look at the Yahoo Fantasy Rankings. What is the probability of a running back gaining 1000 yards (A1) given that he was ranked in the top 10 of the Yahoo rankings (B)?

Known quantities:
Here are the # of 1000 yards rushers (out of the top 50) in the NFL per season since 2002:
2015 7
2014 13
2013 13
2012 16
2011 14
2010 17
2009 15
2008 16
2007 17
2006 23
2005 16
2004 18
2003 18
2002 17

On average there has been 15 per year (although notice the huge downward trend in 2015!). Using 50 running backs per season as our base this means:
-p(A1), probability a running back gains 1000 yards = 15/50= 0.3
Of course, then p(A2), probability a running back does not gain 1000 yards = 0.7
-p(B), probability running back is ranked in top 10 (again based on top 50) = 10/50 = 0.2

For these I used the Yahoo rankings for the past 5 seasons:
-p(B|A1), probability a running back is ranked in top 10 given that they gained 1000 yards = 0.62
-p(B|A2), probability a running back is NOT ranked in top 10 given that they gained 1000 yards = 0.38

What we want to find?
p(A1|B), probability a running back will gain 1000 yards given that they were ranked in the top 10 of the Yahoo Fantasy rankings=?

Formula:
p(A1|B) = [p(A1)*p(B|A1)]/[p(A1)*p(B|A1) + p(A2)*p(B|A2)]
=(0.3*.62)/(0.3*.62 + 0.7*.38)
=0.412

In other words, if Yahoo ranks a running back in the top 10 in the pre-season, there is a 41.2% chance they will gain 1000 yards during the season