Dec 9, 2012; Green Bay, WI, USA; Green Bay Packers quarterback Aaron Rodgers (12) celebrates a touchdown run by jumping into the stands during the third quarter against the Detroit Lions at Lambeau Field. The Packers won 27-20. Mandatory Credit: Jeff Hanisch-USA TODAY Sports

Detroit Lions Road Losing Streak to Packers: Calculating the Odds

Earlier today, I took a general look at the odds of the Detroit Lions losing 22-straight road games to the Green Bay Packers. It was interesting but left me a little unsatisfied as it didn’t say anything about what the actual odds of the streak are.

What I wanted was a better look at the streak based on the win probability of each individual game rather than lumping all the games together as a coin flip or a road game between equal teams or a road game with a representative disparity between the home and road teams.

Fortunately, pro-football-reference was a great help in compiling a list of the closing point spreads for the games going back to the first loss in the streak in 1992. Needing a way to approximate win probability with point spred, I found They have kept track of winning percentages based on closing point spreads for games going back to 2003. Since the data has some unexpected inconsistencies (like 11 point favorites winning more frequently than 13 point favorites), I used the data to produce a trendline rather than relying on the winning percentages on their own.

The line yielded an equation of y = -0.0294x + 0.5, where y is the win probability and x is the point spread. With an r^2 value of 0.95, the line is a 95% fit to the data – good enough for a football blogger.

Putting it all together in table form (point spreads and win probabilities in respect to the Packers):

Year Spread Win Prob
1992 -3.5 60.29%
1993 -3.5 60.29%
1994 -5.5 66.17%
1994* -4.5 63.23%
1995 -3.5 60.29%
1996 -11 82.34%
1997 -10 79.40%
1998 -9.5 77.93%
1999 -4 61.76%
2000 -3.5 60.29%
2001 -5.5 66.17%
2002 -10.5 80.87%
2003 -7 70.58%
2004 -9 76.46%
2005 -5.5 66.17%
2006 -6 67.64%
2007 -3 58.82%
2008 -11 82.34%
2009 -14 91.16%
2010 -14.5 92.63%
2011 +6.5 30.89%
2012 -7 70.58%

Now we are left with a simple probability calculation. Because each game is an independent event, all we have to do is multiply all the individual win probabilities together to determine the overall probability of the Packers winning each of the 22 games in the streak.

The result? Just a shade over 0.02%, that’s a 1 in 4,988.5 chance. In other words, to expect to see this result, the 22 game series would need to be played almost 5,000 times. Yet here we are.

According to a quick google search, the odds of this streak are on par with the odds of an amateur golfer hitting a hole in one or the odds of dying from electrocution.

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