Redskins are lucky to play bad teams, but how lucky?

A recent article in Yahoo Sports pointed out that the Washington Redskins are the first team in history to play six winless teams in a row. Here is their schedule so far (also according to the article cited above):

 

Week 1 — at New York Giants (0-0)

Week 2 — vs. St. Louis Rams (0-1)

Week 3 — at Detroit Lions (0-2)

Week 4 — vs. Tampa Bay Buccaneers (0-3)

Week 5 — at Carolina Panthers (0-3)

Week 6 — vs. Kansas City Chiefs (0-5)

 

The author of the article, Chris Chase (or, as he notes, his dad-let’s call him Mr. Chase), calculates the odds of this as 1 in 32,768. This calculation is incorrect and far too high for several reasons, which I get to below. But first, let me explain how the calculation was likely performed.

 

The calculation assumes, plausibly, that the Redskins have the same chance of playing any given team (unlike some college teams, who purposely make their schedules easy, this is not possible in the NFL).

 

The calculation also assumes, not plausibly, that teams that have thus far won no games have a 50-50 chance of winning each game. The implicit assumption there is that all NFL teams are evenly matched. The fact is that there are a few really good teams, a few really bad teams, and a bunch of teams in the middle. Thus, there are likely to be a bunch of winless teams after 5 games, and not, as the incorrect calculation below implies, only 1 winless team of 32 after 5 games.

 

Finally, the calculation, apparently in a careless error, assumes the chances of playing a winless team the first week are 50-50, when, of course, all teams are winless the first week.

 

So the Mr. Chase’s (incorrect) calculation is
Week 1 chances: 50% ( 1 in 2)
Week 2 chances: 50% (1 in 2)
Week 3 chances: 50%*50%=25% (1 in 4)
Week 4 chances: 50%*50%*50%=12.5% (1 in 8)
Week 5 chances: 50%*50%*50%=12.5% (same as week 3 because the team they played had only played three games)
Week 6 chances: 50%*50%*50%*50%*50%=3.125% (1 in 32)

 

A law of probability is that the chance of two unrelated events happening is the product of their individual chances. Thus, if the chance of rain today is 50% and the chance of rain tomorrow is 50%, the chance of rain both days is 25%, if those chances are unrelated (which, by the way, they probably aren’t). This is why the chances for multiple losses are multiplied together.

 

But back to the football schedule. To calculate the chances of 6 straight games against winless teams, Mr. Chase reasonably multiplied the 6 individual chances (again it assumed the 6 matchups were unrelated):
50% * 50% * 25% * 12.5% * 12.5% * 6.25% = .003%, or 1 in 32,768.
SO, the 32,768 is the number reported in the article.

 

The easy correction is that the chances of playing a winless team in the first game is 100%, so the calculation should be:
100% * 50% * 25% * 12.5% * 12.5% * 6.25% = .006%, or 1 in 16,384.
This error has been pointed out in comments on the article.

 

In addition, other comments point out the other major flaw: teams do not have equal probability of losing. Thus the chance that a team will be, say, 0-2 is not 25% (50%*50%) but something else, depending on the quality of the teams. At the extreme, half the teams lose every game and half win every game (this of course assumes losing teams only play against winning teams, but it is possible).

 

The reality is certainly not this extreme, which would imply a 50-50 chance each week of playing all losing teams (and thus a 1 in 32 chance of playing 6 in a row). So, how do we figure out the reality?

 

The easiest way is to look at, each week the percent of teams that are winless. If we assume the Redskins have an equal chance of playing each team, then we can compute the odds each week (click on the week to see the linked source). Note that everything is out of 31 teams instead of 32 because the Redskins can’t play themselves.

 

Week 1: 31 out of 31 teams winless. Chances: 31/31=100%
Week 2: 15 out of 31 teams winless. Chances 15/31=48% (I am assuming no byes first week and I know redskins lost their first game).
Week 3: 8 out of 31 teams winless. Chances: 8/31=26%
Week 4: 6 out of 31 teams winless. Chances: 6/31 = 19%
Week 5: 6 out of 31 teams winless. Chances 6/31 = 19%
Week 6: 4 of 31 teams winless. Chances: 4/31 = 13%

 

So the actual chances, assuming the Redskins have an equal chance of playing each team each week and cannot play themselves, are: 100%*48%*26%*19%*19%*13% = 0.06%, or 1 in about 1,700. Much more likely than 1 in 32,000 but still pretty unlikely.

 

And after all these easy games, how are they doing? Unluckily for Redskins fans, not too well…they’re 2-3 going into Sunday’s game against the winless Chiefs.

One thought on “Redskins are lucky to play bad teams, but how lucky?

  1. Jonathan Scott

    In 2008 the Detroit Lions set a record by losing all 16 of their games. If they had a 50-50 chance of winning each game, then there was only a 1 in 65,536 them losing all 16. A simpler explanation is that they were just plain lousy! The Yahoo Sports article didn’t factor in that not all NFL teams are created equal, as many fans can attest.

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