From Sam Farmer of the Los Angeles Times comes this quirky tidbit: the New Orleans Saints are 0 for 11 on coin flips (10 to start the game, plus the overtime against Atlanta). Cleveland is also 0-fer on the year, starting the season with 9 coin flip losses.
This is where we can ask, how rare is that? Farmer’s article quotes STATS, inc. as saying the odds are roughly 2,000 to 1. Assuming that the coin isn’t rigged or Jerome Bettis isn’t calling the toss, I suppose that is one correct answer (2,048 to 1 to be exact). The odds that the Browns would have an 0 for 9 stretch, by the way, is a far more reasonable 1 in 512 by the same logic.
Of course, this question can be interpreted different ways. What are the odds that an NFL team would have an 0 for 11 stretch on coin flips? Turns out, probably pretty good that someone will every few years. It’s hard to come up with an exact number, given the randomness of overtime coin flips and how they vary in frequency from team to team, but think about this. There are 32 teams. They each have at least one coin flip for every game. Over the course of 4 seasons (64 games) then that’s at least 2,048 team coin flips (32*64) and more depending on overtimes. Finding an 11 game stretch would be somewhat rare, but not shocking, even though the odds are enormous. There have been a lot of coin flips in the NFL over the years, way more than 2,048.
In fact, it would be shocking not to hear about an NFL team going 0 for 11 every once in a while.
[photo via Getty]
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