What Are The Odds of the NFC Winning the Coin Toss 13 Years in a Row?
by Robert Quigley | 6:51 pm, February 7th, 2010
The NFC just won the coin toss for the 13th time in a row. What are the odds?
Assuming the odds of correctly calling a coin toss are 50/50, the odds of calling one toss correctly are 1/2, the odds of calling 2 in a row are 1/4 (1/[2^2]), the odds of calling three in a row are 1/(2^3)=8, et cetera.
So: 1/(2^13) = 1/8192 = 0.000122070312.
Course, that is assuming the odds are 50/50 and not some other number tweaked by NFC voodoo.
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Hi Robert,
Cute observation, but I want to suggest a slight correction – you’re not being very clear about what probability you are talking about. What you are calculating is the apriori probability that one league would win 13 games in a row *OUT OF 13 GAMES*.
The more relevant probability (I think most statisticians would agree) is “what is the apriori probability, given an event which will run for 44 years, that one league would win the toss 13 or more years in a row?” This is a common probability fallacy. The history of an event is an important selection effect, which can artificially raise the salience of something happening – if the superbowl had gone on for 1000 years, no one would be surprised that there was a streak of 13 at some point along the way.
So I think the better answer is approximately 31 times your figure (44-13 tickets to this lottery), or about .004, assuming there have been 44 superbowls . So good enough to raise suspicion in social science, but hardly a smoking gun.
Perhaps more to the point, the probability of the NFC winning the toss this year was still 0.5, assuming the previous years were a given and the toss’s are independent. Or, now that’s it’s over, you could argue the probability of it happening was 1. :)
Love the blog! Great work by you, Andrew, and the others.
-JB
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