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I did this the obvious way, and my friend came back with a better idea. Can you guys adjudicate or improve on both?

My way:

The Cincinnati Bengals and the Cleveland Browns both won on Sunday for the first time in 46 weeks (says ESPN). That seemed way too improbable...

46 is too high though. If we account for bye weeks, MNF nights, head-to-head games, etc., we get 31 weeks where they each had a chance to win.

Now we can take their respective records since 2009 (CLE: 11-31, CIN: 18-24) to calculate win probabilities for any given week. This gives a probability of 11% that both win in the same week (assume independence).

So...the probability of this 31-week drought? 2.5%...statistically significant but not earth-shattering. For reference, if these teams had even odds of winning any given week, the probability would plummet to 0.01%!

My friend's response:

man this is the most thought provoking post i've ever read…now i've spent 30 minutes thinking about it. anyway…i'm probably making myself sound like an idiot right now, but i'm not sure that independence assumption is correct. i think a more accurate way to think about this problem is the classical jar and ball problem. so if we disregard head to head games and mnf games and all that, the bengals and browns have played 42 games. now let's first put the 11 wins by the browns into separate jars. so if we now put the 18 wins by the bengals one by one into the jars, there is a 31/42 chance that the first win won't end up in a jar with a browns win…the second one has a probability of 30/41 (since it doesn't have the option to end up in the same jar as a previous bengals win)…third one has a 29/40 chance of not ending up in a jar with a bengals win…so on and so forth. if we think about it that way, the chances that a bengals win and browns win does not end up in the same jar after the 18 bengals wins and 11 browns wins have all been put into jars is ~.058%.

anyway…just figured it might be a slightly more accurate way to think about it since the probability of two team with winning percentages above .500 not having wins in the same weekend over the span which they were both above .500 is 0%…i think.

That all makes sense except for the intuitive feeling that the games were independent events (assuming neither team thought about the other). Who's right? Thanks!

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  • $\begingroup$ Your friend's description is incoherent: it sounds like there are only $11$ "jars", one for each Browns win. If you put the Bengals wins into those $11$ jars, the Pigeonhole Principle guarantees there will be jars having wins from both teams. But so what? The burden is on your friend to show how this "jars" metaphor--however s/he manages to fix it up--is an accurate model of how games are actually scheduled and played. For instance, they should be able to tell you what the jars are meant to represent. Are they all Sundays? Sundays when at least one team plays? Both teams? Are they games? $\endgroup$
    – whuber ♦
    Commented Mar 27, 2013 at 14:06

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There is a big selection bias. It would make more sense to calculate the probability of any two teams going 31 weeks without both teams winning during the same week than just these two teams.

Your way of calculating seems better than your friend's. Assuming that probability of winning a game is 11/42 makes more sense than assuming that the team will win exactly 11 out of 42 games (if the team loses their first game they aren't more likely the win their second game).

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