The Three Cap Problem

December 14, 2008, 7:59PM

A couple weeks ago I bought a Mountain Dew and when I looked underneath the cap I saw the word "Bears". So I read the label to see what ridiculous contest they were offering. In brief, you had to match three NFL team names to win a baseball cap with that team's logo on it. That got me thinking, what is the probability of selecting X number of caps and winning? I doesn't matter about the team, just the probability of winning any team's cap. The more I thought about this the harder the problem became.

For simplicity sake, lets assume there are just 30 NFL teams and you had an unlimited number of caps. You need at least three caps to win, so the probability of winning at 1 and 2 caps is zero. At the other extreme, if you pick up 61 caps, you are guaranteed to win. So our effective range is between 3 to 60 caps. (*It is interesting to think of what the probability of winning after the 60th cap, but I am not going to discuss that.)

Assuming we are sampling from an uniform distribution, we have an 1/30th chance of picking a team each time we get a cap. So if the first cap you flipped over was the "Jets" you have an 1/30th chance of picking the "Jets" again on the second flip. If you managed to pick up the "Jets" up for the third cap, your probability is (1/30)^2 or 0.11% for 3 pulls.

But here is where it gets tricky, lets say that you didn't pick the "Jets" three times in a row, but picked this: "Jets", "Jets", "Cowboys". You still have the chance to pick the "Jets" on the fourth pull but what is the probability?

Update: Thanks to Michael Culbertson analysis, he seems to have nailed the solution on the head.

4 Comments

I tried this again this summer and still couldn't get it. It's really rather complicated. Maybe if a took an advanced Prob/Stats course...

- Kenneth on

August 7, 2009, 6:30PM

hey, remember when that time when dos and i were convinced over the answer to that NFL probability problem, and we were dead wrong. yeah, you owned us.

- Gentle Ben on

November 25, 2009, 12:09PM

I'll bite: If there are 30 teams, the chance of winning on the third draw is 0.1%. It takes about 18 draws to have a 50% chance of winning, 27 draws to have a 90% chance of winning, and 38 draws to have a 99.9% chance of winning. Details here.

- Michael Culbertson on

November 27, 2009, 3:19PM

This question is an extension of the classic "brthday problem" but with a match of three (instead of just two) and 30 possibilities instead of 365.

You should see

http://tinyurl.com/yetsutv

See in particular Problem 3 857-858, which gives an example with 30 (just like yours)

Culbertson is right, but there is a simpler way of presenting it.

- Tobin Grant on

December 6, 2009, 8:56PM