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Question: In the NFL, a professional American football league, there are 32 teams, of which 12 make the playoffs. In a typical season, 20 teams (the ones that don’t make the playoffs) play 16 games, 4 teams play 17 games, 6 teams play 18 games, and 2 teams play 19 games. At the beginning of each game, a coin is flipped to determine who gets the football first. You are told that an unknown team won ten of its coin flips last season. Given this information, what is the posterior probability that the team did not make the playoffs (i.e. played 16 games)?

Hint: Conduct both a Bayesian and Frequentist analysis of data to make inferences about a proportion

Correct answer: 0.556

Why might a Bayesian and Frequentist analysis be necessary? This strikes me as a Bayesian problem.

Is there a relationship between whether a football team wins a coin flip and whether they make the playoffs? The two probabilities seem independent.

Below are the probabilities that I'm deducing. Please help me find out the problems with my logic.

  • Posterior = P(did not make playoffs | won 10 out of 16 coin flips)
  • Prior = P(did not make playoffs) = $\frac{20}{32}$ = 0.625
  • Likelihood = P(won 10 out of 16 coin flips | did not make playoffs) = $\binom{16}{10}(1-0.5)^{10}(1-0.5)^{6}$ = 0.122. There are two teams per game, and I assume that each has an equal probability of winning since we have no further information.I am assuming that the coin is fair, so p = 0.5 and q = 0.5.
  • Data = P(won 10 coin out of 16 coin flips) = 0.122

Posterior = $\frac{Likelihood * Prior}{Data}$ = $\frac{0.122*0.625}{0.122}$ = 0.625

Am I calculating the likelihood improperly because there is a relationship between the coin flip and the probability of winning? Any help is appreciated.

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  • $\begingroup$ Why are you dividing by $0.122$? as P(data)? That's $P(data|did not make playoffs).$ $\endgroup$
    – spaceisdarkgreen
    Commented Feb 17, 2017 at 20:13
  • $\begingroup$ @spaceisdarkgreen: I was assuming that the data is won 10 out of 16 coin flips. If winning coin flips is independent of making the playoffs, shouldn't be be dividing by 0.122? $\endgroup$
    – graywarbler
    Commented Feb 17, 2017 at 20:15
  • $\begingroup$ No, you divide by total probability of the data, not the conditional probability of the data (which you did calculate correctly). See answer. $\endgroup$
    – spaceisdarkgreen
    Commented Feb 17, 2017 at 20:19
  • $\begingroup$ @spaceisdarkgreen: Thanks for all your help today. I understand the logic now. :) $\endgroup$
    – graywarbler
    Commented Feb 17, 2017 at 20:23

2 Answers 2

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Let $p_n$ be the probability that our team played exactly $n$ games.

Our prior: $$p_{16}=\frac {20}{32},\,p_{17}=\frac 4{32},\,p_{18}=\frac 6{32},\,p_{19}=\frac 2{32}$$

Now, if there were $n$ coin tosses, the probability that our guys won exactly $10$ of them is $P(W=10\,|\,n)=\binom n{10}2^{-n}$. We can work this out case by case to see that $$n= \{16,17,18,19\}\implies P(W=10\,|\,n)=\{0.122192383,0.148376465,0.166923523,0.176197052\}$$

Of course our revised estimate that our team played only $16$ games is then $$\hat {p_{16}}=\frac {P(W=10\,|\,16)\times p_{16}}{\sum_{n=16}^{19}P(W=10\,|\,n)p_n}\approx 0.556521739$$

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  • $\begingroup$ Thank you so much. I didn't understand that the prior should contain information for all the teams, and not just the team for which we are trying to estimate the parameter. $\endgroup$
    – graywarbler
    Commented Feb 17, 2017 at 20:19
  • $\begingroup$ Yes....your prior is just the Frequentist's prior. After you add new information, that changes your estimates. $\endgroup$
    – lulu
    Commented Feb 17, 2017 at 20:22
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Your problem is in dividing by $P(data).$ $P(data)$ is the total probability of the data but you plugged in the probability conditional on not making the playoffs. (Would it really make sense if that canceled out every single time... why wouldn't the formula just be posterior = prior in that case?)

To get the total probability of the data, you need to sum over all the possible $n$'s: $$ p(data) = p(data|n=16) P(n=16) + p(data|n=17)P(n=17)+p(data|n=18)P(n=18)+p(data|n=19)P(n=19)$$

and compute all those just like you computed $p(data) = p(data|n=16) P(n=16)=0.122*0.625$

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