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Recently my boss has asked me to run some statistical analysis on our office's college football pool. He asked some specific questions and then also left it open ended so I can provide additional analysis if I found anything interesting.

The assumptions are as follows:

  • The pool consists of 60 participants
  • Each week, each participant picks the winners of 10 pre-set football games
  • The pool is 15 weeks long
  • Each person has a 50% chance of correctly picking each game

And the question I am struggling with is:

  • With 60 participants, what is the confidence of the best participant hitting a season average of 55%. 57.5%, 60%? 62.5%

All of the other questions are pretty straightforward normal distribution questions that I was able to solve the answer to. Can anyone help me solve this, please? My issue with the question is I can't think of how to conceptually attack this problem. It seems that the most logical way to solve for this would be to run simulations, but it has been a while since my college stats classes in which they taught this. Any guidance or help would be much appreciated. I have been using excel for this analysis

The rest of the questions he asked, I left below:

  • For an individual in a given week, what is the probability of getting 10 right, 9, 8, etc…
  • Over the season, what can an individual expect out of 15 weeks- __ weeks of 6 right, __ weeks of 5 right, __ weeks of 4 right, etc.
  • For the season, what is the chance an individual might get 55% right? 57.5%? 60%? 62.5%
  • For all participants for all weeks, how many 10 week wins can we expect? 9 wins? 8? For example, could we expect that 2 people hit 10 wins sometime over the course of the season?
  • What is the standard deviation of the group’s overall performance and some statistics on where the group might come out for say 1-standard deviation, 2 sd?
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  • $\begingroup$ Max B: You should see the comment thread beneath my answer. $\endgroup$ – Brian Tung Sep 12 '18 at 6:27
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I think this will work. Someone please let me know if I've had a brain-o. (ETA: It does—kind of. But see the comments.)

With each player picking $150$ games, the distribution of the number of correctly guessed games is approximately normal with a mean of $150/2 = 75$, and a variance of $150/4 = 75/2$, and therefore a standard deviation of $\sqrt{75/2} \approx 6.12$.

What you're looking for is the distribution of the $k$th order statistic, with $k = n = 60$ (that is, the maximum). We can do this by finding this order statistic for a uniform distribution (the percentiles, effectively), and mapping this onto the normal distribution. This random variable $M$ (for Maximum) has a Beta distribution:

$$ M \sim \text{Beta}(60, 1) $$

which has a PDF of

$$ f_M(m) = 60 m^{59} \qquad 0 \leq m \leq 1 $$

So, let's suppose you want to know the probability that the winner gets at least $60$ percent. This is $90$ of the $150$ games, or $15/6.12 \approx 2.45$ standard deviations out. This puts it at the $99.2847$th percentile. Then we integrate

\begin{align} P(\text{winner picks at least $60$ percent}) & = \int_{m=0.992847}^1 60m^{59} \, dm \\ & = \left. m^{60} \right]_{m=0.992847}^1 \\ & \approx 0.35 \end{align}

I don't dare trust that figure very closely until I try with some more significant digits. (ETA: I think it's still pretty good—or it would be, if the normal approximation were good enough.) But you get the idea.

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  • $\begingroup$ Binomial check: Probability of a participant getting $\ge60\%$ correct is: $1-\text{binomcdf}(150,0.5,89) = .0087988$. Probability of $1$ out of $60$ getting $\ge60\%$ correct will be $60\cdot .0087988\cdot .9912012^{59} = .3134$ $\endgroup$ – Phil H Sep 12 '18 at 6:16
  • $\begingroup$ @PhilH: Actually, I think it's off in the other direction, because what we really want is the probability that at least one player gets at least $60$ percent, which would be $1-(1-0.0087988)^{60} \approx 0.411$. Interesting! I didn't think it would be that far off. (I'm trusting your binomial figure, of course.) $\endgroup$ – Brian Tung Sep 12 '18 at 6:24
  • $\begingroup$ Yes, I agree. My binomial figure came from my TI-83 for a cumulative probability of at least 90 correct out of 150 with individual probability of 0.5. It's actually 1 minus the cumulative probability of at most 89. $\endgroup$ – Phil H Sep 12 '18 at 15:44
  • $\begingroup$ Why are we using 89 (I assume it is coming from 90-1) instead of 90? Thanks to all! This has been super helpful. $\endgroup$ – Max B Sep 12 '18 at 16:54
  • $\begingroup$ @MaxB: We want the probability that it's at least $60$ percent—i.e., $90$ correct picks. That's the complement that it's at most $89$ correct picks. You can't have $89.5$ correct picks. $\endgroup$ – Brian Tung Sep 12 '18 at 19:14
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Let's say $X_i$ is the number of games chosen correctly by the $i$th individual, for $i = 1,2,3, \dots ,60$. Then $X_i$ follows a Binomial distribution with $n=150$ and $p=0.5$, which has a mean of $\mu=np = 75$ and a standard deviation of $\sigma=\sqrt{np(1-p)} = 6.12$. We can reasonably approximate this with a Normal distribution with the same mean and standard deviation. A success rate of $60%$ corresponds to $X_i \ge 90$. Consider the complementary event, $X_i < 90$.
$$P(X_i < 90) = P \left( \frac{X_i - \mu}{\sigma} < \frac{90-\mu}{\sigma} =2.45 \right) = \Phi(2.45) = 0.99286$$ where $\Phi$ is the cumulative distribution function of a Normal(0,1) random variable. (By way of comparison, use of the Binomial distribution instead of the Normal yields $P(X_i <90) = P(X_i \le 89) = 0.99120$.)

So using the Normal approximation, the probability that all $60$ individuals have a success rate less than $60$ is $P(X_i < 90)^{60} = 0.651$, and the probability that the highest-scoring individual has a success rate greater then $60$ is $1- 0.651 = 0.349$.

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  • $\begingroup$ Doesn't this have the same issue as my answer? $\endgroup$ – Brian Tung Sep 12 '18 at 16:02
  • $\begingroup$ @BrianTung If by "issue" you mean the use of the normal distribution to approximate the binomial, I don't think that is a problem here because 150 is large enough that the approximation should be very good. I know my solution is very similar to yours, but I thought it should be posted since it avoids some complexity-- the use of the Beta distribution and the need to integrate a function. $\endgroup$ – awkward Sep 12 '18 at 16:52
  • $\begingroup$ @BrianTung P.S. I added a comparison of the Normal approximation with the Binomial so the reader can judge how well the approximation does in this case. $\endgroup$ – awkward Sep 12 '18 at 17:05
  • $\begingroup$ I didn't think so, either, but notice that the error is not insubstantial: It's a few percent. ($0.41$ vs $0.35$) $\endgroup$ – Brian Tung Sep 12 '18 at 19:15

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