# Football Pick 'Em Pool Probability (Normal Distribution)

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?
• Max B: You should see the comment thread beneath my answer. – Brian Tung Sep 12 '18 at 6:27

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.

• 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 $60\cdot .0087988\cdot .9912012^{59} = .3134$ – Phil H Sep 12 '18 at 6:16
• @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.) – Brian Tung Sep 12 '18 at 6:24
• 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. – Phil H Sep 12 '18 at 15:44
• Why are we using 89 (I assume it is coming from 90-1) instead of 90? Thanks to all! This has been super helpful. – Max B Sep 12 '18 at 16:54
• @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. – Brian Tung Sep 12 '18 at 19:14

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 $1- 0.651 = 0.349$.

• Doesn't this have the same issue as my answer? – Brian Tung Sep 12 '18 at 16:02
• @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. – awkward Sep 12 '18 at 16:52
• @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. – awkward Sep 12 '18 at 17:05
• I didn't think so, either, but notice that the error is not insubstantial: It's a few percent. ($0.41$ vs $0.35$) – Brian Tung Sep 12 '18 at 19:15