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?

 A: 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.
A: 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 $1- 0.651 = 0.349$.
