# Just how difficult is it to fill out a perfect bracket in the NCAA March Madness Tournament?

March Madness season is back, and with it are disappointments of a busted bracket. I stumbled upon an article (from the Weather Channel — what?!) about the probabilities of getting a perfect bracket: 1 in 9,223,372,036,854,775,808. But just how small is that? Let’s do some math to find out!

Suppose everyone in the world, population $N$, fills out a random bracket at every time interval $t_0$, at the same time. Let the probability of a person getting the perfect bracket be $p$. The probability that nobody gets the bracket right on any time interval is

$(1-p)^{N}$,

and thus the probability that a person will get the bracket at any time interval is

$1-(1-p)^{N}$.

The probability that someone on Earth gets the bracket right for the first time at the kth time interval is the probability that nobody gets it right in the first $k-1$ intervals and that someone gets it right on the kth interval. Let’s denote a random variable $X$ corresponding to the interval that someone on Earth finally gets the bracket right. Putting this all together, we have

$P(X=k) = \underbrace{(1-p)^{N(k-1)}}_{\text{bracket won't be right before interval } k-1}*\underbrace{(1-(1-p)^{N})}_{\text{bracket is right at interval }k}$.

The expected interval of someone getting it right, i.e. the expectation value of $X=E[X]$ is calculated as,

$E[X] = \sum\limits_{k=1}^{\infty}kP(X=k) = \sum\limits_{k=1}^{\infty} k(1-p)^{N(k-1)}[1-(1-p)^{N}]$.

To simplify this problem, let

$\tilde{p} :=1-(1-p)^{N}$,

or

$1-\tilde{p} =(1-p)^{N}$,

such that

$E[X] = \sum\limits_{k=1}^{\infty} k(1-\tilde{p})^{k-1}\tilde{p}$.

To perform the summation we observe that

$k(1-\tilde{p})^{k-1} = \dfrac{d}{d\tilde{p}} (1-\tilde{p})^{k}$.

So the expectation value calculation becomes,

$E[X] = \tilde{p}\dfrac{d}{d\tilde{p}}\left(\sum\limits_{k=1}^{\infty}(1-\tilde{p})^{k}\right)$,

where the derivative has been taken out of the summation. The quantity in the parenthesis is just a sum of an infinite geometric series with ratio between successive terms equal to $1-\tilde{p}$, so it can be performed easily to yield:

$E[X] = \tilde{p}\dfrac{d}{d\tilde{p}}\left(\dfrac{1-\tilde{p}}{\tilde{p}}\right)$.

After taking the derivative, and subsituting back $p$ for $\tilde{p}$ we get

$E[X] = \dfrac{1}{\tilde{p}} = \dfrac{1}{1-(1-p)^{N}}$.

Now theoretically that does it, but we can make our arithmetic easier by observing that $p \ll 1$. Then, using the binomial expansion

$(1-p)^{N} \simeq 1-Np$,

we obtain

$E[X]\simeq\dfrac{1}{Np}$.

So now to answer our question. From the weather.com article linked earlier, we have

$p = \dfrac{1}{\text{9,223,372,036,854,775,808}} \approx 1.084 \times 10^{-19}$.

According to Wolfram Alpha, the estimated world population in 2013 was

$N\approx7.13\times10^{9}\text{ people}$.

Doing the math, we get the $E[X]\approx 1.29 \times 10^{9}\text{ times}$.

If everyone in the world filled out a bracket every second, we would expect about 40 years before someone gets it right.

If everyone in the world filled out a bracket every year, we would expect 1.29 billion years before someone gets it right, or about 1/10 the age of the universe.

Good luck with that!

EDIT: Just realized, we were not exactly correct in saying $p \ll 1$ alone justifies using the binomial expansion approximation, as $N \gg 1$. However, that the product $Np \ll 1$ does fully justify our approximation. The author regrets this error.