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


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


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},


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


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.


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