What are the exact odds of getting a perfect NCAA bracket?

Question

With the NCAA March Madness Finals nearing, I thought it'd be appropriate to ask this.

From everything that I've read and heard online, there seems to be varying opinions on the exact odds of getting a perfect NCAA bracket, especially from different sources. That seems certainly strange, because I had originally thought there'd be only one way to calculate the odds of that.

For example, this link seems to suggest the odds are 9.2 quintillion. Which seems odd that, compared with this, suggesting it's around 4 quadrillion. Which leads me to being somewhat confused as to the exact number for the odds.

What I'm far more interested is how do I calculate the chances of getting a perfect bracket? With 68 possible winners, I had originally thought it'd be a simple 68!, but I thought since there were a total possible of 68 slots, it could be 68 ^ 68? Is my thinking off, or am I somewhere in the right court?

Your first link assumes only 64 teams. – Santiago Canez Jan 23 '14 at 03:57 — Santiago Canez, Jan 23 '14 at 03:57

Mark McClure · Accepted Answer · 2014-03-20T16:27:48.923

9

I would say there's you have a chance of around 1 in 147 quintillion. More specifically, there are 67 games, since all but one of the 68 must lose. Thus, assuming you pick all your games with the flip of a coin, you've got a probability of $1/2^{67} = 1/147573952589676412928$ of picking everything correctly.

The computation is fairly simple. I'm assuming that the odds of picking a single game correctly are 50-50 or a probability of $1/2$. I'm also assuming that the choices are independent of one another so that their joint probabilities can be computed via multiplication. We then simply multiply 67 one-halves together to get the result.

Your first link points to a USA Today article which in turn points to a page on Geekosystem which shows an image of a Depaul mathematics professor standing in front of a white board with the computation $2^{63} = 9223372036854775808$. That suggests that the 9 quintillion number refers to the smaller 64 team bracket we had prior to the play-in games. My computation arises from the larger bracket that includes four play-in games.

Edit: It might be worth mentioning that most online bracket challenge games start after the play in games and, therefore, use the 64 team bracket.

Of course, some folks might be better than 50-50 at picking basketball games. Let's suppose you can pick a winner three fourths of the time. (Note that, earlier this month, Sports Illustrated's Seth Davis was 36 of 50 or just 72%.) Then, your odds would be $(3/4)^{67}$ or about 1 in 234 million. Still not very good.

edited Mar 20 '14 at 16:27

answered Jan 23 '14 at 03:50

Mark McClure

30,510

3

Of course, one would hope that you have a better method than randomly selecting the winner of each game! Even more so after Nate Silver's use of statistics in predicting winners last year. – Viktor Vaughn Jan 23 '14 at 04:01
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@SpamIAm I was just thinking about that and editing my post accordingly! :) – Mark McClure Jan 23 '14 at 04:06
I think your second paragraph is very informative. +1 – Viktor Vaughn Jan 23 '14 at 04:08
@SpamIAm can you provide a link? – Emmet B Jan 23 '14 at 04:46
So how did you reach that conclusion? I'm just curious in the methodology – yuritsuki Jan 23 '14 at 05:51
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@Retrosaur I expanded my answer a bit. I hope that helps. – Mark McClure Jan 23 '14 at 13:02
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@EmmetB Here's a link to his blog. – Viktor Vaughn Jan 23 '14 at 14:30

What are the exact odds of getting a perfect NCAA bracket?

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