I read somewhere that the odds of picking the winner in every game of the NCAA basketball tournament are 1 in … a lot. So staggering are the odds, in fact, that last year (and possibly this year) some web site offered $1,000,000 to any person who could do it.
This raises two questions:
How many possible outcomes are there?
How many would there be if we assumed that no team seeded 11th or greater would win in the first round?
Well, I’m only moderately geeky, but let’s see.
Try it by region. Fifteen teams are eliminated before one advances to the Final 4. That’s 215 possibilities. Four regions add a factor of 4 and 2 semis plus the finals add a factor of 8. So 215(4)(8) = 2**20 which is 1,048,576.
Question 2 is easier. Since in each region seeds 10 and 9 meet 7 and 8 and we don’t have ties, your conditions can’t be met. The answer is zero.
Actually, jcgmoi, I think you are wrong on both counts.
There are 32 games in the 1st round, each with 2 possible outcomes. That makes 2^32 or 4,294,967,296 combinations in the 1st round! I’m not completely clear how to calculate it past that, but I think you multiply it by the number of rounds (6).
Question 2 simply means that the 2 games you mentioned (10 vs. 7 and 9 vs. 8) would be the only ones with undetermined outcomes, so the total combinations in the 1st round would be 2^8 or 256. I’m too lazy (or too wasted) to think about what that means for the remaining rounds.
If any of the above is incorrect, blame the saturday night soiree 
Let’s see, as you;ve pointed out, there are 32 games in the first round, with (2^32) possible outcomes.
Then there are 16 games in the second round, with (2^16) possible outcomes. Eight in the third round, etc…
If I’m doing my math right, that’s
(2^32) * (2^16) * (2^8) * (2^4) * (2^2) * (2^1) = (2^(32+16+8+4+2+1)) = (2^63)
That’s just my opinion. I could be wrong.
OK, I’m back-tracking. I now think Will is right. My bad was complicating the problem by dealing with regions and then screwing up
when trying to tie the regions together. I used the factor of 4 when I should have used 215 to the 4th power. Now that you guys have cleared my thinking, I see this: 63 games mean 263 possibilities. Thanks for the help.
Yeah, that’s right, for the first question there are
2^63 = 9,223,372,036,854,775,808 possibilities,
and for the second question, that means 24 of the games are predetermined, so there are
2^39 = 549,755,813,888 possibilities.
I think you guys (& gals?) are missing an important factor here. Not only does filling out an NCAA bracket require us to pick the WINNERS of 63 games, but it also requires us to pick the COMPETITORS of 41 of those games.
EXAMPLE: In the first round, 1st-Seeded St. Rastahomie takes on #16 Wattsamotta U. In the 8/9 matchup, #8 Altered State takes on #9 Shamalamadingdong. Now, in order to make accurate predictions for the first game of the second round, one must accurately predict who will win in the first round. In other words, there’s no profit in picking Altered State to beat Wattsamotta U if neither team even plays in that game. Understand?
Therefore, I believe the odds are considerably worse than what we’ve been talking about here.
Wrong thinking is punished, right thinking is just as swiftly rewarded. You’ll find it an effective combination.
But you’re forgetting the fact that picking the winners will automatically pick the competitors, too.