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## Tournament Bracket Question

I hate to expose my ignorance, but...
If you have a 64 team, 4-division tournament (i.e. NCAA Basketball Championship Bracket), how many possible combinations can there be?

2. Originally Posted by SG-1 Fan
I hate to expose my ignorance, but...
If you have a 64 team, 4-division tournament (i.e. NCAA Basketball Championship Bracket), how many possible combinations can there be?
What do you mean by "combinations"? If you mean, how many different ways could the "Final Four" turn out, that would be 16^4 or 65536.

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Originally Posted by SeanF
What do you mean by "combinations"? If you mean, how many different ways could the "Final Four" turn out, that would be 16^4 or 65536.
Thanks SeanF. From tip off of the first game to the crowning of the champion, how many combinations (of winners) could there be?

4. Oh, Lord.

Let's see. There'd be 32 games in the first round, each with two possibilities, so there'd be 32^2 or 1024 possible turn-outs.

Second round would have 16 games, each with two possibilities, so that would be 256 possible turn-outs.

Third round would have 8 games, 64 possibilities.

Fourth round would have 4 games, 16 possibilities.

Fifth round would have 2 games, 4 possibilities.

Final round has 1 game, 2 possibilities.

1024*256*64*16*4*2 = 2,147,483,648 total possible combinations.

Does that seem right, or am I missing something?

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Originally Posted by SeanF
Oh, Lord. ...

1024*256*64*16*4*2 = 2,147,483,648 total possible combinations.

Does that seem right, or am I missing something?
Thanks again SeanF. That's what I was thinking but a co-workers child has a teacher (elementary) that is reporting that there are over a trillion combinations. That seemed unlikely, so I went to mathforum.org looking for an answer. They claim that there are "more than nine quintillion possibilities"!

6. Originally Posted by SeanF
Oh, Lord.

Let's see. There'd be 32 games in the first round, each with two possibilities, so there'd be 32^2 or 1024 possible turn-outs.

Does that seem right, or am I missing something?
You're missing something. There aren't 32^2 possibilities but 2^32 possibilities. All the possibilities can be represented as a list of numbers each 32 binary digits long, going from

0000 0000 0000 0000 0000 0000 0000 0000

to

1111 1111 1111 1111 1111 1111 1111 1111

So you're already up to 4,294,967,296 possible results in the first round alone.

No wonder no one gets them all right.

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Thanks ToSeek. Is 9,223,372,036,854,775,808 the correct answer?

8. Originally Posted by ToSeek
You're missing something. There aren't 32^2 possibilities but 2^32 possibilities.
Uh, yeah, that's what I meant.

Originally Posted by SG-1 Fan
Thanks ToSeek. Is 9,223,372,036,854,775,808 the correct answer?
2^32 * 2^16 * 2^8 * 2^4 * 2^2 * 2^1 = 2^(32+16+8+4+2+1) = 2^63 = 9,223,372,036,854,775,808

9.22 quintillion.

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Originally Posted by SeanF
Uh, yeah, that's what I meant.

2^32 * 2^16 * 2^8 * 2^4 * 2^2 * 2^1 = 2^(32+16+8+4+2+1) = 2^63 = 9,223,372,036,854,775,808

9.22 quintillion.
WOW! I think I'll stick to the ponies...

10. There can be only one.

(Connor Mcleod)

11. Wait, I must be missing something or having a senior moment, but if the question is how many combinations of teams can be in the championship game, wouldn't it be 32x32, since how the teams got there wouldn't matter. There are 32 teams on each bracket, and each one might face each one of the other.

Or are you just looking for the total possible number of combinations throughout?

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Originally Posted by Enzp
Or are you just looking for the total possible number of combinations throughout?
Yes. Just the total number of possible combinations. I just could not get my mind around 9.22 quintillion. It looks like that is the correct answer though....

13. Originally Posted by Enzp
Wait, I must be missing something or having a senior moment, but if the question is how many combinations of teams can be in the championship game, wouldn't it be 32x32, since how the teams got there wouldn't matter. There are 32 teams on each bracket, and each one might face each one of the other.

Or are you just looking for the total possible number of combinations throughout?
I think what prompted the question was the office pools that usually go on for March Madness, where you try to guess the winner in each game. Assuming random chance, the odds that you would get everything right on your bracket sheet are 1 in 9.22 quintillion. Which suggests to me that getting everything right ought to have a bigger payout than it usually does!

But, yes, the odds that you'd get the champion right are 1 in 64, and the odds you'd get the final two right are 1 in 1024 (32^2).

14. Originally Posted by SG-1 Fan
Yes. Just the total number of possible combinations. I just could not get my mind around 9.22 quintillion. It looks like that is the correct answer though....
I don't know if it helps or not, but it might be worth pointing out that it takes a 63-bit binary number to reflect any given completed tournament slate, and that each binary number up to 2^63 is a possible outcome. So there are

111111111111111 1111111111111111 1111111111111111 11111111111111112 possibilities overall.

15. Just bet on Duke.

You'll narrow those odds considerably if you bet on Duke -- from 1 in 9.2 trillion down to about 1 in 5.

16. Not to nitpick, but aren't there 65 teams in the tournament, with teams #65 and #64 playing to see who gets last spot in one bracket?

How would that effect the math? Or should we just forget about it?

17. Originally Posted by Gruesome
Not to nitpick, but aren't there 65 teams in the tournament, with teams #65 and #64 playing to see who gets last spot in one bracket?

How would that effect the math? Or should we just forget about it?
That just doubles the number of possibilities, to 18.4 quintillion.

18. Originally Posted by SG-1 Fan
Thanks again SeanF. That's what I was thinking but a co-workers child has a teacher (elementary) that is reporting that there are over a trillion combinations. That seemed unlikely, so I went to mathforum.org looking for an answer. They claim that there are "more than nine quintillion possibilities"!
Please correct me if I'm wrong, but I'm thinking nine quintillion may be high. From the above link:
If there were four teams, and they played three games, how many
different ways would there be to fill out a bracket?
You can write
them down. There are only eight of them. Where do that eight come
from? Well, there are three games, and you have two possible choices
for each game. Hence, 2^3 = 8 possibilities.

Now back to the real tournament. Since there are 63 games to be
played, and you have two choices at each stage in your bracket, there
are 2^63 different ways to fill out the bracket.

2^63 = 9,223,372,036,854,775,808

That's more than nine quintillion possibilities.
To me, it seems like they are starting with the 64 (or 65) teams and figuring out *every* possibility of fitting them into the brackets.

But--we start out knowing the initial positions of all the teams! Does having that set lower the number of possibilities?

19. Originally Posted by pumpkinpie
But--we start out knowing the initial positions of all the teams! Does having that set lower the number of possibilities?
Nope. If you allowed the 64 teams to be assigned their first-round opponents randomly, that would create even more possibilities. The 9.22 quintillion number is considering those first round opponents are pre-assigned.

20. Originally Posted by SeanF
Nope. If you allowed the 64 teams to be assigned their first-round opponents randomly, that would create even more possibilities. The 9.22 quintillion number is considering those first round opponents are pre-assigned.
'Kay, thanks!
Anyone care to report that number??

21. Originally Posted by pumpkinpie
'Kay, thanks!
Anyone care to report that number??
The other number times 65! ?

That would be times 8.2 x 1090, and we've probably run out of -illions

22. Originally Posted by hhEb09'1
The other number times 65! ?

That would be times 8.2 x 1090, and we've probably run out of -illions
Getting close to a Googol!!

23. Originally Posted by pumpkinpie
Getting close to a Googol!!
More!, since the previous number was 18 x 1018, the new one would be 1.5 x 10110

24. Originally Posted by hhEb09'1
That would be times 8.2 x 1090, and we've probably run out of -illions
Surely not! Just google for them!

Get it? Googol for them .... I'll get me coat.

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