Predicting NCAA Tournament Winners

Date: 03/27/2000 at 13:30:10
From: Asher Pleasants
Subject: Odds and Probability

Hello! What are the odds of one person winning all 63 games in the 
NCAA basketball tournament? I tried using factorial (63 x 62 x ... =) 
but the number is too large for me to multiply after I get to 59. Is 
there a another way to figure it out?

Date: 03/27/2000 at 14:36:18
From: Doctor TWE
Subject: Re: Odds and Probability

Hi Asher! Thanks for writing to Dr. Math!

I assume you mean, "what are the odds of one person PREDICTING THE 
WINNERS of all 63 games in the NCAA basketball tournament," since a 
single person can't win the games [it takes a whole team to play ;-)].

This is not strictly a mathematics question, since a person's 
probability of predicting the winner of each game depends on the 
person's knowledge of the teams' abilities, etc. A person who can 
correctly pick the winner of each individual game 90% of the time will 
have a much better probability of picking all the winners than the 
person who correctly picks the winner of each individual game only 50% 
of the time.

Let's suppose the person in question randomly guesses the outcome of 
each game and has a .5 probability (50% chance) of guessing correctly 
each time. To determine the odds of this person guessing all winners 
correctly, we have to determine how many possible outcomes there are.

To determine the number of possible outcomes, let's take a look at 
some simpler single-elimination tournaments and see if we can find a 
pattern.

With 2 teams, we have 1 game and 2 possible outcomes (A wins or B 
wins).

With 4 teams, we have 3 games in 2 rounds. Let's say that team A plays 
team B, and team C plays team D in the first round. There are 4 
possible outcomes of round one (listing the winners): AC, AD, BC and 
BD. For each of these possible round one outcomes, there are 2 
possible round two outcomes; the first team (A or B) wins, or the 
second team (C or D) wins. Thus we have 4*2 = 8 total outcomes. 
Listing them:

     R1     R2
     --     --
     A,C    A
     A,C    C
     A,D    A
     A,D    D
     B,C    B
     B,C    C
     B,D    B
     B,D    D

With 8 teams, we have 7 games in 3 rounds. Let's say we have the 
following "bracket":

     _A_
        \___
     _B_/   \
             \___
     _C_     /   \
        \___/     \
     _D_/          \
                    \___
     _E_            /
        \___       /
     _F_/   \     /
             \___/
     _G_     /
        \___/
     _H_/


There are 16 possible outcomes of round one (listing the winners): 
ACEG, ACEH, ACFG, ACFH, ADEG, ADEH, ADFG, ADFH, BCEG, BCEH, BCFG, 
BCFH, BDEG, BDEH, BDFG and BDFH. For each of these possible round one 
outcomes, there are 4 possible round two outcomes; the first (A/B) and 
third (E/F) teams win, the first (A/B) and fourth (G/H) teams win, the 
second (C/D) and third (E/F) teams win, or the second (C/D) and fourth 
(G/H) teams win. Finally, in the third round, each second round 
outcome has two possible outcomes; the "top half" bracketed team wins 
or the "bottom half" bracketed team wins. Thus we have 16*4*2 = 128 
total outcomes. Listing them (R1 winners/R2 winners/R3 winner):

     ACEG/AE/A   ACEG/AG/A   ACEG/CE/C   ACEG/CG/C
     ACEG/AE/E   ACEG/AG/G   ACEG/CE/E   ACEG/CG/G
     ACEH/AE/A   ACEH/AH/A   ACEH/CE/C   ACEH/CH/C
     ACEH/AE/E   ACEH/AH/H   ACEH/CE/E   ACEH/CH/H
     ACFG/AF/A   ACFG/AG/A   ACFG/CF/C   ACFG/CG/C
     ACFG/AF/F   ACFG/AG/G   ACFG/CF/F   ACFG/CG/G
     ACFH/AF/A   ACFH/AH/A   ACFH/CF/C   ACFH/CH/C
     ACFH/AF/F   ACFH/AH/H   ACFH/CF/F   ACFH/CH/H
     ADEG/AE/A   ADEG/AG/A   ADEG/DE/D   ADEG/DG/D
     ADEG/AE/E   ADEG/AG/G   ADEG/DE/E   ADEG/DG/G
     ADEH/AE/A   ADEH/AH/A   ADEH/DE/D   ADEH/DH/D
     ADEH/AE/E   ADEH/AH/H   ADEH/DE/E   ADEH/DH/H
     ADFG/AF/A   ADFG/AG/A   ADFG/DF/D   ADFG/DG/D
     ADFG/AF/F   ADFG/AG/G   ADFG/DF/F   ADFG/DG/G
     ADFH/AF/A   ADFH/AH/A   ADFH/DF/D   ADFH/DH/D
     ADFH/AF/F   ADFH/AH/H   ADFH/DF/F   ADFH/DH/H
     BCEG/BE/B   BCEG/BG/B   BCEG/CE/C   BCEG/CG/C
     BCEG/BE/E   BCEG/BG/G   BCEG/CE/E   BCEG/CG/G
     BCEH/BE/B   BCEH/BH/B   BCEH/CE/C   BCEH/CH/C
     BCEH/BE/E   BCEH/BH/H   BCEH/CE/E   BCEH/CH/H
     BCFG/BF/B   BCFG/BG/B   BCFG/CF/C   BCFG/CG/C
     BCFG/BF/F   BCFG/BG/G   BCFG/CF/F   BCFG/CG/G
     BCFH/BF/B   BCFH/BH/B   BCFH/CF/C   BCFH/CH/C
     BCFH/BF/F   BCFH/BH/H   BCFH/CF/F   BCFH/CH/H
     BDEG/BE/B   BDEG/BG/B   BDEG/DE/D   BDEG/DG/D
     BDEG/BE/E   BDEG/BG/G   BDEG/DE/E   BDEG/DG/G
     BDEH/BE/B   BDEH/BH/B   BDEH/DE/D   BDEH/DH/D
     BDEH/BE/E   BDEH/BH/H   BDEH/DE/E   BDEH/DH/H
     BDFG/BF/B   BDFG/BG/B   BDFG/DF/D   BDFG/DG/D
     BDFG/BF/F   BDFG/BG/G   BDFG/DF/F   BDFG/DG/G
     BDFH/BF/B   BDFH/BH/B   BDFH/DF/D   BDFH/DH/D
     BDFH/BF/F   BDFH/BH/H   BDFH/DF/F   BDFH/DH/H

Let's see if we can find the pattern.

     # Teams     # Games     # Outcomes
     -------     -------     ----------
        2           1             2
        4           3             8
        8           7           128

The number of games is always one less than the number of teams in the 
tournament. This makes sense, since every game eliminates one team, 
and all but one team must be eliminated. Can you see a relation 
between the number of games and the number of outcomes? It's not a 
factorial relation, but an exponential one.

In the NCAA tournament there are 64 teams, so there are 63 games 
played. As long as you're only interested in determining the odds of 
predicting ALL the winners, the "rounds" don't matter. Once you're 
considering round two, you're only interested in the situation where 
all the round one winners were correctly predicted. Thus, you don't 
have to concern yourself with situations where one or both of the 
teams in the game were not predicted to advance to round two.

I'll leave it to you to compute the number of possible outcomes. (And 
yes, it is an EXTREMELY large number!) Of course, not all of these 
outcomes are equally likely...

- Doctor TWE, The Math Forum
  http://mathforum.org/dr.math/