Predicting NCAA Tournament WinnersDate: 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/ |
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