4 ms·

I heard before the tourney started that you'd have better odds to win the lottery, twice in one week than to pick a perfect bracket.

machrider 17y ago
Yes, I'm pretty sure there are a total of 63 games to pick (including all rounds), so the odds of picking them all correct are 1 in 2^63. If 300 million Americans filled out brackets, the odds of anyone picking it perfectly are still 1 in 30,744,573,500.
- qeorge 17y ago
  And he picked two rounds in a row. Since all 64 teams could theoretically make it to the second round, the odds are more like: 1 / (2^126) I'll have to leave it to Wolfram Alpha to expand that one: http://www.wolframalpha.com/input/?i=1+%2F+%282^63+*+2^63%29 http://www.wolframalpha.com/input/?i=1+%2F+%282^63+*+2^63%29
  - graywh 17y ago
    First round: 1 / 2^32 since there are 32 winners to pick, and 2 choices for each. Second round: 1 / 4^16 since there are 16 winners to pick and 4 choices for each. But this assumes you can pick a loser from round 1. So, it's actually 1 / 2^16 since there are 16 winners to pick and 2 choices for each. Picking 2 rounds correct is 1 / 2^32 * 1 / 2^16 = 1 / 2^48
    - qeorge 17y ago
      Nice! Thank you. One thing: "this assumes you can pick a loser from round 1" You can pick a loser from Round 1, because all selections are made before any games are played. So it looks like its: 1 / 2^32 * 1 / 4^16 (And we should probably add another 1/2 term to account for the play-in game =)
      - irq11 17y ago
        No. The chance of being incorrect in the first round is factored into the first term (since that term is calculating the probability of picking an entirely correct first bracket). The second term need only represent the chance of picking correctly from the teams in the second round, given that the first round was correctly predicted. In other words, it's a straightforward application of the chain rule for probabilities: p(A,B) = p(A) * p(B|A)
        qeorge 17y ago
        I don't think we disagree. Mind bearing with me for a moment? Probability of A (round 1 / selecting the right urn): 1 / 2^32 (32 games, 2 possible winners for each) Probability of B (round 2: 1 / 4^16 (16 games, 4 possible winners for each) Probability of A and B: (1 / 2^32) * (1 / 4^16) So it would appear the chain rule for probability agrees with my point. What am I missing?
        irq11 17y ago
        Your second term is wrong. There are 16 matches in the second round, and you have a 1/2 chance of picking the winner in each, not a 1/4 chance. The first term (.5^32) is the probability of picking all of the winners in round one (p(A)). The second term (p(B|A)), must be the probability of picking all the winners in round two, given that you picked correctly in round one. Again, that's .5^16, not .25^16.
        qeorge 17y ago
        You're right, thanks for explaining it. I still couldn't see it this morning, but I talked it out with my co-founder, and he got through to me. Thanks for sticking with it.
      - graywh 17y ago
        No, "loser from round 1" means a team you picked to lose. That is, you can't have Team A beat Team Z in round 1 and Team B beat Team Y, then have Team Z beat Team Y in round 2.
- houseabsolute 17y ago
  > 1 in 2^63. Assuming all teams have the same probability of winning. So, it's not nearly as bad as 2^63. Probably in any given game one team has a 0.66 chance of winning. That takes it down to a much more manageable 2^63 in 3^63. Wait . . .

machrider 17y ago

Yes, I'm pretty sure there are a total of 63 games to pick (including all rounds), so the odds of picking them all correct are 1 in 2^63. If 300 million Americans filled out brackets, the odds of anyone picking it perfectly are still 1 in 30,744,573,500.

qeorge 17y ago
And he picked two rounds in a row. Since all 64 teams could theoretically make it to the second round, the odds are more like: 1 / (2^126) I'll have to leave it to Wolfram Alpha to expand that one: http://www.wolframalpha.com/input/?i=1+%2F+%282^63+*+2^63%29 http://www.wolframalpha.com/input/?i=1+%2F+%282^63+*+2^63%29
- graywh 17y ago
  First round: 1 / 2^32 since there are 32 winners to pick, and 2 choices for each. Second round: 1 / 4^16 since there are 16 winners to pick and 4 choices for each. But this assumes you can pick a loser from round 1. So, it's actually 1 / 2^16 since there are 16 winners to pick and 2 choices for each. Picking 2 rounds correct is 1 / 2^32 * 1 / 2^16 = 1 / 2^48
  - qeorge 17y ago
    Nice! Thank you. One thing: "this assumes you can pick a loser from round 1" You can pick a loser from Round 1, because all selections are made before any games are played. So it looks like its: 1 / 2^32 * 1 / 4^16 (And we should probably add another 1/2 term to account for the play-in game =)
    - irq11 17y ago
      No. The chance of being incorrect in the first round is factored into the first term (since that term is calculating the probability of picking an entirely correct first bracket). The second term need only represent the chance of picking correctly from the teams in the second round, given that the first round was correctly predicted. In other words, it's a straightforward application of the chain rule for probabilities: p(A,B) = p(A) * p(B|A)
      - qeorge 17y ago
        I don't think we disagree. Mind bearing with me for a moment? Probability of A (round 1 / selecting the right urn): 1 / 2^32 (32 games, 2 possible winners for each) Probability of B (round 2: 1 / 4^16 (16 games, 4 possible winners for each) Probability of A and B: (1 / 2^32) * (1 / 4^16) So it would appear the chain rule for probability agrees with my point. What am I missing?
        irq11 17y ago
        Your second term is wrong. There are 16 matches in the second round, and you have a 1/2 chance of picking the winner in each, not a 1/4 chance. The first term (.5^32) is the probability of picking all of the winners in round one (p(A)). The second term (p(B|A)), must be the probability of picking all the winners in round two, given that you picked correctly in round one. Again, that's .5^16, not .25^16.
        qeorge 17y ago
        You're right, thanks for explaining it. I still couldn't see it this morning, but I talked it out with my co-founder, and he got through to me. Thanks for sticking with it.
    - graywh 17y ago
      No, "loser from round 1" means a team you picked to lose. That is, you can't have Team A beat Team Z in round 1 and Team B beat Team Y, then have Team Z beat Team Y in round 2.
houseabsolute 17y ago
> 1 in 2^63. Assuming all teams have the same probability of winning. So, it's not nearly as bad as 2^63. Probably in any given game one team has a 0.66 chance of winning. That takes it down to a much more manageable 2^63 in 3^63. Wait . . .