3 ms·

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 e

by 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.

by graywh 17y ago

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.