5 ms·
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 corr
by 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 agoI 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 agoYour 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 agoYou'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.