8 ms·

The odds of winning the lottery are highly improbable yet every almost every week someone wins it. That said if he is that good reliably, there is lots of mone

by fleitz 17y ago

The odds of winning the lottery are highly improbable yet every almost every week someone wins it. That said if he is that good reliably, there is lots of money to be made on sports betting.

hackernews 17y ago
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.
  - 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 . . .
qeorge 17y ago
The lottery is a poor, if often used, comparison for the odds involved in picking the NCAA tournament correctly. Picking 2 rounds of the NCAA tournament perfectly is orders of magnitude harder than picking the winning lottery numbers. For starters, in the lottery: 1) No selection (number) is better than another. i.e., the winners are random 2) Each round has no effect on the next 3) The conditions of the experiment will not change The NCAA tournament is the opposite: 1) Some selections (teams) are clearly better than others. Furthermore, the tournament is structured to give the best teams the best chances of winning. i.e., selection is not random. 2) Each round limits the possible outcomes of the next. To have a chance of picking Round 2 correctly, you must have already aced Round 1. 3) Each game can be affected by an unlimited number of variables. For instance, if John Wall gets hurt it would alter the chances of a Kentucky national championship. More subtly, so could a delayed flight. Even picking the correct lottery numbers twice in a row is not a fair comparison, because the results of the first experiment have no bearing on the second.
- fleitz 17y ago
  Non-random events are generally a lot easier guess correctly at than random events. This is why people will do things like card count or use weighted dice / coins with two heads / tails, etc. This is also why SSL keys are generated from random data rather than weighted data. I'd love to play the lottery if I knew certain balls were weighted more heavily than others. If you think non-random events are difficult to predict, give me your longitude and latitude and I'll predict with in five minutes what time the sun will rise tomorrow. Want to bet $1000 against? The real reason the it is more difficult is because the odds randomly are 1 in 4 billion for the first round, where as a lottery like the 6/49 is 1 in 13,983,816. However, the NCAA is not completely random, it would be difficult to discern what the real odds are but the fact that many players are within a couple games should give you some clue to the decrease in odds given the extra information. If 100 million brackets are filled out and the realistic odds are 1 in 1.5 billion (given the extra information) then someone should fill out a perfect bracket every 15 years. If you look at the prizes being offered for a correct bracket, it should give you some indication as to the greatly reduced odds that one actually faces. If the odds are really 2^48 for a correct 2 round bracket then it would stand to reason that you could offer a 1 trillion dollar prize for a correct bracket with a ticket price of $1 and still have a house margin of 99.64% 100 - ((1 / ((2^48) / 1 trillion)) * 100) = 99.644 This simply isn't the case. (My math may be way off, I was never a stats major)

by fleitz 17y ago

The odds of winning the lottery are highly improbable yet every almost every week someone wins it. That said if he is that good reliably, there is lots of money to be made on sports betting.

hackernews 17y ago

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.
- 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 . . .

qeorge 17y ago

The lottery is a poor, if often used, comparison for the odds involved in picking the NCAA tournament correctly. Picking 2 rounds of the NCAA tournament perfectly is orders of magnitude harder than picking the winning lottery numbers. For starters, in the lottery: 1) No selection (number) is better than another. i.e., the winners are random 2) Each round has no effect on the next 3) The conditions of the experiment will not change The NCAA tournament is the opposite: 1) Some selections (teams) are clearly better than others. Furthermore, the tournament is structured to give the best teams the best chances of winning. i.e., selection is not random. 2) Each round limits the possible outcomes of the next. To have a chance of picking Round 2 correctly, you must have already aced Round 1. 3) Each game can be affected by an unlimited number of variables. For instance, if John Wall gets hurt it would alter the chances of a Kentucky national championship. More subtly, so could a delayed flight. Even picking the correct lottery numbers twice in a row is not a fair comparison, because the results of the first experiment have no bearing on the second.

fleitz 17y ago
Non-random events are generally a lot easier guess correctly at than random events. This is why people will do things like card count or use weighted dice / coins with two heads / tails, etc. This is also why SSL keys are generated from random data rather than weighted data. I'd love to play the lottery if I knew certain balls were weighted more heavily than others. If you think non-random events are difficult to predict, give me your longitude and latitude and I'll predict with in five minutes what time the sun will rise tomorrow. Want to bet $1000 against? The real reason the it is more difficult is because the odds randomly are 1 in 4 billion for the first round, where as a lottery like the 6/49 is 1 in 13,983,816. However, the NCAA is not completely random, it would be difficult to discern what the real odds are but the fact that many players are within a couple games should give you some clue to the decrease in odds given the extra information. If 100 million brackets are filled out and the realistic odds are 1 in 1.5 billion (given the extra information) then someone should fill out a perfect bracket every 15 years. If you look at the prizes being offered for a correct bracket, it should give you some indication as to the greatly reduced odds that one actually faces. If the odds are really 2^48 for a correct 2 round bracket then it would stand to reason that you could offer a 1 trillion dollar prize for a correct bracket with a ticket price of $1 and still have a house margin of 99.64% 100 - ((1 / ((2^48) / 1 trillion)) * 100) = 99.644 This simply isn't the case. (My math may be way off, I was never a stats major)