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Posers and Puzzles

Posers and Puzzles

  1. Standard member uzless
    The So Fist
    09 Mar '09 17:02
    you have four sets of numbers from 1 to 16 for a total of 64 numbers.

    You put them all in a hat.

    12 people draw 5 numbers each out of the hat.

    What are the odds that at least one person will not draw a number from 1-4?
  2. Standard member forkedknight
    Defend the Universe
    09 Mar '09 17:45 / 1 edit
    Originally posted by uzless
    you have four sets of numbers from 1 to 16 for a total of 64 numbers.

    You put them all in a hat.

    12 people draw 5 numbers each out of the hat.

    What are the odds that at least one person will not draw a number from 1-4?
    I didn't take the time to compute the full probability, but I figured for the worst case, being that you have the greatest probability of drawing a low number (1-4) if all of the previous drawers drew a single low number.

    The lower bound for the probability of someone NOT drawing a low number is 84.4%

    First first drawer has a probability of 22.5% of drawing all high numbers.

    Now assume the first drew a single low number. The second person now has a probability of (100-22.5)(21.7) = 16.8% of drawing all high numbers.

    Etc, etc... adding up all the conditional probabilities provides the worst case probability of 84.4%
  3. 09 Mar '09 17:50 / 1 edit
    Is the qeustion about odds, or about probability? It's not the same thing.
    A high odds is a low probability, yet, in betting odds <> 1/probability.
  4. Standard member forkedknight
    Defend the Universe
    09 Mar '09 17:56
    Originally posted by FabianFnas
    Is the qeustion about odds, or about probability? It's not the same thing.
    A high odds is a low probability, yet, in betting odds <> 1/probability.
    I think you are incorrect about a high odds being a low probability. If you have a 1/64 chance of winning, your odds are 63:1 against winning and your probability is 98.4% against winning. In other words, they are entirely the same thing.
  5. Standard member PBE6
    Bananarama
    09 Mar '09 17:58
    Originally posted by uzless
    you have four sets of numbers from 1 to 16 for a total of 64 numbers.

    You put them all in a hat.

    12 people draw 5 numbers each out of the hat.

    What are the odds that at least one person will not draw a number from 1-4?
    Just to clarify, it sounds like there are 16 available top picks, call them 1A-4A, 1B-4B, 1C-4C and 1D-4D. I think the easiest way to solve this is to work backwards and find the probability that all 12 players will get at least one top pick, and then subtract this probability from 1 to find the answer.

    First off, there are 64c60 ways (i.e. "64 choose 60" or 64!/(60!*4!) ways) to choose which picks will get assigned, and then (60c5)*(55c5)*...(10c5)(5c5) ways to assign those picks to the players 5 at a time, so there are (64c60)*(60c5)*(55c5)*...(10c5)*(5c5) assignments in total.

    Now, in order to ensure all 12 players receive a top pick we choose 12 out of the 16 top picks and assign one to each player randomly. Then, of the remaining 52 choices we choose 48 and assign them 4 at a time. The total number of ways to do this is (16c12)*12!*(52c4)*(48c4)*(44c4)*...(8c4)*(4c4). The 12! comes from the random assignment of the 12 top picks, because (12c1)*(11c1)*...(2c1)*(1c1) = 12!.

    The overall probability that each player receives at least one top pick is just the ratio of these:

    P = (16c12)*12!*(52c48)*(48c4)*(44c4)*...(8c4)*(4c4) / (64c60)*(60c5)*(55c5)*...(10c5)*(5c5) = 13.5%

    The probability that at least one player does not receive any top picks is simply 1-P, which is approximately 86.5%.
  6. Standard member forkedknight
    Defend the Universe
    09 Mar '09 18:01
    In terms of "odds", your worst case odds are (100-84.4)/84.4 = better than 5.41:1 that someone won't draw a low number.
  7. 09 Mar '09 18:12
    Originally posted by forkedknight
    I think you are incorrect about a high odds being a low probability. If you have a 1/64 chance of winning, your odds are 63:1 against winning and your probability is 98.4% against winning. In other words, they are entirely the same thing.
    What's the odds of winning on one number in roulette? What is the probability of winning? Hence - not the same. The difference is the house's winnings.
  8. Standard member forkedknight
    Defend the Universe
    09 Mar '09 19:36
    Originally posted by FabianFnas
    What's the odds of winning on one number in roulette? What is the probability of winning? Hence - not the same. The difference is the house's winnings.
    Odds against winning on any one number in roulette: 37:1
    Probability of winning 1/38 = 2.63%
    Probability of not winning = 37/38 = 97.37%
    Probability of not winning / Probability of winning = 97.37/2.63 = 37:1

    I'm failing to follow...
  9. Standard member forkedknight
    Defend the Universe
    09 Mar '09 19:37 / 1 edit
    Originally posted by forkedknight
    Odds against winning on any one number in roulette: 37:1
    Probability of winning 1/38 = 2.63%
    Probability of not winning = 37/38 = 97.37%
    Probability of not winning / Probability of winning = 97.37/2.63 = 37:1

    I'm failing to follow...
    Now, the payout in roulette is 35:1, but that has nothing to do with your odds of winning...

    We have no idea what the payout of drawing no low numbers is, so how is that in any way relevant?
  10. 09 Mar '09 21:22
    Originally posted by forkedknight
    Odds against winning on any one number in roulette: 37:1
    Probability of winning 1/38 = 2.63%
    Probability of not winning = 37/38 = 97.37%
    Probability of not winning / Probability of winning = 97.37/2.63 = 37:1

    I'm failing to follow...
    Put a dollar on number 17. You have one chance of 37 of winning, right. How much do you win if the croupier's ball decides to hit the slot #17? Not 37 dollars, I don't know how much, but the estimated value is not 1.0 for sure.

    But if you chance on red, and red only, the estimated value E(p) is exactly 36/37 or nearly 1.0

    Odds should be 1/p, where p=probability, but it is not. The casino, have to win some too. Estimated value or E(p) is not exactly 1.0
  11. Standard member forkedknight
    Defend the Universe
    10 Mar '09 00:29 / 1 edit
    Originally posted by FabianFnas
    Put a dollar on number 17. You have one chance of 37 of winning, right. How much do you win if the croupier's ball decides to hit the slot #17? Not 37 dollars, I don't know how much, but the estimated value is not 1.0 for sure.

    But if you chance on red, and red only, the estimated value E(p) is exactly 36/37 or nearly 1.0

    Odds should be 1/p, where ...[text shortened]... y, but it is not. The casino, have to win some too. Estimated value or E(p) is not exactly 1.0
    I will resubmit that a casino's payouts have no bearing on the odds or probability of winning.

    The odds of winning back your dollar on a roulette wheel are different than the odds of winning a spin, and I think you are confusing these two different probabilities.

    The odds of winning a spin on a (US) roulette wheel is 1/38, providing the odds against winning of 37:1
    The payout of winning a spin on a roulette wheel is generally 35:1
    This provides a net expected value of:
    (-1 * 37/38) + (35 * 1/38) = -0.0526 on the dollar
  12. 10 Mar '09 08:22
    Originally posted by forkedknight
    I will resubmit that a casino's payouts have no bearing on the odds or probability of winning.

    The odds of winning back your dollar on a roulette wheel are different than the odds of winning a spin, and I think you are confusing these two different probabilities.
    I see the differences between the odds and the probability.

    Firstly - betting in a casino isn't fair. The house are always winning in the long run, always. Using odds as a measure of probability is assuming the betting is fair.

    Let's make an experiment: I flip a fair coin and invites people to bet of the outcome. I say: "You bet ten dollars that you can guess the outcome of one flip. Right? If you lose you lose the ten dollar. But if you win you win your 10 dollar back, and another one dollar on top!"

    This betting is not fair, of course. The odds for the bet is 1 to 10, yet the probability is 0.5. Here the odds are not 1/P = 2, it's 10!

    But if you get two dollars at a win, then the betting is fair and odds are 1/p.

    Betting companies know this, and by tweaking the odds, then they can make a profit out of it. Who will win the next football match? What is the estimated probability and what are the odds for that. The difference is the profit for the betting company.

    If we still disagree, then I cannot explain it further, so this will be the last posting in the matter.
  13. Standard member Palynka
    Upward Spiral
    10 Mar '09 10:56 / 1 edit
    FabianFnas, I think you are mixing two different things.

    In probability theory, the odds of an event are defined as being p/(1-p) where p is the probability of the event.

    Bookmaker's odds, which I think are the ones you are thinking of, are defined using payoffs as primitives (and not probabilities). In that sense you would be correct but, since nothing is said about payoffs, I think using the term as in the above paragraph makes more sense in this context.
  14. 10 Mar '09 11:16
    Originally posted by Palynka
    FabianFnas, I think you are mixing two different things.

    In probability theory, the odds of an event are defined as being p/(1-p) where p is the probability of the event.

    Bookmaker's odds, which I think are the ones you are thinking of, are defined using payoffs as primitives (and not probabilities). In that sense you would be correct but, since nothi ...[text shortened]... out payoffs, I think using the term as in the above paragraph makes more sense in this context.
    You're quite right. I settle with this.