Sloppy Joe

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Posers and Puzzles 15 Apr '21 13:29
  1. R
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    15 Apr '21 13:291 edit
    Sloppy Joe doesn't pair his socks in the drawer. There are currently 6 white socks and 4 black socks in the drawer. If Sloppy Joe reaches into the drawer and randomly pulls 2 socks, what is the probability the socks will be mismatched?
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    15 Apr '21 20:34
    @joe-shmo said
    Sloppy Joe doesn't pair his socks in the drawer. There are currently 6 white socks and 4 black socks in the drawer. If Sloppy Joe reaches into the drawer and randomly pulls 2 socks, what is the probability the socks will be mismatched?
    The way of solving this with the least amount of thinking - I'm sure there is a more fundamental way, but I can't be fundamented to work it out- is:

    - either he draws a white sock first, with probability 6/10. If so, there are now 5 white and 4 black socks left, and the probability of the next sock being a mismatch are 4/9.
    - or he draws a black sock first, with probability 4/10. There are now 3 black socks and 6 white ones left, so the probability of a mismatch is now 6/9.

    So the total is 6/10 * 4/9 + 4/10 * 6/9; that is, (6*4)/90 + (4*6)/90. This is beautifully symmetric, which I rather think I should've realised before. In any case, the end result is 2*(4*6)/90 = 48/90 = 8/15 =0.5(3), so a little more than half.

    Nice problem.
  3. R
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    15 Apr '21 20:42
    @shallow-blue said
    The way of solving this with the least amount of thinking - I'm sure there is a more fundamental way, but I can't be fundamented to work it out- is:

    - either he draws a white sock first, with probability 6/10. If so, there are now 5 white and 4 black socks left, and the probability of the next sock being a mismatch are 4/9.
    - or he draws a black sock first, with prob ...[text shortened]... e, the end result is 2*(4*6)/90 = 48/90 = 8/15 =0.5(3), so a little more than half.

    Nice problem.
    Correct!
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    15 Apr '21 21:21
    @joe-shmo said
    Correct!
    Of course - since I did it arithmetically - but I'd still like to know if there is a solution based on principles rather than mere calculation.

    Unfortunately, when it comes to probabilities, there often isn't. (Knot theory is even worse.)
  5. R
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    15 Apr '21 23:327 edits
    @shallow-blue said
    Of course - since I did it arithmetically - but I'd still like to know if there is a solution based on principles rather than mere calculation.

    Unfortunately, when it comes to probabilities, there often isn't. (Knot theory is even worse.)
    I'm not sure what principles you are looking for, but they are all implicit in your calculation.

    The set can either be matched or not matched. The probabilities of each event are independent, and thus additively form the space of all possible outcomes.

    P ( mismatched ) = 1 - P( matched )

    You can compute the LHS directly ( as you have) or indirectly from the RHS.

    You might instead use combinatorial argument:

    WWWWWW BBBB

    To calculate P ( matched )

         P( matched ) = ( C(6,2) + C(4,2) ) / C(10,2)
          
               = ( 15 + 6 )/45

               = 7/15
    Then it follows that:

    P ( mismatched ) = 1 - P( matched )
           
            = 1 - 7/15

            = 8/15       

    However, there is nothing wrong with treating the draws as chronologically ordered as you have done, as I believe simultaneously drawing two things is a probability zero event anyhow.

    Hope that is what you are looking for, if not your going to have to elaborate a bit further.

    P.S. if you are interested, there are still some unsolved puzzle in the lineup below.

    Topic: Probabilty

    https://www.redhotpawn.com/forum/posers-and-puzzles/joe-should-go-to-bed.188751

    Topic: Algebra

    https://www.redhotpawn.com/forum/posers-and-puzzles/one-train-two-train-red-train-blue-train.188801
  6. RSA
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    16 Apr '21 07:30
    @shallow-blue said
    Of course - since I did it arithmetically - but I'd still like to know if there is a solution based on principles rather than mere calculation.

    Unfortunately, when it comes to probabilities, there often isn't. (Knot theory is even worse.)
    It's not too tedious to do what you did in this case and sum the products of the conditional probabilities. But when you have more draws it helps to find a general solution.

    This is the mass function for this random variable:

    ((KCk)((N-K)C(n-k)))/(NCn)

    Where:

    N is the number of socks
    K is the number of white socks
    n is the number of draws
    k is the number of white socks in your draws

    To meet the conditions for a mismatch, set k = 1.

    ((6!)/(1!)(6-1)!)((4!)/(1!)(4-1)!)/((10!)/(2!)(10-2)!) = (6 x 4)/45 = 8/15.

    So, this is hypergeometrically distributed. It's a useful distribution - change the numbers as necessary.
  7. Subscribervenda
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    16 Apr '21 12:31
    @ashiitaka said
    It's not too tedious to do what you did in this case and sum the products of the conditional probabilities. But when you have more draws it helps to find a general solution.

    This is the mass function for this random variable:

    ((KCk)((N-K)C(n-k)))/(NCn)

    Where:

    N is the number of socks
    K is the number of white socks
    n is the number of draw ...[text shortened]... his is hypergeometrically distributed. It's a useful distribution - change the numbers as necessary.
    First of all,welcome to Joe's problem page.(not reallyJoe's but he is posting most of the problems).
    I asked earlier in the thread about a formula I had which looks to be the same as your mass function example so thanks for that.
    I'll have another look later.
  8. R
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    16 Apr '21 12:48
    @venda said
    First of all,welcome to Joe's problem page.(not reallyJoe's but he is posting most of the problems).
    I asked earlier in the thread about a formula I had which looks to be the same as your mass function example so thanks for that.
    I'll have another look later.
    Most of these are borrowed problems...not my own. I'm just sharing the wealth!
  9. R
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    16 Apr '21 12:53
    @ashiitaka said
    It's not too tedious to do what you did in this case and sum the products of the conditional probabilities. But when you have more draws it helps to find a general solution.

    This is the mass function for this random variable:

    ((KCk)((N-K)C(n-k)))/(NCn)

    Where:

    N is the number of socks
    K is the number of white socks
    n is the number of draw ...[text shortened]... his is hypergeometrically distributed. It's a useful distribution - change the numbers as necessary.
    Yeah, I don't know if I would have ever made it to that generalization. Nice work!
  10. Joined
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    16 Apr '21 17:34
    @ashiitaka said
    It's not too tedious to do what you did in this case and sum the products of the conditional probabilities. But when you have more draws it helps to find a general solution.
    Yup, that's what I meant. But I could never have found that generalisation.
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