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15 Apr 21
@joe-shmo saidThe 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:
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?
- 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.
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15 Apr 21
@shallow-blue saidCorrect!
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.
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15 Apr 21
@joe-shmo saidOf 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.
Correct!
Unfortunately, when it comes to probabilities, there often isn't. (Knot theory is even worse.)
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15 Apr 21
7 edits
@shallow-blue saidI'm not sure what principles you are looking for, but they are all implicit in your calculation.
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.)
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
16 Apr 21
@shallow-blue saidIt'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.
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.)
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.
16 Apr 21
@ashiitaka saidFirst of all,welcome to Joe's problem page.(not reallyJoe's but he is posting most of the problems).
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.
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.
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16 Apr 21
@venda saidMost of these are borrowed problems...not my own. I'm just sharing the wealth!
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.
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16 Apr 21
@ashiitaka saidYeah, I don't know if I would have ever made it to that generalization. Nice work!
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.
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16 Apr 21
@ashiitaka saidYup, that's what I meant. But I could never have found that generalisation.
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.