re: Math guys- Statistics Question

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The answer is 1/11. This is reminiscent of the Monty Hall problem.

The fact that she held one then drew another doesn't matter. It can be treated as one act. She drew two.

The fallacy of 1/21 is you are looking for a specific sock. As if what are the odds of getting THE "left green sock". It's not that. If you drew them one at a time your first odds were not 1/22. They were nothing. There were not odds.

In reality you are looking for ANY TWO SOCKS THAT MATCH

Picking one sock first didn't change that goal. You didn't begin by looking for THE "right green sock". Your criteria don't change just because you put it in two meaningless steps. The division in time between drawing each sock did nothing.

The odds of picking two matching socks is 1/11

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Uhhh.


That is absolutely not right.

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The odds of picking two matching socks is 1/11

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Its asking about the probability of picking the matching sock of those left in the drawer, not the probability that you remove two of the same socks if you picked at random. That is very different

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Its asking about the probability of picking the matching sock of those left in the drawer, not the probability that you remove two of the same socks. That is very different

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That's actually the same act phrased differently. Regardless, the probability is the same.

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Seriously, am I missing something?

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No.

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Why is this so difficult to figure out?

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It's not. The answer is 1/21.

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Its asking about the probability of picking the matching sock of those left in the drawer, not the probability that you remove two of the same socks. That is very different

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Why don't the odds add up then?

There can only be 11 different first socks. Because they are interchangeable pairs.

So Sock 1 has a 1/21 chance of getting its mate
Sock 2 has a 1/21 chance
Etc

To Sock 11


That's only 11/21. What are the other 10 possibilities? Answer that.

The first sock is random and part of an interchangeable pair. Those conditions don't change because it got selected

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Unless I am missing something...

She has 11 pair of socks shuffled in a drawer. That's 22 socks.

She picks out 1 sock, that leaves her with 21 socks in the drawer. If she is trying to pick the sock that matches the first sock she took out then she has 1 and 21 shot at picking that sock..

Seriously, am I missing something? Why is this so difficult to figure out?

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My whole premise for starting the thread. That's exactly how I see it, but for some reason feel like it's too easy and I'm missing something.

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The answer is 1/11. This is reminiscent of the Monty Hall problem.

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Good answer. How sure are you?

Total number of two-sock selections possible, ignoring which sock is picked first, is C(22,2) = 231. There are 11 pairs of socks. 11 / 231 is the same as 21:1.

If we introduce ordering, P(22,2) is 432. There are 22 ordered sock-pairs. 22 / 432 = 21:1 as well.

The Monty Hall problem had an extra wrinkle that is not present here: Monty revealed information after the initial selection. There's nothing like that here.

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Chegg has the answer.

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Ha, I actually just came across that site for the first time asking about another question that I have no idea what the answer is. Of course they have it, but it's $14.95 a month.

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genro

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Maybe this will help you. Instead of doing the whole one-sock-two-sock deal, let's rephrase the problem to just say she grabs two random socks, what is the probability that she grabbed a matching pair?

First we need to figure out the total number of different pairs we can possibly make, matching or not. There is a formula to calculate the number of combinations, and this operation is usually called "n choose k", which in this case with 22 socks and choosing 2 at a time would be "22 choose 2". The formula:

n!/((k)!(n-k)!)

(22)! / ((2)!(22-2)!)

(22 * 21 * 20!) / ((2)!(20)!)
(22 * 21) / 2
462 / 2
231

So there are 231 different total combinations of pairs of socks, and of those 11 are matching. So your odds of blindly choosing a matching pair is 11/231, or 1/21.