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Some green socks, some black socks, their total is four.
Some green socks, some black socks, some socks on the floor.
The chance of getting two black ones, the chance of getting two black ones, that chance is a third.
On the first draw, on the first draw, and there are no more draws, no more draws, no more draws.
Some green socks, some black socks, some socks on the floor.
What's the chance, what's the chance, of getting exactly one green sock, exactly one green sock, I wonder?
And remember, and remember, their total is four!
Some green socks, some black socks, some socks on the floor.

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closed as off-topic by Rand al'Thor, Deusovi♦ Sep 17 '16 at 23:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Have you read Are math-textbook-style problems on topic? on meta? $\endgroup$ – Rand al'Thor Sep 17 '16 at 23:03
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    $\begingroup$ Just for fun, @randal'thor, suppose the probability of 2 blacks socks had been 1/3... that would not occur in a mathematics class and would require lateral-thinking $\endgroup$ – humn Sep 17 '16 at 23:07
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    $\begingroup$ @TheBitByte Yes, and? The "basic maths exercise" close reason still exists. Questions like this aren't what PSE is for. $\endgroup$ – Rand al'Thor Sep 17 '16 at 23:29
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    $\begingroup$ Looking forward to enjoying how lateral the thinking will get! $\endgroup$ – humn Jan 29 '17 at 22:20
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    $\begingroup$ @Close voters, this question has been fixed/modified. Please review it now and let me know if there's anything more I can improve. $\endgroup$ – Buffer Over Read Feb 5 '17 at 3:17
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The latest edit to the question makes this answer invalid. It is in the process of being updated.

No chance of getting 2 green socks(!), if the probability of getting two black socks is 1/2. The latest edit to the question makes it 1/3, so this answer is in the process of being updated.

...because there are...

...3 black socks but only 1 green sock.

This was derived by using $b$ for the number of black socks, so the probability of getting 2 would be...

... $1/2 ~ = ~ b/4 {\small\times} (b{-}1)/3 ~ = ~ b(b{-}1)/12 ~ = ~ 6/12$ ...

...which leaves only one possible value for $b$.

Even better than having 2 green socks, though, is this straightforward way to verify this conclusion, thanks to Peregrine Rook:

Since you are dividing the four socks randomly into two groups of two (the two that you take out and the two that you leave behind), there’s a 50% probability that you select the green sock and a 50% probability that you don’t. But you get two black socks if and only if the green sock is one of the ones you leave in the box.

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    $\begingroup$ Here’s a slightly simpler way of doing the “check”: since you are dividing the four socks randomly into two groups of two (the two that you take out and the two that you leave behind), there’s a 50% probability that you select the green sock and a 50% probability that you don’t. But you get two black socks if and only if the green sock is one of the ones you leave in the box. $\endgroup$ – Peregrine Rook Sep 19 '16 at 5:14

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