I think it's 1/3. Imagine he picks a white sock. There are now 2 pink socks (non-matching) and 1 white sock, leaving a 1 in 3 chance of a matching pair.
Agree. It doesn't matter which he draws first, the chance of drawing a matching sock is 1/3.
Also agreed. Doesn't matter if he draws them simultaneously or not, for the purposes of the problem, you can treat one draw as the "first."
Whichever color he draws first, there are now three socks left. One is the matching color, two are not. Therefore, the probability he will draw a match is 1/3.
That doesn't make any sense. There are 4 possible outcomes, not 3 possible outcomes. The clue isn't about picking one and then the odds of picking one that matches, it's picking TWO at random and whether those two match.
The person is removing a sock after he picks each. He is not replacing them (at least as worded in the opening topic, otherwise he's not ending up w/ a pair of socks in hand anyway). If he takes out a white sock, he cannot select that white sock again on his 2nd pick from what was originally four. His 2nd pick is among a pool of three because one possible option has been removed.
He's being handed two simultaneously. There are four possible outcomes. Ergo 2/4 or 1/2 ---