More dice trick questions

I hope these aren't too hard for the basic statistics forum, but some friends and I are disagreeing on the answers to these two questions:

1) Show that if two dice are loaded with the same probability distribution, then the probability of doubles is at least 1/6.

2) If P2 is the probability of a 2 showing and we are told that at least one 2 has shown, what is the probability that upon looking at the pair of dice the sum will be 4?

Now for each question:

1) My initial solution to this was to show that the probability of doubles on die with even distributions are 1/6, so any variation in the distribution leads to a smaller number. But I can't quite prove this algebraically and I'm stuck. Is my approach wrong?

2) The simple answer (which some are sticking to) is that it's simple: The answer is P2. But I don't believe it could be that easy, and here's the answer I came up with:

Let A = the event where the sum is 4

B = The event where at least one 2 shows up

A is equal to the probability of getting a (1,3), (3,1), or (2,2). Since we know one die is a two, the probability of 1,3 or 3,1 is zero and P[A] = P[(2,2)].

As for B:

$\displaystyle P[B] = 1 - P[not B] = 1 - (1-P_2 )(1-P_2 ) = P_2 (2 - P_2 )$

Now, we want to know P[A|B]:

$\displaystyle P[A|B] = \frac{P[A \cap B]}{P[B]} = \frac{P[(2,2)]}{P[B]} = \frac{P_2^2}{P_2 (2 - P_2 )} = \frac{P_2}{2-P_2}$

I can't explain it...it just *feels* wrong. Like I did too much. Is the simple answer really correct or is this one right?