The Wizard will bank this bet: 1/6 vs 1/11

[kewl]

How is it different if the peeker said one is a 3 instead of a 2. The question is still what is on the other die. That is always going to be 1 of 6.

Let me rephrase.

If the question was:

You roll a pair of dice and no matter what the outcome is the peeker will announce the value of one of the dice (at random). What are the odds that there will be a: pair, a (announced value) and 1, a (announced value) and 2, a (announced value) and 3...whatever),
then the answer will be 1/6.

And if the question was (in your case with 5 and 3):

You roll a pair of dice. The peeker makes announcement when at least one 3 or one 5 is under the cup. What are the odds that there are exactly 5 and 3 under the cup?
then the answer is 1/11.

Now, compare the second question with the one asked here. Keep in mind that there was discussion about whether the peeker announces a value every time or only if there is a 2 under the cup and it was agreed upon, and Alan concurred, that the case in discussion is the one where the peeker only announces if there is a 2. Otherwise he says nothing.