Originally Posted by 1in11 View Post
The craps table example - where one die lands on 2 and the other die is hidden - is not equivalent to the question posed. Once you know for certainty the value of a specific die is 2, then the probability of 2-2 is 1 in 6.

The question posed is "at least one of the dice is a 2." For this question, we do not know if it is the first die that is a 2, or if it is the second die that is a 2 (for how ever you want to define "first die" and "second die"). The person who viewed the dice knows this information, but we do not. This is where the 1 in 11 answer comes from - there are 11 ways to roll at least one 2 (2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2), and each of these has the same probability. Of these 11 ways, only 1 way satisfies the condition that "both dice are showing a 2." Hence, 1 in 11.
No, I am going to stick with the craps table example and once again I am going to post the original question on the Wizard's Forum:

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."

What is the probability that both dice are showing a 2?

It doesn't matter which of the dice is a 2 -- the answer is still 1/6.
Even if both dice showed a 2 -- the answer is still 1/6.

This exercise of yours (and everyone else including the Wizard) where you list the 11 ways to roll at least one 2 (2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2) is meaningless.

When you have a 2 showing on one die -- there is only a 1/6 chance of a 2 showing on the second die. And again, it doesn't matter which die the 2 is showing on. And it doesn't matter if a 2 is showing on both dice because the answer is still 1/6.

I am afraid that all of you "overthought" this question.

I am going to copy this and post it on the Wizard's forum also.

Thanks for joining my forum.