Originally Posted by Alan Mendelson View Post
Kewl you wrote "it's done before the roll." That alters reality.

The original problem specifically states that two dice were rolled and at least one is showing a 2. Whatever you thought prior to that no longer matters. Now you have to deal with at least one die showing a two. The question now centers on the second die in the problem which may or may not be a two and the chances for that second die to be a 2 is 1/6.

By figuring probability before the roll alters the reality of the question. You are changing the facts and the conditions of the problem.

As I have said a hundred times, your 1/11 is "good math" but for a different question -- perhaps for a question that asks: "how many different combinations of two dice have at least one 2 and of those combinations how many would show 2-2?" That would be your 1/11 answer.
This is pretty good Alan.

The only change I would make is to substitute "how many different combinations of two dice involve the number 2 and of those..." for "how many different combinations of two dice have at least one 2 and of those...". Exactly one 2 is (an element) of at least one 2 - which is the same as one 2 or two 2's - but one 2 isn't (defined, or normally referred to, as) at least one 2. {One 2} isn't {one 2 or two 2's}. Those are two different sets.

Alan, you did a good thing by pointing out that this problem isn't written for the 1/11 chance answer in theory. The original problem supplies only a specific roll, or rolls of a specific nature; and nothing has been restricted about the number 2 beyond the "peeker's" specific observation.

To view also a specific roll theoretically - in the probability or "how often" sense (given that the combinations of two dice involve the number 2) - ask how often will the left die involve a 2. Half the time; the other half of the time the right die will involve a 2. Hence, half the time the other die will show a 2 given the number 2 somewhere. When it's the left die, the other die will show a 2 with 1/6 chance; and, when it's the right die, the other die will show a 2 with 1/6 chance. Now, we put this into a calculation as did the 1/11 chance answerers. There are two parts to this calculation:

(1/2 X 1/6) + (1/2 X 1/6) = 1/12 + 1/12 = 1/6.

This is the specific roll or rolls theoretical counterpart to the 1/11 chance answer. The way to perform the calculation if going by a specific roll or rolls in theory. All specific rolls considered, in theory, in terms of which side the roll or rolls involve the number 2.

This calculation may be done also within the 1/11 chance calculation:

(5/11 X 1/6) + (5/11 X 1/6) + {(1/22 X 1/6) + (1/22 X 1/6)} = 1/6.
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Note that the only way to arrive at the 1/6 chance answer in and of itself for a specific roll or rolls is to find a 2 on the only die looked at. This is the exact manner in which the original problem is written.

Again, exactly one 2 is (an element) of at least one 2 - which is the same as one 2 or two 2's - but one 2 isn't (defined, or normally referred to, as) at least one 2. {One 2} isn't {one 2 or two 2's}. Those are two different sets. The reason to require "one die or the other is a 2" - the same as involving the number 2 - instead of "at least one two". "One die or the other is a 2" would allow for the specific roll or rolls to read the same as the general roll or rolls, though waiting on a 2 would still be a requirement in advance. This is not the original problem as written.