Originally Posted by OneHitWonder View Post
A specific roll which is (an element) of a defined set of rolls is different from a roll which is a defined set of rolls. A major difference, but one which may seem an unobvious nuance to someone with virtually no algebraic experience.

The problem as written here doesn't address a specific roll which "is (an element) of" a set of rolls. It addresses the specific roll as, "At least one of the dice is a 2." It does not mention the words "is (an element) of" the "at least one of the dice is a 2" set of rolls of the 1/11 chance answer.

The roll in the problem by which "at least one of the dice is a 2" means a roll by which one or two of the dice is a 2. This definitely doesn't include a non-2 with a 2, or 2-2. Eg, an end roll of 6-2 isn't included in one or two of the dice is a 2, because 6-2 has no possibility of two of the dice is a 2. However, a specific roll of 2-X, where X has yet to be determined, has the possibility of one or two of the dice is a 2.



The set of rolls defined by "at least one 2" definitely does include {a non-2 with 2} or {2-2} but is different from a specific roll defined by "at least one 2". The latter doesn't include a non-2 with 2, or 2-2. It's the set of rolls of 2-X, where the X-die has yet to be looked at: {2-(1), 2-(2), 2-(3), 2-(4), 2-(5), and 2-(6)}. An end roll of this set is either 2-1, 2-2, 2-3, 2-4, 2-5, or 2-6.


You apply the nonsense word as "a crutch" to not make the effort to understand what I and others have written, and then to not write an intelligent reply. Even a nonsense idea deserves an intelligent reply, as I continue to reply to you.

In my opinion, this is, foremost, a math question. Not a gambling forum, or court of law, exercise. To this end, I will post it up at one of the math forums, along with a summary of my own arguments of course, when I have some more free time.

It doesn't surprise me that neither the Wizard nor Alan has chosen such a form of "binding arbitration". They aren't the math and formal logic experts. Aside from redietz, and perhaps Singer, I get the feeling that a lot of this went over everyone's head. Same at the Wizard's.
I'm using "nonsense" as a short and to the point way to express my believe you are barking the wrong tree here with your desire to dismantle a simple logical math question into philosophical argument about the meaning of a phrase, detail twisting and
challenging the very structure of math-semantic-logic up to the point where every effort
to make sense of anything becomes futile.
I can do that as well. You tell me something simple, such as "Good day." and I can start questioning, arguing and wooden philosophizing the "real" meaning of the phrase to the end of all days. But I choose not to, because to me, such exercise is an utter waste of time and demonstrates perfectly how wooden philosophizing on simple things can lead to total paralysis of the mind and breaks all possible ways of communication. That's all it accomplishes.

Why don't you bring the question to a math experts( or semantics experts or both) dedicated forum or blog and see how they interpret it and answer it? I don't think discussion as the one you seek belongs in a gambling forum.

I bet the math experts there will first clear one thing - the intention of the peeker, and then will give you the correct answer in identical manner as this answer:

"There are 36 combinations of two 6-sided dices, only 11 of them show at least a 2, from which one is 2-2. Hence the probability would be 1/11.

However, this would only hold if the player would say nothing when no 2 were spotted.

If (before peeking) your friend is determined to say "at least one of the dice is a X" where X is one of the numbers he may spots, the probability for a pair would be different.

If the friend sees any pair (1/6), probability of "X-X" where X is the named number, is one.
If the friend sees a non-pair (5/6), probability of "X-X" is zero whatever he choses for X.

Hence, like the Monty-Hall-Problem, if your friend must always say "At least one of the dice is a X" (while not lying), probability of a pair is 1/6."