This thread has been brought to the attention of users at http://math.stackexchange.com/ That being said, I am a user there and have come to give an argument in favor of the "1-11'ers." I have multiple degrees in mathematics and my field is in combinatorics. This is precisely the type of question we would ask students in an undergraduate class involving basic probability.

The scenario (as I understand): You have two dice. You roll both dice and someone checks to see if there are any twos. If there are *no* twos, then the person verifying reveals the results of the dice and has the dice cast again.

Eventually the verifier will truthfully state "There is at least one two" at which point you ask the question of "What is the probability that *both* dice are two?"

Let us reimagine this scenario somewhat. To keep track of what is going on a bit more easily, let us imagine that the two dice are *different colors*, one red, and one green. Clearly, adding or removing color to the mix will not in any way change the probability of the situation.

As the dice are *fair*, every result is equally likely. We have the following results:
1-1. 1-2, 1-3, ...
2-1. 2-2, 2-3, ...
...
For a total of 36 equally probable scenarios. As per usual, our definition of probability in an equiprobable setting (such as this one) is Pr(A) = |A|/|S|, where A is the event in question, |A| represents the size of the event in question, and |S| represents the size of the sample space.

This question however is one about "conditional probability," written mathematically as Pr(A|B) "the probability of event A occuring given that we know that B occured" or more simply as "probability of A given B."

Let A represent the event that BOTH dice show a 2. Let B represent the event that "AT LEAST ONE" die shows a 2. Let us go one step further and also let C represent the event that the red die shows a 2, to explain where your answer comes from.

The question being asked is: "What is the probability that both dice show 2 given that at least one die (be it the red one *or* the green one) is confirmed to be a two", i.e. Pr(A|B)=?

The *definition* of conditional probability states that Pr(A|B) = Pr(A and B)/Pr(B). which, since we are in an equiprobable setting can be simplified further to be = (|A and B|/|S|)/(|B|/|S|) = |A and B|/|B|.

We now try to calculate |A and B| and also calculate |B|. How many outcomes have it such that "both dice are 2 AND at least one die shows a 2", the only outcome is 22, so |A and B|=1, and the probability Pr(A and B) = 1/36. Calculating the size of |B|, we look at all outcomes where there is at least one two. Referring to our chart above, we see that occurs when either the red die is 2 or when the green die is 2 (or both), for a total of 11 outcomes. (specifically 2-1. 2-2, 2-3, 2-4. 2-5, 2-6, 1-2. 3-2, 4-2, 5-2, 6-2). Remember: Each of these events is equally likely to be the one that has been tossed.

Thus, we calculate Pr(A|B) = |A and B|/|B| = 1/11.

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Why are you convinced that the probability is instead 1/6? That is explainable as well. You seem to be confused as to the nature of how the dice are acting. You think "The *FIRST* die is a 2, the second die could either be a 2 or not", however how do you define the "first die." With colors on the dice, it makes it easier to see whats going on.

You seem to have mistaken the situation as: You have two dice. You roll both dice and someone checks to see if THE RED DIE IS A TWO. If the red die is not a two, then the person verifying reveals the results of the dice and has the dice cast again.

Eventually the verifier will truthfully state "The red die is a two" at which point you ask the question of "What is the probability that *both* dice are two?" I.e. Pr(A|C)=?

We calculate this the same way as before, noting that |A and C| = 1, as before, and |C| = 6. You get then that Pr(A|C) = Pr(A and C)/Pr(C) = (1/36)/(1/6) = 6/36 = 1/6.

Yes, of course the results of the dice themselves are independent, however an event which uses information about both dice simultaneously ("at least one die is a two") is no longer independent of the results of the dice.