Two Dice Question: If at least one is a $2$, what is the probability both are $2$? There is an internet debate raging about a two dice question. Here it is:
You have two dice in a cup.  You shake them up and slam the cup to the table.  Your partner peeks under the cup and announces "At least one of the dice is a 2." What is the probability that the other die is a two?
This question was originally asked on the WizardofVegas forum.  Two camps emerged, the 1 in 6 camp and the 1 in 11 camp.  I myself originally thought it was 1 in 6 but after doing the math changed to 1 in 11.  Michael Shackleford, of the Wizard of Odds site says it is 1 in 11 and has challenged anyone to gamble with him on this proposition:
They will roll the dice.  Anytime at least one of the dice is a two, if the other die is a 2 his opponent will get 8 points.  If the other die isn't a 2 the the Wizard will be awarded 1 point.  The first person to get to 25 points wins the contest.  
One person, Alan Mendelsen, a B celebrity and businessman accepted the challenge on April 29th but so far has failed to honor his commitment.  Others in the 1 in 6 crowd say the Wizard is wrong but refuse to put their money where their mouth is.  
The debate has subsided on Wizardofvegas but still rages on at Alanbestbuysforum in the Las Vegas section. This forum is owned by Alan Mendelsen. The thread is called "The Wizard will bank this bet."  The motivation for the question here is to provide a correct answer that will convince people in that thread.
Could any of you provide an answer with explanation?
 A: The probability is $1/11$. The easiest way to see this is just to think of the two dice as being distinguishable (a red die and a green die, say). Now there are $36$ equiprobable outcomes from the result of rolling both dice (imagine a $6 \times 6$ table). Saying "at least one die is 2" says that one of the 11 outcomes "Red 2 Green 1, Red 2 Green 3, ..., Red 2 Green 6; Red 1 Green 2, Red 3 Green 2, ..., Red 6 Green 2; Red 2 Green 2" occurred, so with that information, each of those 11 outcomes is equiprobable. In particular, the probability of "Red 2 Green 2" is $1/11$, and that's the probability we're after.
Another way of phrasing this is to explicitly use conditional probability. You are asking for $P[\text{both dice are 2}|\text{at least one die is 2}]$. Since the first event implies the second, this probability is equal to $P[\text{both dice are 2}]/P[\text{at least one die is 2}]$. Since $P[\text{both dice are 2}] = 1/36$ and $P[\text{at least one die is 2}] = 11/36$, we have $P[\text{both dice are 2}|\text{at least one die is 2}] = (1/36)/(11/36) = 1/11$.
(But, nobody is going to go register for some other forum just to wade into a debate there.)
