Most of the folks are saying 1/6. Someone even has a table where the post below corrects the interpretation of the table to count the double 6 properly.
Actually is it stated anywhere that the dice are 6 sided? That is the most common configuration, but not teh only one.
If one die is >= 6 sides, and the other is < 6:
If one die shows 6, the odds are 0 that other one does 
(assuming it is numbered normally, and not 3,4,5,6 or something)
Brian
Good point. D&D dice make probability fun.
I just rolled two dice. One of them shows a 6.
What are the chances the other die shows a 6?
What are the chances the other die shows a 3?
Thinking the answers will be different is nonsensical.
Those who have written simulations giving the 1/11 result have written flawed programs. Musicat appears to be on the right track.
No, I don’t think that’s the problem; everyone seems to agree on that.
The interpretation where 1/6 is the correct answer: Roll two dice. If the red die is a 6, what is the probability that the green die is a 6? Equivalently, if the green die is a 6, what is the probability that the red die is a 6.
The interpretation where 1/11 is the correct answer: Roll two dice. If at least one of the dice is a 6, what is the probability that both are 6s?
In both interpretations, we are not considering rolls where neither is a 6.
No, not wrong track - just a different understanding of the problem (which matches my understanding, as the OP).
Roll two dice. You could get any combination at all. 36 possibilities. If you roll and I said “out of all the things you could have rolled, you rolled at least one six” then the chances the other die is a six is 1/11. By the way, the chance it is a three is 2/11.
Now, if I said “That roll doesn’t count because you didn’t roll a six. Roll again” until you rolled a six, the chance that the other die is also a six is 1/6. The chance of a three is also 1/6.
ETA: Thudlow Boink is also correct. If I said “the red die is a six. What are the odds that the green die is a six?” the answer is 1/6. But that’s not the question.
You’ve counted the (Red=6,Green=6) event twice here.
A nice thing with probability is that you can list out all the mutually exclusive events that have an equal chance of happening, and then you can just count them to determine the probability you’re looking for. There are 36 ways for that pair of red and green dice to be rolled. Eleven of them will have at least one six showing. Those are:
(1,6)
(2,6)
(3,6)
(4,6)
(5,6)
(6,6)
(6,1)
(6,2)
(6,3)
(6,4)
(6,5)
When you did it, you listed the (6,6) event twice.
Please, show us your “unflawed” program, or at least the algorithm it’s based on.
**Let’s return to the OP. I still haven’t run my program yet, but I think I know the answer.
Now break that down:
“Two fair dice are rolled together…” One roll, and one roll only.
“…and you are told that ‘at least one of the dice is a 6.’” We now have a simple odds situation. Of two dice, one is a 6. If it is not, we don’t have a question and this exercise is finito.
“A 6 is removed,…” So this die is no longer involved. Fuggitaboudit. Only one die remaining, in so far, an indeterminate state.
“…and you are shown the other die.” We have a simple problem, which can be reduced to the question, “what are the odds of a single die, in a single throw, being a 6?” Nothing else that happened before matters, *unlike the boy/girl problem, or the Monty Hall door problem. *
“What is the probability of the…die showing a six?” Six possibilities, one possible outcome. Answer: 1/6
If you don’t agree with me, please show me where I have defined the problem wrong, or how the first parts of this problem affect the final question.
The phrase “the other die” reveals a lot: It’s only meaningful in one interpretation of the problem. If I roll a red die and a green die, and both happen to come up 6, and I say “at least one die is a 6”, which die is “the other one”? The one I didn’t look at? But I looked at both. The one that didn’t come up 6? But both did. The green one? Why?
In any setup of the problem where “the other die” is meaningful, the answer is 1/6. In the setups where the answer is 1/11, that phrase doesn’t make sense.
If nothing that happened before matters, then what determines which die is the “other die,” the one that you are shown?
That’s because showing a red 6 and having a green 6 is a different event than being shown a green 6 and having a red. You can make two choices about which 6 to show in the double 6 case.
You are only showing cases where a 6 occurs. Use two different colored dice or denominations of coins and it’s pretty evident they are different events.
In Excel you can do this.
Make a column representing one die =RANDBETWEEN(1,6) do that for another column. Third column (C) is true if either are a 6. Fourth column (D) is true if both are a 6.
(C+D)/(2D) = 1 / 6
Or get a penny and a nickel and flip them 100 times and do the experiment by showing if you count the times you can show two heads by the times you get one head while discarding the events where you get double tails you get a 1/2 ratio which is analogous to 1/6 for the dice.
The reason for two different denominations is just the clarity that two different coins have in the choice of showing one to be heads has on the number of events possible. Whereas with interchangeable pennies or dice that detail is lost easily.
What? What else are you looking at but the die that wasn’t revealed? Aka, the “other die.” And look the green and red is just to illustrate that even though the numbers are the same and they are both fair dice looking at one of them and seeing that it’s a 6 is different than looking at the other and seeing that it’s a 6. It could be looking at numbers in column A vs looking at numbers in column B of a spreadsheet. Those are two different peeks.
But these events aren’t as likely as the others. There are two ways to throw 1&6 (red 1 + green 6, and red 6 + green 1), but only one way to throw 6&6 (red 6 + green 6).
Are you assuming the Axiom of Choice? 
See my formulation quoted below. The “other die” is the one that’s remaining after a 6 has been removed. In the case where two sixes were originally rolled, it’s irrelevant which is the one that is removed.
The way I was interpreting the problem, it’s equivalent to “If at least one of the dice is a 6, what is the probability that both are sixes?” The answer to this question would be D/C, which turns out experimentally to be quite close to 1/11.
What’s the explanation for your (C+D)/(2D) formula?
Yeah but in a chart of events where a die is being a revealed and a die is being concealed there are 12 permutations which I listed above will do so again.
Revealed Red result Green Result
Red 6 1
Red 6 2
Red 6 3
Red 6 4
Red 6 5
Red 6 6
Green 1 6
Green 2 6
Green 3 6
Green 4 6
Green 5 6
Green 6 6
Out of the 12 possibilities of a particular die being revealed and a particular die being concealed there are only 2 that have two 6s. 2/12 = 1/6.
If you do this as a real life experiment with two people. You will see that 1/6 of the time when at least one six is rolled that the concealed die is a 6. Do it with real dice and two people.
Then you must have removed a die from one of the following 11 rolls, all having an equal chance of having occurred:
1,6
2,6
3,6
4,6
5,6
6,6
6,5
6,4
6,3
6,2
6,1
There is a 1/11 chance of the other die being a 6.
When the dice are thrown, there are 11 possible outcomes, all equally likely.
In the case of (6, 6), there are two possible ways that one of the dice could be revealed, but that doesn’t make that outcome more likely to have come up in the first place.
If you do it with real dice, how do you decide which die to remove when you roll two sixes? That is, how do you know which of the following events you’re in?
Red 6 6
Green 6 6
Do you agree with this statement: If you roll two dice, you’ll get a 1 and a 6 twice as often as you’ll get two sixes.