How do you interpret the "dice problem"?

[Zedd]

Besides all of this, one item remains from the start of the original dice problem thread which hasn't been addressed at all with respect to the original problem. A number isn't agreed upon in advance of the roll. I think it was Mango who called it "forcing the number" when all of the rolls are used or none is discarded. He concluded that if a number is forced for each roll, then the chance of two numbers the same on a roll is 1/6.

I did addressed this in the other thread.

But, the original question is indeed incapable of solution without making assumptions on how the condition of 'at least one deuce' was met.

However, after we had cleared up this ambiguity and established that the number IS agreed upon in advance--Alan still believes the answer is 1/6. This is simply wrong.