If you are clearly able to distinguish between the "spinner" and the not spinner, then yes the probability will be 1/6. However, how could you have "chosen" one to be frozen and one to be a "spinner". What if the "frozen" die is not a two and the "spinner" eventually rests on a two? Why would the verifier not have stated "there is at least one two" in that situation. It is as I said in my previous post, you have somehow arbitrarily changed the wording of the question to be "Given that the *red* die lands on a two, what is the probability that both are a two", or "Given that the leftmost die lands on a two, what is the probability that both are a two", or in your wording "Given that only one die is stationary at the time and is a two, what is the probability that after the second comes to rest that both are a two". In each of these alternate wordings, there is some clear distinction being made between the dice, and you are ignoring the possibility that the second die could be a two with the first one not, and still be considered as a scenario where the verifier truthfully states "at least one die is a two."
In the interpretation the 1/11'ers are using, which appears to be the original wording of the question, is that "both dice are tossed" (and both come to rest), and the verifier sees the results of both dice before stating "at least one die is a two."
The two are very different scenarios, and have different probabilities (as explained in my previous post).
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