How do you interpret the "dice problem"?

[Harry Porter]

Unfortunately, in the real physical world we have to decide which die it is.

If you want to flip-flop and double-count be my guest... just don't try it in Vegas. It's against the rules.

Fine, so kill that portion of the post.

The enumeration of candidate throws stands. There are 36 equally probable dice throws. Of these, there are 11 specific throws where someone who looked at the outcome would say, "At least one of the dice is a 2." (as posed in the original problem).

Of these 11 possible outcomes, only 1 satisfies the criteria.

If you wish to discuss this further, run the suggested trial and post the results. (it's a short trial, so "anything can happen", but odds favor a result near 1/11)