The Wizard will bank this bet: 1/6 vs 1/11

[kewl]

This has been a record-setting waste of space. I will say this for Rob -- in entry #10 on this thread, he provided the clearest summary and explanation of the situation. It's obvious Rob had a handle on this from the beginning. I'm no fan of Rob's video poker riff, but if he can sit down and, in a few minutes off the top of his head, summarize the issues in language Alan (and virtually all readers) can grasp, then that highlights how the original question was just a parlor trick.

Nope. Not a parlor trick at all. The question is as clear as it gets (I would've add an additional sentence just to make it crystal clear, but it works fine as it is now). It's just some folks need to learn basic probability before making bold erroneous claims.

If we're at this point, I'll explain why. When two dice are peeked at and we're informed that one of them is a 2, if that die is removed from the scene then we're left with a single die that no longer has a numerical relationship to the die showing a 2. This is the way I interpret the problem with the wording from the OP, and the odds that 2nd die will be a 2 is 6-1.

Alrighty then. Remove the dice with a two from scene at will. Take the bet at 9 for 1 and remove the dice whenever at least one of them is a two and reveal the other.
Lol, what is that even going to accomplish? Alter the odds on the remaining die??

Need to read some Wiki facts on Probability, me thinks.