It was two years ago today...

[a2a3dseddie]

HOWEVER, if I were a court appointed arbitrator I would rule solidly in favor of the "6" crowd, while giving the "11" crowd honorable mention for their deep insight, unforeseen complex problem solving abilities, ingenuity, and overall argumentative tenacity. Why? Because NOWHERE IN THE ORIGINAL STATED PROBLEM DOES IT MENTION ANYTHING OTHER THAN A ONE TIME TRIAL. If the peeker sees only one die and it is a 2, or if he sees both dice and at least one of them is a 2--and because he doesn't identify which is the case--"6" wins as the simple, common sense conclusion.

from the looks of things, the wizard is only paying 9 units for something that has 11-1 odds.

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.

However, if you take the WoV interpretation, simply telling you that one of the dice shows a 2 does not remove it from its numerical relationship with the die with the unknown number showing. Whereas rolling 2-2 is a 36-1 possibility, knowing half the outcome, ie. one of the dice is showing a 2, reduces the odds to 11-1 since it is no longer a single die event. You now have to consider all combinations of the two dice where either one of them shows a 2.