Here's another example of their stubborn thinking that cannot consider ONE die but must consider only two dice answers. This comes form "Dalex64":

I realize that there probably isn't anyone left reading who doesn't believe at this point, but I came up with an explaination using Calculatus Eliminatus, in other words, the process of elimination.

I'll start with the original question:
Quote:
You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."

What is the probability that both dice are showing a 2?


Now, little bits at a time.

Quote:
You have two 6-sided dice in a cup.


I think everyone agrees on that.

Quote:
You shake the dice, and slam the cup down onto the table, hiding the result.


This is our starting condition. Two dice, shaken and rolled, now unmoving, and hidden from view.

Potentially, the first point of contention. I think right here, before you peek, there is an equal 1 in 36 chance that any one of the two dice arrangements could have landed under the cup. namely:

1-1
1-2 2-1
1-3 2-2 3-1
1-4 2-3 3-2 4-1
1-5 2-4 3-3 4-2 5-1
1-6 2-5 3-4 4-3 5-2 6-1
2-6 3-5 4-4 5-3 6-2
3-6 4-5 5-4 6-3
4-6 5-5 6-4
5-6 6-5
6-6

that arrangement should look familiar to craps players. 36 ways the dice can land, with equal probability.

If you don't think all 36 of those can be under the cup before we peek, under the terms of the question, please explain why not.

Quote:
Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."


this is our main point of contention:

At this point, I eliminate all of the possibilities that do not have a 2. They do not meet the condition that "at least one of the dice is a two"

I marked the remaning possibilities in bold. All of the remaining die rolls fulfill the condition that "at least one of the dice is a 2"

With the informaiton given in the question, I don't have any more information that I can use to eliminate any more of the possibilities.

Quote:
What is the probability that both dice are showing a 2?


Of the 11 possibilities that are left, only one of them is 2-2, so the probability that both dice show a 2 is 1 in 11.

To get to 1/6, you need to eliminate 5 more of the possible rolls. Which ones do you eliminate, and why?

The people who say the answer is one in 6 say that under the cup, there must be only: 2-1, 2-2, 2-3, 2-4, 2-5, 2-6.
Or perhaps 1-2, 2-2, 3-2, 4-2, 5-2, 6-2. I'm not sure which.

I'd like to know how you eliminate the remaining possibilities that meet both the condition that at least one of the dice is a two, and both dice started shaken and hidden from view under a cup.

I'll listen to explainations as to why all 36 possibilities couldn't be under the cup before you peek, too. Please don't forget the starting condition that "You shake the dice, and slam the cup down onto the table, hiding the result."

As an aside, I think another way to say that "at least one of the dice is a 2" is "this roll of the dice has resolved the bet" that is to say, the bet where you lose on one two and win on double twos.


Again, their only answer seems to come from the idea that the answer must come from a two dice combination of numbers. They don't even consider the idea that the answer can come from a single die with six faces.

Is this something taught in school? That the easy, simple answer can't be right?

I think this is the most important part of that post on WOV:

I'll listen to explainations as to why all 36 possibilities couldn't be under the cup

Well how about this one: there is one other die with only six faces on it, 1 thru 6?