Originally Posted by RS__ View Post
Originally Posted by Alan Mendelson View Post
My interpretation is simply this:

A peeker looks and sees at least one die is a 2. That die will not change. It will always be a 2.

Now you tell me: if one die is always a two, and there are two dice, what are the odds that the other die is also showing a 2? Is it still 1/11 for a six-sided die?

LOL
I'm trying to give you the benefit of the doubt. I don't think you realize that, though.

I'm not sure what's so funny. If you mean you set one die to a '2', and roll the other one, then it's pretty obvious there's a 1/6 chance it lands on '2'. Why would you even ask that question?
So there is no misunderstanding: when two dice are rolled and you are told that one die has landed on the 2, there is a 1/6 chance that the other die is also a 2.

When you roll two dice, there are 11 combinations on those two dice that include a 2, and one of those 11 combinations is 2-2. That's 1/11.

Do you realize what the difference is? The difference is when you consider TWO dice or you consider only ONE die. When someone tells you that one die has landed on a 2, then there are only six possibilities to consider for the second die.

That's what Ive been telling you for two years. It's what regnis has been telling you. And redietz even acknowledges the difference in the two situations because as he put it, telling someone that a 2 has been rolled is sequential and not simultaneous.

Frankly, as I've said all along, it's a matter of reading comprehension and understanding the words of the problem. Your math isn't wrong, but you are applying the math incorrectly.