Originally Posted by RS__ View Post
Alan, would you accept the results of the following: If I were to roll the dice many many times, record the times where at least one 2 was present, then show the results (ie: frequency of "other" die being a 2).

If the correct answer is 1/6, then we should see the other die show up as a deuce 1/6 of the time at least one die is present, yes?


Edit. Alan, we've showed (shown?) you many times why the answer is 1/11 but you are not willing to accept it. You can lead a horse to water but you can't make him drink.

PS: Did you ever wonder, why just about everyone on the WOV forum was willing to bank the bet I presented a while ago? Why would a bunch of APs be willing to risk their money (thousands of $$$) for somethijg where they don't expect to win?
No I will not accept any number of rolls. And it's because that was not the question that was asked. YOU and many others have said the answer is 1/11 to the original question. Show me how you arrived at that answer given the information in the original problem. And again that information is: there are two dice and at least one is showing a 2. No additional rolling needed. You said the chance of both dice showing 2-2 is 1/11, so go ahead and demonstrate that to me just as the original question lays out the problem.

I look forward to it.