Originally Posted by
a2a3dseddie
So, in the original question, the guy looks under the cup and could see one of the following:
Dice One Dice Two
2...............1
2...............2
2...............3
2...............4
2...............5
2...............6
1...............2
2...............2
3...............2
4...............2
5...............2
6...............2
Your chart is exactly right and you must consider only ONE of the following. If it is Die 1, the answer would be (as your graph shows) 1/6. If it is Die 2, it is also 1/6. YOU CANNOT SWITCH BACK AND FORTH.
In order to switch back and forth, you have to do what the Wizard did on his forum in his video: you must CHANGE the result of a die after it comes to rest. In the real world that doesn't happen.
In fact, the original problem is simply this: a shooter throws two dice on a craps table. One die immediately comes to rest showing a 2. The second die is a spinner and keeps spinning. What are the chances that the spinner will come to rest on a 2 as well? The answer is 1/6.
Guys, that's the same thing as the stated problem. When the "observer" looks and says one of the dice is a 2, what are the odds that the other die is also a 2, it is the same thing as the "spinner" problem.
If you don't interpret it as being the same as being the "spinner" problem you are making up your own conditions. The only way to interpret the original question is to consider it to be the same as the spinner problem. Because we know at least one of the two dice is showing a 2, just like with the spinner.
Sorry, but you are going to have to deal with this reading comprehension issue.