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  1. #11
    Originally Posted by Alan Mendelson View Post
    Now let's talk about two red dice that are shaken in a cup and slammed down on a table. One person peeks and sees that at least one of the two red dice is showing a 2.

    Both of you are unable to show me 11 possible options. You say they are there and you ask me to trust that they are there but you can't demonstrate that they are there.
    I watched the Wizard's video, and I don't think I would have used the wording "double count" even though I know exactly what he means. At the end of the video, he has 36 sets of dice laid out with all 36 possible combinations 2 dice can be arranged.

    Why didn't you accept that?

    You know the 11 possible 2 dice combinations that contain at least one 2.

    red die 1=2 red die 2=1
    red die 1=2 red die 2=2
    red die 1=2 red die 2=3
    red die 1=2 red die 2=4
    red die 1=2 red die 2=5
    red die 1=2 red die 2=6

    OR

    red die 1=1 red die 2=2
    red die 1=2 red die 2=2
    red die 1=3 red die 2=2
    red die 1=4 red die 2=2
    red die 1=5 red die 2=2
    red die 1=6 red die 2=2

    How can you exclude any of these combinations the way the question is worded?

    Once again, take it from your friend:
    Originally Posted by Rob.Singer View Post
    Originally Posted by Alan Mendelson View Post
    Rob I am 1000% honest here. I haven't got a clue as to what you're saying. So let me be point blank with some questions:

    1. If wo dice are rolled at the same time and one die shows a 2 do you think the chance that the second die rolled at the same time is 1/6 for also showing needa 2 ?

    2. If you said "yes" to question #1 why won't you accept the wager?

    And leave out all of your criticism about foreigners, WOV, the Wizard, the WOV forum members, Obama, and anybody else. Answer the two questions.
    No and I never have, not under that premise, which is the same way wizard interpreted the problem and not my interpretation of it. You need to understand that if one die is a 2, it's not a function of simply how many faces that 2nd die has. They work TOGETHER to form one of the possible 11 combinations remaining where a 2 is on at least one of the dice. Arci laid them out for you. I don't know how else better to show it to you.
    Last edited by a2a3dseddie; 05-23-2017 at 08:23 PM.

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