RE: Probability question: names in hats
March 14, 2016 at 11:25 am
(This post was last modified: March 14, 2016 at 11:26 am by robvalue.)
Say player 1 takes number 2 out the hat.
This leaves the following numbers: 1 3 4 5 6 7 8 9
9 numbers, and 8 of them keep the game going. So 8/9
It's the same deal as for player 1, except he had a redundant extra number (his own).
Now say player 1 takes another number, but not 2:
1 2 x x x x x x x
9 numbers, but one of them is a 2. Which he can't pick. So it's 7/8 to continue.
This leaves the following numbers: 1 3 4 5 6 7 8 9
9 numbers, and 8 of them keep the game going. So 8/9
It's the same deal as for player 1, except he had a redundant extra number (his own).
Now say player 1 takes another number, but not 2:
1 2 x x x x x x x
9 numbers, but one of them is a 2. Which he can't pick. So it's 7/8 to continue.
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