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I first read this problem here.

I am having trouble finding the expected value of the game. I understand for the first round,but on the 2nd round assuming that no group is killed, does just one person role the dices and if not (6,6), then someone roles for the other 89?

Or is it 10 people role the dice and if none of them die, then one role for the other 80?

I am having trouble viewing a copy of the primary paper.

Dan Brumleve
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yiyi
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  • @danbrumleve What did you edit? (just want to know so I can improve my question writting ability. ) – yiyi Jan 20 '13 at 08:04
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    I changed the link to point at the permalink of the article instead of the front page of the blog; I had to scroll down past a few posts to find the one you are asking about. – Dan Brumleve Jan 20 '13 at 08:14

1 Answers1

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In the shooting room paradox, there is one roll of the dice per group. So, there is one roll for the first person; one roll for the next nine, assuming that the first person is not shot; one roll for the next 90, assuming that the last nine people were not shot; and so on.

The point of the paradox is that each person who enters the room is killed iff the roll is double sixes, which, according to the usual laws of probability, occurs with probability 1/36. However, since the group sizes increase geometrically until a group is killed, at least 90% of the people who ever enter the shooting room are killed.

The paper referred to is here, but it's behind a pay wall.

David Moews
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