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 Russian Roulette (Posted on 2010-06-08)
Eight men, including Colonel Mustard, sit at a round table, for a modified game of Russian roulette. They are using a six chamber revolver which has been loaded with 5 bullets.

The game begins by one of the men reaching into a hat, and randomly drawing the name of the first player.

If the first player survives his turn, the gun is handed to his adjacent clockwise neighbor, and his name is immediately returned to the hat.

If the first player loses, his name is thrown away, and the men pull from the hat, and choose the name of the next player.

The game is continued in such a way until either all five bullets have fired, OR a player survives his turn, but no longer has an adjacent clockwise neighbor to pass the gun to.

What is the probability that the Colonel will survive the game?

(Note that the chamber is spun every time a player takes his turn).

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 First Thoughts | Comment 5 of 12 |
I definitely think simulation is the way to go on this one.

If done analytically, I would analyze the game in rounds, where each round ends when a bullet is fired or when the game ends because a player survives his turn but no longer has an adjacent clockwise neighbor to pass the gun to.

Round 1 is simple.  Someone eventually dies, and there are 4 bullets left, and one group of 7 consecutive seats.

Round 2, however, is already complicated.  It could end with:
a) Nobody dies. Game over, or
b) a 2nd person dies, 3 bullets remaining, and
i) 6 consecutive seats, or
ii) 5 consecutive seats and an isolated seat, or
iii) two groups of seats, a 4-group and a 2-group, or
iv) two groups of seats, each one a 3-group
In effect, 5 different states, where the probability of transitioning from the round 1 end-state is a little complicated.

Round 3 (if we get that far) is even worse. It could end with:
a) Nobody dies. Game over, or
b) a 3rd person dies, 2 bullets remaining and
i) 5 consecutive seats, or
ii) 4 consecutive seats and an isolated seat, or
iii) two groups of seats, a 3-group and a 2 group, or
iv) three groups of seats, a 3-group and two 1-groups, or
v) three groups of seats, a 1-group and two 2-groups
In effect, 6 different states, where the probability of transitioning from the round 2 end-states requires tracking a lot of different cases.

I might calculate it all at the end of the day while watching a basketball game, but the probability is low.

 Posted by Steve Herman on 2010-06-08 14:46:17

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