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Russian Roulette (Posted on 2010-06-08) Difficulty: 4 of 5
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).

No Solution Yet Submitted by John zadeh    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts More complicated than I thought. How to do it. | Comment 10 of 12 |
I wasn't thinking about the fact that each time a person survives the gun is passed.  This means it could be passed up to 6 times and still not end the game if the person clockwise of the first empty chair is pulled.

This is completely solvable in principle but way too much work.

I will find the exactly one person dies. 
Number the players 1 through 8 and wlog let 8 be the first to die.  Passing the gun means passing it to the next higher number and if player 7 survives the game ends.  There are 4 bullets left so a 1/3 chance the gun will be passed instead of killing a second person.

If player 7 is drawn there is a 1/3 chance the game will end.
If player 6 is drawn there is a (1/3)^2 chance players 6 and 7 will both survive and the game will end.
If player 5 is drawn there is a (1/3)^3 chance players 5, 6 and 7 will all survive and the game will end.
etc.
Add (1/3)+(1/3)^2+...+(1/3)^7 = 7651/15309
since each player was equally likely to be drawn divide this by 7 to get 1093/15309 for the probability of only one death.

The problem is if there is a second death it can leave one string of 6 players in a row
or 1 person between chairs and 5 in a row
or 2 in a row and 4 in a row
or 3 in a row and 3 in a row
these each have a different probability of occurring based on arguments above.  Then for each of these we would have to find the probability of the game ending again based on arguments above.

The third death makes things even messier.  I think it is best to trust Charlies program.

  Posted by Jer on 2010-06-09 15:56:32
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