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Topic Title: Russian roulette problem
Created On Fri Apr 08, 05 10:48 PM
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chengwf
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Fri Apr 08, 05 10:48 PM
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Assuming both players take turns, what is the probability the player who goes first will lose at Russian roulette using a gun with six chambers?
 
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Tomfr
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Sat Apr 09, 05 03:42 PM
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that's a GS brainteaser, right?

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Cat pow‚!
 
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Aaron
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Sat Apr 09, 05 05:18 PM
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It depends on the rules.

If there is one bullet, you never spin and always play until one person loses, then both players have the same 50% chance. The bullet can be in either an odd or even numbered chamber, the first person loses if it is odd (calling the first chamber 1), the second person loses if it is even.

If there is one bullet, you spin after every trigger pull, and play until one person loses then call the probability of the first person losing p. If she doesn't lose, then clearly her probability goes to 1-p, since the second person is now first and has p chance of losing. That gives us:

p = 1/6 + 5*(1-p)/6
11p/6 = 1
p = 6/11.

There are other variants. One bullet removed instead of one bullet put in. Spin after every other trigger pull. Play one one, or some set number, of rounds. Person doesn't have to pull the trigger if it is certain there is a bullet in the chamber. Person who dies wins.

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Aaron Brown
 
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andyk
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Fri Apr 15, 05 01:12 PM
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In a very similar vein, and easy to solve with Aaron's answers,
but still entertaining for friends and family alike...

Dr Evil has you captured in his underground bunker, with
his heavily armed flunkeys ready to kill you instantly if you
try anything funny. He loads a 6 chamber revolver with a
single live bullet and hands it to you. He offers you a choice
between two games of Russian Roullete.

Game 1
Spin the gun barrel once.
Put the gun to your head.
Pull the trigger 4 times.

Game 2
Spin the barrel. Put the gun to your head. Pull the trigger.
Do this sequence 5 times in total.

Which game offers you the best chance of surviving ?
 
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Wilbur
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Mon Apr 18, 05 02:25 PM
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game one:

You live if bullet is in chamber 5 or 6.

You live 2/6 of the time. =33%

game two:

Each time you have a 5/6 chance of living.

after 5 times you have a (5/6)^5 of still living = approx 40%

I pick game 2.

-------------------------
Most things I worry about, never happen anyway.

Edited: Mon Apr 18, 05 at 02:30 PM by Wilbur
 
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Singlestrand
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Tue Oct 04, 05 07:58 PM
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what if there were two bullets in adjacent chambers? you fire once, nothing happens. next you have a choice of firing or spinning the barrel and then firing. Which would you choose?
 
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friesenjung
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Wed Oct 05, 05 04:16 PM
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If you have two adjacent chambers filled, you have a "revolver with three chambers" problem. Let chamber 3 be filled. If you survived the first "shot" you have either been in chamber 1 or 2, which gives you 2 or 3 for the current try. That gives you a 50% survival chance.

If you spin again you end up in one of the three states, with one chamber filled, which leaves you with a survival probability of 66.6%.

Hence spin again.

[typo]

Edited: Wed Oct 05, 05 at 04:18 PM by friesenjung
 
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Aaron
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Wed Oct 05, 05 09:17 PM
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I disagree. Suppose you pull the trigger first on chamber 1. If it's empty, the lower number of the two loaded chambers is equally likely to be 2, 3, 4 or 5. So there's 1 chance in 4 that it will be 2 and you'll die if you pull without spinning.

If you spin, there are two loaded chambers among six, so there's one chance in 3 of dying.

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Aaron Brown
 
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Singlestrand
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Wed Oct 05, 05 11:23 PM
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yup, don't spin. chance of dying is 1/4 if you don't spin. 1/3 if you spin.
 
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friesenjung
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Thu Oct 06, 05 10:39 AM
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you guys are right. stupid mistake on my side. sorry.
 
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