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Two men plays Russian roulette. In revolver there are 2 bullets in consecutive chambers( 2 bullets are in 2 chambers next to each other).

One man spun the cylinder, pulled the trigger and he is fine, so the chamber was empty. What should the second man do: pull the trigger or spin the cylinder.

My answer: Pull the trigger

Explanation:

If the man spins the cylinder there is 2 full chambers with the bullet, 4 empty chambers. The probability of drawing empty chamber is $\frac{4}{6} = \frac{2}{3}$

If the man just pull the trigger we have to remember that after 3 empty chambers there is empty chamber and after 1 empty chamber there is full chamber so the probability of that the next chamber is empty is $\frac{3}{4}$

$\frac{3}{4} \gt \frac{2}{3}$

Question: Am I right?

miracle173
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Yoda
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3 Answers3

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Let's model the gun has a list of 6 letters ABCDEF where EF have the bullets. The first person survives, so we must be in positions ABCD. Thus, there is a 1/4 chance of death if we pull the trigger because we can end up in positions BCDE and E is the only one that kills us. If we spin the wheel, then we have a 2/6=1/3 chance of death.

However, the strategic part (given that one of you must die, hurting your opponent is just as good as surviving for yourself) is that suppose I pull the trigger and survive. Now, I am in position CDE, so my opponent can now pick between a 1/3 chance (spinning) and a 1/3 chance (triggering). If he chooses to trigger again, the gun can only be in positions CD, so I have a 1/2 with trigger, and 1/3 with spinning.

Assuming optimal play for both parties, player one survives (otherwise, nothing interesting), player two triggers*, player one triggers (because spinning helps player two), player two spins (because 1/3 is better than 1/2), then player one triggers*. Note that the * parts are the same with opposite players. So basically the roles reverse. I am also of the opinion that player one has an advantage in long term, but is probably overshadowed by immediate chance to die. I will think about that some other time.

Thoth19
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Your logic is correct, but your computation is incorrect; $\frac{4}{6}$ is not equal to $\frac{1}{3}$.

CaptainCodeman
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  • Thank you, could you have a glance at this stuff I don't know any technique to solve this thing http://math.stackexchange.com/questions/849296/puzzle-about-phographers-in-africa-taking-pictures-of-2-animals-which-name-corre?noredirect=1#comment1751313_849296 – Yoda Jun 27 '14 at 11:45
  • @Yoda, I looked at that link, as far as I can tell there is no real method, you just have to consider the possibilities and eliminate the ones that don't work.. like a sudoku. – CaptainCodeman Jun 27 '14 at 21:50
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Just another way to look at this problem: when you pull the trigger and nothing happened you can conclude 2 things:

  1. You lost one 'good' spot (empty chamber)
  2. You know that the 2nd consecutive chamber is irrelevant

We know that if you spin the roulette your death probability is 1/3 And if we factor 'pull the trigger' approach factored with the 2 items above you have 6-2=4 spots. So death probability is 1/4.

Since you want a lower probability of death you should choose 'pull the trigger approach'

adhg
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