Russian Roulette Part 2

Probability Level 3

After meeting some unsavory characters playing CS:GO, the Russian mafia forces you to play Russian Roulette. You have survived Part 1 , and they now pit you against another survivor.

A gun, which can hold 6 bullets, is chambered with a bullet. Both of you will take turns to shoot the gun (without spinning the barrel between turns), until one of you gets shot. By sheer blink luck, the mafia allows you to choose between going first or second.

To increase your chance of survival, should you:

A. Go first

B. Go second

Equal (non-zero) chance of surviving either way B A Dead either way

3 solutions

Brian Charlesworth
Jan 14, 2016

Edited solution (reflecting intended interpretation and Calvin's hint)

The person who goes first will die if the bullet is in chamber $1, 3$ or $5$ , and the person who goes second will die if the bullet is in chamber $2, 4$ or $6$ . Since the bullet is equally likely to be in any of the chambers, you will have the same chance of dying, namely $3*\dfrac{1}{6} = \dfrac{1}{2}$ , whether you go first or go second.

(Note that if there were an odd number of chambers in the barrel then it would better to go second.)

Initial solution (for scenario where each person takes just one shot)

If you go first, since $5$ of the $6$ chambers are empty, there is a probability of $\dfrac{5}{6}$ that you will survive.

If you go second, then there are two (mutually disjoint) survival scenarios:

(i) the other person goes first and shoots themselves, guaranteeing that the chamber you fire is empty. This happens with probability $\dfrac{1}{6}$ .
(ii) the other person goes first and fires an empty chamber, which will happen with probability $\dfrac{5}{6}$ . Since initially each chamber has an equal probability of housing the bullet, there is a $\dfrac{1}{5}$ chance that the next chamber has the bullet, so you subsequently have a $\dfrac{4}{5}$ chance of surviving your turn.

As these scenarios are mutually disjoint we add the respective probabilities to find that going second gives you a survival probability of $\dfrac{1}{6} + \dfrac{5}{6}*\dfrac{4}{5} = \dfrac{5}{6}$ , which is exactly the same as the survival probability of going first.

There's a "one-line" solution to explain why the probabilities are equal.

Hint: Continue playing the game.

Calvin Lin Staff - 5 years, 5 months ago

Oh, right. Since there is a $\dfrac{1}{6}$ chance of the bullet being in any given chamber, there is the same $\dfrac{1}{6}$ chance of shooting yourself whether you shoot first or second, (or for that matter third, fourth, fifth or six if there were six participants). So either way there is a probability of $\dfrac{5}{6}$ of surviving, and a $\dfrac{2}{3}$ chance that both participants will survive.

If there were just two chambers it wouldn't matter statistically whether you went first or second, but psychologically there might be an individual preference. If you go first then at least you could hope for the best when pulling the trigger regardless, but if you went second and the first person survived then the terror would be that much more profound. Mind you, at least you would then be certain that you have the loaded chamber and you quickly turn the gun on the Mafia dude before they can draw their weapon, and then afterwards try to escape.

Brian Charlesworth - 5 years, 5 months ago

Oh wait, I realized that you interpreted the question in a different way. Let me clarify that we play till one person dies.

Calvin Lin Staff - 5 years, 5 months ago

@Calvin Lin – Oh, haha, yeah ... I guess that's the whole point of Russian Roulette; someone must die. :( I'll edit my solution to reflect the intended interpretation.

Brian Charlesworth - 5 years, 5 months ago

@Calvin Lin – Do you play CS:GO ? :P

Nihar Mahajan - 5 years, 5 months ago

Michael Mendrin
Jan 14, 2016

The person that goes first gets shot if the bullet is in an odd barrel, otherwise if the bullet is in an even barrel, then person that goes second gets shot--ergo, equal probability.

Moderator note:

Great! This is the explanation that I was looking for. Having the suitable interpretation makes the problem much easier to work with.

Yes! I did the same way rather than applying great mathematics.

Department 8 - 5 years, 5 months ago

Patrick Heebels
Jan 14, 2016

Let $k$ be the number of chambers that haven't been fired yet. One chamber holds the bullet, so the change that person A (the one to pull the trigger first) won't survive this round is $\frac{1}{k}$ . If person A is to survive this round and end the 'game' in this round he has to fire one of the empty chambers himself and after that person B (the one to pull the trigger second) has to die. The chance that this will happen is also $\frac{k-1}{k} \cdot \frac{1}{k-1} = \frac{1}{k}$ .

So in each round the chance for person A to survive or not is the same.

Russian Roulette Part 2

3 solutions

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