### Brain Teasers

# Russian Roulette

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?

### Answer

Label the chambers 1 through 6. Chambers 1 through 3 have bullets and chambers 4 through 6 are empty. After you spin the cylinder there are six possible outcomes:1. Chamber 1 is fired first: Player 1 loses

2. Chamber 2 is fired first: Player 1 loses

3. Chamber 3 is fired first: Player 1 loses

4. Chamber 4 is fired first: Player 2 loses (First shot, player 1, chamber 4 empty. Second shot player 2, chamber 5, empty. Third shot player 1, chamber 6 empty. Fourth shot player 2, chamber 1 not empty.)

5. Chamber 5 is fired first: Player 1 loses (First shot, player 1, chamber 5 empty. Second shot player 2, chamber 6, empty. Third shot player 1, chamber 1 not empty.)

6. Chamber 6 is fired first: Player 2 loses (First shot, player 1, chamber 6 empty. Second shot, player 2, chamber 1, not empty)

Therefore player 2 has an 4/6 or 2/3 chance of winning.

Hide Answer Show Answer

## What Next?

**Solve a Similar Brain Teaser...**

Or, get a random brain teaser.

If you become a registered user you can vote on this brain teaser, keep track of

which ones you have seen, and even make your own.

## Comments

By winning you mean "not dying" so I presume you prefer to go second unless you have a large insurance policy and lots of creditors.

I would not play this kind of game ! But since we are just talking about your teaser, it is a great one. Not hard (just imagine the possibilities as a drawing or a table) yet it is a good one.

I wish you had told me in the teaser itself how many chambers a revolver has....I liked it though!

Why not just shoot the other guy and you win?

Hey robertbbr! I wished the same thing, but then thought about it a bit more and realised I didn't really need to know unless you want to work out the exact probability. The thing to remember is that with only one bullet, it would be evens. Then notice that after the first shot has been fired, and no bullet comes out, the only bullet you need to worry about is the first one, so it becomes similar to the one bullet problem. Hence it's just the first person to have a go that suffers a disadvantage as a result of the increased number of bullets.

fishmed,,,LOL.

i guess right but i dunno the reason.. haha

I think your answer is too complicated. You already have a better chance of living by going second because no mater what happens, you will shoot the same amount of times or less than the other person. FUN RIDDLE

It's scary either way.

In the first round your opponent has a 50% chance of living/dying if he takes the first shot. If your opponent takes the first shot and lives, you have a 2/3 chance of dying!

But there is only a 1/2 * 2/3 = 2/6 or 1/3 chance of that exact scenario happening.

In the first round your opponent has a 50% chance of living/dying if he takes the first shot. If your opponent takes the first shot and lives, you have a 2/3 chance of dying!

But there is only a 1/2 * 2/3 = 2/6 or 1/3 chance of that exact scenario happening.

Too easy.

Hey!

I worked out a formula for the probabilities given an unspecified number of chambers...as previously mentioned, it is always best to go second. As there are 3 bullets there must be at least 3 chambers. Lets do the trivial ones first: P1, P2 = Player 1, Player 2

3 Chambers: P1 Loss (Trivial)

4 Chambers: 3/4 Chance P1 Loss (3/4 Chance of P1 dying in first shot. Otherwise, P2 definitely loses.)

5 Chambers, lets draw the chambers, the next chamber fired is one up from the previous one and I have written the result for P1 firing that chamber first. B means chamber has bullet, E means it is empty.

B P1 Loss

B P1 Loss

B P1 Loss

E P2 Loss (P1 shoots empty, P2 gets the bullet)

E P1 Loss (P1 shoots empty, P2 shoots empty, P1 gets bullet)

So, P1 has 4/5 chance of losing.

We can just extend the above, For example 8 chambers: (The author has already did 6 )

(Nc = Number of chambers)

B P1 Loss

B P1 Loss

B P1 Loss

E P2 Loss

E P1 Loss

E P2 Loss

E P1 Loss

E P2 Loss

(The reader should be able to check that...)

There is a pattern here which will quite clearly continue. P1 Loses if they pick one of the 1st 3 chambers first, then it alternates.

So the probability of P1 losing is 3 plus half the number of remaining chambers, if the number of remaining chambers is odd we round down.

The number of remaining chambers is Nc-3, so half the number of remaining chambers is (Nc-3)/2.

Rounding down can be expressed by the floor() function.

SO Probability of P1 Losing is (For Nc > 2):

(3 + Floor((Nc-3)/2)))/2

This will always be greater than 0.5*, so P1 is always more likely to lose (Note: as the number of chambers increases, the probability will approach 0.5)

*To prove this, simplify the equation the odd and even Nc and it should be clear.

I worked out a formula for the probabilities given an unspecified number of chambers...as previously mentioned, it is always best to go second. As there are 3 bullets there must be at least 3 chambers. Lets do the trivial ones first: P1, P2 = Player 1, Player 2

3 Chambers: P1 Loss (Trivial)

4 Chambers: 3/4 Chance P1 Loss (3/4 Chance of P1 dying in first shot. Otherwise, P2 definitely loses.)

5 Chambers, lets draw the chambers, the next chamber fired is one up from the previous one and I have written the result for P1 firing that chamber first. B means chamber has bullet, E means it is empty.

B P1 Loss

B P1 Loss

B P1 Loss

E P2 Loss (P1 shoots empty, P2 gets the bullet)

E P1 Loss (P1 shoots empty, P2 shoots empty, P1 gets bullet)

So, P1 has 4/5 chance of losing.

We can just extend the above, For example 8 chambers: (The author has already did 6 )

(Nc = Number of chambers)

B P1 Loss

B P1 Loss

B P1 Loss

E P2 Loss

E P1 Loss

E P2 Loss

E P1 Loss

E P2 Loss

(The reader should be able to check that...)

There is a pattern here which will quite clearly continue. P1 Loses if they pick one of the 1st 3 chambers first, then it alternates.

So the probability of P1 losing is 3 plus half the number of remaining chambers, if the number of remaining chambers is odd we round down.

The number of remaining chambers is Nc-3, so half the number of remaining chambers is (Nc-3)/2.

Rounding down can be expressed by the floor() function.

SO Probability of P1 Losing is (For Nc > 2):

(3 + Floor((Nc-3)/2)))/2

This will always be greater than 0.5*, so P1 is always more likely to lose (Note: as the number of chambers increases, the probability will approach 0.5)

*To prove this, simplify the equation the odd and even Nc and it should be clear.

## Follow Braingle!