Wednesday, November 11, 2009

Russian Roulette

Here's a nice little probability problem that I first saw in David Darling's Universal Book of Mathematics:

You're in a game of Russian Roulette (with only one other player), but instead of one bullet the gun (a six-shooter revolver) is loaded with three bullets in three consecutive chambers. The barrel is spun only once. Each player points the gun at his head and pulls the trigger. If he is still alive, the gun is passed to the other player. The game ends when one player dies. Would you stand a better chance of surviving if you shot first or second, or does it make a difference?

This puzzle, or a variation of it, is often used as a brain teaser interview question. To avoid spoiling it for anyone, I'll add the solution as a comment to this post (so don't scroll down until you think you've worked it out).

Bill the Lizard said...

Solution:

After the chamber is spun, there are six possible bullet configurations (B = bullet, E = empty):

BBBEEE -> player 1 dies

EBBBEE -> player 2 dies

EEBBBE -> player 1 dies

EEEBBB -> player 2 dies

BEEEBB -> player 1 dies

BBEEEB -> player 1 dies

It's easy to see that the odds are in your favor if you elect to shoot second. Player 2 has a 2/3 chance of surviving the game, and consequently a 1/3 chance of dying.

Colin said...

You need to specify how many players there are.

Also, you need to simplify real world logic. I imagine three consecutive bullets in a barrel would end up closer to the ground than the empty chambers; what with gravity and all.

Bill the Lizard said...

Colin,
I clarified the description to specify that there are only two players.

As for gravity, you can negate the effects of the weight of the bullets by pointing the gun straight up in the air when you spin the chamber. Naturally, in problems like this we want to assume that each configuration is equally likely. Otherwise there's really no way to answer the question definitively.

Vrungel said...

There is a variation of this puzzle with only two bullets in adjacent chambers.

(there also variation with only one player that gets to choose whether to skip first chamber for example by pointing gun at the air first).

Bill the Lizard said...

Vrungel,
I'd rather have the option to skip the last chamber. :)

Vishal said...

Good one! But it's easy to figure it out once you surpass the initial (false) intuition that both choices would yield the same (50/50) survival rates!

Eric Hsu said...

Nice one. Another way to think of this: the first player has a 3/6 chance to live. The second player only goes if the first player gets an empty slot. Two of the empties are followed by empty, only one is followed by a bullet, so player 2 lives 2/3 times.

The benefit of this approach is that it generalizes to other situations. E.g. A twenty bullet chamber with ten consecutive bullets. In that case, P1 still has 10/20 chance to live, but P2 has a 9/10 chance to live.

Anonymous said...

The logic isnt completely correct....if the first player dies on the first shot....the game is over....

Bill the Lizard said...

Anonymous said "The logic isnt completely correct....if the first player dies on the first shot....the game is over...."

That is the point. In four of the different arrangements of bullets, player 1 dies and player 2 is the winner. In the other 2 arrangements, player 2 dies and player 1 is the winner.

As others have pointed out, this method of enumerating all the different possible outcomes can be used to solve other variations of this problem, as well as many other probability problems.

bryan said...

Well, my thought process is - (1) if one bullet is loaded, it's 50% chance for the first guy to survive; (2) if six bullets are loaded, it's 0% chance for the first guy to survive; (3) between one bullet and six, the chance is between 0% and 50%; (4) so.. pick the second.

chipuni said...

A strange game, Professor Falken. The only way to win is not to play. How about a nice game of chess?

BeMeCollective said...

You guys are very smart, really. I am humbled....
But if you were playing against me, and I went second, AND you survived the 1st and 3rd clicks, I would just shoot you in the head with the "sure death" next click.

Sorry, it's Mr Hyde, I can't control him sometimes....

Best Regards
Ari

Bill the Lizard said...

BeMeCollective,
Sometimes it's the simplest strategies that we overlook. :)

betcode said...

I completely agree that in the other two arrangements, when the player 2 dies, then the player 1 obviously wins. I think that's simply logical. Joseph

Fergal Daly said...

This question is a bit odd. You don't need to do any enumerations etc, #1 has a 50% chance of dying on the first bullet and a non-zero chance of dying later, so #1 has > 50% chance of dying. It's trivial.

Is there any Russian roulette game where going first is actually an advantage?

Evan Browns said...

I like Russian roulette, but only with real gambler!

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