In Russian roulette, is it best to go first? | Hacker News Reader

In Russian roulette, is it best to go first?

math.stackexchange.com

18 pointssrand_23 comments

Assume that we are playing a game of Russian roulette (6 chambers). Assume that there is no shuffling after the shot is fired.

I was wondering if you have an advantage in going first?

If so, how big of an advantage?

I was just debating this with friends, and I wouldn't know what probability to use to prove it. I'm thinking binomial distribution or something like that.

If n=2, then there's no advantage. Just 50/50 if the person survives or dies.

If n=3, then maybe the other guy has an advantage. The person who goes second should have an advantage.

Or maybe I'm wrong.

Interesting discussion Math.SE about this question.

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A rarely mentioned factor is that one can apparently cheat at roulette. If you've got a gun with a barrel that really spins freely - which is a precondition for the first shot being at all "random" - then since the chamber with a bullet in it is heavier than the empty chambers, it's quite likely to end up on or near the bottom of the gun (with "bottom" defined relative to whatever way you're holding it) and quite unlikely to end up on top.

Thus if the rules were that one spins the barrel every time, the game can actually go on indefinitely. And if the rule is that you spin once, the first round is the safest.

If the chamber is ratcheted, and thus can spin in only one direction is this still true? None of my meager math nor experience can suss this one out.

This question reminded me of a sketch in an old Argentinean comedy TV show under the name "A morir por la casetera". The show was hosted by one of the most famous Argentinean comedians, Tato Bores.

The sketch consisted of two contestants playing Russian roulette in order to win a VCR recorder. But one of the contestants wants to commit suicide, but every week he cannot achieve his goal and wins the VCR recorder, and the other one gets killed.

I specially remember one time when the suicidal player triggers the gun five times without success and then the second player asks not to fire the gun, only to be coerced to fire by social pressure.

Sorry, I looked after videos of the sketch but a quick googling didn't show anything relevant.

PS: AFAIK, no person was actually killed while filming this.

This isn't really a question of probabilities: In classic Russian roulette, the single bullet is either going to be in an odd or even-numbered chamber with equal likelihood. Hence, the odds are the same for either player.

> This isn't really a question of probabilities: …

Didn't you just solve this question using basic probability theory?

With this assumption, you are right. However in Russian Roulette you spin every turn, thus player 2 has an advantage.

First turn gets 1/6, second 1/6*5/6 and so on.

Don't confuse "overall odds" with "initial advantage" in a game that can stop on the first move.

The game begins with player #1 pulling the trigger (1/6 chance dying), and player #2 waiting (0/6 chance dying).

So player #2 has the initial advantage (yet same overall odds) in this game.

It's best to be player #2.

I'm arguing my case here:

http://math.stackexchange.com/questions/97396/in-russian-rou... (in the comments)

There seems to be a strange relationship and independence between initial advantage and overall (to the last move) odds in this game that's difficult to define.

If you don't cherry-pick conditions like some are doing in stackexchange then it's best to go last. That is, assuming that it will go on until one player dies. This is a first-to-lose game and by the moment of expected outcome, the first player will have lost with higher probability than the second. That said, the difference is so small and the volatility so high (loss of everything including life) that it makes no sense to engage in such game unless you are willing to die.

EDIT: refer to http://news.ycombinator.com/item?id=3439595 for further explanation before downvoting away rollseyes

I'll assume that there are two players, and that you spin the cylinder at the beginning, but not after each turn. If this is the case, then at the start, the die is cast; the bullet is in one of the chambers, each with equal probability. It therefore has an equal probability of being in an even or odd chamber. Depending on if you go first or second, you either take odd-numbered or even-numbered turns. Either way, you have the same probability of losing.

On the other hand, if you spin the cylinder after each turn, it's best not to go first.

Very very interresting, you make me doubt of my answer...

But I finally continue to think that a player don't have to go first (if the barrel is spined once).

First: for n=1 chamber, it is obvious - the clue.

If we extend the n=1 case to n>1, we can feel what will happen: there is intrinsically the time in the game rule: you have to be the last. Then acting on this rule shall make a difference.

Let's take n=2 chambers. The payer don't know where the bullet is. Then, from her point of view, she still have to let her opponent try the first one XD from the player point of view, the die is not cast, the fact that the barrel has been spined is virtual: the fact that the variable's drawing has been pre-made or not for each probability system is not relevant

So, in short, when the player press the trigger, the die is casted once like in each probability system - but becareful, here, the die is changing after each try since the rule is that the barrel is spinned once=) Great great problem, I like it=)

And so on for n>2.

I'm pretty sure your first answer was wrong, but I don't get why would anyone downvote instead of correcting or giving a different answer.

It seems that I'm victim of some kind of ennemy's downvotes

Ok no, I think now I'm wrong=) this is even, the die is cast once when the players choose to start or not and after, the game is made.

If someone asked me this on an interview, I'd assume it is one of those "thinking outside the box" questions, and say "go first, and quickly squeeze off 6 shot attempts toward the second player".

I stay away from probability questions as my mind is still a bit warped from reading about the Monty Hall problem[1].

[1]: http://en.wikipedia.org/wiki/Monty_Hall_Problem

This paradox looks easily resolvable to me: keeping the same door can be seen as choosing a new door - but the same.

The host eliminates an "incorrect" door only after you first choose one. The monty hall problem is about using the new information that the host is obligated to give you. if you use it you increase your chances to 1/2, if you do nothing with the new information you remain at 1/3. Consider an alternative scenario with the same concept: you have 100 doors, and after you choose a door the host eliminates 98 wrong doors. you can now choose the 1 other door or keep your original choice. Its obvious the other one is a better choice

Yes you have to do something: choosing a new door - but it can be the same. Do you get me? The action of making a new choice is either visible if you are choosing a new door, but you can also say that you are making a new choice by choosing the same door.

By the way, it is simpler: the probability the door you've choosen at first hide the car is just increasing when the host is eliminating one. You don't have to do something for that.

> if you use it you increase your chances to 1/2,

I may be misreading you here, but if I'm not, then you are incorrect. If you switch, you win 2/3 of the time, not 1/2. This is one of the many "doesn't make sense" conclusions of this problem, which makes it so great.

Consider this:

There are 2 people playing simultaneously, and you both have to agree on the same door initially. Monty then reveals one of the other losing doors. Because of your personalities, you switch every time, the other person never switches.

Since Monty always reveals a loser, one of you HAS to win. Since the other person never switches, his chance of winning remains at 1/3. Since someone ALWAYS wins, and he wins 1/3 of the time, you, the switcher, are left with the remaining 2/3.

Another alternative scenario that sometimes helps people have their "ah-ha" moment is to change it so that Monty doesn't actually open a door, but instead says "Do you want to keep your door, or would you like to switch and take BOTH remaining doors?".

Another ennemy's downvote :V

Ask PG: is it fair? Shouldn't be the downvote ability attributed at a case by case basis - like some kindof community moderators?... the upvote ability already serving as quality mesurement...

The probability the shot is fired is more and more important after each try for n>=2 (with n=1, the first try is the good one).

So now, the question is: is it best to go first? In terms of probability, yes you have to go first.

But.

This is like going all-in in a tournament of poker: the winner is the last man, so you have to avoid at maximum each situation which can make you out even if the probability are in your favor: once you are out, you've loose. The simple fact of going all-in makes you far in danger: you are just playing your life. If you aren't all in, for the moment, you are not playing your life and if you are not playing your life, you can't loose it.

Knowing that you can't repeat the game infinitely, don't play first because the theory of probabilities become nul and void in this case: the bullet is in one chamber, the latest you press the trigger, the latest you die.

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