

In Russian roulette, is it best to go first? - srand_
http://math.stackexchange.com/q/96331/12205
Assume that we are playing a game of Russian roulette (6 chambers). Assume that there is no shuffling after the shot is fired.<p>I was wondering if you have an advantage in going first?<p>If so, how big of an advantage?<p>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.<p>If n=2, then there's no advantage. Just 50/50 if the person survives or dies.<p>If n=3, then maybe the other guy has an advantage. The person who goes second should have an advantage.<p>Or maybe I'm wrong.<p>Interesting discussion Math.SE about this question.
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glenra
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.

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michaelcampbell
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.

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aurelianito
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.

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drcode
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.

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

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

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powertower
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.

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powertower
I'm arguing my case here:

[http://math.stackexchange.com/questions/97396/in-russian-
rou...](http://math.stackexchange.com/questions/97396/in-russian-roulette-
does-the-initial-advantage-of-player-2-count) (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.

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muyuu
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_

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wch
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.

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FredBrach
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.

~~~
muyuu
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.

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FredBrach
It seems that I'm victim of some kind of _ennemy's downvotes_

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tzs
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".

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waffle_ss
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>

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

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funthree
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

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michaelcampbell
> 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.

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FredBrach
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_.

