# Cool Math ConundrumsMay 14, 2012 9:27 AM   Subscribe

In Russian roulette, is it best to go first?

So far, yes.
posted by Thorzdad at 9:33 AM on May 14, 2012 [4 favorites]

It's certainly not best to go last.
posted by mazola at 9:34 AM on May 14, 2012 [15 favorites]

Next to last however, is pretty good.
posted by samsara at 9:39 AM on May 14, 2012 [3 favorites]

It's certainly not best to go last.

posted by empath at 9:50 AM on May 14, 2012

It's certainly not best to go last.

Well, no:

Now, the idea in a 2 player game is that it is best to be player 2, because in the event you end up on turn six, you KNOW you have a chambered round, and can use it to shoot player 1 (or your captor), thus winning, changing your total odds of losing to P1 - 3/6, P2 - 2/6, Captor - 1/6
posted by eugenen at 9:51 AM on May 14, 2012 [3 favorites]

It's best to go first, because the bullets aren't weightless. The revolver spin is effectively a weighted dice throw due to the metal of the bullets, so the first player is less likely to die as empty chambers are favoured by the weight differential. The benefit goes up with the amount of players.

If not playing with a revolver, make sure not to go first.
posted by jaduncan at 9:55 AM on May 14, 2012 [5 favorites]

I think you missed the joke. The person who goes last in Russian roulette is, by definition, either dead or seriously injured.
posted by jedicus at 9:58 AM on May 14, 2012

Are there any other sites treating the same topics in a similar way, but with less... StackExchangeness? I don't want to complain, the links are fascinating, but there's something about StackExchange that doesn't quite work for me personally.
posted by Talkie Toaster at 10:00 AM on May 14, 2012 [1 favorite]

The advantage in Russian roulette is clearly the guy that gets to go home to young Meryl Streep.
posted by The 10th Regiment of Foot at 10:03 AM on May 14, 2012 [1 favorite]

In Russian roulette, is it best to go first?

Depends what you mean by "go".
posted by chavenet at 10:03 AM on May 14, 2012 [7 favorites]

Is it really considered good form at Stack Exchange to pose sixth-grade brainteasers looking for answers from professional mathematicians?
posted by RogerB at 10:06 AM on May 14, 2012

How many sides does a circle have?

Didn't anyone here see the trailer to the action movie version of Flatland? Remember when the Jason Statham character -- who played the circle -- was asked that very question by the evil septagon and the circle replied "I've got two sides, and you're on THE BAD ONE!" just before he drove a protractor through a hexagon's skull?
posted by Harvey Jerkwater at 10:06 AM on May 14, 2012 [12 favorites]

Talkie Toaster: "Are there any other sites treating the same topics in a similar way, but with less... StackExchangeness?"

I didn't understand this question until I went and read the links. OMG you are so right, that format is awful.
posted by I am the Walrus at 10:26 AM on May 14, 2012

in the event you end up on turn six, you KNOW you have a chambered round, and can use it to shoot player 1

By that logic, just stand far enough away to pull the trigger six times. Spoilsport.

This is winning at Russian roulette: palm the bullet.
posted by justsomebodythatyouusedtoknow at 10:36 AM on May 14, 2012

In Russian roulette, is it best to go first?

Who cares?

*Click*
posted by Skygazer at 10:37 AM on May 14, 2012 [3 favorites]

Hm.
posted by justsomebodythatyouusedtoknow at 10:40 AM on May 14, 2012

The circle question apparently originated from the poster's son's 2nd-grade math teacher, and has an interesting twist:

I ran into this teacher recently and mentioned this quiz problem. She said she thought my son had written "8" and didn't know that a sideways "8" means infinity.
posted by swift at 10:58 AM on May 14, 2012 [3 favorites]

Now, the idea in a 2 player game is that it is best to be player 2, because in the event you end up on turn six, you KNOW you have a chambered round,

Player one knows this after the click of turn 5. Why would they then pass you the gun they know is loaded?
posted by fightorflight at 11:29 AM on May 14, 2012 [3 favorites]

I give a version of the Russian roulette problem to my calculus II students. Instead of just passing the gun back and forth, you spin the chamber every time before you pull the trigger. Under these conditions, is it "better" to go first or second? How much "better"; i.e., what is the probability that you win if you go second?
posted by King Bee at 11:32 AM on May 14, 2012

If you're down to the final chamber, you shoot the other guy, obviously.
posted by Ironmouth at 11:33 AM on May 14, 2012

MetaFilter: I didn't understand this question until I went and read the links.
posted by Pruitt-Igoe at 12:21 PM on May 14, 2012 [3 favorites]

Russian Roulette is never totally serious until the first person pulls the trigger. Then it's suddenly deadly fucking serious. "Okay, this is really happening." By this measure, you should always avoid being the first to go, if you care about survival. Most of us operate by this thinking, which is why it's very rare that a Russian Roulette game ever actually happens. So regardless of the probabilities involved in the actual game, don't be the one to initiate. Unless someone, like, holds a gun to your head.
posted by naju at 12:37 PM on May 14, 2012

Russian Roulette: Play 'til you win!
posted by double block and bleed at 1:19 PM on May 14, 2012

If you are willing to use the last bullet to shoot someone, if someone gives you the gun first why, not shoot them pulling the trigger twice?

Assume you're in the jungle playing high stakes for lost Nazi gold.
posted by ersatz at 4:13 PM on May 14, 2012

Or playing for an edit window.
posted by ersatz at 4:14 PM on May 14, 2012 [1 favorite]

If after the first guy fires a blank, the host offers you to swap turns, should you?
posted by Obscure Reference at 4:54 PM on May 14, 2012

RogerB: "Is it really considered good form at Stack Exchange to pose sixth-grade brainteasers looking for answers from professional mathematicians"

There are actually two Stack Exchange sites for maths questions. This one, math.stackexchange.com, is aimed at a general audience (undergrads and below) and is pretty liberal with what questions it accepts. Similar to AskMe, the only hard and fast rule is that the the question must be answerable. Professional mathematicians who want to discuss serious, research level, questions use the other site, mathoverflow.net, which is much stricter about the kinds of questions it accepts.
posted by Proofs and Refutations at 9:06 PM on May 14, 2012 [2 favorites]

I've always had the same question about musical notes, and have had the same difficulty in finding the answer. Everyone with enough musical knowledge gives some answer that completely misses the point, like "if that key wasn't there, you wouldn't be able to play the note." While everyone with enough technical expertise somehow misses the point in a completely different way, like "integers make easy ratios."

There are really twelve notes in an octave, but 5 of them are delegated to a sub- or semi- note status. C#, D#, F#, G# and A# are all treated like they are shadows of other notes, but they are not. If the distance from E to F is one integer, then the distance from F to G would be two integers, F# being the integer in between them. But F# gets treated as if it's not an integer; It's treated like it's F.5 (or F + 0.5), when it is really F + 1. In fact, people often start off wondering why there is no B.5 (B#) or E.5 (E#), until they are taught that all twelve notes are equidistant or integer separated,

So, to illustrate the problem, let temporarily assign new names to the notes: rather than the labels C, C#, D, D#, E, F, F#, G, G#, A, A#, and B, instead we shall label them C, C+1, C+2, C+3, C+4, C+5, C+6, C+7, C+8, C+9, C+10, and C+11. So why do the notes C, C+2, C+4, C+5, C+7, C+9, and C+11 get to be part of the octave, while C+1, C+3, C+6, C+8, and C+10 get excluded, with the reasoning that they aren't really notes, just sub- or half-notes.

So why do 1, 3, 6, 8, and 10 not get counted as integers, while 0, 2, 4, 5, 7, 9, and 11 do?
posted by BurnChao at 2:50 AM on May 15, 2012

Try writing some music. If you use just the black notes, it's a song in a pentatonic scale, and it pretty much all sounds good. If you use just the white notes, you're writing in a minor or c major, and you aren't going to hit a lot of sour notes, either.

If you mix them, though, you really need to know what you're doing, to not create an off key mess.

You can shift the positions of the black keys around, btw, it doesn't really matter. It's the relationship that matters.
posted by empath at 4:14 AM on May 15, 2012

The answer given was the correct one. Given 11 notes, break them up into 5 and 6 based on closeness by perfect 5ths.
posted by Obscure Reference at 9:12 AM on May 15, 2012

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