Russian roulette


OK, I got to watching the HBO mini-series Epitafios (awesome show, IMHO), where one of the main characters is in the habit of playing Russian Roulette. My instinct was to agree with the odds initially posted on the wikipedia article here and discussed here , that with no respin, the odds go up on each subsequent trigger pull: 1/6, 1/5, 1/4…. etc. to 1/1 on the sixth. I realized that this was a misreprentation of the statistics, because only one "selection" was made at the beginning of the game (the spin). So I wrote a program to simulate the game, and to my pleasant surprise, I was right, each pull is 1/6 up until the sixth pull which is 1/1.

Long story short, Russian Roulette doesn’t become successively more dangerous as you play. Don’t get me wrong, I’ll never play, but this was kind of interesting to me to do the math (and a interesting example of how the self-correcting nature of Wikipedia is accessible to everybody.)

By Dave

He was born in Canada, but currently lives in Boulder, CO up in Boulder Heights.

9 replies on “Russian roulette”

Dave, I think it all depents on how you frame the problem.  All comments below assume NO respin after each trigger pull. First, when I run your program, the result that I see is that the probability is 1/6th for each pull INCLUDING the sixth one.  That is, I don’t see it going to 1/1 on the sixth pull. This is probably because your program continues to pull the trigger even if the gun fired before the sixth try.  In reality, this is not the way the game is played (with one bullet, anyway). I think the correct way to frame the problem is to stop each game when the gun fires.  Then figure out for what percentage of first tries that the gun fires, what percentage of second tries that the gun fires, etc.   The probability will build up to 1/1 for the sixth try.

Without running simulation, it seems to me that the probability that the gun will fire on the first pull will be 1/6th…on the second pull, 1/5th…on the third pull, 1/4th…etc.

The trick is deciding whether the game begins anew on each trigger pull. I contend that only one selection is made (at the beginning, when the chamber is spun), and that revealing empty chambers in order does not change the initial selection.

Yes, that is true.  AT THE BEGINNING OF THE GAME the odds are equally likely that the gun will fire on the first pull or the sixth. But every time that the trigger is pulled the probabilities change.  That is because some of the possible outcomes have been removed from consideration.  By the time that you get to trigger pull #5, you now know that the bullet was not in chambers 1 through 4 for CERTAIN. So….if you are handed the gun after four trigger pulls (without the gun firing) you know that the bullet is in chamber number 5 or 6 for CERTAIN and your odds are now 50/50 that it will fire on the next pull. The odds could not still be 1/6th for pull #5 and 1/6th for pull #6.  Nor could they be 1/6th for pull #5 and 1 for pull #6.  For starters, neither set of probabilities adds up to 1.

A quick counter example should suffice. If I show you three face up cards, one of which is, say, a Queen, and the other two are Jacks, then place them face down. You choose a card (at random), and I reveal a card to be a Jack (not the one you picked). The chances the card you have under your finger are still 1/3. Now let’s reveal another card, it’s a Jack too! Now you KNOW the card under your finger is a Queen. So now we see the probabilities goes 1/3, 1/3, 1/1. These don’t add up to 1 either. 

Think about the trigger pull as your selection, after a rotation of the chamber. So at the beginning of the game, the order of play determines which trigger pull will be yours (let’s assume there are 6 players for the sake of this). Thus, at the beginning of play, the cylinder was rotated so that the bullet is 2 off the active chamber. You’re fourth to pull the trigger. All the selections have already been made…. Revealing empty chamber doesn’t change that…

At each point in time, the sum of probability of the remaining outcomes has to be equal to 1.  In the example you just gave, before any of the cards are turned over the chances that each of them are a queen are 1/3.  So, since there are three cards, the probabilities are 1/3, 1/3, and 1/3 for each of them to be a queen (and the sum equals 1).  After you reveal one of the cards and it is a jack, we know for sure that it is not a queen.  The two remaining face-down cards each now have a probability of 1/2 (not 1/3 as you state) and the sum still equals 1. One of us isn’t getting it, and I think it is in the statement of the problem. Let’s say there are six people and they determine before the first trigger pull in which order they will go, then they all will have equal odds of getting killed during the course of the game. However…if you are number 5, say, and the game has actually proceeded to where are handed the gun for your turn, I maintain that you better start sweating, because your odds of being killed have just increased to 1/2.

Actually, the example of the three cards is from Marilyn Vos Savant’s column. Please see the description of the Monty Hall Problem (wikipedia has a good write-up) for why revealing a DIFFERENT card doesn’t change the picked card’s chances to 1/2 (it indeed stays at 1/3). 

I assume you’ll grant me that if I’m right about the card example, then I’m also right about Russian Roulette. So then all I have to do is convince you of the card example, and we’re set….. 😎

You are right about one thing in the last post… The odds that the picked card remain at 1/3.  But the odds that it is the remaining card have increased to 2/3.  Still summing to unity.  I am familiar with the Monty Hall problem, but in my haste assumed that it made a difference because Monty knows where the prize is. The Monty Hall problem is not the same as Russian roulette.  For one thing, Monty isn’t allowed to choose the same door that you did. Where the "trigger pull" might very well do so. Also, you have shown by bringing the Monty Hall problem up that revealing things changes the odds.  If I choose door number 1 and you reveal door number 2 as being a booby prize, you have increased the odds that door number 3 will have the real prize from 1/3 to 2/3.

Yes, yes…. but… The card/MH problem are actually one and the same. The key is that if you do NOT change selection, the odds remain the same (1/3 for MH, 1/6, for RR). 

The selection as to the order of the trigger pulls happened BEFORE the game started (i.e., you had your chamber all picked out, shined up and dudded up before anyone else pulled the trigger). You had and have no aforeknowledge when the selection was made, meaning the probability cannot change as a result of other events that took place in the time between when you picked out your chamber and when you pulled the trigger.

Interestingly enough, I believe that on the 5th pull of the trigger, if you shouted WAIT, and changed to the last chamber, the chance of that chamber containing the bullet go to 5/6 (how do you like them apples?)

Here’s the odds before each trigger pull (the first number is the odds of the chamber under the pin killing you, subsequent odds refer to SKIPPING that chamber in favor of another chamber). Note that the odds sum to 1 as required in each case:

Before 1st:     1/6, 1/6, 1/6, 1/6, 1/6, 1/6 (i.e., spinning again won’t change anything!)

Before 2nd:    1/6, 5/24, 5/24, 5/24, 5/24 (skipping chambers increases odds)

Before 3rd:     1/6, 5/18, 5/18, 5/18

Before 4th:     1/6, 5/12, 5/12 (nearly 1/2 on each of the other two chambers)

Before 5th:     1/6, 5/6 (ouch, don’t switch chambers here!)

Before 6th:     1

It’s very similar to another Vos Savant problem: If there’s a 1 in a million chance of there being 1 bomb on a plane, and a 1 in a billion chance of there being 2 bombs on a plane, then shouldn’t you bring a bomb on the plane every time?

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