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        riddles >> easy >> Conditional Russian Roulette
        
(Message started by: markr on Apr 18^th, 2015, 12:49am)

Title: Conditional Russian Roulette
Post by markr on Apr 18^th, 2015, 12:49am

I was playing around with a simple conditional probability problem (pick up a revolver (six-shooter), pull the trigger and the cylinder is empty - what's the probability that there are n=0 to 5 bullets in the remaining cylinders?) that I extended to become:
What's the probability that if you pull the trigger again, a shot will be fired?

I came up with an answer (1/3), then solved the problem for a two-shooter, three-shooter, ..., five-shooter. Surprisingly, to me, I got the same answer for all cases, and I assume that for n>1 an n-shooter will have the same answer. Maybe I did this wrong.

I'm looking for confirmation that I did this correctly, and if so, is there an intuitive explanation for the fact that the probability is constant?

The statement of the extended problem would go something like this:
- You pick up a gun with N>1 cylinders that you know nothing about with respect to which cylinders are loaded.
- You pull the trigger and no shot is fired (i.e. the cylinder is empty).
- if you pull the trigger again, what is the probability that a shot is fired (i.e. the next cylinder is loaded)?

Thanks in advance.

Title: Re: Conditional Russian Roulette
Post by towr on Apr 18^th, 2015, 3:59am

On the one hand, there's no a priori distribution given. On the other hand whether one chamber is empty or not needn't have any bearing on whether the next is. So Bayesian statistics gets us nowhere.

If I pick up an N-shooter, and each chamber has an equal random chance of containing a bullet, then there's a 50% chance the next chamber contains a bullet if I didn't just shoot myself (and also if I did).

The story is different if we're picking (with equal probability) from one of N+1 N-shooter guns, where the K-th gun has K bullets randomly distributed in the chambers (K running from 0 to N+1). Then I get 1/3rd (at least for a 2- and 3-shooter). So I'm guessing this is what you used as a priori distribution.

Title: Re: Conditional Russian Roulette
Post by markr on Apr 18^th, 2015, 11:02am

What I intended to do (thought I did) was use Bayes to determine the probabilities of there being 0, 1, ..., N-1 bullets in the gun based on the fact that a random sample (the first trigger pull) was empty. Then, for each of those results, I multiplied them by the probability of the next chamber being full and summed it all up.

The first step did not result in equal probabilities for the different loading options. For example, for the 3-shooter, I got:
P(0 bullets in the gun) = 1/2, P(next chamber full) = 0
P(1 bullet in the gun) = 1/3, P(next chamber full) = 1/2
P(2 bullets in the gun) = 1/6, P(next chamber full) = 1
P(3 bullets in the gun) = 0

(1/2 * 0) + (1/3 * 1/2) + (1/6 * 1) = 1/3

Title: Re: Conditional Russian Roulette
Post by towr on Apr 19^th, 2015, 9:44am

That's only correct under the assumption that each of 0 to N bullets are equally likely. But there's no reason to assume that's the case, because having a 50% chance for each chamber to be empty or filled is an equally fair interpretation of a randomly filled gun.

As for a possible intuitive approach why the answer doesn't change as N increases: An N-shooter is equivalent to a K-shooter plus an M-shooter with N=M+K; so for all intents and purposes we might as well only be dealing with 2-shooters (and (N-2)-shooters that we'll never touch).

Title: Re: Conditional Russian Roulette
Post by markr on Apr 19^th, 2015, 10:56am

OK - thanks.