Russian roulette (with a light bulb)

[PrayingMantis]

It might be a simple and obvious question, but I'd still like to get some feedback from the math experts here.

Someone plays "Russian roulette" with a gun and a light bulb, i.e, the gun is pointed to the light bulb. Before starting, he puts 2 bullets into the magazine, which has place for 5. Now he starts pushing the trigger, time after time. He does so until a bullet is shot and explodes the bulb. Each time he pushes the trigger, is "a trial".

Well, The bulb explodes. He tells you that it took no more than 3 trials.

What is the probability the bulb exploded in each of the three first trials?

[Nottom]

Does the gun get reset after each trial?

If they are dependant trials then the only way he would miss would be if he fired the 3 empty chambers first. So the chances of him hitting would be 1-(60%*50%*33.3%)=90%

If they are independant then you just have to miss 3 times where you have a 60% chance of missing each time. So its 1-(60%^3)=78.4%

[PrayingMantis]

I think you read the question a bit differently from what I meant.

The man didn't reset, or reload. He loaded once, with 2 bullets, in random order, into this 5-chambers magazine. Then started pushing the trigger, one time after the other.

You know that the bulb exploded, and you know it took *no more* than 3 trials, i.e, 3 times of pushing the trigger.
The bulb could have exploded in the first, second, or third push. That's it.

The question is: what is the probability the bulb exploded in each of the different three trials?

I'm looking for answer that will look like this:

Prob. of the bulb exploding on the first push (trial): X

Prob. of exploding on the second trial: Y

Prob. of exploding on the third trial: Z

[Bozeman]

Asssuming no respinning of the chambers (dependent prob.), P(1)= .4, P(2)=(1-P(1))*.5=.3, P(3)= (1-P(2|!1)-P(1))*2/3=.2. To find the conditional probabilites of these events knowing that the gun went off in 3 or less, the ratios of the probabilities will remain the same, but their sum will be one, so P(1|<4)=.4/.9=4/9, P(2|<4)=1/3, P(3|<4)=2/9.

For the respinning problem, P(1)=.4, P(2)=.4*.6=.24, P(3)=.4*.36=.144, so P(1|<4)=.4/.784, etc.

Craig

[PrayingMantis]

Thanks, Craig. Your first answer is basically what I was looking for, since I didn't mean it to be a "respinning" problem.

Now, I have a question regarding this.

[ QUOTE ]
Asssuming no respinning of the chambers (dependent prob.), P(1)= .4, P(2)=(1-P(1))*.5=.3, P(3)= (1-P(2|!1)-P(1))*2/3=.2.

[/ QUOTE ]

You treat this, as if the the bullets are randomly spread in the magazine (that's how the question was phrased: the man simply puts in the 2 bullets). However, we already "know", in retrospect, for example, that there was *at least* one bullet in the first three chambers (that were "shot"). In other words: the probability of 2 consecutive bullets in the last 2 chambers is 0.

Shouldn't this have some implication on the calculation?

[Jezebel]

[ QUOTE ]
It might be a simple and obvious question, but I'd still like to get some feedback from the math experts here.

[/ QUOTE ]
I believe that there is a very simple answer to this question.

[ QUOTE ]
Before starting, he puts 2 bullets into the magazine, which has place for 5.

[/ QUOTE ]
Since we are putting bullets into a magazine we must assume that the weapon is not a revolver (since a revolver does not have a magazine, but a cylinder). Handguns that use a magazine are always automatics. Therefore the likelyhood of the bulb being shot on the first trigger pull is 100%, absent a misfire.

This response is only half meant as a smart ass response. You did not state why you wanted to know the information. If a "buddy" asked you this question, it may have been posed in this manner to give you the run around and him a cheap laugh. Just a thought.

[Dan S]

[ QUOTE ]
Thanks, Craig. Your first answer is basically what I was looking for, since I didn't mean it to be a "respinning" problem.

Now, I have a question regarding this.

[ QUOTE ]
Asssuming no respinning of the chambers (dependent prob.), P(1)= .4, P(2)=(1-P(1))*.5=.3, P(3)= (1-P(2|!1)-P(1))*2/3=.2.

[/ QUOTE ]

You treat this, as if the the bullets are randomly spread in the magazine (that's how the question was phrased: the man simply puts in the 2 bullets). However, we already "know", in retrospect, for example, that there was *at least* one bullet in the first three chambers (that were "shot"). In other words: the probability of 2 consecutive bullets in the last 2 chambers is 0.

Shouldn't this have some implication on the calculation?

[/ QUOTE ]

Yes. We can throw out one of the ten possibilities of bullet distribution (4th and 5th chambers). So the probabilities are now 4/9, 3/9, and 2/9, not 4/10, 3/10, and 2/10.

Note that the probabilities have to add up to 1 or the answer makes no sense.

[Dan S]

Oops, I see that Craig already addressed this in his original post that you replied to.

[PrayingMantis]

[ QUOTE ]
Since we are putting bullets into a magazine we must assume that the weapon is not a revolver (since a revolver does not have a magazine, but a cylinder). Handguns that use a magazine are always automatics. Therefore the likelyhood of the bulb being shot on the first trigger pull is 100%, absent a misfire.

This response is only half meant as a smart ass response. You did not state why you wanted to know the information. If a "buddy" asked you this question, it may have been posed in this manner to give you the run around and him a cheap laugh. Just a thought.

[/ QUOTE ]

Very impressive. there are two reasons why I used magazine instead of revolver: 1) I know nothing about this stuff (except from TV), 2) If I knew the terminology, it probably wasn't in English, but in Hebrew.

Yeah, I know there's some link between the state of Israel and all kinds of weapons, but I don't really have anything to do with it.

I used gun and bullets, instead of cards, for example, only to make my point as clear as possible (first bullet=end of story), and also to make it more attractive, and less boring. I guess it worked...

Thanks for giving it a thought... /images/graemlins/grin.gif

[BruceZ]

The number of ways to load 2 bullets in 5 chambers is C(5,2) = 5*4/2 = 10. We can rule out 1 arrangement where the bullets are in chambers 4 and 5 since we know there was one in the first 3. So there are 9 possible arrangements. Now just put a bullet in chamber 1, count the number of ways to put a bullet in the other chambers, and divide by 9. Repeat for chambers 2 and 3.

1: (1,2) (1,3) (1,4) (1,5) = 4/9
2: (2,3) (2,4) (2,5) = 3/9
3: (3,4) (3,5) = 2/9

These add to 1 as they must.

[PrayingMantis]