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 Posted: Fri Apr 27, 2007 11:28 am
Hi guys.
My 8th grade class had a test on probability and straight lines this Thursday. It was quite simple until I reached the last question on Russian roulette, which I simply had no idea how to solve. We'll get the tests back (and as such the answers) next Thursday but I can't wait for that, I need the answer or I won't be able to sleep.
A guy named Kid is playing Russian roulette. He is alone, not playing with someone else ((sick b*****d)). The revolver he is using has eight chambers, and he places two rounds in it. He plays three games. What is the risk of him dying?
In the first game, the risk of him dying is of course 2/8 = 1/4, but I have no idea how to go from there.
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Posted: Sun Apr 29, 2007 8:52 pm
Assuming that 1 game is pulling the trigger once and you never spin the barrel after each game:
First Game:
2/8 = 1/4 = 25%
Second Game(assuming you live):
2/7 = 28.5%
Third Game(assuming you survive the second):
2/6 = 1/3 = 33%
Your overall chance in dying is:
2/8 + 2/7 + 2/6 = 73/84
Which is about 87% chance you will die.
If you happen to respin after each game of 1 shot each:
1st:
2/8 = 1/4
2nd:
2/8 = 1/4
3rd:
2/8 = 1/4
Final = 3/4
75% Chance you die there.
In both cases, you better write a decent will.
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Posted: Wed May 02, 2007 1:37 pm
Buga Assuming that 1 game is pulling the trigger once and you never spin the barrel after each game: Whereas it wasn't explicitly stated whether he spins the barrel or not after each game, we had to assume he didn't as no one probably would have managed to solve it if he did. Should have written that in my original post. Buga If you happen to respin after each game of 1 shot each: ... Final = 3/4 This would mean that he'd be certain of dying the fourth time he plays. : -) Anyways, I figured the answer out after some trial-and-error. I got it confirmed by one of the teachers. In the first game, he has a 1/4 chance of dying. In the second game, he first has to survive in order to play it and die. As such, 3/4 * 1/4. In the third, he has a 3/4 * 3/4 * 1/4 chance of dying, following the above. As such, 1/4 + (3/4 * 1/4) + (3/4 * 3/4 * 1/4) = 37/64 chance of dying. I came pretty close to thinking of this on the test, shame I somehow decided to multiply instead of adding... oh well.
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Posted: Thu May 03, 2007 10:17 am
The way I see it, there are, essentially, four ways to put the bullets in the gun. They can be in adjacent chambers, have one chamber between them, two chambers between them, or three chambers between them (a gap of four is the same as a gap of two, five is the same as 1 and six is the same as adjacent chambers). Call these case0, case1, case2 and case3.
Assuming no respins, you can work out that given case0, you have three bad starting chambers and five good ones (ones that you survive from); Given case1, there's four bad and four good; case2, five bad and 3 good; case3 six bad and two good.
Now, the odds of being in case0, case1 and case2 is 2/7 because there are two ways that they can come about out of 7 and 1/7 of a chance of being in case3. So, the odds of surviving are
(2/7)*(5/8 + 4/8 + 3/8 ) + (1/7)*(2/8 )= 12/28 + 1/28 = 13/28.
I think Buga got it wrong when he straight added them up. For instance, whats the chance of throwing a 6 in two throws of a dice? Its 11/36 not 2/6 and the latter is what buga's method would get you I think.
I don't really like the question. Its difficult to see how to distribute the probabilities. Perhaps a tree would help, but I doubt it. Probability questions should have clear distributions, or else its like asking what chance a triangle has of being acute.
Edit: your answer assumes respins and is correct for that.
Edit2: I made some errors with the odds of survival from the different cases: for case0, there are 4 bad and 4 good places to start from; for case1 there are five bad and 3 good; for case2 there are 6 bad and 2 good; and for case3 there are 6 bad and 2 good.
This gives a survival probability of
(2/7)*(4/8 + 3/8 + 2/8 ) + (1/7)*(2/8 )= (2*(4+3+2) + 2)/56 = 20/56 = 5/14.
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Posted: Mon May 14, 2007 8:35 am
It's easy actually. You just have to live through the 3 shots
Respin:
First: 3/4 he will survive this one
Second: 3/4 he will survive this one
Third: 3/4 he will survive this one
Total: 3/4 * 3/4 * 3/4 = 27/64 he will survive
This means 1 - 27/64 = 37/64 chance of dying, exactly the same as your teacher said it.
Without Respin:
First: 3/4 he will survive this one
Second: 5/7 he will survive this one
Third: 2/3 he will survive this one
Total: 3/4 * 5/7 * 2/3 = 5/14 he will survive
This means: 1 - 5/14 = 9/14 chance of dying.
The difference why this solution is shorter than the other is because I use the easiest way: calculating the chance he lives and then taking the complement of that. (you either live or die, so the sum of those chances are 1.)
If you want to know the chance of living, he has to live through all 3 shots, which is an easy calculation.
However, if you want to know the chance of dying, you have to calculate what is the chance he dies the 1st, 2nd, 3rd shot, assuming he lived through the previous ones. You can't die twice or more, can you? So, that gives more difficult calculations.
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Posted: Wed Jun 27, 2007 10:04 pm
This reminds me too much of the story I read about Russian Roulette, only that in that story,
A) there were two people playing Russian Roulette
B) they put an extra bullet in after each round
C) they respun it every time
Anyway! The point I thought up was the one which The_Bartner said in his last post. If he dies he has to stop playing.
It's like those funny tee diagrams were you have a really short barnch on one end (he dies first try), and an infinetly long one on the other one (he keeps shouting but never dies)
You just have to look at the first 3 levels though.
First shot
He dies 1/4 he lives 3/4
Second shot
he dies 1/4 he lives 3/4
Third shot
he dies 1/4 he lives3/4
thus the probability he dies equals
(1/4) + (3/4*1/4) + (3/4*3/4*1/4)=37/64
That's how I would have figured out this problem anyway.
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Posted: Mon Jul 02, 2007 9:53 am
The probability of him dying is 0 or 1. ninja
[Really, there is an important distinction that needs to be made unless you are a Bayesian <_<]
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