An Uncomplicated Kill
On a top secret assignment in Azerbaijan, you and your partners are rather bored, and Vodka is in short supply. Being the daredevils that you all are, you decide to play a modified game of Russian roulette to eliminate one of you. A six shooter revolver is purchased from the neighborhood arms dealer and three bullets are loaded in a row in three consecutive chambers. The barrel is spun only once. Then, each player points the gun at their head and pulls the trigger. The game stops when one person is dead.
Would you rather be first or second to shoot?
Would you rather be first or second to shoot?
posted by littlecow at 10/03/2006 01:50:00 AM
17 Comments:
For the first person p(death) = 1/2.
For the 2nd person, we are interested in p(death|first person is not dead). This is 1/3. So choose to go second.
For the first person p(death) = 1/2.
For the 2nd person, we are interested in p(death|first person is not dead). This is 1/3. So choose to go second.
Would go second:
P(death_of_A) = 1/2;
P(death_of_B) = P(survival_of_A due to his encountering the bullet from the last empty chamber)= 1/6
sorry, the last tline should read
P(death_of_B) = P(survival_of_A due to the barrel being positioned at the last empty chamber during his shot)= 1/6
Nice retake on sholay :)
Coming to the problem:
For A- 3 chances in 6 that the bullet is present in the chamber loaded.
For B- Since the guy before is not already dead,it means one safe chamber has been passed.Now,we have the following situations:
a
f b
--- the chambers
e c
d
where a,b,c are empty and d,e,f are loaded.
1.Suppose A gets chamber a.Then B is also safe.
2.A gets chamber b,then B is safe again.
3.A gets c,then C is not safe anymore.
These are the only possibilities that can occur for given constraint that the barrels are consecutively loaded and A doesnt die.
Thus probability of B's survival is only 1 in 3.
Of course,it doesn't matter,all of them will die eventually cos of driniking alcohol...
Seems that the chambers haven't come out exactly as expected,sorry about that.
Sorry,I meant that the probability of B's death is only 1 in 3.
looks like there is an interesting math problem goin on here - i jus wanted to say thanks for the comment... i will certainly check out the vellore tmeple the next time i am close by....
I dont think i have ever enjoyed any other temple visits other than the ones in mayavaram - mostly because, the others I go cos my mom thinks i should and my mom always has 101 dieties to see and pujas to do - sort of takes the pleasure out of the whole thing.
And yes, its quite a nice feeling to try and go back in history and visualise the whole place in a different era -
And more than anything, there is this feeling of majesty one gets in some forts / temples - and a feeling of greatness from those places which is incomparable - one of the reasons i loved hampi too... and Golconda too!
And i have to say, i never thought i wud get a comment for that post of mine :)
>grenade:The barrel is held in place by a pin,and the chambers weight must far exceed the weight of the three bullets combined,thus making the effect of gravity very small.If gravity was a factor,then I think the equilibrium position would be with the three bullets in the lower three chambers.The gun empties the topmost barrel first.Thus the probability in this case would be zero.
A simpler assumption to negate gravity would be to spin the barrel with the gun facing downward.
Clarification:The probability that either person should die would be zero.
Is this assumption feasible:The torque required to spin the barrel will far exceed the moment of friction between the pin and barrel,so that we stop the barrel by hand,and when the mechanism of the gun is involved,the friction becomes sufficient.
(I think revovlers work on the basis of a lock holding them in place when being fired,so that it is not friction which is stopping the barrel,but the lock.)
A new puzzle:
In how many ways can two queens be placed on a chessboard so that their firing lines are separate?
Regarding the coin toss,even with ordinary coins,one may have observed that if we start with a given face up,then more often than not,it will show the same face up if it is caught at approx. the same level as it was thrown from.
[It must involve a lot of physics to understand its dynamics,though.]
Sorry guys,but I dont have the answer to the question....from where I checked the problem,he never considered the diagonal firing lines.Since the answer is obtained by brute force as well as a mathematical procedure,it must be right.
Here is another problem:
If you remember your solid state chemistry,the close packed structures are said to be the most efficient in terms of volume packed by the spheres. So what are the structures that enables maximum packing and what is their efficiency?
It doesn't matter. The odds are the same!
Assume chambers are labeled ABCDEF. Wlog, assume the bullets are loaded in ABC. Then there are only 6 possibilities, ABC, BCD, CDE, DEF, EFA, FAB. ABC means first person tries A, second tries B, etc. +1 if you are alive, -1 if you die. The scores come out, -1, -1, -1, 0, -1, +1. So the first one to pull the trigger dies with 2/3 chance,, stays alive with 1/6 chance, both stay alive with 1/6 chance.
If they get to turn the chambers randomly before trying every time, the muubers might come out different.
The actual probabilities depend on the number of players, which wasn't specified.
Let [p(1)...p(n)] be the probability that the kth player is the first to die.
For n>4, p(k)=0 for k>4.
For n>=4, p(4)=1/6, p(3)=1/6, p(2)=1/6, and p(1)=1/2
For n=3, p(3)=1/6, p(2)=1/6, p(3)=2/3
For n=2, p(2)=1/3, p(1)=2/3
In any case p(2) < p(1). If there's a trick I don't see it. But it is interesting to note that there's no difference between being in positions 2 through n, unless n>4.
Post a Comment
<< Home