Probablity

Assuming the other two guys are shooting at the guy most likely to shoot them... then the chances of DeadEye Dick being the first to boot hill is pretty much zero.

Half will shoot at Third, Third will shoot at Half until one or both of them is toast.

I dont think you understood me. I want probability. As in fraction. REMEMBER: i'm looking for the probability of the 1/4 guy being one of the first to be killed.

Asumption: Each of the guys knows the chance they have of hitting and the chances of the other two.

Answer: 0

As SealPoet said...

pardon my language, but goddamn you people. This is a math question.

Lazydude, the answer is zero. Is that not mathematical enough for you?
Can you please explain why you want God to damn us?

Sorry, golazy, I feel for you, but their answers are funny. You should rephrase the question so that the target of each mans choice is expressed in probability. In other words there is a fifty percent chance they will DECIDE to aim at either one of their rivals. Otherwise we can just choose their decisions.

forget it nooone will get it right. the answer is 5/12. Here is why: 1/2 half guy has a 1/2 chance to shoot at the 1/4 guy. That means that his chance of killing the 1/4 guy is 1/4 becuause you multiply his accuracy by the chance of him shooting a the guy. The 1/3 guy also has a 1/2 chance of shooting at the 1/4 guy, so his chance of hitting the 1/4 guy is 1/6. Then, all you simply do is add 1/4 and 1/6 to get your answer.

I had gotten that far, but isn't that only for the first round of combat? For the second round of combat you have to factor whether or not 1/2 shot and killed 1/3, or 1/3 shot and killed 1/2, or both. Then you have to do the same for the third round of combat and so on. I was going to assume that they were using six-shooters, and there were only six rounds of combat total, but the math would still be beyond me.

But you are assuming independence when independence cannot be demonstrated -- and no statistical analysis is better than statistical analysis founded in bad assumptions.

i am saying on the first round. And patiodog you complicating things

Well, it's been too long for this to be a homework problem, so I'll tell you the way I figure it.

The first thing to consider is who is shooting at 1/4. There are four possible outcomes here, all of equal probability:
(a) P(1/2 and 1/3 shoot at 1/4) = 1/4
(b) P(only 1/2 shoots at 1/4) = 1/4
(c) P(only 1/3 shoots at 1/4) = 1/4
(d) P(no one shoots at 1/4) = 1/4

Then we have to figure the probability of 1/4 being shot in instances a, b, and c. (Clearly the probability of 1/4 being shot in instance c is zero.)

P(1/4 is shot in instance a) = (1/2) + (1/3) = 5/6
P(1/4 is shot in instance b) = 1/2
P(1/4 is shot in instance c) = 1/3

So,
P(1/4 is shot by someone) = (1/4)*((5/6)+(1/2)+(1/3)) = 5/12

Which, unfortunately, is the same answer I got the first time when I figured it in the way I though was oversimplistic, which is to calculate the sum of the probability that 1/4 gets shot by 1/2 and the probability that he gets shot by 1/3:
(1/2)*(1/2) + (1/2)*(1/3) = 5/12

The reasoning in this latter case seems flawed to me, since it doesn't seem to count out the possibility that 1/4 is shot by both of his adversaries at the same time. Hmmm...

You REALLY need to rephrase the question if everyone only fires once.

If each shooter chooses his target entirely at random, our 1/4 accuracy marksman will die in the first round with a probability of 3/8.

It would be 5/12, but the probability of him taking two bullets is 1/24, which must be subtracted... leaving 3/8

I don't think that needs to be subtracted. Either way he dies, so you don't discount one event.

See, the first time I did it I was dissatisfied with the 5/12 answer for the reason that apostrophe states, but when I used another method to account for this instance, I got the same answer. I think you'll find the way I put it above rules that out as a problem.

Another way of doing it:

Probability that Mr. 1/2 will not kill Mr. 1/4 = 3/4

Probability that Mr. 1/3 will not kill Mr. 1/4 = 5/6

Probability that neither will kill Mr. 1/4 = (3/4) * (5/6) = 15/24 = 5/8

And if there is a 5/8 chance of him not dying, there is a 3/8 chance of him dying.

Damn it, you're right. I didn't fix anything with my way of doing it.

And so simple, too.

i don't understand why the riddle says it is a shoot out between two people.. yet there are three people described... that just makes it a little more confusing

my bad, i dont look while i type