Quick logic puzzle help!

[Tomec]

I'm in a class where the instructor is encouraging us to cheat... So I'm hoping you all can help me really quick with this quiz item

Here is the problem


3 people are having a shoot out. They are taking turns shooting in a random order standing in a triangle shooting whoever they want. Guy A has a 100% chance to hit his target. Guy B has a 80% chance to hit. Guy C has a 50% chance to hit.

What is each of their chances to be the last one standing.

[Blueobelisk]

I can't do it quickly lol.

What's the time limit?

[Tomec]

Well it isn't due for another 15m or so... and I have a couple general ideas going on for how it should be solved just in case. But I HATE logic problems and this is the hardest one he gave lol

[Blueobelisk]

Just draw a chart of the scenarios and add up the math.

Tier 1 is the chance each person starts
Tier 2 is the next person after (and one after)

Then continue including if they hit or miss by %s.

[Tomec]

Just draw a chart of the scenarios and add up the math.

Tier 1 is the chance each person starts
Tier 2 is the next person after (and one after)

Then continue including if they hit or miss by %s.

Basically what I'm doing. Thanks for that, I'm on the right track at least lol. Now just to sit down and grunt it out.

THough with encouragement to cheat if anyone gives a solution (and it matches mine or is close enough) before the deadline... One internetz cookie!

[MrKnubbles]

Assuming that Guy A always starts first...

Guy A starts off and shoots Guy B, then Guy A would have a 50% chance to survive.
Guy A starts off and shoots Guy C, then Guy A would have a 20% chance to survive.

If Guy A shoots Guy B then Guy C would have a 50% chance to survive. But then he has to hit his shot which is a 50% chance to hit so I think that's a 25% chance to survive.
If Guy A shoots Guy C then Guy B would have a 50% chance to survive. But then he has to hit his shot which is an 80% chance to hit so I think that's a 40% chance to survive.

Am I on the right track here?

[Blueobelisk]

It's hard to answer for C and B since they could potentially keep missing each other lol...

[Machismo]

If the shooting order is random, then you are not going to get it solved in 15 minutes.

[Reeve]

I'm in a class where the instructor is encouraging us to cheat... So I'm hoping you all can help me really quick with this quiz item

Here is the problem


3 people are having a shoot out. They are taking turns shooting in a random order standing in a triangle shooting whoever they want. Guy A has a 100% chance to hit his target. Guy B has a 80% chance to hit. Guy C has a 50% chance to hit.

What is each of their chances to be the last one standing.

What order do they shoot in? Are their targets fixed, or is there a 50-50 chance they'll choose one of the other two gunners?

[Machismo]

Shooter A will always choose to shoot at B, and B will always choose to shoot at A. The issue is the randomness of the second and third sets. (the first set is A=50% and C=50%.

[Blueobelisk]

10% chance A wins

---------- Post added 2013-02-06 at 10:34 PM ----------

Wait no lol. That's be if we only considered if he started. Then we have to consider if he won with someone else starting...

This is too complicated for a short amount of time.

[Machismo]

This is a ballpark figure, as the magnitude can get rather daunting in the 2nd and third sets:

A=21.1, B=24 C=54.9

That is with a fully random order, and each one gets to choose their target.

[Reeve]

10% chance A wins

---------- Post added 2013-02-06 at 10:34 PM ----------

Wait no lol. That's be if we only considered if he started. Then we have to consider if he won with someone else starting...

This is too complicated for a short amount of time.

If A starts, and he chooses B as his target, then A and C would both have a 50% chance of being the last standing. Unfortunately we have no way of knowing who starts, how they choose their targets, or anything.

[Machismo]

There are three sets to this, determined by who randomly starts first. The first set is easy, as A would immediately shoot B, thus giving A and C and equal 50-50 chance of winning. The second set would involve B shooting first (at A). The person who started this did not state if they were all allowed a turn, or if the order was completely random ie. shooter C could get two shots in a row.

[Tomec]

Yea that whole "who starts first" thing was the biggest kicker. The problem didn't specify who starts, in fact it even went so far as to say it was a random order. But like if I'm remembering the question right it was like if the order was BCA then B shot and unless C was dead C shot then A shot (unless dead) repeat until 1 stands. I think A will always have the highest chance, and C the lowest though. Just due to the %. A always kills, and B and C can miss (even if B rarely does), so as soon as he's up you're in a world of trouble. But if it wasn't for that random thing it'd be a lot easier.

Anyways, I'll know next week (I believe) what the answers were and I'll be really interested to know how we were actually supposed to solve it and stuff lol. Thanks for the answers/help.

[Twoddle]

If I remember this puzzle right the order is C fires first then B then A.

Semi-Spoiler:

C's best chance is to fire his first shot straight in the air, B will then shoot at A as his greatest threat.

I would go on to solve it but I'm running late. Cheers.

[Catta]

I saw this in QI...

[Wildtree]

I'm in a class where the instructor is encouraging us to cheat... So I'm hoping you all can help me really quick with this quiz item.

Cheating could very well mean, to go and google the answers..
There's leads out there for it...
F.e. http://vishal12.wordpress.com/2011/0...fight-puzzles/