Tripli puzzle

Re: Tripli puzzle

Hopefully, A's first shot doesn't take out B by mistake.

Clarification

A and B may occasionally miss who they are aiming for, but would never hit someone they are not aiming for.

Re: Tripli puzzle

A's best chance shooting at someone else is to shoot at B and miss. B's shot almost certainly will target C. If he hits then C is out and then A will get the next shot. If he misses then C will likely play the percentages and shoot at B. B will certainly be out. A then gets the next shot and since he missed the first time his second effort percentage wise is pretty good. If he misses then C is the last stander.

On the right track

Just convert those observations to a strategy.

For bonus points, given the optimal strategy by all three, who is most likely to survive? Least likely?

For any fellow probability geeks: what is each participant's probability of survival?

Re: On the right track

I don't grasp on how to make it a mathematical strategy. The parameters seem pretty strict. If I'm "A" I figure the others will play percentages so I'm likely to get shots one and three. That means in a round I likely get 50% more shots than the others. I do know that B is a gonner if C is standing after shot two. Here's my strategy. After shot #1 my strategy is to offer the least amount of target area to the other two participants. I'll watch for the probability people to reply to find how it's done mathematically.

For sure C

has to go because as long as hes alive someone will get shot. It's possible if A and B go at it they both can live--for awhile. If they don't continually miss then B has the edge because of accuracy.

Re: Tripli puzzle

I would say that A should shoot at C. If he were to shoot at B and hit, then he is dead. If He shoots at C and hits, he still has a chance. If he shoots at C and misses, the odds are that C will shoot B, so A still has a chance. Statistically, if A always shoots at C and the hit probabilities hold, then he has the best chance to end up the last one standing.

PS, I am not smart enough at statistics to determine the odds for each one. ;-)

Re: Tripli puzzle

Also, B has to shoot at C because, C will most likely shoot at him and he would then be dead. Again, If he hits C, then A still has a chance, and if he misses, he (B) is likely dead.

Think outside the box

Are we making this more difficult than

it has to be? "A" needs to miss/or fire into the air. If "A" unloads at the second-worse shot and kills "B", then "A" is dead. If "A" unloads at "C" and kills him, "A" then has only a one-third chance of living. By "A" initially killing one of the others then "A" would only make his/her chances worse, because then "B" or "C" would then shoot at "A" instead of one of the others. By missing, "A" gets another shot, with better percentages.

We have a winner!

You were so close before when you talked about what would happen if he aimed at B and missed. But it's hard to jump to that next part, omitting the "aim at B and". Good job

For the followups, with this strategy, A has the best chance of survival (59/110 by my calculation), and the best shot, C, has the worst chance of survival (10% = 20% chance of surviving the first shot from B (after which he will shoot B) times a 50% chance of surviving the next shot from A)

PS

In your response, you said

"If "A" unloads at "C" and kills him, "A" then has only a one-third chance of living."

Your reasoning is right, but the math is a little off. He only has a 20% chance of surviving the first shot from B, and even then, he is not home-free--he has only a 50% chance of killing B, and if he misses, he stands only a 20% chance of surviving the next shot...

The math gets tricky here, and I am doubting the numeric answer I gave in the previous post. Maybe someone can chime in with some help.

Re: PS

Alex, it has been a long time since Probability 101, but this is how I think it can be done:

Consider a simpler problem:

Just A and B, with same probabilites of hit/miss. There are an infinite number of ways (with decreasing probabilites of happening) for A to win (last one standing). Construct the tree of all possible outcomes for A to win; note that there are two trees, depending on who has first shot.

If A has first shot, then A wins if:

A hits B (prob = .5), or if

A misses, B misses, A hits B (prob = .5x.2x.5), or if

A misses, B misses, A misses, B misses, A hits B (prob = .5x.2x.5x.2x.5), or if .... (infinite series).

This series is .5(1+ .1*1 + .1*2 + .1*3 + ...) = .6111..., so

If A has first shot, then A wins with prob = .6111...

(and B wins with prob = 1- .6111... = .39999)

If B has first shot, similar reasoning shows A wins if

B misses, A hits B (prob = .2x.5), or if

B misses, A misses, B misses, A hits B (prob = .2x.5x.2x.5), or if ...

This series is .1*1 + .1*2 + .1*3 + ... =.111..., so

If B has first shot, A wins with prob = .111...

(and B wins with prob = 1 - .111... = .8999...).

Similar logic can be applied to the original problem; this exercise is left to Alex ...

Re: PS

Thanks, Larry. You took a slightly different approach than I did, and with a couple of minor arithmetic corrections, your solutions avoid the inconsistencies mine were raising. Going back to f1rst principles is usually the right approach! And I now see the flaw in my reasoning.

Here are the corrections I noted:

"This series is .5(1+ .1*1 + .1*2 + .1*3 + ...) = .6111..., "

S/b .55555...

And B wins with probability .44444

"If B has first shot, similar reasoning shows A wins if

...

(and B wins with prob = 1 - .111... = .8999...)."

s/b 1-.111... = .8888....

Now I get the probabilities of survival for each of the participants as

A: 49/90 = .5444...

B: 16/45 = .3555... and

C: 1/10 = .1

That removed some cobwebs!!!

Boing!

What an awakening. As in the song �Wonderful World� by Sam Cooke, the lyrics �Don�t know nothin bout his-to-ry, Don�t know nothin bout bi-ol-o-gy�, well, I don�t know nothin bout prob-a-bil-a-ty. But it occurs to me that� Is this what they teach at War College? Does the Pentagon have math guys on staff that labor in their service to project the probability of entering fire fights, battles, wars? Do the senior, senior leaders of such things determine their participation by mathematically predicting victory, defeat, losses, based on mathematical probability rather than field theaters? It looks painfully similar to setting a line on a sporting event�I�ll take Notre Dame by six over Ohio State. I�ll take India by eleven million over Pakistan. I�m too na�ve. Of course they do, it would be dumb to not assume so.

Re: PS

Alex

Thanks for correcting the arithmetic errors.

Gary - yes indeed the Pentagon has mathemeticians/statisticians who perform analyses like this. The Generals may not pay much attention to the actual numerical results since the probability of various events occuring is usually difficult to quantify. But they might pay attention to the general outlines suggested by an analysis like this. In this case, one conclusion might be that it is important to have the first shot regardless of accuracy. Ready, Fire, Aim!

Larry