[bushindo]

Two gunmen are standing 200 paces apart. Each man has only 1 bullet in his gun. Every second, the two men take 1 step towards each other. If the first man (call him A) were to shoot, his chance of killing B is k/100, where k is the number of steps he already took. So when k = 0, the two men are 200 paces apart, A's chance of killing is 0. When k=100, which is when the two men are right next to each other, A's chance of killing is 1.

B's chance of killing is 1.25*k/100 when k is less than 80. If k is more than 80, B's chance of killing is 1. Essentially, B's chance of killing increases between 0 and 80 paces, and stays constant at 1 after 80 paces.

Assume that the two men are in a locked room, and there's nowhere to run. Therefore, if a man were to shoot first and miss, the other man would kill him for sure.

1) Let's say that A and B know about their accuracy functions as well as their opponent's, at which step should A shoot for an optimal chance of survival? At what step should B shoot?

2) Let's say that A and B know their accuracy function, but not their opponent's? Where is the optimal shooting step for A? Where is the optimal step for B?

[Joe's Student]

Two gunmen are standing 200 paces apart. Each man has only 1 bullet in his gun. Every second, the two men take 1 step towards each other. If the first man (call him A) were to shoot, his chance of killing B is k/100, where k is the number of steps he already took. So when k = 0, the two men are 200 paces apart, A's chance of killing is 0. When k=100, which is when the two men are right next to each other, A's chance of killing is 1.

B's chance of killing is 1.25*k/100 when k is less than 80. If k is more than 80, B's chance of killing is 1. Essentially, B's chance of killing increases between 0 and 80 paces, and stays constant at 1 after 80 paces.

Assume that the two men are in a locked room, and there's nowhere to run. Therefore, if a man were to shoot first and miss, the other man would kill him for sure.

1) Let's say that A and B know about their accuracy functions as well as their opponent's, at which step should A shoot for an optimal chance of survival? At what step should B shoot?

2) Let's say that A and B know their accuracy function, but not their opponent's? Where is the optimal shooting step for A? Where is the optimal step for B?

Could you clarify the bit in red for me please? If A takes 100 paces k=100. Seeing as the distance between them is 200, is there not still 100 paces between them? Therefore meaning the two men are not right next to each other? Or do both A & B each take a step at the same time?

Thanks.

[Guest]

as i understand there is two hundred paces between them not to the center. So if each take 100 steps then 2 * 100 they have gone 200 steps.

[bushindo]

as i understand there is two hundred paces between them not to the center. So if each take 100 steps then 2 * 100 they have gone 200 steps.

Final's understanding is correct. There are 200 paces between the two men. After 100 paces from each men, the two will be right next to each other.

[Joe's Student]

as i understand there is two hundred paces between them not to the center. So if each take 100 steps then 2 * 100 they have gone 200 steps.

Yeah so they each take a step together? ok thanks.

[Joe's Student]

Ok well for the first premise I think it's a type of WIFOM situation:

I'm not sure if this is right but I'll give it a go anyway, maybe someone can point me in the right direction.

1)

For any value of k B's chance of hitting A is higher than A's chance of hitting B. A knows that B is guaranteed to shoot at A when k=80 (if he hasn't before then). So therefore to give A the best possible chance of hitting B, he should shoot when k=79. But b/c B knows A's accuracy functions B should shoot before A i.e when k=78.

Following on from this, A should also realise that B will shoot at k=78 so A should shoot at k=77. You can see where this is going. I think this is more of a psyching-out-the-opponent type set-up.

[Guest]

Ok well for the first premise I think it's a type of WIFOM situation:

I'm not sure if this is right but I'll give it a go anyway, maybe someone can point me in the right direction.

1)

For any value of k B's chance of hitting A is higher than A's chance of hitting B. A knows that B is guaranteed to shoot at A when k=80 (if he hasn't before then). So therefore to give A the best possible chance of hitting B, he should shoot when k=79. But b/c B knows A's accuracy functions B should shoot before A i.e when k=78.

Following on from this, A should also realise that B will shoot at k=78 so A should shoot at k=77. You can see where this is going. I think this is more of a psyching-out-the-opponent type set-up.

I was thinking along the same lines, but with a small modification:

1)

as joe's student said its a game of psych out, but for optimal survival B will not shoot b4 his chance of a hit is at least 50%, i.e. 40 steps. Now, B knows that A is thinking the same thing, but for A it is 50 steps. Since A also knows that B will have > than 50% accuracy after 40 steps, A's best chance is to shoot at 40 steps. B has a better chance if he waits for his opponent to reach 50 steps (50%), and then shoot, i.e on the 50th step.

Ans: A would shoot at 40 steps, B would shoot at 50 steps (if A has not jumped the gun)

2)

A knows his accuracy is at a 50% at 50 steps. He does not know what B's accuracy is, but if he were to take an avg., it would be 50% at 50 steps too. So the ideal time to shoot would be 50 steps. B, however, his knows he is at an advantage to the avg. (50% at 50 steps), so B would play his advantage and shot at the 50th step.

Ans: Both will shoot at the 50th step (B would win)

[Joe's Student]

I was thinking along the same lines, but with a small modification:

1)

as joe's student said its a game of psych out, but for optimal survival B will not shoot b4 his chance of a hit is at least 50%, i.e. 40 steps. Now, B knows that A is thinking the same thing, but for A it is 50 steps. Since A also knows that B will have > than 50% accuracy after 40 steps, A's best chance is to shoot at 40 steps. B has a better chance if he waits for his opponent to reach 50 steps (50%), and then shoot, i.e on the 50th step.

Ans: A would shoot at 40 steps, B would shoot at 50 steps (if A has not jumped the gun)

But b/c they know each others functions, going by your idea, would B not shoot before 40 steps. He wouldn't want to take the chance of A shooting and killing him. You want to be the one to shoot first, especially if your B, as you have a higher chance than A earlier on. So therefore would B not shoot at 39 steps?

I replied in your spolier .

[Guest]

Couldn't get the attachment into the spoiler... but:

1) Ans: A and B both shoot on the 44th step.

2) Ans: A shots on the 50th step, B shoots on the 44th step.

(44th could be 45th... the graph lines intersect between 44 and 45, so lets say 'or')

Edited May 30, 2009 by adiace

[bushindo]

Couldn't get the attachment into the spoiler... but:

1) Ans: A and B both shoot on the 44th step.

2) Ans: A shots on the 50th step, B shoots on the 44th step.

(44th could be 45th... the graph lines intersect between 44 and 45, so lets say 'or')

Good job, adiace. Your response for part I is entirely satisfactory. I would like a bit more justification and elaboration for part II, though.

Edited May 30, 2009 by bushindo

[Guest]

i would argue for 2 that the optimal shooting time would be the same as in number one or are you asking from their point of view what is there best guess at when they should shoot?

[Guest]

First: I want to ask; have you (anyone) ever been shot at in an hostile situation?

Second: It is already too late;

[Guest]

Ok well for the first premise I think it's a type of WIFOM situation:

I'm not sure if this is right but I'll give it a go anyway, maybe someone can point me in the right direction.

1)

For any value of k B's chance of hitting A is higher than A's chance of hitting B. A knows that B is guaranteed to shoot at A when k=80 (if he hasn't before then). So therefore to give A the best possible chance of hitting B, he should shoot when k=79. But b/c B knows A's accuracy functions B should shoot before A i.e when k=78.

Following on from this, A should also realise that B will shoot at k=78 so A should shoot at k=77. You can see where this is going. I think this is more of a psyching-out-the-opponent type set-up.

lol!

Have any of you seen "The Princess Bride"? This is totally Vizzini logic. Going by that, what they should do is each wait to 100 step and shoot eachother after switching guns, but B will have spent the last several years building up an immunity to bullets.

[Guest]

Using math brings back memories... aah..

Now in my solution for the 1st part I used the intersection of lines to determine the step to shoot on for ideal survival.

For A: His chance if killing B is defined by the line y=x/100, x = number of steps taken.

If he does not know B's ability, then his chance of being killed by B is y = 1 - p*x/100, where p = B's multiplication factor (it was 1.25 in the 1st part)

We an solve the two lines and find its intersection.

we get: x/100 = 1- p*x/100

> x = 100/(p+1)

Now this gives the ideal step on which A should shoot. But since he does not know the ability of B (i.e. 'p'). He must take the average on the entire range of possible 'p''s.

p = 0 is the minimum, i.e. no matter what B can't shoot him. ( we can't take negative values for p, or the lines will never intersect)

p= 100 is the maximum, i.e., B has 100% accuracy after the 1st step itself.

To find the Average Value of a Function over a range [a,b]

our function is x = 100/(p+1)

so the Average Value of x for A is:

1/(a-b) * 100 * [ln(a+1) - ln(b+1)]

a = 0

b = 100

> 1/100 * 100 * [ln(101) - ln(1)]

= 4.615 rounded = 5.

Now for B the equation for x is:

x = 100(p+1.25)

so the Average Value of x for B is:

> 1/100 * 100 * [ln(101.25) - ln(1.25)]

= 4.394 rounded = 4.

2nd Part Ans: A shoots on 5th step, B shoots on 4th step.

Edited May 31, 2009 by adiace

[Guest]

in the last post when i say 'our function is x = 100/(p+1)'

i.e. x= f(p) = 100(p+1)

so in the integral formula it is f(p) and dp not f(x) and dx

[Guest]

For A:

Chance of surviving = Chance of not being shot * Chance of killing

P = x/100 * (1 - x/80)

dP/dx = 1/100 - x/4000

1/100 - x/4000 = 0

x = 40 steps.

For B:

P = x/80 * (1 - x/100)

dP/dx = 1/80 * x/4000

1/80 * x/4000 = 0

x = 50 steps.

[Guest]

For A:

Chance of surviving = Chance of not being shot * Chance of killing

P = x/100 * (1 - x/80)

dP/dx = 1/100 - x/4000

1/100 - x/4000 = 0

x = 40 steps.

For B:

P = x/80 * (1 - x/100)

dP/dx = 1/80 * x/4000

1/80 * x/4000 = 0

x = 50 steps.

Hey,

If A were smart he would know that B would (using ur logic) shoot on the 50th step, so why would he not wait till the 49th step and improve his chances.

I think it is wrong to multiply the probability functions, it would be right to solve for when they intersect.

x/80 = 1 - x/100

[Guest]

Hey,

If A were smart he would know that B would (using ur logic) shoot on the 50th step, so why would he not wait till the 49th step and improve his chances.

I think it is wrong to multiply the probability functions, it would be right to solve for when they intersect.

x/80 = 1 - x/100

Well how does he know that the other person will shoot? Obviously it is a mind game and really pointless to work out a theoretical value.

But I stand by my method.

[Guest]

Well how does he know that the other person will shoot? Obviously it is a mind game and really pointless to work out a theoretical value.

But I stand by my method.

Umm... Ok...

A's survival = (x/100)*(1-x/80)

B's survival = (x/80)*(1-x/100)

A shoots at 40

B shoots at 50

Thinking of A & B:

*A knows that B will shoot at 50, A also knows that B will think that A will shoot on 40. A can also shoot on the 49th step to improve his chances, because B will not shoot b4 the 50th step, because he is better off facing A's inaccurate bullet.

I know this seems circuitous but the point I am driving at is: B will not shoot b4 step 50 no matter what.

* This means that A can shoot anywhere between 40 and 50 to improve his chances. In fact the calculation of A's survival from 0 to 50 steps is independent of B:

That means A and B should both shoot on the 50th step.

But I still think the equation has something off about it.

If A's prob of kill B is x/100 and increases every step, and A's prob of being safe from B is 1-x/80 and this decreases every step. The best time for A to shoot is when A's prob to kill B is = to A's prob of being safe from B.

Its almost like A and B are running towards each other and B is running faster than A, and they will collide on the 44th step.

Edited May 31, 2009 by adiace

[bushindo]

For A:

Chance of surviving = Chance of not being shot * Chance of killing

P = x/100 * (1 - x/80)

dP/dx = 1/100 - x/4000

1/100 - x/4000 = 0

x = 40 steps.

For B:

P = x/80 * (1 - x/100)

dP/dx = 1/80 * x/4000

1/80 * x/4000 = 0

x = 50 steps.

I'll just make a point that if A were to shoot at 40 paces,his chance of surviving is 40%, regardless of where B shoots. If he were to shoot at 44 paces, his chances of living is now 44% percent, regardless of B's strategy. So an improvement is possible there.

Using math brings back memories... aah..

Now in my solution for the 1st part I used the intersection of lines to determine the step to shoot on for ideal survival.

For A: His chance if killing B is defined by the line y=x/100, x = number of steps taken.

If he does not know B's ability, then his chance of being killed by B is y = 1 - p*x/100, where p = B's multiplication factor (it was 1.25 in the 1st part)

We an solve the two lines and find its intersection.

we get: x/100 = 1- p*x/100

> x = 100/(p+1)

Now this gives the ideal step on which A should shoot. But since he does not know the ability of B (i.e. 'p'). He must take the average on the entire range of possible 'p''s.

p = 0 is the minimum, i.e. no matter what B can't shoot him. ( we can't take negative values for p, or the lines will never intersect)

p= 100 is the maximum, i.e., B has 100% accuracy after the 1st step itself.

To find the Average Value of a Function over a range [a,b]

our function is x = 100/(p+1)

so the Average Value of x for A is:

1/(a-b) * 100 * [ln(a+1) - ln(b+1)]

a = 0

b = 100

> 1/100 * 100 * [ln(101) - ln(1)]

= 4.615 rounded = 5.

Now for B the equation for x is:

x = 100(p+1.25)

so the Average Value of x for B is:

> 1/100 * 100 * [ln(101.25) - ln(1.25)]

= 4.394 rounded = 4.

2nd Part Ans: A shoots on 5th step, B shoots on 4th step.

I like the way you think,adice, but there must be a math error somewhere. Your method is roughly equivalent to guessing what your opponent's accuracy function would be using some guesses about the population of gun fighters. Obviously if A were to shoot on the 5th step, his chance of hitting is only 5%. Normally,you would only do that if you think that if your opponent is vastly superior to you in accuracy, and that in 5 paces his accuracy already surpassed 95%.

[Guest]

I'll just make a point that if A were to shoot at 40 paces,his chance of surviving is 40%, regardless of where B shoots. If he were to shoot at 44 paces, his chances of living is now 44% percent, regardless of B's strategy. So an improvement is possible there.

I like the way you think,adice, but there must be a math error somewhere. Your method is roughly equivalent to guessing what your opponent's accuracy function would be using some guesses about the population of gun fighters. Obviously if A were to shoot on the 5th step, his chance of hitting is only 5%. Normally,you would only do that if you think that if your opponent is vastly superior to you in accuracy, and that in 5 paces his accuracy already surpassed 95%.

I don't think there is an error in my math, but my range for the opponent's ability is between 0 to 100 times the ability of A. So for the range 0 to 1, A is at an advantage but from 1 to 100, he is out gunned by his opponent.

If you add some information such as "A is 1/2 as good as the best gun slinger in the world" or something like that, then the range for the opponent's ability will change to 0 to 2.

And the ans would be:

for A: 55th step

for B: 48th step

Edited May 31, 2009 by adiace

[Guest]

I'll just make a point that if A were to shoot at 40 paces,his chance of surviving is 40%, regardless of where B shoots. If he were to shoot at 44 paces, his chances of living is now 44% percent, regardless of B's strategy. So an improvement is possible there.

Well I have to disagree there. It does matter what strategy the other person plays. If A was shot before he chose to fire his own gun then his chance of survival would be 0 obviously. Also I don’t know why you say that after 44 paces he has a 44% chance of survival. That sounds as though you are assuming that B will shoot after A has shot.

I realized that i missed the chance that B fires before A in my calculations. But since we don’t know what this is I dont think anything can be worked out.

Anyway, maybe I'm missing something. But I'd be glad if you could explain the logic here.

This reminds me of paradox I heard where a teacher says that the class will be given a surprise test sometime next week. The test couldn’t be on a Friday because on Thursday everyone would know that they haven’t had the test yet so it must be tomorrow. So it wouldn't be a surprise test if it was on a Friday. So we know that the teacher wouldn’t do the test on that day. But the same logic can be applied to Thursday. Since the test can’t be on a Friday, on Wednesday everyone would know the test would be on a Thursday. Following this logic shows that the test cannot be on any day.

[bushindo]

For A:

Chance of surviving = Chance of not being shot * Chance of killing

P = x/100 * (1 - x/80)

dP/dx = 1/100 - x/4000

1/100 - x/4000 = 0

x = 40 steps.

For B:

P = x/80 * (1 - x/100)

dP/dx = 1/80 * x/4000

1/80 * x/4000 = 0

x = 50 steps.

Sorry for the double post, i waited too long and the edit function went away. This approach is the answer to this question, "If A and B were to shoot simultaneously at step k, what is the best place for each person to shoot so that he lives and the other one dies?"

Now, let me clarify my statement earlier. If A were to shoot at 44 paces, he is guaranteed minimum chance of living of 44%. If he were to shoot at 40, his minimum chance of living is 40%. See adiace's graph for the proof.

I think you might be missing the key info that there is only 1 bullet in each person's gun, and that if the person who shoots first misses, the second person will have a guaranteed 100% chance of killing, since we assume that it's a closed room and there's nowhere to run.

Edited May 31, 2009 by bushindo

[Guest]

@bushindo:

Actually I have limited my range for the the ability of the opponent from 0 to 100. Technically B could be infinitely better than A. And your equation is defined as k*p/100 and for the zeroth step, you will have infinity*0/100. i.e an undefined number And A can never know when to shoot... better off shooting himself

Edited May 31, 2009 by adiace

[Guest]

@bushindo:

Actually I have limited my range for the the ability of the opponent from 0 to 100. Technically B could by infinitely better than A. And your equation is defined as k*p/100 and for the zeroth step, you will have infinity*0/100. i.e an undefined number And A can never know when to shoot... better off shooting himself

Anyway, Bushindo, I'm afraid I don't see the significance of where the graph crosses. I can infer from that, that at 44 steps his chance of killing B is equal to his chance of surviving B. Is that useful?

Edited May 31, 2009 by psychic_mind

[Guest]

Hey psychic_mind and bushindo,

Here is a graph of all the functions being discussed.