Number crunchers

[Slasher]

Here's a job for all the number crunchers here - if there is an unknown intruder on the other side of a closed door, and you have no idea where in that room the intruder is positioned - what are the chances of firing a killer shot through that closed door?

[CaptainD]

Do we know how big the room is, what else is in it, what the door's made of, what type of firearm we're using?  All those would come into it but even without knowing them those things it would seem an infinitesimal chance.  How do we even know there's someone on the other side?  If you can hear them, how good your locational hearing is would come into it!  (If it was me... mine's rubbish, so it would be of not help.)

[Stupot]

Why don't you ask Oscar Pistorius?

[Khris]

If we make some basic assumptions, we can put a rough number on it.

Let's say the intruder is standing, and we're firing at chest height. Say they are three meters away, and we can fire at an angle of +/- 60° through the door.
The diameter of that circle is six meters, thus the circumference is roughly 19. 120° is one third, so the arc is 19/3 meters long.
Let's say we wound him deadly hitting anywhere in a 33cm wide area.
The probability is now 1/3m / 19/3m = 1/19 ~ 5%.

This is proportional to the distance, so if they are six meters away, it drops to 2.5%.
If they are closer, about 1m away (from the muzzle), it's 15%.

[KBlaney]

1 in 2*pi*r if we know the distance based on  Khris' s math. If we don't know the distance, but we have an upper bound that we call m, the odds become 1 in Integral(2*pi*r,r,0,m)/(m). Fittingly, the limit of the odds as m approaches infinity is 0.

[Adam H]

A little late, but I have a much more intuitive answer if we tweak our assumptions a little: keep the projectile the same while making the location of the intruder variable. This makes more sense to me because you have no reason to fire at different angles. Assuming the room is rectangular, you are firing chest height across the length of the room, the intruder is equally likely to be anywhere in the room, and iff you hit the intruder within 6 inches of the center of his chest he will die, then:

There is about a 1/(W-1) chance that the intruder will die, where W is the width of the room in feet.

If the room is 2 feet wide, then the only place for the intruder to stand is in the path of the bullet (assume he has to stand with his center of his chest about 6 inches from the wall) so there's a 100% chance he'll be hit, no matter how far away he is. If the room is 11 feet wide, then the intruder has a 1/10 chance of standing in that middle 1-foot-wide section.

[Ryan Timothy B]

Of course upward or downward angles matter. Midget or tall person. That's another variable.

[Khris]

Adam H:
You are making unrealistic simplifications that give a faulty result. Why?

[Adam H]

Not to be antagonistic, but I believe my assumptions are more in line with the original problem than yours. Your solution doesn't take into account the width of the room or the fact that we have no clue where the intruder is.

Since we have no idea where the man is, the shooter should not care where he aims.

What other assumptions do you think are faulty? The shape of the room? The height of the shot? Meh.

[Khris]

My solution is all about not knowing where the intruder is, I am only assuming that they are somewhere within a sector of ±60° from a shot straight through the door, which is pretty reasonable, unless they're pressing themselves against the wall near the door (in which case the chances of hitting them drop to 0, which is certainly not what slasher was asking about).
The width of the room doesn't matter, only the distance between gun and intruder. Unless of course the room is tiny or very narrow, but there's no reason to assume that based on the initial question.

The big problem with your proposed solution is that the door isn't as wide as the room. The shooter isn't going to change position, just angling the gun. Just imagine the room on the other side is really big, but the intruder is standing relatively close to the door. In that case it's obvious that your solution's result is going to be way off base, while mine is still pretty accurate.

[frenzykitty]

Whoa - MATHS FIGHT! *throws calculator and hides under desk*

[budgerigar]

Due to faulty memory of an aging brain, I expected the conversation to be about this thing:

Oh well. Back to the home for me.

[Adam H]

My solution is all about not knowing where the intruder is, I am only assuming that they are somewhere within a sector of ±60° from a shot straight through the door, which is pretty reasonable, unless they're pressing themselves against the wall near the door (in which case the chances of hitting them drop to 0, which is certainly not what slasher was asking about).
The width of the room doesn't matter, only the distance between gun and intruder. Unless of course the room is tiny or very narrow, but there's no reason to assume that based on the initial question.

The big problem with your proposed solution is that the door isn't as wide as the room. The shooter isn't going to change position, just angling the gun. Just imagine the room on the other side is really big, but the intruder is standing relatively close to the door. In that case it's obvious that your solution's result is going to be way off base, while mine is still pretty accurate.

Sure... but we don't know where the intruder is standing!

"I am only assuming that they are somewhere within a sector of ±60° from a shot straight through the door" - but slasher said "you have no idea where in that room the intruder is positioned". Why couldn't the intruder be in the corner of the room next to the door???

If the room is very wide, then the intruder is PROBABLY not going to be standing in the shooter's path. And if the room is very narrow, then the intruder is PROBABLY going to be standing in the gun's path - assuming the shooter knows the layout of the room he'll shoot along the length of the room and not into the side wall, like he might in your interpretation.

My solution doesn't involve the shooter changing positions. If we were to test our solutions by having 100 intruders in 100 rooms, my shooter would stand in the center of the door and fire straight along the length of the room 100 times. This is because the shooter has no reason not to - there's no point in angling the gun or moving around, because the shooter doesn't know where the intruder is.

The fun thing is, both our solutions are correct. Slasher can pick the one that he was looking for. My answer fits MY interpretation of the problem better than yours, but I wouldn't be too surprised if my mind works differently from slasher's.


Anyways... NUMBER MUNCHERS!!! Yes! I loved that game! Or else I hated it and I was just forced to play it. I love it now, anyways.

[Khris]

I reread your posts and I realized that there was a misunderstanding due to syntactic ambiguity.

This:

Your solution doesn't take into account [...] the fact that we have no clue where the intruder is.

made me shake my head in disbelief at first, but I get now that you were talking about 120° not covering the entire room.

So yeah, the main problem here is that slasher never got back to this, and didn't give sufficient info in the first place.

[Adam H]

Cheers!

I did some playing around with your solution and got a simple reduction just by tweaking your angle from 120 to 115 (which is 360/pi) and using my 1-foot-death-zone approximation instead of your 33 cm: the odds are approximately 1 in 2r, where r is the distance in feet (3 ft ~ 1 m). Then if you don't know the distance, the solution is 1 over the integral of 2r from 0 to M (upper bound), which is 1/M^2.

So if you know the distance between you and the intruder then the solution is 1/(2r),if you have a cone shaped area the intruder could be in then the solution is 1/(M^2), and if you have a rectangular area then the solution is 1/(W-1). And I think we should leave it at that.

[EchosofNezhyt]

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