Math: A Monstrous Problem

[Monte314]

I'm a big fan of vintage sci-fi. From time-to-time I've watched old monster movies where the hero knows the "creature" is in some particular area, but instead of just firing a bunch of bullets randomly into the area trying for a "lucky shot", he always waits until the thing is rushing out toward him in the dead of night just as his gun jams. Are bullets that expensive!?

Anyway, I began to wonder what the chances of a "lucky shot" in this kind of scenario are. I did some pencil and paper work and some computer modeling, but for our purposes here I've boiled it down to a simple scenario.

Suppose I draw a large circle of radius L>0 in the plane. Concentric with it (i.e., centered at the same point) I draw another, smaller circle of radius S>0 (so that 0<S<L... small less than large... get it?).

I now have two nested, concentric circles: the larger outer circle is the boundary of the monster's domain, and the smaller inner circle is the monster itself. (Notice that I've put the monster at the exact center of its domain... this simplifies the problem a LOT!)

Thinking of a line in the plane as representing the path of a fired bullet in the "lucky shot" scenario above, I pose the following problem:

Let the outer circle have radius L=10, and the inner circle have radius S=1. If a line intersects the outer circle, what is the probability that it also intersects the inner circle?

(That is, with the given sizes, what's the probability that a random shot fired into the creature's domain will be a "lucky shot"?)

You can give an approximate answer by writing a computer simulation. If you are clever, you can you develop a general formula for any pair of concentric nested circles having radii 0<S<L as above. As a check, I can tell you that if the outer circle has twice the radius of the inner, the probability of a "lucky shot" is exactly 1/3. Less than you would think, huh?

Note: It's not too hard to develop a simulation that produces estimates for dimensions three and above, as well. The results are very, very surprising. (I'd tell you about it, but the last time I discussed this with someone, there was a head-explosion incident... uh, theirs, not mine.)

[ssrprotege]

Let the outer circle have radius L=10, and the inner circle have radius S=1. If a line intersects the outer circle, what is the probability that it also intersects the inner circle?


I need clarification. Intersect = the line should go through more than one point of the inner circle? Or just going through one point (i.e. the line becomes a tangent) is okay?

[Monte314]

I need clarification. Intersect = the line should go through more than one point of the inner circle? Or just going through one point (i.e. the line becomes a tangent) is okay?

Excellent question.

It doesn't make any difference; you'll get the same answer either way. For definiteness, let's include tangent lines, since they fulfill the technical requirements of the problem.

As with all these problems, this one requires no advanced techniques... nothing more than high-school mathematics (and in the United States, that's not asking for much).

[SirJac]

Unless I'm missing something, the chances of hitting your target is simply the ratio between the size of the monster (S) and the size of its domain (L).

Area of a circle is Pi(r)^2

Ratio of the difference in area between S and L would be (Pi(S)^2)/(Pi(L)^2)

Simplified: (S^2)/(L^2)

Given S = 1 and L = 10, then the solution would be 1/100 or 1% chance to hit.

However using this formula, a domain being twice the size of the target would give a chance to hit as 25%, not 33.333%. I'm not sure if I made a mistake or not.

[Monte314]

Unless I'm missing something, the chances of hitting your target is simply the ratio between the size of the monster (S) and the size of its domain (L).

Area of a circle is Pi(r)^2

Ratio of the difference in area between S and L would be (Pi(S)^2)/(Pi(L)^2)

Simplified: (S^2)/(L^2)

Given S = 1 and L = 10, then the solution would be 1/100 or 1% chance to hit.

However using this formula, a domain being twice the size of the target would give a chance to hit as 25%, not 33.333%. I'm not sure if I made a mistake or not.



Good idea as a starting point; the answer is going to be some kind of ratio, and sets with greater areas would seem to be easier to "hit". (In fact, if we were "bombing from above", thereby selecting *points* inside the larger circle rather than lines, your answer would be correct.)

Yours is a tempting approach, but I think it is pretty easy to show that looking at the ratio of areas is not going to work here; for this problem, the shape of the area also matters.

For example, if I include as part of my monster all the points on the circumference of the outer circle, I don't change the ratio of areas at all (since the circumference is a set having area zero)... but the probability of a "lucky hit" is clearly now 100%.

I suggest that a better approach is to consider how you might group all the lines that hit the outer circle into two sets: those that are "lucky hits", and those that aren't. It's ratios built from these quantities that need to be considered.





Monte314 added to this post, 1215 minutes and 24 seconds later...

This picture will tell you how to compute the proportion of the lines that hit the outer circle will be "lucky shots"...

To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

[Antisocialite]

I'm a big fan of vintage sci-fi. From time-to-time I've watched old monster movies where the hero knows the "creature" is in some particular area, but instead of just firing a bunch of bullets randomly into the area trying for a "lucky shot", he always waits until the thing is rushing out toward him in the dead of night just as his gun jams. Are bullets that expensive!?

Anyway, I began to wonder what the chances of a "lucky shot" in this kind of scenario are. I did some pencil and paper work and some computer modeling, but for our purposes here I've boiled it down to a simple scenario.

Suppose I draw a large circle of radius L>0 in the plane. Concentric with it (i.e., centered at the same point) I draw another, smaller circle of radius S>0 (so that 0<S<L... small less than large... get it?).

I now have two nested, concentric circles: the larger outer circle is the boundary of the monster's domain, and the smaller inner circle is the monster itself. (Notice that I've put the monster at the exact center of its domain... this simplifies the problem a LOT!)

Thinking of a line in the plane as representing the path of a fired bullet in the "lucky shot" scenario above, I pose the following problem:

Let the outer circle have radius L=10, and the inner circle have radius S=1. If a line intersects the outer circle, what is the probability that it also intersects the inner circle?

(That is, with the given sizes, what's the probability that a random shot fired into the creature's domain will be a "lucky shot"?)

You can give an approximate answer by writing a computer simulation. If you are clever, you can you develop a general formula for any pair of concentric nested circles having radii 0<S<L as above. As a check, I can tell you that if the outer circle has twice the radius of the inner, the probability of a "lucky shot" is exactly 1/3. Less than you would think, huh?

Note: It's not too hard to develop a simulation that produces estimates for dimensions three and above, as well. The results are very, very surprising. (I'd tell you about it, but the last time I discussed this with someone, there was a head-explosion incident... uh, theirs, not mine.)


I think these questions can be solved with learning gun kata (To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.)!

[Monte314]

I think these questions can be solved with learning gun kata (To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.)!

Yes... if you threaten me with a gun, I'll show you the answer!

*brandishes trout menacingly*

[PHS Philip]

Crap, I think I just broke the sine function.

If you have sin(x)=o/h, it's inverse sin(o/h)=x, right? So for that diagram, sin(x)=1/10, so inverse sin(1/10)=x and then 180/2x? Wait, no, 2x/180. Then *100, so 0.1%?

EDIT: Wait, no, that doesn't work for the 1/3 check. Never mind. I guess I'm off on a tangent (bad pun fully intended)

[Monte314]

Crap, I think I just broke the sine function.

If you have sin(x)=o/h, it's inverse sin(o/h)=x, right? So for that diagram, sin(x)=1/10, so inverse sin(1/10)=x and then 180/2x? Wait, no, 2x/180. Then *100, so 0.1%?

EDIT: Wait, no, that doesn't work for the 1/3 check. Never mind. I guess I'm off on a tangent (bad pun fully intended)


You are very close... just a few details to wrap up.

I'll post a solution in a couple of days.

[Antisocialite]

Yes... if you threaten me with a gun, I'll show you the answer!

*brandishes trout menacingly*

:p :laugh:

[thod]

sin-1(S/L) / 90 ,using degrees.

I guess I am misssing something since this is pre teen trig.

[Monte314]

sin-1(S/L) / 90 ,using degrees.

I guess I am misssing something since this is pre teen trig.


Thod's answer is formally correct, as usual. It needs to be converted to radians to be complete (that is, 90 degrees = pi/2 radians).

[WiredBrain]

I developed this simple formula thinking that the shooter must stay away from the Monster's Domain and assuming the shots wil always reach that zone, no matter how unexperienced as a shooter he is. I had already done this when I saw Monte's picture, then said to myself "Man, this is what the domain was meant for..." :thinking:

anyways here it goes for those who are interested:

X: distance between the shooter and the center of both circles.
S: small circle radius
L: big circle radius

lucky shot% = 100*Arcsin(S/X)/Arcsin(L/X)

This answers the stated problem if X=L

[Monte314]

Refer to the picture: To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts. (To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.)

By definition, sin(theta/2) is S/L. Therefore, theta = 2*Arcsin(S/L)

The red line gives the extremes that constitute a "hit" of the inner circle.

The total amount of angular freedom is pi radians (180 degrees, not 360), since once the line turns 180 degrees, its "other end" starts sweeping out the same space over again.

To compute the probability of a hit, then, we just need to compute theta and divide
by pi radians:

Probability(S,L) = theta/pi = (2*Arcsin(S/L))/pi

For S=1 and L=2, this gives 1/3.

For S=1 and L=10, this gives 0.06376856, which is about 1/16.

{For those of you who are picky mathematicians: because of symmetry, every point around the circle will produce the same distribution, so the expected value need not be computed by integration. The problem with the inner circle off-center is nightmarishly hard.}

The results above have been subjected to empirical validation by a monte carlo simulation which agrees closely with the theoretical result.


WiredBrains's answer is correct.