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Suicide Bomber Problem


bloodhound

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So yeah, was just thinking about this on a bus.

 

Suppose you have a bus length l and a suicide bomber intent to kill as many people as he/her can. Assume the position where the bomber sits/stands has pdf 1/l (inside the bus). Also take injury/death which has a strictly decreasing pdf symmetrical around the bomber (like the bell curve or something linear, does it matter?).

 

What is the safest place to be in the bus?[minimise the prob of death/injury] Does it even exist?

 

The solution is probably trivial or something, but I cannot think of an intuitive answer

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Probably but with chlorine gas, a rebreather, some kewl explosives, and also something on the bottom of the bus to blow a hole so you can jump out with your leather coat on and some head protection.

 

The passengers will be screwed.

 

But that's only if you have a bomber who isn't a stupid newb screwing around or else a bomber who knows wtf they are doing.

 

I mean you have to think about how well this terrorist is going to go through with things. Many buses have a type of air hole thing that will have air pushed on a person.

 

It all depends on what the terrorist uses I guess.

 

I think you have to consider ballistics and OTC chemical creations for the rent-a-bomber. I'm guessing you will have shrapnel and the physical properties of chemicals and how fast they can leave an area through a hole.

 

With the idea of shrapnel though, if you hit the ground usually it might fly over you. Gases that are dense however tend to hover around the ground.

 

I'm not going to go into this anymore than I should be.

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The safest place would be at the ends. The "unsafest" would be in the middle. This can be intuited from a symmetry argument, but can also be reasoned out pretty unmessily (just cumbersome to LaTeX it all out). Particularly, notice that contributions from separations greater than L/2 are smaller than those from separations smaller than L/2. This will demonstrate that x=L/2 provides the maximum probability of getting hit. A similar argument tells you that the ends are the safest. The trick is in adding amplitudes - that's all.

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So yeah' date=' was just thinking about this on a bus.

 

Suppose you have a bus length l and a suicide bomber intent to kill as many people as he/her can. Assume the position where the bomber sits/stands has pdf 1/l (inside the bus). Also take injury/death which has a strictly decreasing pdf symmetrical around the bomber (like the bell curve or something linear, does it matter?).

 

What is the safest place to be in the bus?[minimise the prob of death/injury'] Does it even exist?

 

The solution is probably trivial or something, but I cannot think of an intuitive answer

The one behind!!

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take [0,L] == the bus, and suposse that you are sit in x=x0. the expectation of injury in that place is int_0^L D(x,x0)P(x) dx... P(x) is the pdf of the bomber-location (uniform by hipotesis) then E(x0)=1/L int_0^L D(x,x0) dx.

 

D_x0 > 0 if x0<x and D_x0 < 0 if x0 >x then by simmetry

E'(x0) = 0 when x0 is in the middle but in that case E(x0) is maximized.

in fact, D_x0x0 is always negative because its more dangerous if x0->x

 

Then, the solution its seem to be x0 =0 & x0=L

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So there you have it... I'm going to always sit in the very front or back now.
Maybe give greater weight to the probability the terrorist sits at the front because he has a heavy backpack, which means he'll want to minimize walking. If he gets on at the front then there more chance he'll sit at the front than at the back, as opposed to that rectangular R(0,l) assumption, perhaps use exponential pdf.
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