# Bullet Richocheting in a Room

1. May 24, 2012

### lugita15

Here is an interesting puzzle from Steve Landsburg's excellent blog:
http://www.thebigquestions.com/2012/05/22/tuesday-puzzle/

"You’re in a rectangular room. Elsewhere in the room is a man with a gun, who shoots a bullet in a random direction. The bullet careens around the room, bouncing off walls, until it hits either you or one of the various punching bags you’ve placed around the room for purposes of absorbing the bullet. The punching bags must be positioned before you know the random direction of the bullet (though you do know both your own location and the bad guy’s location, neither of which you can change). How many punching bags do you need to guarantee your survival?

This being a math problem, you should treat the room as two dimensional, and yourself, the bullet and the punching bags as points."

2. May 24, 2012

### Whovian

None. As you're a point, and so is the bullet, the probability that, when shot in a random direction, the bullet hits you, for any finite amount of time, is 0.

3. May 24, 2012

### lugita15

Of course, the problem is asking you to survive with certainty, not just with 100% probability.

4. May 24, 2012

### collinsmark

From the comments section of the website having the original riddle:

That was from the author of the riddle. So apparently a probability approaching 100% isn't good enough for the answer. It needs to be 100% guaranteed.

5. May 24, 2012

### collinsmark

1 punching bag. Place the bag at the exact location of the shooter. (The exact location of the shooter is one of the few things that you do know exactly.)

6. May 24, 2012

### lugita15

Even a probability equalling 100% is not good enough. Certainty is required.

7. May 24, 2012

### lugita15

Both the shooter and the punching bag are points, so it will just be as if the bullet is leaving the punching bag.

8. May 25, 2012

### Topher925

Zero. Point the shooter perpendicular to the wall so the bullet bounces back and hits him.

9. May 25, 2012

### Whovian

But the problem did say he shot in a random direction.

10. May 25, 2012

### Topher925

The direction can be random. It just needs to be perpendicular to a wall.

11. May 25, 2012

### Whovian

So you know the direction beforehand, then?

And I don't understand how an angle of, say 30˚ relative to a random wall in a rectangular room could be perpendicular to the wall.

12. May 25, 2012

### Topher925

No, you don't know the direction the bullet will travel so its still random. The room is rectangular, so 4 sides. There is a random probability of 25% that the bullet will hit one of the 4 walls. Just don't stand between the wall and the bullet and you'll survive as long as the shooter is standing perpendicular to the wall.

13. May 25, 2012

### Whovian

But what if it ricochets, bouncing off one wall, so it hits another wall, and ricochets back towards you? We were never told the bullet's direction would be perpendicular to the wall, and the only situation in which it would just bounce back and hit the shooter is if it's perpendicular to the wall it hits.

14. May 25, 2012

### lugita15

It doesn't matter what direction the shooter is faced in, regardless he will still shoot at a random angle, so the probability that he will fire perpendicular to a wall is zero.

15. May 25, 2012

### Staff: Mentor

No matter where I put the bag and myself, if we are points and we are not in the same place, it is always possible that the guy with the gun will stay between me and the bag and will aim exactly at me. So I don't see how it can be solved in a general case. It is not clear if putting bag and myself in the same position makes me safe or not.

16. May 25, 2012

### collinsmark

The riddle statement specifically said, "This being a math problem, you should treat the room as two dimensional, and yourself, the bullet and the punching bags as points."

What that implies is that the gun shooter and bullet (before the bullet is shot, if it's not already inside a bag) can be treated as a single point. The bullet stops the moment the bullet location is exactly equal to to the location of a bag or you. Placing a punching bag at the location of the shooter satisfies the mathematical requirement for stopping the bullet.

Think of it as placing the shooter, gun, and bullet inside a single punching bag, if that helps.

[Edit: or alternately, think of it as sticking a single punching bag inside the bullet (which happens to be at the exact location as the shooter and gun). In either case, the bullet and bag intercept, from a mathematical perspective. The requirement is satisfied. The bullet is stopped.]

Last edited: May 25, 2012
17. May 25, 2012

### lugita15

Remember, your location and his location are fixed, and you can put the punching bags wherever you want. If you want to prevent him from shooting directly at you, just put a bag between you and him.

18. May 25, 2012

### Staff: Mentor

OK, I misread the problem. Positions are known beforehand and fixed, somehow I missed that part [PLAIN]http://www.bpp.com.pl/IMG/grumpy_Borek.png. [Broken]

Still, there is an infinite number of directions in which the gunner can shoot to hit me, so I don't see how it can solved with a finite number of bags.

But I don't have to be right.

Last edited by a moderator: May 6, 2017
19. May 25, 2012

### lugita15

The thing is, there may be a finite set of points through which all of the infinitely many ricocheting paths go.

Last edited by a moderator: May 6, 2017
20. May 25, 2012

### lugita15

The bullet stops when it HITS a bag, not when it starts from or leaves a bag.

21. May 25, 2012

### Jimmy Snyder

I will assume that the location of the shooter and my location are off limits for punching bags.

In that case, it is easy to prove that countably many bags suffice. If the bullet is to strike me without ricochet, then it must be aimed at me and I can stop it with a single bag. If it is to strike me after a single ricochet, then there are no more than 4 directions that I need to protect against, one for each wall. I can protect myself with 4 bags. For two ricochets, there are 12 directions, first hitting one wall, then a different wall, then me. 12 bags suffice. For each subsequent number of ricochets, multiply the number of bags by 3. That is to say, a bullet can ricochet off of any wall except the wall it just ricocheted off of, and then hit me. So for any finite number of ricochets, I need a finite number of bags. And since there can only be a countable number of ricochets, there can only be a countable number of directions I need to protect against. That is to say, a countable union of finite sets is countable.

22. May 25, 2012

### lugita15

That's clear enough, but the hard part would presumably be to prove that finitely many bags suffice.

23. May 25, 2012

### lugita15

The thing is, the paths can intersect. So just because you have distinct paths doesn't mean you need distinct punching bags for each path. It may be that there is a finite set X of points such that for any path P, among the infinitely many ricocheting paths, there exists a point in X such that P passes through that point.

24. May 25, 2012

### Jimmy Snyder

I know. That's why I deleted it. Not soon enough though.

25. May 25, 2012

### Jimmy Snyder

I think it can be done with
only 7
bags.

Just let me know if I'm right and if so I'll provide the proof. Otherwise it's back to the drawing board. Literally.

Last edited: May 25, 2012