Junebugs, Darkness, and You

Originally posted by crsmithers
Ok. Let me try to put this another way. IF they are moving at the same speed, then A will occupy a square on any given move. This square could be A1 or B2 or whatever. We assume that since he is in the room he must occupy one square. If the Bug moves to a square every move as well, and is completely ignorant as to the position of the person, then there i ...[text shortened]... changing your numbers every week doesn't decrease your chances of winning, for the same reason.

So there is no difference between moving and not moving then?

As far as i can tell. If we ignore several factors which make it extremly complicated.. (speed, size of person when moving compared to not moving, the room isn't actually a gird etc.)

Yeah, I think I see where you got the 250 multiplier from, there's 250 possible squares to be in. That's 1/250 possibilities for the bug, same for the human, and 250 potential spots where they can be in the same spot, so you get (1/250)*(1/250)*250=1/250, same odds should apply if one or the other is stationary.
My earlier back of the envelope thing didn't take into account the 250 'good' results, it assumed only 1.
Well, I think that's been satisfactorily solved!

My guess is that the chance of you getting hit is greater when standing still. I have no proof, just ideas

- A bug tends to avoid moving things if it can see the moving objects

- The air rushing past your face while running is sometimes enough to swipe the bug out of the way

- You might end up running after the bug, thus never getting hit :p

Originally posted by agryson
Yeah, I think I see where you got the 250 multiplier from, there's 250 possible squares to be in. That's 1/250 possibilities for the bug, same for the human, and 250 potential spots where they can be in the same spot, so you get (1/250)*(1/250)*250=1/250, same odds should apply if one or the other is stationary.
My earlier back of the envelope thing didn't ...[text shortened]... e 250 'good' results, it assumed only 1.
Well, I think that's been satisfactorily solved!

Sounds reasonable :p

Originally posted by Choreant
how can I run into something while standing still???
And,seriously, i think that while moving one is more prone to the crash.

THe thing is running into you while standing still.


Einstein, edit 1: And thank you for your answers.

I kept running into Junebugs int our kitchen at night and it was scaring the living hell out of me. I'm only 16, a faint-hearted young man when left in the dark with unknown buzzing noises, turning out to be about 5 or 10 junebugs in the bloody kitchen.

I hate Junebugs...

Thanks Again!

---Einstein---

Einstein, edit 2: Junebugs fly at up to about 2 maybe 3 mph possibly 4.

Einstein, edit 3: So from these results, it is more likely to occur if I am standing still?

Of course. That makes sense. If I stand still, the beetle will eventually find me in the dark and hit me (most likely in the face) through random paths of aviation.

If I move however, and the beetle is moving along with me, both of our paths are somewhat randomized in respect to the other's. Therefore, it seems logical that the beetle is less likely to hit me when both our paths are randomized and not just one of us.

Thank you. I just needed a confirmation on that.

Everyone who contributed was a great help!!!

Originally posted by EinsteinMind
THe thing is running into you while standing still.


Einstein, edit 1: And thank you for your answers.

I kept running into Junebugs int our kitchen at night and it was scaring the living hell out of me. I'm only 16, a faint-hearted young man when left in the dark with unknown buzzing noises, turning out to be about 5 or 10 junebugs in the bloody nk you. I just needed a confirmation on that.

Everyone who contributed was a great help!!!

Not necessarily, it all depends on the motion properties of both. I believe that if both follow an unrestricted random movement*, the chances would be the same. However, the walls may change this significantly. Once you hit a wall, your next direction is restricted and therefore your movement is not unrestrictedly random.

So I would guess that standing still in a corner (or perhaps the midpoint of the largest wall) would maximize your chances, at least it is better than standing still in the center. Emphasis on guess, I'm feeling too lazy to work this out entirely.

*For the purpose of the room, this would require (as someone said) teleportation to a random point around the room.

** Edit 2 - Which wouldn't then be a brownian motion, obviously, for the pedants out there. 😉

Well, hey, everything has to depend on something eh?

Consider 2 particles in abox of fixed size both moving randomly with an average velocity of V (Brownian motion, random-walk, etc scenario). Lets say chances of collision in 1 hour is N.
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Now lets speed them up x2. Isnt chances of collision now 2N?

Now velocity is just a vector quantity so we can 'freeze' one of our particles (from initial setup) so that the chances of collision is still N but one particle is stationery and the other is moving at 2v.
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So we have deduced that with one particle stationery and the other moving at 2V the chances of collision is N.

And we have deduced that with both particles moving at average velocity of 2V the chance of collision is 2N.
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Therefore the chances of a collision are increased when both particles are moving.

QED (I think)

Originally posted by wolfgang59
Now velocity is just a vector quantity so we can 'freeze' one of our particles (from initial setup) so that the chances of collision is still N but one particle is stationery and the other is moving at 2v.

I think this part is false. You can't simply sum the velocities because we're supposing the each direction changes randomly and these changes are uncorrelated.

So you'd have a random velocity of approach (sorry, I don't know the precise term, I hope you understand) that can ranges anywhere between -2v and 2v. My point was that this interval is reduced whenever one of the two is sufficiently near a wall.

Tough one.

In chemistry, we learn of the particle model of gases which leads to the Ideal Gas Law. Gases are basically lots of particles which move around randomly and at random velocities which average out to a velocity that is mathematically calculated from the temperature of the gas. That is, temperature represents particle movement to a good approximation.

Collisions between particles lead to reactions if the gases are reactive with one another. Higher temperature increases reaction rate, which means indirectly that higher average velocity of the particles leads to more collisions.

Thus I think moving means you're more likely to hit the bug.

Originally posted by AThousandYoung
Tough one.

In chemistry, we learn of the particle model of gases which leads to the Ideal Gas Law. Gases are basically lots of particles which move around randomly and at random velocities which average out to a velocity that is mathematically calculated from the temperature of the gas. That is, temperature represents particle movement to a good ...[text shortened]... les leads to more collisions.

Thus I think moving means you're more likely to hit the bug.

Just read your post but I think that's maybe because, on average, for each particle that distances itself away, another one closes in. So I'm not convinced that higher velocity means that you'd have a higher probability for one particle to hit each specific particle but you just increase the number of particles you might hit per unit of time. This cannot then apply when the particles are only two.

I think my answer correct but I now know my argument is flawed.

Adding two random velocities cannot be equivalent to adding them.

Simple analogy: Take two random numbers both between 1 & 6 and add them. They will not produce a random number between 2 & 12.

(eg two dice)

Originally posted by wolfgang59
I think my answer correct but I now know my argument is flawed.

Adding two random velocities cannot be equivalent to adding them.

Simple analogy: Take two random numbers both between 1 & 6 and add them. They will not produce a random number between 2 & 12.

(eg two dice)

Of course they will produce a random number between 2 and 12, it's just that the probability distribution changes. It will still be random and between 2 and 12, though.

OK bad wording.