Question

Four snails are travelling in uniform rectilinear motion on a plane.

The directions of their paths are random but not parallel and no more than 2 paths can cross at any point.

Five of 6 possible encounters have already happened.

What is the probability of the sixth encounter occurring?

Answer 1

Surely you need more information, because I can come up with an example where the sixth can never happen, and also ones where the sixth must always happen.

Answer 2

Explain me the example where the sixth can never happen.

Answer 3

If we assume that one snail is travelling in 'essentially' the opposite direction to the other three. This snail can easily cross two of those snails but never the third, whereas the other three snails can all cross each other. I hope that the snails in paint make sense. \[x\] (wanted to see if I'd get an achievement, lol)

Answer 4

I think you might get one if you do a pic, editor and text response all at once, dunno, I haven't done it myself yet..

Answer 5

You do :). Does it make sense, or have I made an assumption that I shouldn't have?

Answer 6

Your pic is ignoring the fact that 5 of 6 possible encounters have already happened.

Answer 7

True, but the three black ones can and will meet much further down the page. So after that has happened, we are in the situation you described and the red will never meet another.

Answer 8

How many encounters are you portraying in your pic (label the snails a,b,c,d,e).

Answer 9

Hold on...

Answer 10

a,b,c,d I mean

Answer 11

The blue circles are the meetings. Snail D is the 'issue'.

Answer 12

Yes, but wait a moment.

You are treating the situation statically, not dynamically, you need a diagram of space vs time.

Leaving that aside for a minute, one snail, say a, must have met the other 3 (because 5 encounters).

Imagine him just sat at (0,0) (put him at rest in that frame, no loss of generality there).
So the other 3 have been to (0,0).

One of the other three must have met the remaining 2 (five encounters).

Only if the 3 others are moving on the same line is that possible.

So the sixth encounter is certain.

Answer 13

Let me see if I can come up with some sort of diagram....

Answer 14

That last conclusion is not true. You've fixed the frame at (0,0) and so we have the three lines going through (0,0), since lines meet at one point only (unless they are the same line) the others will never meet and so fixing the frame at one point tells you nothing other than one snail meets three others. Whilst I haven't indicated specifically time into my diagram it is clear that it is a contraction to all six meetings occurring.

Answer 15

I disagree, the problem is with the way you have crossed the lines, all four lines must intersect in pairs at 5 points, I will put up a diagram in a bit.

Answer 16

You have to take time into account; if not, on your diagram,for example, you could have c having passed the intersection points before a and b got there (or many other such scenarios).

On my diagram, you have the 5 encounters (A to E) and 4 snails (a to d)

If you take A,B,C it is a plane in an x,y,t world and the 4th and 5th encounters have to be IN that plane as well therefore so is snail d path.

So you can see from this why the last encounter (c and d) must occur.

Answer 17

If the lines were infinitely extended in both directions then what you say it true. However, even when one of the directions terminates you can construct examples which contain 5 intersections but not 6 (like the one above), since clearly two lines don't have to meet when they are not extended infinitely in both directions (without being parallel), consider the y-axis and the equation y=x for x>1. It even fails in the projective plane if the lines have a starting point.

Answer 18

I agree with your analysis per se but that does not take into account the conditions of the problem (ie there have been 5 encounters already, so sooner or later...)