Question

Fool's problem of the day:

There are two ants on opposite corners of a cube. On each move, they can travel along an edge to an adjacent vertex. If the probability that they both return to their starting position after \( 4 \) moves is \( \frac mn\) , where \( (m,n)=1 \), can you find \( m+n \)?

(NOTE:They do not stop if they collide.)

[The right answer was posted by @experimentX]

My apologies I made the wrong assumption as @Arnab09 did.



PS: If you Google it might spoil the problem for you.

Answer 1

@nikvist, please check it out

Answer 2

@callisto, what do u think?

Answer 3

Can they travel in the way like go forward by one side length, then go backward to the original position, then go forward by one side length, and then go backward to the original position?

Answer 4

I guess so..

Answer 5

So, they have like 3 ways of travelling?
1. travel in a square form
2. forward -> forwarrd -> backward ->backward
3. forward -> backward -> forward -> backward

Answer 6

square form.. there are 2 types

Answer 7

I meant like that..

Answer 8

i got it..

Answer 9

that square path can be in anti clockwise direction too..

Answer 10

but i think if i doesn't specify the first step, it doesn't matter?

Answer 11

Sight..* I don't, i need to do more grammar exercises :S

Answer 12

@experimentX I think so

Answer 13

*sigh

Answer 14

well, i solved it this way:
we assume the sides of the cube along the co-ordinate axes, one ant is at (0,0,0) and the other is at (1,1,1)
from the beginning, each ant has 3 options to choose a path. after one move, the next move can be chosen is 3 ways. so, an ant can follow 3^4=81 different paths.
to come back to home, within 4 moves, 2 are along chosen from +x, +y, +z
other 2 are automatically selected.. they are among -x,-y,-z.
we have to choose 2 moves from x, y, z.. this can be done in 3C2=6ways
other moves are automatically selected..
now, among these 4 moves, we have to divide them in 2 groups.. for instance, from the point (0,0,0), positive moves have to come first, and from (1,1,1), negative moves have to come first..
so, there will be total 3C2*4!/(2!*2!)=36 moves possible for each ant to come back home..
so, probability to come back home for each ant= 36/81=4/9

as the moves of the ants are independent to each other, net probability to come back home for each ant= (4/9)^2=16/81
so, m+n= 16+81= 97
is it correct??

Answer 15

sorry, i made a mistake earlier..

Answer 16

ans is either 10 or 778

Answer 17

it's 778

Answer 18

Nobody has posted the right answer. Albeit @Arnab09 is very close! :)

Answer 19

I wish I could do your problems, It would be fun being among you,trying to solve them , but I don't know , I'm dumb at these ! :( @FoolForMath

Answer 20

i cant find whats wrong in my method :|
if the ants collide, there path change?

Answer 21

Is it safe to assume ants won't travel back to where they came from unless they collide with eachother?

Answer 22

what a silly mistake by me!!! 3 C 2= 3

Answer 23

i am making it correct again..

Answer 24

well, i solved it this way:
we assume the sides of the cube along the co-ordinate axes, one ant is at (0,0,0) and the other is at (1,1,1)
from the beginning, each ant has 3 options to choose a path. after one move, the next move can be chosen is 3 ways. so, an ant can follow 3^4=81 different paths.
to come back to home, within 4 moves, 2 are along chosen from +x, +y, +z
other 2 are automatically selected.. they are among -x,-y,-z.
we have to choose 2 moves from x, y, z.. this can be done in 3C2=3ways
other moves are automatically selected..
now, among these 4 moves, we have to divide them in 2 groups.. for instance, from the point (0,0,0), positive moves have to come first, and from (1,1,1), negative moves have to come first..
so, there will be total 3C2*4!/(2!*2!)=18 moves possible for each ant to come back home..
so, probability to come back home for each ant= 18/81=2/9

as the moves of the ants are independent to each other, net probability to come back home for each ant= (2/9)^2=4/81
so, m+n= 4+81= 85
is it correct??

Answer 25

@Foolformath, check this out..

Answer 26

I think Arnab's right.

Answer 27

thanx :D, @Foolformath

Answer 28

Did you Googled? ;)

Answer 29

nope.. honestly
why? is it solved somewhere in same process..??

Answer 30

Anyways good for you! :)

Answer 31

where is the solution? its only a question sheet :O

Answer 32

and if i would google it.. i wouldnt do mistake in 1st chance :)

Answer 33

solution is not here

Answer 34

variant=??

Answer 35

There is some problem about my understanding with about problem I will address this issue sometime later.

Answer 36

It's fixed now. Congrats to @experimentX!!

Answer 37

post the solution link please?

Answer 38

oh.. ok.. i got it now.. congrats to @experimentx

Answer 39

sorry people, i couldn't find a better solution than to determine by counting the possible paths. it's was 21