ant can move around triangle sides 3! times. With every additional time the ant moves around the triangle sides it has decreased probability assigned to moving towards vertexes. Hence C>B>A

b

john123 wrote:An ant decides to move around an equilateral triangle. At each of the vertex the ant has a probability of 1/2 to move to either of the remaining two vertices.
Given.
A: The ant crawls around the edges 10 times and comes to the starting point.
B: The ant crawls around the edges 8 times and comes to the starting point.
C: The ant crawls around the edges 6 times and comes to the starting point.

Which of the following is true --

a)B>A>C b) C>B>A c) A>B>C d) C>A>B e) A>C>B f) A=B=C g) B>C>A

A, B, C are the probabilities that each of the described outcome happens?

john123 wrote:An ant decides to move around an equilateral triangle. At each of the vertex the ant has a probability of 1/2 to move to either of the remaining two vertices.
Given.
A: The ant crawls around the edges 10 times and comes to the starting point.
B: The ant crawls around the edges 8 times and comes to the starting point.
C: The ant crawls around the edges 6 times and comes to the starting point.

Which of the following is true --

a)B>A>C b) C>B>A c) A>B>C d) C>A>B e) A>C>B f) A=B=C g) B>C>A

B IMO too.

Slightly different explanation. Assume the triangle is XYZ. Let's say the ant start from X.

A: Comes back to X after 10 moves.
Move 1: It can go to any of Y or Z. Prob 1. Let's say it went to Y.
Move 2: Now it has to go to Z, otherwise it would return to starting point X. Prob = 1 * 1/2
Move 3: It has to go back to Y, otherwise it would return to X. Prob = 1 * 1/2 * 1/2
...
In fact It has to pick the same edge YZ for each of the moves until move 10.
Move 10: Whichever vertex the ant is at, it has to pick X to return to starting point. Prob = 1 * 1/2 *...* 1/2 (eight 1/2s, one each for each of the 8 preceding moves. Leave out move 1, it has prob 1) * 1/2

P(A) = (1/2)^9

In general if the ant has to return to starting position in n moves, the probability is (1/2)^(n-1) (Assuming n > 1).

P(B) = (1/2)^7
P(C) = (1/2)^5

Hence, P(C) > P(B) > P(A)