# Math Problem (one-to-one correspondences)

Alex the ant starts at $(0,0)$. Each minute, he flips a fair coin. If he flips heads, he moves one unit up; if he flips tails, he moves one unit right. Betty the beetle starts at $(2,4)$. Each minute, she flips a fair coin. If she flips heads, she moves one unit down; if she flips tails, she moves one unit left. If the two start at the same time, what is the probability that they meet while walking on the grid?

• Why is the ant just "Ant $A$" while Betty the beetle gets a name? The lack of consistency here is quite troubling. – Mike Pierce Apr 2 '15 at 20:15
• Um fine use Alex or something...does it really matter? – user228320 Apr 2 '15 at 20:20
• Have you tried making a list of the different possible combinations of moves that will result in the two insects meeting? There are not very many possibilities. – David K Apr 2 '15 at 20:21
• Yes I have noted the 3 places where A and B could meet. But how would that help me find the probability? – user228320 Apr 2 '15 at 20:24

They must meet at one of those points: $(0,3),(1,2),(2,1)$. each ant has a probability of $1/2^3$ to go through a specific path to one of the points, since they are 3-step from the start point.

Separate into complementary events:

meeting at $(0,3)$: "A" has 1 path, "B" has 3, we get: $P(0,3)=(1/8)*(3/8)$

meeting at $(1,2)$: Both has 3 paths: $P(1,2)=(3/8)*(3/8)$

meeting at $(2,1)$: same as the first

Summing the cases: $P=3/64 + 9/64 + 3/64 = 15/64$

• isn't the probability of going through a particular path 1/7? – Tyler Hilton Apr 2 '15 at 20:36
• @Tyler Hilton why? – avim Apr 2 '15 at 20:38
• consider ant A. It has three ways to get to $(2, 1)$ and $(1, 2)$ but only 1 way to get to $(0, 3)$ which is a total of 7 ways. – Tyler Hilton Apr 2 '15 at 20:43
• This is the number of ways it can walk to any point, not to a particular one. In order to walk in a particular path, say "up up right", it has to see heads,heads,tails, which has the probability of $0.5^3$ – avim Apr 2 '15 at 20:49