Probability of a Mutual Meeting!

For each person, there are $\dfrac{7!}{3!4!} = 35$ "shortest" paths, each of length $3 + 4 = 7,$ ( $3$ blocks up and $4$ blocks right for $P,$ $3$ blocks down and $4$ blocks left for $Q.$ ) So if $P$ and $Q$ are to meet, it will be as they are each moving from their respective third to fourth grid points, i.e., after they have each traveled $3.5$ blocks.

Now label the grid points with $A$ being $(0,0)$ and $B$ at $(4,3).$ Then after traveling $3$ blocks the probabilities that $P$ and $Q$ will be at point $(a,b)$ will be as follows:

• $P_{P}(0,3) = \dfrac{1}{8}, P_{P}(1,2) = \dfrac{3}{8}, P_{P}(2,1) = \dfrac{3}{8}, P_{P}(3,0) = \dfrac{1}{8};$

• $P_{Q}(1,3) = \dfrac{1}{8}, P_{Q}(2,2) = \dfrac{3}{8}, P_{Q}(3,1) = \dfrac{3}{8}, P_{Q}(4,0) = \dfrac{1}{8}.$

Now there are $7$ block midpoints at which $P$ and $Q$ can meet. For example, for them to meet at the midpoint of $(1,2)$ and $(2,2)$ person $P$ will have had to travel to $(1,2)$ and then choose to travel right with probability $\frac{1}{2},$ and person $Q$ will have had to travel to $(2,2)$ and then choose to travel left with probability $\frac{1}{2}.$ This event would then occur with probability

$P_{P}(1,2) * \dfrac{1}{2} * P_{Q}(2,2) * \dfrac{1}{2} = \dfrac{3}{8}*\dfrac{1}{2}*\dfrac{3}{8}*\dfrac{1}{2} = \dfrac{9}{256}.$

This will be the same probability for $2$ other midpoints, namely between $(2,1)$ and $(2,2),$ and between $(2,1)$ and $(3,1).$ As for the other $4$ potential meeting points, the respective probabilities are as follows:

• $P_{P}(0,3) * 1 * P_{Q}(1,3) * \dfrac{1}{2} = \dfrac{1}{128} = P_{P}(3,0) * \dfrac{1}{2} * P_{Q}(4,0) * 1,$ and

• $P_{P}(1,2) * \dfrac{1}{2} * P_{Q}(1,3) * \dfrac{1}{2} = \dfrac{3}{256} = P_{P}(3,0) * \dfrac{1}{2} * P_{Q}(3,1) * \dfrac{1}{2}.$

Adding together the probabilities of these $7$ mutually exclusive events, we reach a final probability of

$3*\dfrac{9}{256} + 2*\dfrac{1}{128} + 2*\dfrac{3}{256} = \dfrac{37}{256},$ and thus $\alpha + \beta = 37 + 256 = \boxed{293}.$