jmc

This can be solved quickly and easily with generating functions. Let $x^n$ represent flipping $n$ tails. The generating functions for these coins are $(1+x)$,$(1+x)$,and $(4+3x)$ in order. The product is $4+11x+10x^2+3x^3$. ($a{x}^n$ means there are $a$ ways to get $n$ heads, eg there are $10$ ways to get $2$ heads, and therefore $1$ tail, here.) The sum of the coefficients squared (total number of possible outcomes, squared because the event is occurring twice) is $(4+11+10+3{)}^2=28^2=784$ and the sum of the squares of each coefficient (the sum of the number of ways that each coefficient can be chosen by the two people) is $4^2+11^2+10^2+3^2=246$. The probability is then $\frac{4^2+11^2+10^2+3^2}{28^2}=\frac{246}{784}=\frac{123}{392}$. (Notice the relationship between the addends of the numerator here and the cases in the following solution.) $123+392=\boxed{515}$