\(Let\ K(n)\ be\ the\ probability\ that\ Keiko\ gets\ n\ heads,\ and\ let\ E(n)\ be\ the\ probability\ that\ Ephriam\ gets\\ \\ n\ heads. \\ \\ K(0) = 1/2 \\ \\K(1) = 1/2 \\ \\K(2) = 0\ (Keiko\ only\ has\ one\ penny!) \\ \\E(0) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4} \\ \\E(1) = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} \\ \ \ \ \ \ \ \ \ = 2\cdot\frac{1}{4} \ \ \ \ \ \ \ \ = \frac{1}{2} \\ (because\ Ephraim\ can\ get\ HT\ or\ TH) \\ \\E(2) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4} \\ \\The\ probability\ that\ Keiko\ gets\ 0\ heads \\ \\ and\ Ephriam\ gets\ 0\ heads\ is\ K(0)\cdot E(0). \\ \\ Similarly\ for\ 1\ head\ and\ 2\ heads.\ Thus,\ we\ have: \\ \\P = K(0)\cdot E(0) + K(1)\cdot E(1) + K(2)\cdot E(2) \\ \\P = \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2} + 0 \\ \\P = \frac{3}{8} \\ \\Thus\ the\ answer\ is\ 3/8.\)

\(-tommarvoloriddle\)

EDIT:

SUMMARY

- we used a equation K(0)*E(0)+K(1)+E(1)+K(2)*E(2)

- we found the values by using probability and logic.

- we plugged the numbers in

- we got the answer 3/8.

Apologises: 

-quality of the answer

-all the work is in latex

-fact that it may not be super clear

If you have a question, just let me know.

Thanks  Tom.....riddle 

I'd just do it with a probablilty tree.  Which is the same just more visual.

The first toss is Ephran1 , the second is Ephan2 and the last is Keiko's toss.

Which set of three branches gives the desired outcome?

Hint, there are 3 desirable branches ends and 8 ends altogether.