Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
$\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4}$
Divide it into $2$ cases:
1) Keiko and Ephriam both get $0$ heads:
This means that they both roll all tails, so there is only $1$ way for this to happen.
2) Keiko and Ephriam both get $1$ head:
For Keiko, there is only $1$ way for this to happen because he is only flipping 1 penny, but for Ephriam, there are 2 ways since there are $2$ choices for when he can flip the head. So, in total there are $2 \cdot 1 = 2$ ways for this case.
Thus, in total there are $3$ ways that work. Since there are $2$ choices for each coin flip (Heads or Tails), there are $2^3 = 8$ total ways of flipping 3 coins.
Thus, since all possible coin flips of 3 coins are equally likely, the probability is $\boxed{(B) \frac38}$.