Showing that one entropy is greater than another

Question

There is two fair (identical) coins. Heads is worth one point and tails is worth two points. We flip two coins at a single time.

Lets consider the two experiments $X$ & $Y$ on the set $S=$ {2, 3, 4}. Experiment $X$ the set $P_x(x)$ is the probability that the sums of the points on the two coins is $x$. Experiment $Y$ $P_y(y)$ is the probability that the max number of points is $\frac{1}{2}y$ (if we have a tie the max number is the common number).

Determine, with proof wich one of the entropies $H(X)$ & $H(Y)$ is greater.

So I'm having a bit trouble getting started on this problem, first off I don't understand when they say "probability that the max number of points is $\frac{1}{2}y$"

Now I can compute the entropy for the first part of the problem and I would appreciate if I could get some feedback if I did it correctly.
$H(X) = -(\frac{1}{3}log(\frac{1}{3})+\frac{1}{3}log(\frac{1}{3})+\frac{1}{3}log(\frac{1}{3}))$

Answer 1

If we get $(1,1)$, the maximum number of points is $1$. Let $\frac{y}{2}=1$, the corresponding $Y$ would be $2$.

Otherwise, the maximum number of points is $2$. The corresponding $Y$ would be $4$.

Hence we just have to compute $P_Y(2)$ and $P_Y(4)$ and then we can compute the entropy.

Answer 2

Just in case someone wondered: We have $P_x(2)=P_x(4)=1/4$ and $P_x(3)=1/2$, so using tge definition we get $H(X)=0.25\log(64)$. Further, we have $P_y(2)=1/4$ and $P_y(4)=3/4$, so $H(Y)=-0.25\log(0.25)-0.75\log(0.75)=0.25\log(256/27)<H(X)$.