Say there is two fair (identical) coins. Let Heads worth one point and tails is worth two points. We flip the two coins at a single instance .
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 am able to compute the entropy for the first part of the problem. I would appreciate if I could get some feedback if I did it correctly.
$H(X) = -(\frac{1}{3}log(\frac{1}{3})+\frac{1}{3}log(\frac{1}{3})+\frac{1}{3}log(\frac{1}{3}))$
Let the outcomes of the coins be denoted by $A$ and $B$, where $A,B \in \{1,2\}$. Note that $X=A+B$. Therefore, $$\Pr(X=2)=\Pr(A=1)\cdot\Pr(B=1)=1/4,$$ $$\Pr(X=3)=\Pr(A=1)\cdot\Pr(B=2) + \Pr(A=2)\cdot\Pr(B=1)=1/2,$$ and $$\Pr(X=4)=\Pr(A=2)\cdot\Pr(B=2)=1/4.$$
For the experiment $Y$, we compute $P_Y(y)$ by noting that $Y = 2\max\{A,B\}$. Observe that $$\Pr(Y=2)=\Pr(A=1)\cdot\Pr(B=1)=1/4,$$ and $$\Pr(Y=4)=1-\Pr(Y=2)=3/4.$$ Therefore, $$H(Y)=\frac{1}{4}\log(4)+\frac{3}{4}\log(4/3).$$ Now compare $H(Y)$ with $H(X)$ to determine which one is greater.