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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.
$H(X) = -(\frac{1}{3}log(\frac{1}{3})+\frac{1}{3}log(\frac{1}{3})+\frac{1}{3}log(\frac{1}{3}))$

Temirzhan
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2 Answers2

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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.

Siong Thye Goh
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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 $H(Y)=-0.25\log(0.25)-0.75\log(0.75)=0.25\log(256/27)<H(X)$.

Matija
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