C3. Yana and Zahid are playing a game. Yana rolls her pair of fair six-sided dice and draws a rectangle whose length and width are the two numbers she rolled. Zahid rolls his pair of fair six-sided dice, and draws a square with side length according to the rule specified below.
a. Suppose that Zahid always uses the number from the first of his two dice as the side length of his square, and ignores the second. Whose shape has the larger average area, and by how much?
b. Suppose now that Zahid draws a square with the side length equal to the minimum of his two dice results. What is the probability that Yana's and Zahid's shapes will have the same area?
c. Suppose once again that Zahid draws a square with the side length equal to the minimum Yana's figure and the area of Zahid's figure. Find the expected value of $D$.
Solution:
a. Zahid's average area is:
$$
\frac{1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}}{6}=\frac{1+4+9+16+25+36}{6}=\frac{91}{6} .
$$

Yana's average area is:
$$
\frac{(1+2+3+4+5+6)^{2}}{36}=\frac{21^{2}}{6^{2}}=\frac{7^{2}}{2^{2}}=\frac{49}{4} .
$$

Therefore, Zahid's average area is larger by $\frac{182-147}{12}=35 / 12$ (square units).
b. All possible outcomes for two dice are:
$$
\left[\begin{array}{llllll}
(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\
(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\
(3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\
(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\
(5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\
(6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6)
\end{array}\right]
$$

The only way for them to have the same area is if (i) Zahid's minimum value is equal to that on Yana's two rolled dice or (ii) Zahid's minimum is 2 and Yana rolls a 1 and 4 , in some order.
The probability that Zahid's minimum number is 1 is $P_{z}(1)=11 / 36$. Similarly, $P_{z}(2)=9 / 36, P_{z}(3)=7 / 36, P_{z}(4)=5 / 36, P_{z}(5)=3 / 36, P_{z}(6)=1 / 36$.
In general, the probability that Zahid's minimum number is $n$ is $P_{z}(n)=\frac{13-2 n}{36}$.
The probability for Yana to roll the same specific number on both dice is $1 / 36$. The probability to roll either $(1,4)$ or $(4,1)$ is $2 / 36$.
The total probability is $\frac{1}{36} \times\left(P_{z}(1)+\cdots+P_{z}(6)\right)+\frac{2}{36} \times P_{z}(2)=\frac{1}{36}+\frac{1}{72}=1 / 24$.
c. As calculated in part (a), the expected value of Yana's area is $\frac{49}{4}$. From part (b), the probabilities that Zahid's minimum number is $n$ is $\frac{13-2 n}{36}$ for $n=1,2, \ldots, 6$, and in these cases, his area is $n^{2}$. Thus, the expected value of Zahid's area is
$$
\sum_{n=1}^{6} \frac{13-2 n}{36}\left(n^{2}\right)=\frac{11}{36}+\frac{36}{36}+\frac{63}{36}+\frac{80}{36}+\frac{75}{36}+\frac{36}{36}=\frac{301}{36} .
$$

The expected value of the difference of the areas is the difference of the expected values, which is
$$
\frac{49}{4}-\frac{301}{36}=\frac{140}{36}=\frac{35}{9} \text {. }
$$

Answer: $35 / 9$.