Answer

To answer this question, we'll compute the absolute difference between two dice (discrete uniform random variables $[1,6]$). Denote the rolls of the two dice X and Y. Then $E(\lvert X-Y \rvert)$ can be split between three cases: $$X=Y \Rightarrow E(\lvert X-Y\rvert)=0$$ $$X\gt Y \Rightarrow E(\lvert X-Y \rvert)=E(X-Y)$$ $$X\lt Y \Rightarrow E(\lvert X-Y\rvert)=E(Y-X)$$Now note that $E(X-Y)=E(Y-X)$ since X and Y are discrete uniform random variables from the same law. Therefore we can express: $$E(\lvert X-Y \rvert)=2E(X-Y)1_{X\gt Y}$$Where $1_{X\gt Y}$ is the indicator function - it takes value 1 if $X \gt Y$ and 0 otherwise. This allows us to only consider the case where X is greater than Y and halves the number of cases we need to consider. Now note the difference between two different rolls can be at most 5 and at least 1. Let's evaluate a few to see the pattern: $$X-Y=5 \text{ - This only happens if X=6 and Y=1 - so 1 way only}$$ $$X-Y=4 \text{ - This can happen if X=5, Y=1 or X=6, Y=2 - so in 2 ways}$$ $$X-Y=3 \text{ - This happens if X=4, Y=1 or X=5, Y=2 or X=6, Y=3 - so 3 ways}$$ Notice the pattern - $k$ can be the difference between the two dice in $6-k$ ways for $k$ between 1 and 5. There's a total of $6\cdot 6=36$ possibile outcomes when rolling the two dice, so we can find the expected difference as the sum of possible differences weighted by their probabilities: $$\text{Assuming X>Y: }E(X-Y) = \sum_{k=1}^5 k \cdot \frac{6-k}{36}=\frac{35}{36}$$Therefore, the expected absolute difference is: $$E(\lvert X-Y\rvert)=2\cdot \frac{35}{26}=\frac{70}{36}$$Now, let's move on to the smallest between the two rolls. To find the expected minimum, let's again evaluate a few possibilities to notice the pattern. $min(X,Y)=1$ in the following three ways: $$X=Y=1 \text{ - Happens in 1 way only}$$ $$X=1,\, Y\gt 1 \text{5 possible values for Y, so 5 ways}$$ $$X\gt 1, \, Y=1 \text{5 possible values for X, so 5 ways}$$In total, there are $2\cdot 5 + 1=11$ out of 36 possible outcomes where 1 is the minimum between the two. What about for $min(X,Y)=2$? $$X=Y=2 \text{ - Happens in 1 way only}$$ $$X=2, \, Y\gt 2 \text{4 possible values for Y, so 4 ways}$$ $$X\gt 2, \, Y=2 \text{4 possible values for X, so 4 ways}$$In total, that's $2\cdot 4 + 1=9$ out of 36 ways where 2 is the minimum. Notice the pattern: there's always one way to have them both be the same, and $6-k$ ways that $k$ can be the strictly smaller of the two (as you always have $6-k$ values to choose above $k$). That means probability of $k$ being the smallest of the two is $\frac{2\cdot (6-k) + 1}{36}$. We can then compute the expected value by weighing the possible outcomes with their probabilities: $$E(min(X,Y))=\sum_{k=1}^6 k \cdot \frac{2\cdot (6-k) + 1}{36}=\frac{91}{36}$$In conclusion, we found that $E(\lvert X-Y\rvert)=\frac{70}{36}$ and $E(min(X,Y))=\frac{91}{36}$ and therefore minimum of the two is larger on average.