I am trying to understand part c of the following question (taken from Pitman's Probability, Chapter 3.1). I understand the solution using the difference of the max probabilities but I don't get the symmetry argument.
More specifically where is the 7 coming from? Intuitively it makes sense that the minimum and maximum would be symmetrical but I just can't figure out why it's this particular relation.
A hint would be much appreciated.
On
The $7$ comes because of the symmetry of probabilities. The probability of rolling $X$ is the same as the probability of rolling $7-X$. If you want to compute the expectation of the minimum, you could follow through the calculation for the maximum but changing the direction of the $\le$ signs. An alternate approach is to calculate the expectation of the maximum of $7-\text{ roll}$ for each die and subtract from $7$. This only requires that $P(1)=P(6), P(2)=P(5), P(3)=P(4)$, but the three probabilities could be different and the calculation would still work.
\begin{align} 7-1 & = 6 \\ 7-2 & = 5 \\ 7-3 & = 4 \\ 7-4 & = 3 \\ 7-5 & = 2 \\ 7-6 & = 1 \\ \uparrow & \phantom{={}} \uparrow \\ a & \phantom{{}={}} b \end{align} In column $a$ you see the numbers $1,2,3,4,5,6$ and in column $b$ you see those same six numbers.
If the outcome $X$ is one of the numbers $1,2,3,4,5,6,$ all with equal probabilities, then $7-X$ is also one of those six numbers, all with equal probabilities.
And if the maximum in five trials $X_1,X_2,X_3,X_4,X_5$ is $M$, then the minimum of $7-X_1,7-X_2, 7-X_3,7-X_4, 7-X_5,7-X_6$ is $7-M.$