Q:

Calculating mean of histogram of dice rolls

The table in the picture shows a frequency distribution of a weighted dice (the 6th side is twice as likely to occur) and it is followed by a histogram and a calculation of the mean value of the distribution. I am a little confused about the mean calculation because they multiplied each probability with the no. of dice's side. It doesn't make sense because these numbers on the horizontal axis (1 through to 6) aren't real numbers in the sense that "3" means something has happened three times or there are 3 elements there. It's just a way of denoting a side. We might as well have named each side A,B,C,D,E,F or something. If it were a histogram where we looked at no. of people belonging to a certain age (like 5 people are aged 10 years) then it would've made more sense to me. I am pretty sure I am making a wrong assumption somewhere can you please help me?

A:

In the expected value calculation, they are treating the face number on the die as a number, not a label. It is true, as you say, if the sides of the die were labelled A, B, C, D, E, F, we couldn't make an expected value calculation.
Perhaps it helps to think of it as: When you roll this weighted die, you win the number of dollars that shows on the face of the die.  The expected value of $\frac{27}7$ gives your expected winnings when you roll the die once.

