Three standard $6$ sided dice are rolled and you get paid the face value of the highest number rolled.
What is the expected winnings from this game?
My approach is ignoring the two dice that don't count, and summing the possibilities x outcomes for the highest dice
That is
There is $1$ way for $1$ to be the highest value, when all dice are $1$
There are $(2^3-1^3)$ ways for $2$ to be the highest (Let this value $ = a$)
There are $(3^3-2^3)$ ways for $3$ to be the highest (Let $= b$)
...
$(6^3-5^3)$ ways for $6$ to be the highest (Let $= f$)
So the expected winnings is $$ \displaystyle \frac{1 \cdot a + 2 \cdot b + 3 \cdot c + 4 \cdot d + 5 \cdot e + 6 \cdot f}{216}.$$ Is that correct? Or am I misunderstanding something?
