I was wondering about finding expected value of dice throw. I know that in case of single dice throw,the expected value is 3.5 = ((1+2+3+4+5+6)/6). So for purpose of betting the base price could be 3.5 to ensure no profit and no loss scenario. But how can I extend this logic for double dice throw or N dice throw. I though that for case of double dice out of the 36 cases, I could take the maximum value for each pair which would give 6 eleven times, 5 nine times, 4 seven times, 3 five times , 2 three times and 1 once. But this gave expected value to be 4.47, whereas the correct answer in book is 4.25. How can I determine the fair value in this case?
[Math] Expectation of Double Dice Throw
expectationprobability
Best Answer
The number of rolls where $h$ is the highest of $n$ dice is the number of ways for $k$ dice to show less than $h$ and $n-k$ to show $h$ $$ \begin{align} \sum_{k=0}^{n-1}\binom{n}{k}(h-1)^k &=\overbrace{\color{#C00}{(1+(h-1))^n}\vphantom{\binom{n}{n}}}^{\text{sum for $k=0\dots n$}}-\overbrace{\color{#090}{\binom{n}{n}(h-1)^n}}^{k=n}\\ &=\color{#C00}{h^n}-\color{#090}{(h-1)^n}\tag1 \end{align} $$ which makes sense since it is the number of possible rolls showing at most $h$ minus the number of rolls showing at most $h-1$ with $n$ dice.
Assuming $n\gt0$, the total number of possible rolls of $n\times d$-sided dice is the sum of $(1)$ for all $h$ $$ \sum_{h=1}^d\left(h^n-(h-1)^n\right)=d^n\tag2 $$ The expected high die is then $$ \begin{align}\newcommand{\stirtwo}[2]{\left\{#1\atop#2\right\}} \frac1{d^n}\sum_{h=1}^dh\left(h^n-(h-1)^n\right) &=\frac1{d^n}\sum_{h=1}^d\left(h^{n+1}-(h-1)^{n+1}-(h-1)^n\right)\\ &=d-\frac1{d^n}\sum_{h=1}^d(h-1)^n\\ &=d-\frac1{d^n}\sum_{h=1}^{d-1}h^n\tag{3a}\\ &=d-\frac1{d^n}\sum_{j=0}^{n}\binom{d}{j+1}\stirtwo{n}{j}j!\tag{3b} \end{align} $$ where $\stirtwo{n}{j}$ is a Stirling Number of the Second Kind.