Expectation of Double Dice Throw

Question

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?

Answer 1

Mean of a Random Variable: A quantity equal to the average result of an experiment after a large number of trials. For example, if a fair 6-sided die is rolled, the expected value of the number rolled is 3.5.

The expectation of the sum of two (independent) dice is the sum of expectations of each die, which is 3.5 + 3.5 = 7. Similarly, for N dice throws, the expectation of the sum should be N * 3.5.

If you're taking only the maximum value of the two dice throws, then your answer 4.47 is correct. This has been proven here in multiple ways.

Presumably, the answer given in the book is wrong or the question is different than you've interpreted it to be.

Answer 2

Let $X_1$ be the value on the first die and let $X_2$ be the value on the second die. Since $X = X1 + X2$ we have $E(X) = E(X_1) + E(X_2)$. Earlier we computed that $E(X_1) = E(X_2) = 3.5$, therefore $E(X) = 7$.

Answer 3

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.

Answer 4

I guess the rule of the game is that you would roll the dice then decide whether to take the amount you rolled as reward or not. If not, you would roll again and take the result of the 2nd roll as reward no matter what.

If the question is interpretted in this way, the strategy that you should follow is: Stop at the 1st roll if you do better than your expected return if you roll twice. Meaning that you should stop at the 1st roll if you get 4 ,5 or 6.

Hence, your expected return overall is: E(x) = 4*1/6 + 5*1/6 + 6*1/6 + (3.5)*3/6 = 4.25 (where 3.5 is the expected return of the 2nd roll)

Hope this helps.