Best value of $X$ dice thrown: expected value

diceexpected valueprobability

It has been a long time I didn't practice probabilities. If I made notation mistakes, please let me know, thus I can fix it.

History

I play a game where I use ten-sided dice. Depending on my bonus, I have the possibility to throw two dices and keep the highest value from these two dices. The bonus allows me to throw up to 10 dice, and keep the highest value. I want to know the expected value depending of the number of dice thrown.

What I tried

I tried to guess probability based on lower number of dices, recursively. Thus, I got this equation:

$$P_{d}(R) = P_{1}(R)\cdot P_{d-1}(\le R) + P_{1}(\lt R)\cdot P_{d-1}(R)$$

Notation

$d$ : number of thrown dice
$R$ : result (highest value from all dice)
$P_{d}(R)$ : probability to get result $R$ with $d$ dice throws
$P_{1}(\le R) ; P_{1}(\lt R)$ : probability to get result lower-or-equal or lower-than $R$

Known values

$P_{1}(R) = \frac{1}{10}$
$P_{d}(\le R) = \Bigl(\frac{R}{10}\Bigr)^{d}$
$P_{d}(\lt R) = P_{d}(\le R-1) = \Bigl(\frac{R-1}{10}\Bigr)^{d}$

Explanation

There are two ways to get a result $R$ :

  • the first die equals $R$ and the others are lower-or-equal to $R$
    $$P_{1}(R)\cdot P_{d-1}(\le R)$$
  • the first die is lower than $R$ and the best of other dice is equals to $R$
    $$P_{1}(\lt R)\cdot P_{d-1}(R)$$

As these two possibilities are disjointed (no possibility in common), we can sum the two equations to get the final equation.

Results

I implemented these equations using Excel, here are my results:

Equations results

The vertical axis represent the number $d$ of thrown dice. The horizontal axis represent the result $R$. The Esp column stands for the expected value (named Esperance in french).

Question

Might you please check that my reasoning is correct? If yes, do my results seem correct too? If not, what is wrong?

Best Answer

The column with expected values provide the correct values. However, I would suggest simplifying your argument in the following way.

\begin{equation} P_d(R) = P_d(\leq R)-P_d(\leq R-1) = \Bigl(\frac{R}{10}\Bigr)^{d} - \Bigl(\frac{R-1}{10}\Bigr)^{d} \end{equation}

This simplifies calculating the expected value.
\begin{equation} \mathop{\mathbb{E}}[R] = \sum_{R=1}^{10} R * P_d(R) = \sum_{R=1}^{10} R * (P_d(\leq R)-P_d(\leq R-1)) \end{equation}

\begin{equation} = \sum_{R=1}^{10} R * P_d(\leq R)-(R-1)*P_d(\leq R-1)) - P_d(\leq R-1)) \end{equation}

\begin{equation} = 10 * P_d(\leq 10) - \sum_{R=1}^{9} P_d(\leq R) = 10 - \sum_{R=1}^{9} \Bigl(\frac{R}{10}\Bigr)^{d} \end{equation}

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