Dice games, part 1

First, recall that for a discrete random variable, $X$ , we have: $E[X] = \sum_{x=1}^{\infty} x P(Y=x) ...........(i)$ Also note that the dice are independent, from which: $P(max(X,Y)\leq z) = P(X\leq z)P(Y\leq z)\\ \Rightarrow P(max(X,Y) = z_i) = P(max(X,Y)\leq z_i) - P(max(X,Y)\leq z_{i-1})\\ = P(X\leq z_i)P(Y\leq z_i) - P(X\leq z_{i-1})P(X\leq z_{i-1}).....(ii)$ Now, the dice have a uniform distribution between the values of $1$ and $n$ , so $(ii)$ can be rewritten as: $P(max(X,Y)=z) = \frac{z}{n} \cdot \frac{z}{n} - \frac{z-1}{n} \cdot \frac{z-1}{n}$ So, $(i)$ can be rewritten to get: $E[max(X,Y)] = \sum_{z=1}^{\infty} z \: P(max(X,Y) = z)\\ = \sum_{z=1}^{n} z \left(\frac{z^2}{n^2} - \frac{(z-1)^2}{n^2}\right)\\ = \frac{1}{n^2} \sum_{z=1}^{n}2z^2 - z\\ = \frac{(n+1)(4n-1)}{6n}......(iii)$ Bavid gets to roll a third die, and take that into account, so he gets a slightly different version of $(ii)$ : $P(max(U,V, W)=z) = \frac{z}{n} \cdot \frac{z}{n} \cdot \frac{z}{n}- \frac{z-1}{n} \cdot \frac{z-1}{n}\cdot \frac{z-1}{n}$ Plugging that into $(i)$ : $E[max(U, V, W)] = \sum_{z=1}^{\infty} z \: P(max(U, V, Y) = z)\\ = \sum_{z=1}^{n} z \left(\frac{z^3}{n^3} - \frac{(z-1)^3}{n^3}\right)\\ = \frac{1}{n^3} \sum_{z=1}^{n}3z^3 - 3z^2 + z\\ = \frac{(n+1)(3n-1)}{4n}......(iv)$ Now we turn to Alivia and Bavid's game: $E[Bavid]>E[Alivia]\\ E[max(U, V, W)] > E[max(X, Y) + 5]$ (using $(iii)$ and $(iv)$ ) $\frac{(n+1)(3n-1)}{4n} > \frac{(n+1)(4n-1)}{6n} + 5\\ \frac{3(n+1)(3n-1)}{12n} > \frac{2(n+1)(4n-1) + 60n}{12n}$ (since n\neq 0) $3(n+1)(3n-1) > 2(n+1)(4n-1) + 60n\\ 9n^2+6n-3 > 8n^2 + 6n - 2 + 60n\\ n^2 - 60n - 1 > 0$ (quadratic equation and constraint of $n \in \mathbb{N}$ yields) $n>\frac{60 + \sqrt{3600+4}}{2} = 30 + \sqrt{901}\\ \therefore \boxed{n = 61}$ (Good luck getting $61$ -sided dice, though)

X numbers randomly picked will always split the max and minimum evenly (picking a random number between 0-1 will average .5, the max of 2 numbers will average .333, the max of 3 .25, etc.)

Based on this, Alivia's average should be

x*2/3+5 (where x is the number of sides on the die)

And Bavid's average should be

x*3/4

Now, simply solve for x

(x * 2/3+5)-x * 3/4=0

60=x

In other words, when the die is 60-sided, they will both have the same average. However, Bavid needs to have a higher average, so go up one more to |61|

Let's consider a $n$ -sided dice thrown $p$ times. Let be $\overline{x}=(x_1,x_2,...,x_p)$ , $1\leq x_i\leq n$ , $1\leq i \leq p$ the vector containing the result of each roll. Let be $M_m$ the set of all $\overline{x}$ such that $\max{ \overline{x}}=m$ .

Now, let's find $|M_m|$ . For a $\overline{x}$ string long $p$ , you can place the maximum $m$ one time in $\displaystyle \binom{p}{1}$ ways. Than, fill the remaining places with the $m-1$ number smaller than $m$ in $(m-1)^{p-1}$ ways. Or, you can place $m$ two times in $\displaystyle \binom{p}{2}$ ways, than fill the remaining spaces with the $m-1$ smaller numbers in $(m-1)^{p-2}$ ways. Hence $|M_m|$ is

$\displaystyle |M_m|=\binom{p}{1}(m-1)^{p-1}+\binom{p}{2}(m-1)^{p-2}+...+\binom{p}{p}=\sum_{h=1}^{p} \binom{p}{h}(m-1)^{p-h}$ .

So, we can write $\displaystyle \mathbb{P}(\max{\overline{x}=m})=\frac{|M_m|}{n^p}$ , where $n^p$ is the number of all possible $\overline{x}$ .

For the first player, we have that $p=2$ . Let's define the random variable $A=\max(x_1,x_2,x_3)+5=m+5$ , $6\leq A\leq n+5$ so that $m=A-5$ . So, we find that

$\displaystyle \mathbb{P}(A)=\frac{1}{n^2}\sum_{h=1}^{2} \binom{2}{h}(A-6)^{2-h}=\frac{1}{n^2} \bigg(\binom{2}{1}(A-6)+\binom{2}{2}\bigg)=\frac{2A-11}{n^2}$ .

So,

$\displaystyle \mathbb{E}[A]=\sum_{A=6}^{n+5} \frac{2A-11}{n^2}A=\frac{1}{6} \bigg(4n-\frac{1}{n}+33\bigg)$

For the second player, we have that $p=3$ . Let's define the random variable $B=\max(x_1,x_2)$ , $1\leq B \leq n$ . Again,

$\displaystyle \mathbb{P}(B)=\frac{1}{n^3}\sum_{h=1}^{3} \binom{3}{h}(B-1)^{3-h}=\frac{3(B-1)^2+3(B-1)+1}{n^3}$

So

$\displaystyle \mathbb{E}[B]=\sum_{B=1}^{n} \frac{3(B-1)^2+3(B-1)+1}{n^3}B=\frac{(n+1)(3n-1)}{4n}$

Eventually

$\displaystyle \mathbb{E}[B]>\mathbb{E}[A]\ \hspace{5pt} \Longrightarrow \hspace{5pt} \frac{(n+1)(3n-1)}{4n}>\frac{1}{6} \bigg(4n-\frac{1}{n}+33\bigg) \hspace{5pt} \Longrightarrow \hspace{5pt} n>30+\sqrt{901}\approx 60.017 \hspace{5pt} \Longrightarrow \hspace{5pt} n=\boxed{61}$