I roll three dice and keep only the largest two

Question

I know that rolling two 6-sided dice I will get 7 as medium result. But what if I roll three dice and get only the largest two? How much is that going to increase the medium result? And what about with 4 dice, 5 etc...?

Thank you!

Answer 1

Let's suppose we have 3 dice and pick the largest two. Then if $X$ is the value of the largest die result, and $Y$ is the value of the next-largest, we want $p(x,y)=P(X=x,Y=y)$. If we observe 3 distinct die results, there are 6 ways the dice could be assigned to those results. If we only observe 2 distinct die results, there are 3 ways the dice could be assigned to those results, and if we only observe 1 distinct die result, there is only 1 way for the dice to be assigned to those results. So we have $$p(x,y)=\begin{cases}\frac{6(y-1)+3}{6^3}&x>y\\\frac{3(y-1)+1}{6^3}&x=y\\0&x<y\end{cases}=\begin{cases}\frac{6y-3}{6^3}&x>y\\\frac{3y-2}{6^3}&x=y\\0&x<y\end{cases}.$$

And we note that $\sum_{x,y}p(x,y)=\sum_{y=1}^{6}(6-y)\frac{6y-3}{6^3}+\frac{3y-2}{6^3}=1$, as of course it should. Now if we let $T=X+Y$, we can find $p(t)=P(T=t)$, for $2\leq t\leq 12$. This gives $$p(t)=\begin{cases}\sum_{y=\min\{1,t-6\}}^{(t-1)/2}p(t-y,y)&t\text{ odd}\\\sum_{y=\min\{1,t-6\}}^{t/2}p(t-y,y)&t\text{ even}\end{cases}=\begin{cases}\sum_{y=\min\{1,t-6\}}^{(t-1)/2}\frac{6y-3}{6^{3}}&t\text{ odd}\\\left(\sum_{y=\min\{1,t-6\}}^{t/2-1}\frac{6y-3}{6^3}\right)+\frac{3(t/2)-2}{6^3}&t\text{ even}\end{cases}$$ This equals $$p(t)=\begin{cases}\frac{3(t-1)^{2}}{4\cdot 6^3}-\frac{3(t-7)^2}{6^3}\mathbf{1}_{\{>7\}}(t)&t\text{ odd}\\\frac{3(t-2)^{2}}{4\cdot 6^3}+\frac{3t-4}{2\cdot 6^3}-\frac{3(t-7)^2}{6^3}\mathbf{1}_{\{>7\}}(t)&t\text{ even}\end{cases}=\begin{cases}\frac{3(t-1)^{2}-12(t-7)^2\mathbf{1}_{\{>7\}}(t)}{4\cdot 6^3}&t\text{ odd}\\\frac{3(t-1)^{2}+1-12(t-7)^2\mathbf{1}_{\{>7\}}(t)}{4\cdot 6^3}&t\text{ even}\end{cases}.$$ Once again, it's easy to check that this defines a probability distribution, noting that $\mathbf{1}_{\{>7\}}(t)$ is the function which equals 1 if $t>7$ and is 0 otherwise.

Then the mean of this distribution is just $\sum_{t=2}^{12}t*p(t)=\frac{203}{24}\approx 8.46.$

I believe that the approach I've just described is probably the best for extending to more dice; one could, in the case of 3 dice, view the variable of interest as $X+Y+Z-M,$ where $M=\min\{X,Y,Z\},$ and $X,Y,Z$ are the three dice results, and use linearity of expectation to compute the mean. However, with more dice, this will get a bit more... dicey.

Answer 2

If the question is to determine the median, rather than the mean, one approach is to list, in descending order, the number of ways the two largest numbers can sum to $S=12$, $11$, $10$, etc., and keep a running total, stopping when it gets to $108=6^3/2$.

$$\begin{align} N_3(S=12)&=1+3\cdot5=16&T=16\\ N_3(S=11)&=3+6\cdot4=27&T=43\\ N_3(S=10)&=(3+6\cdot3)+(1+3\cdot4)=34&T=77\\ N_3(S=9)&=(3+6\cdot2)+(3+6\cdot3)=36&T=113 \end{align}$$

so the median score is $S=9$.

Remark: The scratchwork above is organized according to the various ways you can get each value of $S$. For example, for $S=10$, you either have a $6$ and a $4$, or you have two $5$'s. In first first case you either a $6$ and two $4$'s, which is possible in three ways, or you have a $6$, a $4$, and something smaller than $4$, which is possible in $6\cdot3$ ways. In the second case, you either have three $5$'s, which is possible in only one way, or you have two $5$'s and something smaller than a $5$, which is possible in $3\cdot4$ ways.

If you roll four dice and keep the largest two, the goal for the running total is $6^4/2=648$ and the counts, if I've done things correctly, become

$$\begin{align} N_4(S=12)&=1+4\cdot5+6\cdot5^2=181&T=181\\ N_4(S=11)&=4+4\cdot3\cdot4+4\cdot3\cdot4^2=244&T=425\\ N_4(S=10)&=(4+4\cdot3\cdot3+4\cdot3\cdot3^2)+(1+4\cdot4+6\cdot4^2)=261&T=686 \end{align}$$

so the median is $S=10$ For five dice, the goal is $6^5/2=3888$ and we have

$$\begin{align} N_5(S=12)&=1+5\cdot5+10\cdot5^2+10\cdot5^3=1526&T=1526\\ N_5(S=11)&=5+5\cdot4\cdot4+6\cdot6\cdot4^2+6\cdot4\cdot4^3=2197&T=3723\\ N_5(S=10)&=\cdots&T\gt3888 \end{align}$$

so the median is still $S=10$. (I didn't bother doing the precise count, since it's clear the median won't go down if you roll more dice.)

One can clearly keep going, at the expense of doing more and more arithmetic. It looks likely, or at least possible, that the median will be $S=11$ when rolling six dice, since the running total came fairly close to the goal with five. I'm hoping someone will check if it does, and doublecheck my numbers along the way.

It's actually easier to determine when the median becomes $S=12$: If you roll $n$ dice, you can avoid getting two $6$'s in only $5^n+n5^{n-1}$ ways, and this is less than $6^n/2$ as soon as $n=10$.