Probability related to launch of dice

Question

Consider the following problem:

Two dice are launched two times. First I launch the first die two times and I order the results from the smallest to the biggest. Then I do the same for the second die. What is the probability that one of the two couples has values bigger than the other.

I would define the random vectors $X=[x_1,x_2]$, $Y=[y_1,y_2]$ where $x_1, x_2$ are the first and the second results of the first die respectively and $y_1, y_2$ are the first and the second results of the first die respectively. What I would do is to count the couples of couples ((x_1,x_2),(y_1,y_2)) for which $\min(x_1,x_2) < \min(y_1,y_2) $ and at the same time $\max(x_1,x_2) < \max(y_1,y_2) $. At the end of course multiply the result by 2 (reverting the roles of the dice) and dividing by the cardinality of the set of all couples. Is it correct? is there a more intelligent way to approach it?

Answer 1

Suppose the dice are $n$-sided. You can break down the type of roll into these subcategories: $(aaaa)$, $(aaab)$, $(aabb)$, $(aabc)$, $(abcd)$. Here $(aabc)$, for instance, means that two dice showed the same value and the other two showed unique values... so far we're not assuming anything about which values are larger or which rolls are which. The probabilities for these categories are: $$ \begin{array} {|r|r|}\hline \textbf{Cat} & \textbf{Prob} \\ \hline (aaaa) & 1/n^3 \\ \hline (aaab) & 4/n^2-4/n^3 \\ \hline (aabb) & 3/n^2-3/n^3 \\ \hline (aabc) & 6/n-18/n^2+12/n^3 \\ \hline (abcd) & 1-6/n+11/n^2-6/n^3 \\ \hline \end{array} $$ The desired outcome can't occur in $(aaaa)$ or $(aaab)$. In $(aabb)$, it can only occur if one die rolled $aa$ and the other rolled $bb$, which is the case with probability $1/3$. For $(aabc)$, we can assume $b<c$; and $a<b$, $b<a<c$, or $c<a$ with probability $1/3$ each. The middle case can produce the desired outcome if the two $a$'s are rolled by different dice (probability $2/3$). The first and last cases can produce the desired outcome if the $a$'s are rolled by the same die (probability $1/3$). So case $(aabc)$ gives the desired outcome with probability $4/9$. Finally, for $(abcd)$, we can assume $a<b<c<d$. The desired outcome happens when the die that rolls $a$ does not also roll $d$... probability $2/3$. Putting it all together, we find $$ \frac{1}{3}\left(\frac{3}{n^2}-\frac{3}{n^3}\right)+\frac{4}{9}\left(\frac{6}{n}-\frac{18}{n^2}+\frac{12}{n^3}\right)+\frac{2}{3}\left(1-\frac{6}{n}+\frac{11}{n^2}-\frac{6}{n^3}\right)=\frac{2}{3}-\frac{4}{3n}+\frac{1}{3n^2}+\frac{1}{3n^3}. $$ For $n=6$, this comes out to $295 / 648\approx 0.4552$.

Answer 2

Update: This answer was for a (mis-)interpretation of the problem where the goal was to have both $x_i$ less than both $y_i$ or vice versa. OP's intended question wants to take $a_1=\min(x_1,x_2)$, $a_2=\max(x_1,x_2)$, $b_1=\min(y_1,y_2)$, $b_2=\max(y_1,y_2)$, and ask for the probability that $(a_1 < b_1 \wedge a_2 < b_2)$ or $(a_1 > b_1 \wedge a_2 > b_2)$.

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Suppose the dice are $n$-sided. Let $P_k$ be the probability that one roll (of the two from a single die) is equal to $k$ and the other is $\le k$. This is equal to $1/n^2 + 2(k-1)/n^2=(2k-1)/n^2$. And let $Q_k$ be the probability that both rolls are $>k$. This is equal to $(1-k/n)^2.$ The desired result is $$ f(n)=\sum_{k=1}^{n-1}2P_kQ_k=\frac{2}{n^4}\sum_{k=1}^{n-1}(2k-1)(n-k)^2=\frac{1}{3}\left(1 - \frac{2}{n} + \frac{2}{n^2} - \frac{1}{n^3}\right). $$ For $n=6$, this evaluates to $155/648 \approx 0.2392$. (As a useful check, you can verify that $f(1)=0$, and $f(2)=1/8$, and $\lim_{n\rightarrow\infty}f(n)=1/3$. It should be clear why all of these are correct.)