A game of dominance

Question

I have $n$ dices corresponding to variables $D_1,\ldots,D_n$. I throw the dices $k$ times and record the outcomes, i.e. $k$ realizations of the variables. Suppose that I throw the dices one more time.

I wonder if it is possible to compute or bound from below the probability that the $k+1$-th realization will be dominated by one of the previous $k$ realizations. Realization $d_1,\ldots,d_n$ of the variables is dominated by realization $d'_1,\ldots,d'_n$ when $d'_i \geq d_i$ for all $i=1,\ldots,n$.

I tried but couldn't determine an efficient way to compute or bound this probability, either considering the actual values of the given $k$ realizations, or more generally, as a function of $k$. Has this problem even been tackled in existing literature?

Answer 1

If we want to consider the $k$ realizations, we can just list the dominated rolls and count them up. For example, suppose we have two dice and have rolled $(2,5)$ and $(4,3)$. There are $10$ rolls dominated by the first and an additional $6$ ($3$ or $4$ on the first die and $1$ to $3$ on the second) dominated by the second, so there is a $\frac {16}{36}$ chance a third roll will be dominated by at least one of the first two.

If we don't consider the existing rolls, there is a $\frac 7{12}$ chance that one die dominates another, so the chance that one roll of $n$ dice dominates another is $\left(\frac 7{12}\right)^n$ I don't see an easy way to account for the correlations to say the chance that one roll of $n$ dice is dominated by one of $k$ previous rolls.

Answer 2

Let $r_i = (x_1,\ldots x_n)$ be the outcome of the $i$-th roll and use the notation $r_i\leq r_j$ to denote dominance.

You can start with the negation that it's not dominated by any of the previous dice rolls and break it up conditional on the outcome of the $k+1$-th roll $$ \sum_{x_1\ldots x_n}P(r_{k+1}\nleq r_1, r_{k+1} \nleq r_2,\ldots, r_{k+1}\nleq r_k|r_{k+1}=\{x_1\ldots x_n\})P(r_{k+1}=\{x_1\ldots x_n\})$$

This is nice cause the events $r_{k+1} \nleq r_i$ are independent and identical when conditioned on the outcome of $r_{k+1},$ so the thing factors $$ \sum_{x_1\ldots x_n}P(r_{k+1}\nleq r_1|r_{k+1}=\{x_1\ldots x_n\})^k P(r_{k+1}=\{x_1\ldots x_n\})$$

For a die in roll k+1 showing $x,$ the probability it will be dominated by its corresponding die in roll $r_1$ is $(7-x)/6$ so the probability that $r_{k+1}=(x_1\ldots x_n)$ is dominated is $\prod_{i=1}^n\frac{(7-x_i)}{6}.$ So we have $$ \sum_{x_1\ldots x_n}\left(1-\prod_{i=1}^n\frac{(7-x_i)}{6}\right)^k P(r_{k+1}=\{x_1\ldots x_n\})$$ for the probability of the negation.

$P(r_{k+1} = \{x_1,\ldots x_n)\})$ is just constant $1/6^n$ for any outcome, so the end result is

$$ P(\mbox{roll k+1 is dominated}) = 1-\frac{1}{6^n}\sum_{x_1\ldots x_n}\left(1-\prod_{i=1}^n\frac{(7-x_i)}{6}\right)^k $$ where each $x_i$ ranges over $1\le x_i \le 6.$

Whether I've made a mistake or how to tame this further if I haven't, I don't know.

Answer 3

In a slight change of notation from the original post, let's say we have a $k$ by $n$ matrix $D$ containing the die rolls, and we are interested in the probability that at least one of the rows of the matrix dominates the other rows, where we say row $i$ dominates row $j$ if $D_{i,m} \ge D_{j,m}$ for all $m=1,2,3,\dots ,n$. It is possible for there to be more than one dominant row. There are $6^{kn}$ possible arrangements of $D$ with $1 \le D_{i,j} \le 6$, all of which we assume are equally likely. We will use the Principle of Inclusion and Exclusion (PIE) to count the number of arrangements in which at least one row dominates all the other rows.

Let's warm up with an example. What are the number of arrangements in which row 1 dominates the other rows? In column 1, if $D_{1,1}=m$ then there are $m$ possible choices for each of the other rolls in that column. Therefore there are $\sum_{m=1}^6 m^{k-1}$ possible arrangements for the rolls in column 1. Since the $n$ columns are independent, there are $(\sum_{m=1}^6 m^{k-1})^n$ possible arrangements of $D$ in which row 1 dominates the other rows.

Suppose rows 1 and 2 both dominate the other rows. Then the situation is similar to the previous example, but now there are only $k-2$ other rows to consider. So the number of arrangements in which rows 1 and 2 are both dominant is $(\sum_{m=1}^6 m^{k-2})^n$.

Now that we are warmed up, let $S_j$ be the number of arrangements in which there are $j$ dominant rows. There are $\binom{k}{j}$ ways to pick the dominant rows, so based on the previous examples, $$S_j = \binom{k}{j} \left( \sum_{m=1}^6 m^{k-j} \right) ^n$$ for $j=1,2,3, \dots ,k$. Since there are $6^{kn}$ possible arrangements, by PIE the probability that there is at least one dominant row is

$$(S_1 - S_2 + S_3 - \dots \pm S_k) / 6^{kn}$$

For a specific example, suppose there are $n=4$ dice and $k=6$ rolls. Then the computation above yields a probability of about 0.0279763, which agrees well with a Monte Carlo simulation in which there were 28,120 successes in $10^6$ trials.