Problem Statement

Two players have a set of dice each. The first player has nine dice with four faces each, with numbers one to four. The second player has six normal dice with six faces each, each face has the usual numbers from one to six. They roll their dice and sum the totals of the faces. The player with the highest total wins (it's a draw if the totals are the same). What's the probability of the first player beating the second player? Later the two players use a different set of dice each. Now the first player has five dice with ten faces each, and the second player has six dice with seven faces each. Now what's the probability of the first player beating the second player? This task was adapted from the Project Euler Problem n.205: https://projecteuler.net/problem=205

Let's start with the solution:

sub likelihoods ($roll) {
    my ($dice, $faces) = $roll.comb(/\d+/);
    my @counts;
    @counts[$_]++ for [X+] |(1..$faces,) xx $dice;
    return [@counts[]:p], $faces ** $dice;
}
 
sub beating-probability ([$roll1, $roll2]) {
    my (@c1, $p1) := likelihoods $roll1;
    my (@c2, $p2) := likelihoods $roll2;
    my $p12 = $p1 * $p2;
 
    [+] gather for flat @c1 X @c2 -> $p, $q {
	take $p.value * $q.value / $p12 if $p.key > $q.key;
    }
}

# We're using standard DnD notation for dice rolls here.
say .gist, "\t", .raku given beating-probability < 9d4 6d6 >;
say .gist, "\t", .raku given beating-probability < 5d10 6d7 >;