Nontransitive Sets of Dice

Consider the following set of dice with nonstandard pips:

Die $A$: $1$ $4$ $4$ $4$ $4$ $4$
Die $B$: $2$ $2$ $2$ $5$ $5$ $5$
Die $C$: $3$ $3$ $3$ $3$ $3$ $6$

A game is played by two players picking a die in turn and rolling it. The player who rolls the highest value wins.

If the first player picks die $A$ and the second player picks die $B$ we get
$P(\text{second player wins}) = 7/12 \gt 1/2$.

If the first player picks die $B$ and the second player picks die $C$ we get
$P(\text{second player wins}) = 7/12 \gt 1/2$.

If the first player picks die $C$ and the second player picks die $A$ we get
$P(\text{second player wins}) = 25/36 \gt 1/2$.

So whatever die the first player picks, the second player can pick another die and have a larger than $50\%$ chance of winning.
A set of dice having this property is called a nontransitive set of dice.

We wish to investigate how many sets of nontransitive dice exist. We will assume the following conditions:There are three six-sided dice with each side having between $1$ and $N$ pips, inclusive.
Dice with the same set of pips are equal, regardless of which side on the die the pips are located.
The same pip value may appear on multiple dice; if both players roll the same value neither player wins.
The sets of dice $\{A,B,C\}$, $\{B,C,A\}$ and $\{C,A,B\}$ are the same set.

For $N = 7$ we find there are $9780$ such sets.
How many are there for $N = 30$?

This problem is a complex combinatorial problem and needs a programming approach to solve it. The conditions clearly state that we need to consider permutations and combinations of dice results and conduct a probability analysis.

The idea behind the non-transitive property indicates that the probabilities are not directly proportional to the face values of the dice. Therefore, direct mathematical computation seems a bit tricky and intensive programming is required.

Procedure to follow:

1. First, we create a collection of all possible dice, as per the provided constraints.
2. Next, we iterate over all possible sets of $3$ dice.
3. We exclude sets that contain duplicate dice.
4. We check if the dice set is non-transitive and if yes, we increment the counter.

This needs efficient programming, which will involve iterations, condition checks, and tracking. Involving the dice’s face values ranging up to $30$ will increase the problem’s complexity.

Such a problem might be solved with more advanced computational mathematics software or high-level programming languages like Python or R, which can handle complex numbers and probability structures.

The Python code you might use to address this problem could start like this:

“`python
import itertools as it
# Firstly, create a function to check transitivity

def check_transitivity(dice):
perms = list(it.permutations(dice))

for perm in perms:
if not (win_probability(perm[0], perm[1]) > 0.5 and win_probability(perm[1], perm[2]) > 0.5 and win_probability(perm[2], perm[0]) > 0.5):
return False
return True

# Function to calculate the win probability
def win_probability(die1, die2):
win = 0
total = 0

# Then iterate over all possible throws
for result1 in die1:
for result2 in die2:
total += 1
if result1 > result2:
win += 1

return win/total

# Complete the code accordingly…
“`

As it becomes clear, this problem makes a deep dive into combinatorial probability and complexity can increase substantially with increasing $N$. Hence, you might require a computational mathematics background and programming skills to solve it.

Note: The implementation provided here is just a starting point. You will need to complete it accordingly and handle any edge cases that may arise due to the conditions of the problem statement.

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