12 sided dice probability question

Question

I need help with determining the probabilities of rolling four 12-sided dice: Red, White, Blue, and Yellow.

I tried to program a simulation in BASIC on Windows, and the language on Windows 10 isn't like my old and long-gone Atari 800XL, so I cannot figure out a simulation.

Basically, if all of the dice combinations are done (should be 12 * 12 * 12 * 12 or 20,736), in what number of occasions would Red "win" with the highest outright number, and how many would Red "tie" in a tie for the highest number, two-way, three-way, and four-way ties are "ties?" (A "loss" would be any other scenario or the remainder.)

Thank you.

Answer 1

Red wins with $12$ in $11 \times 11 \times 11$ ways.

Red wins with $11$ in $10 \times 10 \times 10$ ways.

... and so on.

Two way ties are a little harder.

Red is in a two way tie with $12$ in $3 \times 11 \times 11$ ways. The $3$ picks the die that ties.

Red is in a two way tie with $11$ in $3 \times 10 \times 10$ ways. The $3$ picks the die that ties.

... and so on.

Red is in a three way tie with $12$ in $3 \times 11$ ways. The $3$ picks the die that does not tie.

... and so on.

Counting the four way ties is easy. There are $12$ of them.

Answer 2

In order for red to win, all the other dice rolls must be strictly less than the outcome rolled by red. So if red rolls a number $r$, there are $(r-1)^3$ outcomes of the other three dice that will satisfy this condition. We then sum this over all possible outcomes of $r$ to obtain $$\sum_{r=1}^{12} (r-1)^3 = 4356$$ outcomes where red wins outright. Since each outcome is equally likely, and there are $12^4 = 20736$ possible outcomes, the desired probability of an outright win is $\frac{4356}{20736} = \frac{121}{576}$.

If by a "tie" we mean at least one other die rolls a value equal to that of red's roll, and this value is the highest value among the four dice, and we are not concerned with distinguishing whether such a tie is "two-way," "three-way," or "four-way," then we can modify the above calculation to count the number of outcomes in which the other three dice can be less than or equal to red's roll. Of course, this means if red rolls $r$, then the other three dice may roll any value between $1$ and $r$, so the number of such outcomes is $$\sum_{r=1}^{12} r^3 = 6084.$$ So if we now subtract from this the number of outcomes where red wins outright, we get the number of outcomes where there is some kind of tie with red's roll: $$6084 - 4356 = 1278.$$ Then the probability of a tie is $\frac{1278}{20736} = \frac{1}{12}$.

In fact, it is not difficult to see from the above reasoning that even if there are an arbitrary number of competing dice (in your case, you have three competitors), the probability of red being in a tie for the win is still $1/12$. This seems like a surprising result. Why would it be independent of the number of competitors? Can you think of an explanation as to why?