Rolling $N$ indistinct dice and $1$ distinct die, what is the chance that the distinct die is part of the Top 3?

Question

Rolling $N$ indistinct dice and $1$ distinct die, what is the chance that the distinct die is part of the Top 3? Ties favor the distinct die being included over an indistinct die.

I'm at the point where I think I know the answer, but I definitely haven't convinced myself that my argument is solid. I'm having a block of some sort, making it hard to think further on the problem.

For any roll of all $N+1$ dice, the chance of the distinct die being in the Top 3 is the chance of it being greater than or equal to each of $N-2$ dice. As for the other $2$ indistinct dice, their order in the Top 3 does not matter.

So, the problem then becomes what is the chance of a die rolling greater than or equal to $N-2$ other dice. So this should become an ordering problem, the number of orderings with our distinct die affixed to the top of the order divided by all the possible orderings of $(N-2)+1$ dice.

$$ \frac{(N-2)!}{(N-1)!} = \frac{1}{N-1} $$

The only reason I'm not satisfied with this conclusion is that in the original case, there were $2$ additional dice that need to be woven into the ranking that was generated without them. I've got this gut feeling that I've broken something by removing them from the model and putting them back into the problem will cause issues.

Is it safe to say that regardless of where these dice are added the result is the same? If this is the wrong approach, how would I solve this problem correctly?

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Revisiting this question again a few days later, I've realized what was wrong with my logic.

I presented a problem in which ties favored the distinct die, however in my ordering argument, I considered only the cases in which the distinct die appeared at the TOP of the rankings (when affixed to the top). With ties involved, this doesn't work. The distinct die could be all the way at the bottom, but still qualify for Top 3 in the event of a $(N-1)$-way tie. There's also some nebulous ground on ordering indistinct dice about each other, but I think that is still forgivable if handled properly.

Another issue with the formulation above was that it neglected the other two dice. The distinct die was only tested against the $N-2$ other dice, which meant that it had fewer chances to be in the Top 3. In the case of $N+1=4$, the distinct die should have $3$ chances to be part of the Top 3, as it only needs to surpass one of the other dice. However, it is only tested against one die, netting it a much lower chance than it should have of being in the Top 3. Perhaps starting from that and then weaving the remaining two dice back into the model would have worked, but already seems like it would have been far less elegant than Christian Blatter's answer.

Answer 1

Its more complicated than that. Condition on the number $j\in[6]$ obtained with the special die. The result $j$ will be sufficient for a success iff at most $2$, i.e., $r\in \{0,1,2\}$ of the $N$ standard dices have shown a number $x_k>j$, i.e., a number $x_k\in[(j+1)..6]$. The probability $p$ that a given $x_k$ lies in this interval is $p={6-j\over 6}$. Furthermore the number of dices showing a result in this interval is binomially distributed. The conditional probability of a success therefore is $$P\bigl[{\rm success}\bigm| j\ \bigr]=\sum_{r=0}^2{N\choose r}\left({6-j\over6}\right)^r\left({j\over6}\right)^{N-r}\quad(1\leq j\leq 5)\ , $$ and $P\bigl[{\rm success}\bigm| 6\, \bigr]=1$. Now sum over $j$ and divide by $6$. In the case $N=2$ the final result is $1$, as expected, and in the case $N=3$ it is ${119\over144}\doteq0.826$.