What are the odds of any role of a 24 sided die occurring 4 or more times in 10 rolls? Note that I am not asking about the odds of a chosen roll happening 4 times in 10 rolls, (this has a probability of 0.000517 according to a binomial calculator), rather, the odds of ANY roll happening 4 or more times in 10 rolls.
Another way to ask it is: What are the odds that the mode of the list of 10 rolls of the die will be 4 or more? 
In general how is this calculated?
Thank you.  
 A: Suppose $i$ is an integer between $1$ and $24$, and let $A_i$ be the event that you roll an $i$ at least $4$ times in ten rolls. Notice that two of the $A_i$ could occur, but not three, since that would be twelve different rolls.
By the inclusion/exclusion principle, you get that $$\mathbb{P}(\cup A_i) = \sum \mathbb{P}(A_i) - \sum_{i\ \neq j}\mathbb{P}(A_i \cap A_j).$$ You have already calculated $\mathbb{P}(A_i)$ for each $i$. Now you have to calculate $\mathbb{P}(A_i \cap A_j)$ for $i \neq j$, which is a little trickier (unless I'm missing an easier way).
To do this, you should add up 


*

*the number of ways of rolling (exactly) four $i$s and exactly four $j$s,

*...of rolling four $i$s and five $j$s, 

*...of rolling five $i$s and four $j$s (same as
above), 

*...of rolling five $i$s and five $j$s, 

*...of rolling six $i$s and four $j$s, and 

*...of rolling six $j$s and four $i$s (same as
above).


This is $$22^2 \binom{10}{8}\binom{8}{4} + 2 \cdot 22 \cdot \binom{10}{9}\binom{9}{5} + \binom{10}{5} + 2 \cdot\binom{10}{6}.$$
Call that number $n$ for brevity. Dividing $n$ by $24^{10}$ will give you $\mathbb{P}(A_i \cap A_j)$, given $i$ and $j$. Now add up all the different choices of $i$ and $j$ to give you (assuming you did your previous calculation right) $$\mathbb{P}(\cup A_i)= 24 \cdot (.000517) - \binom{24}{2}\frac{n}{24^{10}}$$ 
A: Outline: We use Inclusion/Exclusion.
Find in one of the usual ways the probability of say $1$ coming up $4$ or more times.  Multiply by $24$, and call the result $a$.
However, this double-counts the ways in which $2$ faces come up $4$ or more times. (We cannot have more than $2$ faces coming up $4$ or more times.)
So we will have to subtract the probability of $2$ faces coming up $4$ or more times.
Which $2$ faces? They can be chosen in $\binom{24}{2}$ ways. So let us find  the probability that (say) faces $1$ and $2$ can come up $4$ or more times, multiply by $\binom{24}{2}$, and subtract the result from $a$.
The face $1$ can come up $4$, $5$, or $6$ times. The simplest case is $6$, for then we find the probability of $6$ $1$'s and $4$ $2$'s. 
For face $1$ coming up $5$ times, we can have face $2$ coming up $4$ or $5$ times. The probabilities can be calculated. The same is true for face $1$ coming up $4$ times, but then there are three cases for face $2$.
Unpleasant, but one can push it through. 
