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.
Answer
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) $$\mathbb{P}(\cup A_i)= 24 \cdot (.000517) - \binom{24}{2}\frac{n}{24^{10}}$$
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