Probability that the third-largest of 39 nine-sided dice will be equal to 8

Question

How can I calculate the probability that the third-largest of 39 nine-sided dice will be equal to 8? How would this be done mathematically?

The following is possible for example:

(two nines, one eight, rest is below 8)

(three eights, rest is below 8)

(one nine, two eights, rest is below 8)

Answer 1

$$\binom {39} 2 \cdot 37 \frac{7^{36}}{9^{39}}$$ choose 2 from 39 that have the value 9. That's $\binom{39}2$. Now choose one that holds 8, 37 dice left, hence 37. The rest can be anything from 1,..,7, hence $7^{36}$. Total number of cases is $9^{39}$.

Sorry for the typo, I'm typing from my phone.

Answer 2

EDIT I began formulatng this answer before the clarification that the OP wanted exactly one eight (i.e., two nines, one eight, all others at most seven). The below is about a too general case then.

---

This amounts to: At least three dice are $\ge 8$, but at most two dice are $>8$.

We count outcomes:

• No nine: $8^{39}$
• No nine and no eight: $7^{39}$
• No nine and one eight: $39\cdot 7^{38}$
• No nine and two eights: ${39\choose 2}\cdot 7^{37}$

Hence

No nine and at least three eights: $8^{39}-7^{39}-39\cdot7^{38}-{39\choose 2}\cdot 7^{37}$

Use a similar method to count

One nine and at least two eights: ...

and

Two nines and at least one eight: ...

Finally, add these three counts of outcomes.