Probability of ordered dice Suppose you throw five dice, order them from lowest to highest number, then select the third die. What is the probability that the third die was number 2?
Or, in general, when throwing $n$ dice and selecting the $i$th die (sorted from low to high), what is the probability that the number of that die is $k_i$? So the above example was $n=5,i=3,k_i=2$. How do you determine a general formula for finding the probability?

We can use the multinomial law to compute this probability. If we split the probability on the number of 2 you draw we have
$$p=\mathbb P(\textrm{exactly one 2 and two 1}) + \mathbb P(\textrm{exactly two 2 and (one or two) 1}) + \mathbb P(\textrm{exactly three 2}) + \mathbb P(\textrm{exactly four 2}) + \mathbb P(\textrm{exactly five 2})$$
$$p=\left(5 \cdot 6 \cdot \left( \dfrac{1}{6} \right)\left( \dfrac{1}{6} \right)^2\left( \dfrac{4}{6} \right)^2 \right) + \left(10 \cdot 3 \cdot \left( \dfrac{1}{6} \right)^2\left(\left( \dfrac{1}{6} \right)^2\left( \dfrac{4}{6} \right)+\left( \dfrac{1}{6} \right)\left( \dfrac{4}{6} \right)^2 \right) \right) + \left(10 \cdot \left( \dfrac{1}{6} \right)^3\left( \dfrac{5}{6} \right)^2 \right) + \left( 5\left( \dfrac{1}{6} \right)^4\left( \dfrac{5}{6} \right)\right) + \left( \left( \dfrac{1}{6} \right)^5\right)$$
$$p=\left( \dfrac{30 \cdot 16}{6^5}\right) + \left( \dfrac{30 \cdot 20}{6^5}\right) + \left( \dfrac{10 \cdot 25}{6^5}\right) + \left( \dfrac{5 \cdot 5}{6^5}\right)+\left( \dfrac{1}{6^5}\right)$$
$$p=\dfrac{1356}{6^5} \approx 0.17438$$
And we get the probability that @Henry found with R. We can also verify it with a simple Python program that gives the same result
from random import randint

ok = 0
nb_draws = 100000
for j in range(nb_draws):
    l = [randint(1,6) for i in range(5)]
    l.sort()
    if l[2] == 2:
        ok += 1

print(ok * 1.0 / nb_draws)

