Probability of getting all faces of a die an equal number of times

I have a question:

A die is rolled 36 times.

What is the probability of getting each number 6 times?

I think the answer is: $6\cdot\left(\frac16\right)^6$

Am I wrong?

Solution 1:

The answer you give is incorrect. There are 6 events you want to have happen, rolling each number 6 times, so you multiply their probabilities. Additionally, there are many ways this could happen, so we multiply by the number of ways of arranging the die rolls. This gives:

$$ \left(\frac{1}{6}\right)^{36} \frac{36!}{(6!)^6}$$

Solution 2:

It's multinomially distributed with $n=36$ ; $x_i=6$ and $p_i=1/6$ for $i=1,2,..,6$, see https://en.wikipedia.org/wiki/Multinomial_distribution . So the probability is $$ P=\frac{36!}{6!6!6!6!6!6!}\cdot \left( \frac{1}{6} \right)^6\left( \frac{1}{6} \right)^6\left( \frac{1}{6} \right)^6\left( \frac{1}{6} \right)^6\left( \frac{1}{6} \right)^6\left( \frac{1}{6} \right)^6$$ which can be simplified to $$ P=\frac{36!}{6!6!6!6!6!6!}\cdot \left( \frac{1}{6^{36}} \right)$$ I don't know if this can be simplified further but it doesn't look like your result, sorry!

Solution 3:

Total number of outcomes:

$$6^{36}=10314424798490535546171949056$$

Number of desired outcomes:

$$\frac{36!}{6!^{6}}=2670177736637149247308800$$

Probability:

$$\frac{2670177736637149247308800}{10314424798490535546171949056}\approx0.02589\%$$

Solution 4:

Among the 36 results, we need each number appear 6 times. Thus this can happen in $$\frac{36!}{(6!)^6}$$ ways. For each of this sequence, the probability is $\frac{1}{6^{36}}$. Thus the required probability is $$\frac{36!}{(6!)^6}\frac{1}{6^{36}}$$