What is the probability that three of six dice show the number five?

Question

When I try to think it through I get $$ \frac{6 \choose 3 } { 6^6} = \frac{5}{11~664} $$ because $6^6$ is every possibility and the number of ways you can get 3 five's is $6 \choose 3$ which is $20$

If this is wrong why?

In contrast when I use the binomial distribution and I do  $$ {6 \choose 3} (1/6)^3 (5/6)^3 = 0.05358. $$ What am I misinterpreting in the question? I'm tempted to lean to my binomial distribution one being wrong because doesn't binomial imply independent trials one after another and doesn't the fact that all the die I'm throwing all the die at once deny that notion?

Answer 1

You should consider the remaining choices in the first attempt.

There are each $5^3$ choices for the remaining non-$6$ values. Multiplying the answer you got in first attempt with $5^3$ yields the correct answer, which is your answer in the second attempt.

Answer 2

As you already know, you have to use the Bernoulli distribution to calculate the probability that something happens $k$ times in $n$ trials. Let's analyze it to understand why the Bernoulli is not in contradiction with the basic fact that:

"probability = #cases for which the event is realized/#total cases".

If you want exactly $k=3$ times "5" out of $n=6$ times that you roll the dice (or $6$ dices), you have: $$ B(n,k) \, p^k (1-p)^{n-k} = 20 (1/6)^{3} (5/6)^3 = 20 \frac{5^3}{6^6} \approx 0.053 $$ where $B(n,k) = n!/(k!(n-k)!) = 20$ is the binomial coefficient. Clearly $p=1/6$ is the probability of "getting 5" and $1-p = 5/6$ is the probability of "getting something that is not 5"). In this way you are asking that the event with probability $p$ ("getting 5") is realized k times and the event "not 5" is realized exactly $n-k$ times.

Interpretation: we obtained $20 \times 5^3/6^6$, which is almost your first claim, apart from that additional factor $5^3$. The total number of possible events is $6^6$, and in fact it is at the denominator (as expected!)... but you can convince yourself that the number of events "with three 5" is $20 \times 5^3$ and not simply $20$... just try to enumerate few of them: (3,3,3,1,1,1)...(3,3,3,5,5,5) are already 5, then (1,3,3,3,1,1)...(5,3,3,3,5,5) are other 5, and still many of them are missing!

Additional note: The coefficient $B$ tells you that you are NOT interested in the temporal order of the events... or, if you roll $n$ dices at the same time, that you are not assigning any "color" to the dices and asking that "5 is realized in the yellow one". In practice you can think of $B$ as the guy that takes into account for the fact that you do not distinguish between the $n$ dices... or that you don't care about the temporal ordering of the $n$ events. In fact, if you are asking for a precise sequence, let's say $\{(not 5),(not 5),5,5,(not 5),5\}$, then the probability is simply $(1/6)^k (5/6)^{n-k}$.