# 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?

• @b00nheT: No, that's completely wrong. That would be the calculation if you were rolling a die 6 times and calculating the probability of getting the same number all 6 times. Nov 1, 2016 at 17:28
• saw this question on the hot network list and came here expecting to see a question about the probability that the die was actually thrown... Nov 1, 2016 at 18:47

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:

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

• That was my solution as well (as you can probably see in the deleted answer below). However, I had some doubts on it, in particularly with the numerator. Nov 1, 2016 at 11:30

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!

Total number of outcomes:

$$6^{36}=10314424798490535546171949056$$

Number of desired outcomes:

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

Probability:

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

• How did you get desired outcome as that value? Possible for you to add some more details, like how did you get 36! Along with 6!^6? Nov 2, 2016 at 4:09

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

• That was my solution as well (as you can probably see in the deleted answer below). However, I had some doubts on it, in particularly with the numerator. Nov 1, 2016 at 11:29
• @barak manos I'm answering on mobile which doesn't seem to show deleted answers. What doubts do you have? Nov 1, 2016 at 11:32
• Once the author deletes a solution, others will not be able to see it (even though to the author it is visible).
– user348749
Nov 1, 2016 at 11:34
• You're doing LaTex on mobile??? Wow! Anyways... Everybody here's giving the same answer, so it's probably correct, I suppose... Nov 1, 2016 at 11:34
• It is correct, @barakmanos , as is your answer of $\approx0.02589\%$. Nov 1, 2016 at 12:39

A combinatorial interpretation of the multinomial coefficient is the # of ways of putting n distinct objects into b bins, with k_1 objects in the first bin, k_2 objects in the second bin, and so on.

Thus the simplest expression would appear to be

$$\binom{36}{6,6,6,6,6,6}\over 6^{36}$$

• Sixes. Sixes everywhere. Nov 2, 2016 at 19:23

Since you need all six outcomes exactly six times, this can happen in $C(36,6)$ ways for outcome $1$, $C(30,6)$ ways for outcome $2$, $C(24,6)$ ways for outcome $3$, and so on.

Req. Probability$$=\frac{C(36,6).C(30,6).C(24,6).C(18,6).C(12,6).C(6,6)}{6^6}$$

Fun fact, if you toss a fair die $6n$ times, the probability that each number comes up $n$ times is, asymptotically, $(\sqrt{3}/4)(n\pi)^{-2.5}$. This approximation gives, for $n=6$, a approximate probability of 0.028%, and the approximation gets better for larger $n$.

You get this result by taking the actual probability, ${6n\choose n,n,n,n,n,n}(1/6)^{6n}$ and using Sterling's approximation $n! \approx \sqrt{2n\pi}n^ne^{-n}$.

• astounding fact! Nov 2, 2016 at 10:35

The probability of the events you are interested in is simply the number of them divided by the number of all possible outcomes.

We can model a run of $36$ dice rolls as a word: $$w \in W = \{1, \dotsc, 6 \}^{36}$$ the events you are interested in is "each possible dice number shows up 6 times" which are from this set $$A = \{ w \in W \mid \#_1(w) = \#_2(w) = \dotsb = \#_6(w) = 6 \}$$ where $\#_a(w)$ is the number of symbols $a$ in $w$.

The sought probability is $$p = \frac{\lvert A \rvert}{\lvert W \rvert}$$ The denominator is $\lvert W \rvert = 6^{36}$. The nominator is trickier. We can generate $A$ from the permutations of the word $w_0 = 1^6 2^6 3^6 4^6 5^6 6^6$ as $$A = \{ w \in W \mid w = \pi(w_0), \pi \in S_{36} \}$$ where $S_{36}$ is the set of all permutations of $36$ elements. We can not simply use $\lvert S_{36}\rvert = 36!$, as many permutations will map to the same word. So we rather have a partition of $A$ into equivalence classes $$\lvert A \rvert = \lvert \{ [ \pi ] \mid \pi \in S_{36} \} \rvert = \lvert S_{36} \rvert / \lvert [ \pi ] \rvert \,$$ where $\pi_2 \in [ \pi_1 ] \iff \pi_1(w_0) = \pi_2(w_0)$. We have $$\lvert [ \pi ] \rvert = (6!)^6$$ because for any permutation $\pi$ acting on the word $w_0$ we can apply $6!$ permutations to the first $6$ symbols $1$ of $w_0$ and do not change the result of the permutation. The same for the other $5$ parts of the same $6$ symbols.

So we get \begin{align} p &= \frac{36! / (6!)^6}{6^{36}} \\ &= \frac{36!}{(6!)^6 \, 6^{36}} \\ &= \frac{371993326789901217467999448150835200000000}{139314069504000000 \cdot 10314424798490535546171949056} \\ &= \frac{371993326789901217467999448150835200000000}{1436944493270691663365581511922731188224000000} \\ &= \frac{371993326789901217467999448150835200}{1436944493270691663365581511922731188224} \\ &= \frac{10730845456521465275}{41451359947637504606208} \\ &= 0.0002588780071408264 \end{align}

This sort of question is something that could easily be set by someone with no knowledge of nor understanding of probability or asked carelessly while dealing with something else. If you are 8(ish) then the person asking the question probably wont be expecting an answer that uses "factorials" = the "!", of course;

2! = 2 x 1

3! = 3 x 2 x 1

4! = 4 x 3 x 2 x 1

If, however, you are taking a university degree in mathematics then factorials (or even more interesting) are likely to be expected. It is difficult to know what level of accuracy is expected. Sometimes questions like this get asked to create social friction with more accurate answers getting derided for "over-thinking it" or "over-complicating it" or "taking too long" whereas giving a quicker answer might be considered too flippant, or hasty or ill-thought out. If your boss has already given an answer then giving something different might be considered "insubordinate".

Most of the answers in this thread give a 'reasonable' approximation. Some veer towards what most non-mathematicians might find easier to understand whereas others veer towards being far more accurate. The answer that is given in the question is perfectly accurate enough for almost all cases where this question might arise. There are complexities it doesn't address but the use-case might not need those to be taken into account. So it's not a 'wrong' answer as such, just as many other answers in this thread are also not really wrong = it's just that they are all approximations with a varying level of accuracy.

To understand why this is a non-trivial question first think of a simpler case where you get 6 rolls and need to get each number but only once each.

1. On the first throw you need any number 1-6, so it's fairly certain = 1. Some people might say you don't even need to throw the die because a good outcome is so certain - unless the die lands on it's edge, or corner, or disappears into an alternate reality or some other thing we can disregard as being "not a proper throw".
2. On the 2nd throw one of the numbers has become unsuitable because it was already thrown in the first throw. So you've got 5/6 chance of getting something suitable.
3. On the 3rd throw there are now 2 numbers that would disqualify you. So that leaves 4 'good' results = 4/6

So it's only the 6th throw where you have a 1/6 chance of getting the right number. I have a nagging feeling about 2 more complexities that make this not quite true but it's a good approximation.

Ok, so now another slightly less simple experiment, but still simpler than the original question. Say 36 throws but you need to get 6 results that are 6.

1. On the 1st throw there is an approximately 6/36 chance of getting a 6. Lets say this does happen, that you do get a 6 on the 1st throw ...
2. On the 2nd throw (given that you got a 6 already on the first throw) the chance of getting a 6 would be approximately 5/35. Lets say this throw is also a 6.
3. Ok, so given the first 2 throws were 6 then we only need 4 more 6s but we only have 34 more throws so the chance of getting a 6 is 4/34.

But doesn't this contradict the fact that each time a die is rolled it has a 1/6 chance of getting a particular number? (assuming a perfect world for dice-rolling) Not really because we are looking at the probability of all 36 dice rolls. Not just the first throw. So we have to consider all the 36 rolls as being 'one' "event".

In the last experiment we could see a factorial pattern forming;

6/36 x 5/35 x 4/34

So on the top its 6 x 5 x 4 x 3 x 2 x 1, which is written "6!" as you probably know.

I think the permutations (or is it combinations) work out pretty well. I think the extra complexities, to do with which order the 6s appear in, obligingly cancel each other out.

So all in all i think the initial answer was close enough for most cases. It's not really a case of being "right or wrong" because all answers have to make some assumptions such as generally ignoring "edge cases". I think the more accurate answers tend towards it being more unlikely.

I really like the answers that use factorials but my fav is the one that starts "fun fact". I'm not keen on results shown as percentages but it gets rid of a couple of zeroes and i guess it does make it more human-readable.

Good luck and regards from Tom :)