0
$\begingroup$

I am working on a fairly simple problem:

"When rolling a fair dice 12 times, what is the probability of getting 2 of each number"

My immediate instinct was to calculate as follows, in this fashion

$P(\text{two of each}) = P(\text{fixed order}) \times \text{permutations of pairs}= P(\text{Roll 1}=\text{anything}) \times P(\text{Roll 2}=\text{Roll 1}) \times P(\text{Roll 3}\neq \text{Roll 2}) \times P(\text{Roll 4}=\text{Roll 3}) \times\cdots \times P(\text{Roll 11}\neq\text{Roll 1,3,5,7,9}) \times P(\text{Roll 12}=\text{Roll 11}) \times \frac{12!}{2!^6}$

Where $\frac{12!}{2!^6}$ is the number of ways of ordering pairs of 2 within a block of 12.

The resulting calculation is: $P(\text{two of each}) = \frac{6}{6} \cdot \frac{1}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} \cdot \frac{4}{6} \cdot \frac{1}{6} \cdot \frac{3}{6} \cdot \frac{1}{6} \cdot \frac{2}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{12!}{2!^6} = \frac{720}{6^{12}} \cdot \frac{12!}{2!^6} = 2.4755... $ Which is clearly incorrect. This question has made me realise that I am lacking some important intuition when it comes to probability.

My issue is I don't understand what I've done wrong here, why is this product of independent probabilities wrong?

My thoughts: I am somehow overcounting the 6! in the numerator of the sequence by multiplying with 12!, which is erroneous.

Thanks in advance!

Note: I have put a conditional probability tag on this question as I understand that $P(\text{Roll k}=\text{Roll k-1}) $ is conditional on $P(\text{Roll k-1})$, feel free to comment if this isn't appropriate.

$\endgroup$
2
  • 4
    $\begingroup$ Not following your computation. You are correct about the number of ways to get exactly $2$ of each possible value. The probability of getting any exact sequence of $12$ rolls is $\frac 1{6^{12}}$, so just multiply and you get $0.003438286$ $\endgroup$ Commented Nov 23 at 23:06
  • $\begingroup$ Take notice of multinomial distribution. $\endgroup$ Commented Nov 24 at 9:34

3 Answers 3

Reset to default
4
$\begingroup$

You correctly compute the probability of tossing a sequence of the form $AABBCCDDEEFF$ But you don't specify the $6$ distinct values. Thus you are off by a factor of $6!$ as that's the number of ways of specifying the values. Accordingly, your result is off by a factor of $6!$ so you should have arrived at $$\frac 1{6!}\times \frac {720}{6^{12}}\times \frac {12!}{2^6}=.003438\cdots$$

bringing it in line with the more straightforward calculation.

$\endgroup$
5
  • $\begingroup$ Or, more simply, set $A=1, B=2$ and so forth, and conclude the probability is $\frac{12!}{(2!)^6 6^{12}}$. $\endgroup$ Commented Nov 24 at 1:40
  • $\begingroup$ @KyanCheung Well, yes. The OP's method is not the most efficient. $\endgroup$ Commented Nov 24 at 1:46
  • $\begingroup$ This defintely clears it up, but I'm still a little bit confused as to what my additional 6! represents, it is clear to me, now, that the $\frac{12!}{(2!)^6}$ double counts $6!$? Do you mind elaborating on what that 6! actually is here, and how it implicitly drops out through multiplying the individual probabilities? $\endgroup$ Commented Nov 24 at 8:54
  • 2
    $\begingroup$ Not really following. You are (correctly) trying to compute the probability of getting a specified string. You then (sensibly) look at a particular form, the $AABB\cdots$ form and compute the probability of that. However that form does not specify a string until you assign values to the variables. There are $6!$ ways to assign those values. That's the factor. Not sure I answered your question though... $\endgroup$ Commented Nov 24 at 10:17
  • 1
    $\begingroup$ As a general rule: in cases like this, where it's obvious that there is overcounting, work the same problem for a smaller number of cases. $6$ is a lot...why not do $2$? That's small enough to work by hand, the only good strings are $1122, 1212, 1221, 2211, 2121,2211$. And yet it reveals the problem just as well as the case of an ordinary die. $\endgroup$ Commented Nov 24 at 10:19
1
$\begingroup$

You seem to have gotten mixed up at a few points.

For a start, the probability of any fixed order is $(\frac{1}{6})^{12}$. This has to be multiplied by the number of ways to get our pairs.

To find the number of ways to get our pairs, I recommend against multiplying out the ways that one number matches another number and nothing else. That computation gets very messy, very quickly.

Instead pick which two of the $12$ will be $1$. There are $\binom{12}{2}$ ways to do that. Then pick which two of the $10$ remaining will be $2$. There are $\binom{10}{2}$ ways to do that. This leads to: $$ \begin{align} p(\text{two of each} &= (\frac{1}{6})^{12} \binom{12}{2} \binom{10}{2} \binom{8}{2} \binom{6}{2} \binom{4}{2} \binom{2}{2} \\ &= \frac{1*(12*11)*(10*9)*(8*7)*(6*5)*(4*3)*(2*1)}{6^{12}*2*2*2*2*2*2} \\ &= \frac{12!}{6^{12}2^6} \\ &= 0.00343828589391861 \end{align} $$

$\endgroup$
1
  • $\begingroup$ I appreciate your answer. My main question is where did I go wrong with the "one number matches another number and nothing else method"? Because it goes against my intuition. And to extend this question, why is my computation not the probability of any fixed order, what has it over counted? And how can I convert it into the correct solution, as it were. $\endgroup$ Commented Nov 23 at 23:28
0
$\begingroup$

So lulu's answer gave me the guidance I need to understand this, I've marked his as the answer but I thought I'd post a little extra information of what I've realised, in case anyone is having a bad day at the office with this kind of probability.

So, as is usually the case, the Maths is doing exactly what I want it to do: $\frac{6}{6} \cdot \frac{1}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} \cdot \frac{4}{6} \cdot \frac{1}{6} \cdot \frac{3}{6} \cdot \frac{1}{6} \cdot \frac{2}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}$

correctly calculates the probability that we get a sequence $AABBCCDD\cdots NN$ and all permutations of this sequence where we treat each pair as a single unit, therefore, by multiplying with $\frac{12!}{(2!)^6}$, which includes all permutations $AABBCCDD\cdots NN$ as well, for example: $BBAADDCC\cdots ZZ$ etc. We multiplicatively overcount the permutations in which each pair is treated as a single unit, hence, as lulu says, we must divide the product by $6!$. This gives the correct answer of $0.00344$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.