In 'An Introduction to Probability Theory and Applications' by W. Feller I encountered this apparently innocuous problem.
A throw of twelve dice can result in $6^{12}$ different outcomes, to all of which we attribute equal probabilities. The event that each face appears twice can occur in as many ways as twelve dice can be arranged in six groups of two each. Hence the probability of the event is $\displaystyle \frac{12!}{2^{6}6^{12}}=0.003438$.
The reasoning is that by doing that you're grouping two 1's, two 2's, ..., two 6's (each group using one partition in the multinomial) and the result is the number of different partitions that can be found with that particular characteristic. However, I had doubts about that answer. To understand better that problem, I did a simpler example with $4$ dice instead of $12$ (in this case, the event is the number of ways in which two faces appear twice).
Using the same result I get as the probability $\displaystyle \frac{4!}{2^{2}6^{4}}=$ 0.46 $0.0046$. Then, to see if that's true I ran a little simulation of this case in Mathematica:
dice = Table[Table[Random[Integer, {1, 6}], {i, 1, 4}],
{j, 1, 1000000}]; i=0;
f[{a_, b_, c_, d_}] := Which[a === b && c === d && a != c,
i++, a === c && b === d && a != b,
i++, a === d && b === c && a != b, i++;];
Map[f, dice, {1}];
After $1000000$ steps I got $69687$ cases in which two faces appear twice. This is equivalent to a probability of $0.069687$. Far smaller than what I expected based on the calculation above.
Since this latter example is much more manageable than the one with twelve dice, I did the following
With four dice we have the following:
- Partition $r_{1}$ contains the first and second dice and partition $r_{2}$ contains the third and fourth dice.
- Partition $r_{1}$ contains the second and fourth dice and partition $r_{2}$ contains the first and fourth third.
- Partition $r_{1}$ contains the first and fourth dice and partition $r_{2}$ contains the second and third dice.
- Partition $r_{2}$ contains the first and second dice and partition $r_{1}$ contains the third and fourth dice.
- Partition $r_{2}$ contains the second and fourth dice and partition $r_{1}$ contains the first and fourth third.
- Partition $r_{2}$ contains the first and fourth dice and partition $r_{1}$ contains the second and third dice.
For each case, we can have $30$ outcomes in which two faces appear two times. For example, for the first case we have $1122, 1133, 1144, 1 155, 1166, 2211, 2233, 2244,2255, 2266,... 6611, 6622, 6633, 6644, 6655$. However, such outcomes are repeated twice (particularly, the outcomes of case 1 repeat the outcomes of case 4, etc). Therefore, the number of different outcomes which produce two faces appearing two times are $\frac{6}{2}6*5=90$. Since there are $6^{4}$ outcomes, we have a probability of $\frac{90}{6^{4}}=0.0694444$, which is the result that produces the simulation in Mathematica.
Is it wrong the first reasoning? If so, is there a general approach to use the multinomial coefficient to solve this kind of problems. For instance, it appears that this only happens for $r_{1}=r_{2}=r_{k}$. Otherwise, there are not repeated outcomes.