Why should these sixpences be considered as different when calculating the probability? I was trying to solve this problem:
A cook makes three plum puddings for Christmas. 6 silver sixpences are being stirred thoroughly into the pudding mixture before dividing it into three equal portions.
Find the probability that there are at least 2 sixpences in each puddings.
From my opinion, I believe that there is only one way to have at least 2 sixpences in each puddings, and it is have only 2 sixpences in each pudding. In this instance, we just need to find the probability of this arrangement, and I thought it should be:
$$(\frac{1}{3})^{2*3}=(\frac{1}{3})^6$$.
This is because I think that each silver sixpence has a probability of $\frac{1}{3}$ to be stirred in one of the portion.
However, when I looked at the mark scheme, the probability of it turns out to be:$$ \frac{6C2 \times 4C2}{3^6}$$. 
In this instance, I thought the answer has treated the identical silver sixpences to be different, may I know why? And May I know when should I treat the identical things to be different, it is quite confusing.
Thank you so much for you guy’s replies. 
 A: In most word problems such as this in probability, words such as "identical" or "indistinguishable" (which do not appear in your problem statement, by the way) really mean only that when we look at the result in the end to determine "success" or "failure," we do not use any distinguishing marks among the "indistinguishable" objects to determine "success" or "failure".
So even if the six coins were all minted in different years with different (distinguishable) markings, we only care that two ended up in each pudding, not whether two particular coins ended up together.
Another way to look at it is, if the coins really were initially indistinguishable,  if you etch the numbers $1$ through $6$ into the surfaces of the coins, it is not going to change how the coins are stirred nor how many coins end up in each pudding.
So any correct solution had better give the same answer as a correct solution for the case where you can distinguish the coins.

As another example, consider the problem where there are $99$ identical red balls and $1$ black ball in an urn. We mix the balls thoroughly, then reach in without looking and pull out one ball at random. What is the probability that the ball is black?
Reasoning from the number of configurations we can have at the end,
there is one configuration with $1$ red ball outside the urn, $98$ red balls inside, and one black ball inside; there is one configuration with $99$ red balls inside the urn and one black ball outside. Since there are only two configurations, each occurs with probability $\frac12,$ therefore there is $\frac12$ probability that we draw the black ball from the urn.
It should be clear that this is not a correct answer. 

In addition to those objections, your solution to the problem does not even consistently count the outcomes. If you truly cannot count one outcome as different than another because different coins ended up in a particular pudding, then there are not $3^6$ possible outcomes.
The number of distinct ways the coins can be distributed according to that counting is far less than that.
(The outcomes also are not equally likely, but that is another matter.)
A: A solution of the problem might also clarify:

You can think of it as selecting one of the functions $f:\left\{ 1,2,3,4,5,6\right\} \to\left\{ 1,2,3\right\} $.
This "at random" in the sense that the functions have equal probability to be selected.
The domain can be identified with the set of sixpences and the codomain with the set of puddings.
Then to be found is the probability on selecting a function $f$ satisfy
$\left|f^{-1}\left(\left\{ i\right\} \right)\right|=2$ for $i=1,2,3$
The probability on selecting such function equals the number of favourable
outcomes divided by the number of possible outcomes which is $3^{6}$.
First we select $2$ elements that are sent to $1$ (there are $\binom{6}{2}$
possibilities).
After that from the $4$ remaining elements we select $2$ elements
that are sent to $2$ (there are $\binom{4}{2}$ possibilities).
The remaining  are sent to $3$. 
This shows that there are $\binom{6}{2}\binom{4}{2}$ outcomes so
that the probability equals: $$\frac{\binom{6}{2}\binom{4}{2}}{3^{6}}$$
A: We calculate $3^6$, as the total number elementary events by assuming that all the sixpences have equal chances of being inside any of the 3 puddings. This is because,the process of mixing makes the location of each sixpence random.First sixpence can go in any of 3 portions,2nd sixpence in any of the 3 portions and so on. So total number of ways this can happen is  $3\times3\times3\times3\times3\times3$ which is $3^6$. If we put a sticker on all the sixpences to denote the pudding's where it ends up(where A,B and C represents any of the three puddings) along with it's number.AAAAAA happens when all the sixpences goes in first part of the portion,AAAAAB is when the last sixpences happens to be in second portion. It can also be in the third portion  (AAAAAC). Thus the chance of the last sixpence being in any of the three portions gives us three different elementary events.If we were to count all the possible 6 letter words with just three letters we will get $3^6$ words. So if we assume AAAAAC and CAAAAA as same, then the total number of elementary events is defintely not $3^6$ in the first place. If we if ignore AAAAAC because we  already accounted for CAAAAA, then we are assuming that last coin cannot  be in pudding C,because first coin already took the place,which is wrong. For such a situation, we have to modify this question, i will explain this at last part.


We know that since there are only 6 coins,there should be two in all the puddings to have a pair in every pudding.We can select the first pair in $^6c_2$ ways(We are using combination because order of the sixpences does not matter, first and second, is same as second and first). We assume that they went to a pudding. We can choose the next pair in $^4C_2$ ways, let it be inside another pudding. Then the last pair can only be in last pudding.If i had to write down one such random  combination,i would number the coins 1 to 6.There are $^6c_2$ ways in which i can chose a pair numbers(without considering order).I chose (1,6), and wrote down A----A ,then i chose (2,3) from remaining four numbers($^4C_2$ ways) and wrote(ABB--A). Then i can easily fill remaining as ABBCCA. So probability that there are at least 2 sixpences in each puddings $$= \frac{Number\ of\ elementary\ events\ with\ atleast\ 2\ in\ each\ pudding}{Total\ number\ of\ elementary\ events}\\
= \frac{^6c_2\times^4C_2}{3^6}$$


An example question, where there is only one combination with all the three portions having 2 coins.
If the cook was trying to make different kinds of pudding by putting sixpences in it. In how many ways can he do this, if a  set of 3 pudding  is only distinct,if it has more or less number of coins in any of the three puddings?. So the three pudding's can have (6,0,0) coins,where one pudding gets all the coins (same as (0,0,6) as the pudding's are indistinguishable).(5,1,0),(4,2,0),(3,3,0)(3,2,1)and (2,2,2) are the other different kinds of pudding.Here(2,2,2) is the only situation in which the three puddings having at least two sixpences.  This not a question directly related  probability, but Enumerative combinatorics. If the question was,  what is the probability of a (2,2,2) set of pudding, when the sixpences are mixed with the pudding mixture?, the answer is still $\frac{^6c_2\times^4C_2}{3^6}$ as calculated above.

If we actually mix six coins in a pudding mixture and do this experiment a very large number of times (say 1 million times),record the results and take the average.We will get an answer close to the calculated value of probabilty.   
