# Why do you need to distinguish between apparently identical objects in probability?

In the math class I taught today I was asked a question, and I was unable to give a good answer.

The problem was as follows: A certain factory produces throat tablets. In each pack, there are from 48 to 52 tablets, with a given probability distribution in a table (I don't remember the exact probabilities; it was far from uniform). If we take two of these packs, what is the probability that there are 102 tablets in total?

The provided solution (as well as how I would've solved it myself) went on to say that there are three ways that this can happen:

• The first box has 50 and the second box has 52
• The first box has 51 and the second box has 51
• The first box has 52 and the second box has 50

For each of these three probabilities, the probability is calculated by multiplying the relevant probabilities. Then the three results are added up and that's the answer.

One student asked me why there are three cases and not two. Why are the first and third cases separate? Why is there a first box and a second box, and not just two boxes?

Another, more standard, example is when working with the sum of two thrown dice. Same thing here: Why can we declare that there is a first die and a second die (or a blue die and a red die)?

I know that we can "pretend to change" the problem text to say that we pick first one box and count the tablets, and then pick the second box and count the tablets. But the problem as stated doesn't say that. According to the statement the two boxes are entirely equal. How can I justify breaking this symmetry? (Apart form saying "See, it gives the answer you're supposed to get.")

Note that this is not about the similar question in combinatorics. When asking "how many ways are there to select three students from the class to form a committee", or something like that, I think many of them have understood (and I feel more confident in how to explain) when order matters and when it doesn't in such questions. And I am similarily confident in the probability questions that spring directly out form such scenarios ("What is the probability that Alice and Bob are both chosen as part of the committee?").

• Returning back to the title of the question, the problem is not actually about order but about the sheer number of outcomes, and the reply below with dice of different colors explains it nicely. – Rusty Core Sep 6 '19 at 17:42
• There isn't a right number of cases. But there is a set of cases that it is easier to calculate the correct probabilities for. – Jessica B Sep 12 '19 at 6:43

The answers provided here so far give lots of good tips but I think they're not addressing a key part of the question, which is "why do we need to count two events (50,52) and (52,50), instead of one event (50,52)?"

The answer is that you can do it either way. The textbook way is counting $$P(X=50 \cap Y=52) + P(X=52 \cap Y=50) + P(X=51 \cap Y=51) = P(X=50)P(Y=52) + P(X=52)P(Y=50) + P(X=51)P(Y=51) = P(X=50)P(X=52)\times 2 + P(X=50)^2$$ This approach assumes that the boxes are iid to express the task in terms of univariate distribution of a single box.

Instead, the student's alternative is $$P(X+Y = 102) = P(one~box=50, another=52) + P(both=51)$$ This is entirely valid so far, and if you had a distribution defining probabilities of these events, you could continue. But the point of such tasks is usually that you are given a univariate distribution, or need to define it from a text-based setting ("fair dice"). It is this step that introduces the factor of 2, because there are two elementary outcomes that result in $$(one~box=50, another=52)$$.
I suppose this could be demonstrated more clearly by a task where an initial part provides the joint distribution, but a second part requires to do the same but derive the distribution first.

• Perhaps it's useful to explicitly analyze the "student's alternative" further - split 𝑃(𝑜𝑛𝑒 𝑏𝑜𝑥=50,𝑎𝑛𝑜𝑡ℎ𝑒𝑟=52) into P(𝑜𝑛𝑒 𝑏𝑜𝑥=50)*P(𝑎𝑛𝑜𝑡ℎ𝑒𝑟=52 | 𝑜𝑛𝑒 𝑏𝑜𝑥=50); the "two elementary outcomes" result appears as a consequence of the fact that P(𝑜𝑛𝑒 𝑏𝑜𝑥=50) is bigger than P(X=50) because there are two boxes. – Peteris Sep 7 '19 at 15:05

This is a very good question. The issue comes up frequently. I explain this using a toy model: throw two regular six-sided die. What is the probability that the sum is 3? With some physical modeling, you can become convinced that the answer is 2/36=5.5%, This corresponds to the two possibilities 1+2 and 2+1.

But why are these two possibilities distinct? We could run the experiment where one dice is red, the other blue. Now it becomes clear why {red->1,blue->2} is not the same outcome as {blue->1,red->2}.

With the boxes of throat tablets in the original post, the boxes are different because they are composed of different atoms, which obey Fermi-Dirac statistics, not Bose-Einstein!

• Atoms may obey either Fermi-Dirac or Bose-Einstein statistics, depending on their total spin. In a quantum state with 10 electrons over here and 11 electrons over there, the state will be antisymmetric under interchange of any two of the 21 electrons. There is, however, no necessary connection of this state and a state with 11 electrons over here and 10 electrons over there. The same will be true if I replace "electron" with "tritium atom" (which is a boson) and replace "antisymmetric" with "symmetric". – Will Orrick Sep 7 '19 at 21:05
• Importantly, merely coloring two existing, otherwise indistinguishable dice does not influence their probabilistic characteristics, so any result observed when they are distinctly colored must hold when they're not. – BallpointBen Sep 8 '19 at 18:19
• +1 for even mentioning F-D versus B-E statistics – kcrisman Sep 12 '19 at 1:23
• @kcrisman I'm really trying to understand what the relevance of bosons vs. fermions is to a discussion of classical objects. In quantum mechanics it is true of both bosons and fermions that particles of the same species are truly identical, so that the wave function describing a multi-particle system must have a symmetry property reflecting this identicality. This can give rise to striking phenomena that we would not expect to see for classical objects, such as Bose-Einstein condensation. The distinction between the two classes of particle is in the nature of the symmetry: for bosons the... – Will Orrick Sep 12 '19 at 15:26
• @kcrisman Thank you for your reply. I've long thought it would be fun to include some calculations related to Bose-Einstein condensates in a probability course as an approachable example demonstrating the dramatic effects that true indistinguishability can have. Unfortunately I think there's just too much extraneous background that students would have to learn (and probably take on faith) to make this practical in an elementary course. – Will Orrick Sep 13 '19 at 14:09

I don't think I can improve on juod's excellent answer, but there are a few points worth elaborating.

First, physical objects are always distinguishable. (I am here ignoring the phenomenon of identical particles in quantum mechanics, the issue there being that we really have to rethink the concept of physical object when talking about the electrons in a two-electron quantum state or the photons in a two-photon quantum state.) So it's not actually "pretending" when we artificially introduce a distinction between the "first pack" and the "second pack"; a natural distinction already exists between physically different ojects.

Second, one may, if one chooses, ignore the distinction between the two packs by forgetting which is which, or by not recording this information in the first place. This requires that one frame the situation in terms of sets of two objects, and that one modify the mathematical description accordingly. Making this choice will not affect the answer you get.

Third, there is an implicit assumption in the problem as stated that ought to be made explicit. This assumption is that the joint probabilities, that is, the probabilities associated with various configurations of two packs, can be computed from the one-pack probabilities. This really is an assumption, one which may or may not hold, depending on the details of the production process and the method of sampling. For example, suppose that the the packaging equipment is such that if one pack is overfull, the next pack produced will have a higher probability of also being overfull. And suppose that the sampling method is to examine two successive packs coming off the assembly line. In this situation the probability of two overfull packs is not simply related to the probability of single packs being overfull. In this problem, you are meant to assume that this isn't an issue, or that the sampling is done in such a way that the problem is ameliorated.

Fourth, how is it that even the single-pack probabilities are assigned? In principle one might deduce these probabilities from a detailed inspection of the packaging equipment, but in practice the probabilities will be empirically determined. Any real production process will only ever produce finitely many packs. If the question is concerned with the packs produced in a particular time period, the probabilities boil down to ratios, $$p_{48}=N_{48}/N$$, $$p_{49}=N_{49}/N$$, and so on, where $$N=N_{48}+N_{49}+\ldots+N_{52}$$ and $$N_i$$ is the number of packs produced with $$i$$ tablets.

Let's now do the computation with and without distinguishing the packs.

(with distinguishing) The probability the pack I open first has 52 tablets and the pack I open second has 50 tablets is $$(N_{52}/N)(N_{50}/(N-1))$$, and similarly for the other configurations. The total probability is $$\Pr(\text{102 tablets})=\frac{N_{52}N_{50}+N_{51}(N_{51}-1)+N_{50}N_{52}}{N(N-1)}\approx p_{52}p_{50}+p_{51}^2+p_{50}p_{52},$$ where the approximation is valid when the numbers of packs are large.

(without distinguishing) The probability that the set of two packs contains a pack with 52 and a pack with 50 tablets is $$\frac{\binom{N_{52}}{1}\binom{N_{50}}{1}}{\binom{N}{2}}=\frac{N_{52}N_{50}}{N(N-1)/2},$$ while the probability that the set contains two packs of 51 tablets is $$\frac{\binom{N_{51}}{2}}{\binom{N}{2}}=\frac{N_{51}(N_{51}-1)/2}{N(N-1)/2}.$$ Combining gives an answer algebraically equivalent to the one before.

Your student's proposed method (combining the first and third cases; not distinguishing a first and second pack) gets the probability for case 2 right essentially "by accident": there is a cancellation of two factors of $$1/2$$ with the result that the joint probability of getting two packs of 51 equals the product of two one-pack probabilities. No similar "accident" occurs in the case of one pack of 50 and one pack of 52: there is a factor of $$1/2$$ in the denominator, but no corresponding factor of $$1/2$$ in the numerator to cancel it.

The succinct answer is that there are two ways to choose $$50$$ and $$52$$ - if two boxes are chosen one could have $$50$$ and other $$52$$ or vice-versa - but this will be more apparent to a student if you write out the space of all $$25$$ possible outcomes as a list of ordered pairs.

"Indistinguishable" does not mean "identical", just as "isomorphic" does not mean "equal".

Each pick of a box is the same as choosing a number from $$\{1, 2, 3, 4, 5\}$$ (with some probabilities). So the problem is equivalent to the problem: what is the probability that two numbers chosen (repetition allowed) from $$\{1, 2, 3, 4, 5\}$$ sum to $$8$$? Or equivalent to: what is the probability that the digits of a number with digits from $$\{1, 2, 3, 4, 5\}$$ sum to $$8$$? But two indistinguishable boxes, each containing $$\{1, 2, 3, 4, 5\}$$ are still not the same box - there are two boxes sitting on the table. The first digit chosen and second digit chosen are different digits. I could have called them the second digit and the first digit - that's a question of how I label them - but they are different.

The terminology "indistinguishable" is misleading - it does not mean that one cannot physically distinguish the boxes - it means only that their contents are identified in some established way ("isomorphism").

The key thing here is really exchangeability. Let $$X^{k}$$ be the set of $$k$$-tuples of elements of the set $$S$$. A probability distribution pn $$X^{k}$$ is exchangeable if it is invariant under permutations. In particular, any distribution that arises from $$k$$ independent samples from $$X$$ is exchangeable. This exchangeability is the key thing. I have a canonical way of identifying the faces of two different dice, but they are still two different dice, and I can tell them apart. There are two ways to put them in order and I can call one first or second as I like - and nothing about probabilities depends on this labeling. However I decide to do it, I can get $$3$$ on one and $$5$$ on the other - and there are two ways this can happen because while indistinguishable, the dice are different.

For convincing a student one can simulate the exercise - just write the numbers on pieces of paper - and make two selections with replacement - or better fill two shoeboxes with numbered papers and choose one from each. Better yet, one can do it on a computer.

• Just subtracting 47 from the contents of each box doesn't really do much in my opinion. The question remains: why aren't there 15 possible choices, with two of them adding to 8? And you say that there are two ways to choose $3+5 = 5+3$, but then you say "order" is a red herring. Which is it? (I'm trying to be difficult here, because I know that some of my students will be. As they should, in my opinion.) – Arthur Sep 6 '19 at 14:02
• @Arthur: I also have a lot of trouble with these kinds of problems, and unfortunately (for me) they arise quite often in my "day-job" work. Here's something that might help, at least in this case. We are assuming the boxes are distinguishable. So instead of "two boxes", think of this as Box A and Box B, and one number is associated with Box A and another (possibly the same) number is associated with Box B. To count the possible outcomes, count ordered pairs of numbers such that the first coordinate is a number associated with Box A and the second coordinate is a number associated with Box B. – Dave L Renfro Sep 6 '19 at 14:33
• @DaveLRenfro But the boxes aren't distinguishable, according to the problem statement. How can I justify to my students that we are allowed to break that symmetry? – Arthur Sep 6 '19 at 14:36
• @Arthur: They are not "distinguishable" in the sense that they have the same contents, but they are "distinguishable" in the sense that one can label them and tell them apart, although it won't matter which gets called Alice and which Bob. – Dan Fox Sep 6 '19 at 14:40
• I don't think this addresses the question. You said "write out the space of all 25 possible outcomes as a list of ordered pairs" as though that was clear, but in fact that is the core of the question: why ordered pairs, and not 2-element subsets? – Chris Cunningham Sep 6 '19 at 14:58

I don't know from what principles you're teaching probability, but the answer I would give is the following:

When calculating probabilities, it is important to remember the basic principles. We have a set of possible outcomes $$S$$ and an event $$E \subseteq S$$. The probability that an outcome in our event is chosen is precisely $$\frac{E}{S}$$, *given that all outcomes in $$S$$ are equiprobable.

Now when your student is unclear about something, ask them to write out all the possible outcomes. Depending on whether they choose to consider order they may have fewer or more. Then ask them to write out all possible outcomes in your event. Depending on whether they considered order they will have $$2$$ or $$3$$. Finally, ask them to tell you why they believe all events are equiprobable. If they chose not to consider order, they're in trouble.

The advantage of this system is that it gives them a hard rule that they can always fall back on.

Note: I do not necessarily advocate for the usage of the terminology I suggest here. If the idea of sets is new to them, then maybe explaining this with Venn diagrams is more helpful.

You didn't disclose the level you teach. When I first ran into this issue (i.e. the need to explain this), I went to the example...

You have 2 coins. There are 3 possible outcomes, 2 heads, One Head One Tail, 2 Tails. Are they each equally likely, each 1 in 3? I then walk them through the physical experiment. Even though we flip two coins at once (as another member stated) the coins are different, and a 'tree diagram' would show that the first flip, whether H or T, eliminates HH or TT, but still leaves us the 50/50 chance of one of each.

The 2 coin model makes both an experiment, or diagrams on the board, easy to tackle. From that, the expansion to higher numbers is easier to explain.

You could explain that the first box always goes to Alice, while the second goes to Bob. Then it becomes obvious that Alice getting 50 and Bob getting 52 is different then Alice getting 52 and Bob getting 50, so much so that one of them could die if they get the wrong amount.

On the other hand, Alice and Bob can both get 51 in only one possible way, and the effects on their throats remain the same even if the packages get switched by mistake in transport. That's why 51-51 isn't counted twice.

Challenge him with the following: let‘s think it through with a soccer match. In how many ways can the match end, such that the sum of the goals is 3? According to his logic only two possible ways (1, 2) or (0, 3) - I‘m pretty sure the fans won‘t agree.

Same goes for the colored dice - how many ways can they sum up to 4? Thus what’s the probability for those 2 dice to sum up to 4? Does he think that removing the color changes the probabilty to get 4?

I think there are a lot of good answers here, but one point of view that might convince some students is to use a simulation - not of the toy problems e.g. with dice suggested, but with this one.

Apologies in advance for my terrible R code - live version - here I assume a uniform discrete distribution on 48 to 52.

counter = 0
for (i in 1:100000)
{
if ( sum(ceiling(runif(2, min=47, max=52))) == 102)
{ counter = counter +1 }
}
counter/100000


Occasionally I get .122 or .118, but usually it's very, very close to 12 percent, which is indeed $$3\times .2^2$$. Add another zero to make it better (but then it actually takes a few seconds).

(This sort of argument seems to help some people accept the correct solution to the Monty Hall problem, for what it's worth.)

• I hinted at this in my answer, but with actual coins. Thinking about random number generators, I felt it really depended on the student's level. I can imagine my example showing .26 as a result, for HH, and the students still not quite understanding how that supports my statement the odds are 1 in 4. (+1, by the way) – JTP - Apologise to Monica Sep 12 '19 at 9:45
• Yes, the sophistication of the student definitely will impact the usefulness of this. If it's just a black box that spits out a "nearly correct" answer to them, it's not so great. – kcrisman Sep 12 '19 at 19:50