I know that such questions may be discouraged (they are on physics.SE), feel free to close.

If we have 11 students and 8 identical chocolates, find the probability that a particular student gets 5 chocolates.

Looks pretty tame. Until I decided to solve it in two different ways.

In probablity problems, we can always 'label' identical objects and get the same answer.

Well, here the above principle seems to not work.

Solving the problem by considering identical chocolates: Here, I take 8 identical chocolates and 10 identical partitions. Shuffling them will get me the total number of ways to distribute the chocolates amongst 11 people ($\binom{18}{8}$). Then I can remove 5 chocolates and one student, I'll get the same thing. The final probability comes to be $\frac{\binom{12}{3}}{\binom{18}{8}}$. I can elaborate how I got here if required.

Solving with labelled chocolates: First, for each chocolate, choose a student to give it to. That gives us $11^8$ total ways. Now, choose five chocolates, give them to the 'special' student, and distribute the rest in the same way. We get $\frac{\binom{8}{5}10^3}{11^8}$

Due to the large power of 11, I doubt that the two answers are equal. So have I made a mistake (double-counting, etc), or are the answers actually supposed to be different? If it's the latter, please explain why the 'labellimg' trchnique doesn't work.


We have $11$ students, one of them George. We find the probability that George gets exactly $5$ chocolates, on the assumption that the chocolates are given out to students, one at a time, independently and "at random." A success, at least from George's point of view, is that George gets the chocolate.

The probability of success on a given trial is $1/11$, the probability of failure is $10/11$. The probability of $5$ successes )and therefore $3$ failures) in $8$ trials is $$\binom{8}{5}\left(\frac{1}{11}\right)^5 \left(\frac{10}{11}\right)^3.$$ That is precisely one of your solutions.

If the distribution of chocolates happens as described in the solution above, then the other solution is not correct. It is easiest to see that under that probability model, the answer is not correct. If we express the answer as a reduced fraction, it will have a high power of $11$ in the denominator. Your first answer, when expressed as a reduced fraction, does not have an $11$ in the denominator. (An $11$ comes from the top of the expression you gave, and one from the bottom; they cancel.)

As @joriki has written in his solution, the probability model was left unmentioned by the question setter. That is a serious error. For then, in principle, we can choose any model we like. Assume, for example, that the chocolate giver promised George that he would get $5$ chocolates. Then the required probability is $1$. Or else assume "randomness," except that someone must get $5$. then the probability is $1/11$. But presumably the question setter had some default model in mind. The most likely default assumption is the one described in my solution above. It is the same as the model used in your second solution.

Now we need to worry about why the other solution is not correct under our probability model. That is a little difficult, since I do not fully understand that solution. It looks like a variant of a "stars and bars" argument, though the particular numerical details are not explained. Stars and bars arguments are very good for counting the number of ways that $k$ identical objects can be distributed into $n$ bins. But that is often not useful for probability calculations, since the ways are in general not equally likely.

Imagine, in our stars and bars picture, that George will get all the chocolates up to the first separator. We can count the number $S$ of ways in which George can get $5$. We can also count the total number $N$ of ways to distribute the chocolates. The answer to our probability problem, under the default assumption, is not $S/N$, since the $N$ ways are not all equally likely. We are so accustomed to equally likely situations that it is easy to forget to check whether our sample space consists of equally likely outcomes.

  • $\begingroup$ What makes the one distribution correct and the other incorrect? The problem statement doesn't say anything about the distribution, and it doesn't say that the chocolates are given out "at random" one at a time. While this interpretation seems to make more real-world sense than a somewhat artificial uniform distribution over all partitions, the word "identical" in the problem statement doesn't make any sense under this interpretation. $\endgroup$ – joriki Mar 3 '12 at 8:09
  • $\begingroup$ In my view this answer as it stands is rather misleading, and it's unfortunate that it got accepted. It's a very common problem that people think that saying something like "we have $n$ Xs and $m$ Ys" has anything to do with probabilities, and you're reinforcing that misconception when you label one solution of the ill-posed problem as correct and another as incorrect instead of pointing out what's missing in the problem statement. All the more so since the word "identical" in the problem doesn't support the interpretation you label as the correct one. $\endgroup$ – joriki Mar 3 '12 at 8:30
  • $\begingroup$ joriki: Good point. I edited the assertion so that it now says that under the probability model that led to the binomial, the solution is not correct. The "identical" part is indeed irrelevant, but for many students there is an important distinction, since they classify problems according to methods or formulas. It looks to me as if the OP saw intuitively that the "identical method" and "distinct method" should yield the same answer, and was puzzled when they didn't. $\endgroup$ – André Nicolas Mar 3 '12 at 8:33
  • $\begingroup$ Your edit hasn't really improved things, I think, since you're still talking about "correct" and "not correct", which implies that the problem is well-posed and there's a single correct answer. The main point of the answer should be that the problem is ill-posed and the answer is undefined; you could then go on to speculate that the intended interpretation may, despite the word "identical" indicating otherwise, have been the one where all assignments of chocolates are equally likely, since this distribution is more likely to occur in a real-world situation. $\endgroup$ – joriki Mar 3 '12 at 8:37
  • $\begingroup$ @joriki: I have re-edited, mentioning the model repeatedly. As happens all too often, the problem is indeed not well-posed. But the ill-poser ordinarily has a "method" implicitly in mind, namely the technique of the chapter in which the problem occurs! $\endgroup$ – André Nicolas Mar 3 '12 at 8:58

The problem isn't in your correct solutions but in the problem statement. The problem is ill-posed; the existence of some students and some chocolates doesn't imply any probabilities. Often when people forget to specify a distribution, one can guess that they had a uniform distribution in mind and forgot to mention it. In this case, however, even assuming the authors had a uniform distribution in mind, it's not clear which uniform distribution this might be. As your different results show, the probability of a particular student getting $5$ chocolates is different if the distribution is uniform over all possible assignments of chocolates to students than if the distribution is uniform over all possible assignments of numbers of chocolates to students.

You can easily see this from a smaller example. (Small examples are often helpful in understanding things in combinatorics.) If there are two students and two chocolates, then there are four possible assignments of chocolates to students and three possible assignments of numbers of chocolates to students. Without any specification of a distribution, there's no reason to believe that the distribution should be uniform over either of these two sets, so there's no such thing as "the probability that a particular student gets $n$ chocolates" without such a specification.


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