# Why does "order matter" when calculating the probability that no people out of seven were born in winter?

SUMMARY

Here's a problem from Harvard's Stats 110 course.

For a group of 7 people, find the probability that all 4 seasons (winter, spring, summer, fall) occur at least once each among their birthdays, assuming that all seasons are equally likely.

The problem and its solution can be found here.

The hard part for me comes when calculating the probability that no people out of seven were born in winter. I don't understand why "order matters" here. (I understand the inclusion-exclusion part of the main problem, just not the "order matters" part.)

WHAT I TRIED

As one part of this problem, I need to calculate the probability that none of the seven people has a birthday in the winter. I tried to calculate this by considering the people as indistinguishable and lumping them into 4 different categories: birthdays in winter, spring, fall, or summer. Thinking like this, I used the "stars and bars" formula to calculate the probability that none of the birthdays fell in the winter category. This can be calculated as

$$P(A) = \dbinom{7 + 3 - 1}{3} / \dbinom{7+4-1}{4}$$

WHAT HARVARD DID

Harvard says this is wrong and that since order matters, the calculation is (the much easier)

$$P(A) = (3/4)^7$$

It seems like it could go either way, and I can actually model either in R. So what am I doing wrong?

• Your approach assumes that every way of choosing how many people have birthdays in each season is equally likely. This is not the same as assuming that each person is equally likely to have his birthday in each season, and indeed the two answers aren't compatible. For instance, if you flip 50 coins, each choice for how many heads isn't equally likely... $N=25$ heads is much more likely than $N=5$ heads. Jun 9, 2015 at 21:38

The "Stars and bars" formula counts the number of ways to arrange a list of stars and bars. It gives the wrong answer for your problem because these arrangements are not equally probable, so you can't simply divide the number of "favorable" arrangements by the total possible number of arrangements.

For example, when assigning birth seasons (call them A, B, C, D) to four people, the arrangement ****||| is less likely than the arrangement *|*|*|* . Why? Think of lining up these four people in a row and having each person pick a season. The first arrangement of stars and bars corresponds to AAAA, while the second arrangement corresponds to ABCD, ACBD, ADBC, ADCB, and many more, for a total of $4!$ possibilities. This is the sense in which order matters---the $4!$ possibilities are distinct, and as a whole don't have the same prob as the the single possibility AAAA.

The various Stars and Bars possibilities are not all equally likely. To take a simpler example, toss a fair coin $10$ times. The Stars and Bars possibilities "(Number of heads, Number of tails)" are $(0,10)$, $(1,9)$, and so on up to $(10,0)$. There are $11$ such possibilities.

However, $10$ straight tails has probability $\frac{1}{1024}$, while $9$ tails and a head (somewhere) has probability $\frac{10}{1024}$. The "middle" case $(5,5)$ is the most likely.

Using a probability model that assigned equal probabilities to the $11$ Stars and Bars possibilities would give results contrary to the actual behaviour of tossed fair coins.

Remark: There are situations in which one gets good agreement with reality by assuming the Stars and Bars possibilities are all equally likely. For an example, please see Bose-Einstein Statistics.

The reason why "order matters" is that you get a different answer. In probability problems where order does not matter, you can do the calculations under the assumption that order does matter, and then find out that by counting all possible sequences of the $7$ people you have simply multiplied both the numerator and denominator of each fraction by a factor of $7!$. Often in such cases you could have treated the $7$ people as an unordered set and arrived at the same answer with far less work.

You're not going to be able to do that here. The fact that $7!$ divides neither $3^7$ nor $4^7$ is a clue.