Harvard Stat 110 Strategic Practice 2, Fall 2011 - Inclusion Exclusion - Problem 1.1 Harvard Stat 110 Strategic Practice 2, Fall 2011 - Inclusion Exclusion - Problem 1.1
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.
I tried to solve it using another method (which came into my mind at that moment) but I must be doing a mistake. Maybe someone could help me finding the error and also explaining WHY I make the error, so I can avoid it in the future.
I tried to apply the naive definition. So, I have $7$ people and $4$ seasons to choose.
1st person can have $4$ picks, the 2nd $4$ picks, etc etc. All are independent, so I have a total of $4^7$ possibilities.
The number of favourable outcomes are when from the $7$ people, $4$ have each Spring, Summer, Fall, Winter and the other $3$ might get any choice.
From $7$ people, I chose $4$ to fill Spring, Summer, Fall, Winter and the other $3$ can have whatever choice.
So, result should be
$$\binom{7}{4} \cdot \frac{4^3}{4^7}$$
which yields $0.546$ which is clearly different from the practice answer of $0.513$.
Could somebody, please, point out what I am doing wrong?
Thank you!

There are multiple mistakes here. First, once you choose the four people, you neglected to designate how the four different seasons are assigned to them, since in the denominator you are counting all possible sequences of seven season assignments. It should be $7 \cdot 6 \cdot 5 \cdot 4$ rather than $\binom{7}{4}$. However, this leads to a nonsensical probability larger than $1$, which I explain below.

You are counting some outcomes multiple times. For example, if the seven people are ordered and their picks are

Fall, Spring, Summer, Summer, Winter, Summer, Summer

in that order,
then this is counted four times too many in your computation, because there are four different ways this could have arisen from choosing the set of "four different seasons" people first, before letting the rest be summer. Specifically, it could have been the 1st, 2nd, 3rd, and 5th people that were chosen first, or it could have been the 1st, 2nd, 4th, and 5th people that were chosen first.
The difficulty is that there isn't a uniform way to correct for this overcounting (like just dividing by 4 or something). For instance, for

Fall, Spring, Summer, Summer, Winter, Winter, Winter

this would be counted six multiple times due to the different ways this could have arisen from a selection of "four different seasons" people first.
If you want to handle this case by case, you have to go through all possible frequencies of exactly how many occurrences of each season there are, which is a bit difficult to compute. Admittedly, the standard approach (inclusion-exclusion) also doesn't have a simple expression either, but is slightly easier to write down.

The complement event is "at least one season does not appear." Letting $A_1, A_2, A_3, A_4$ be the events that "season $i$ does not appear" respectively, we want $1-P(A_1 \cup A_2 \cup A_3 \cup A_4)$. Using inclusion-exclusion, this is
$$1-\left(4 (3/4)^7 - \binom{4}{2} (2/4)^7 + \binom{4}{3} (1/4)^7\right) \approx 0.5126953125 $$
which matches BruceET's simulation.
