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!

Comment: Based in simulation of a million performances of this experiment, which should give 2 or 3 place accuracy, I'm betting on the answer $0.513.$
The simulated result is $0.5135 \pm 0.001.$
set.seed(536)
nr.s = replicate(10^6,
           length(unique(sample(1:4,7,rep=T))))
mean(nr.s == 4)
[1] 0.513456       # aprx P(All seasons)
2*sd(nr.s == 4)/1000
[1] 0.0009996383   # aprx 95% margin of sim error

Note: Your method in clever and comes close, but it's approximate.
