The birthday paradox is a surprising result of probability. Suppose you randomly chose $23$ people and put them in a room. Then there would be a good chance $\big($greater than $\frac{1}{2}\big)$ that two of those people share a birthday (even though there are $365$ days in the year).
What about birth seasons (spring, summer, fall, winter)? Suppose you randomly chose $3$ people and put them in a room. Then is it true that there would be a greater than $\frac{1}{2}$ chance that two of them share a birth season?
Note: Birth seasons do not all have the exact same likelihood. However, their likelihoods are close enough that you can assume they are equal for this problem.
Consider just $1$ person in the room, and then a $2nd$ person enters. The probability that the second person has a different birth season than the first is $\frac{3}{4}.$ The probability that these people share a birth season is therefore $1-\frac{3}{4}=\frac{1}{4}.$
Now suppose that a 3rd person entered the room. He would have a $\frac{2}{4}=\frac{1}{2}$ chance to have a different birth season from the first two.
Now consider the probability of both these events happening (first two people's birth seasons different, $3rd$ person's birth season different as well): $ \frac{3}{4} \times \frac{1}{2}=\frac{3}{8}.$
Since the probability that the first three people's birth seasons are different is $\frac{3}{8},$ the probability that the first three people have at least a pair that share a birth season is $1-\frac{3}{8}=\frac{5}{8}.$