Birth season paradox

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 $1-\frac{3}{8}=\frac{5}{8}.$

There are 4C3 ways of choosing the 3 seasons in which the 3 people have their birthdays. There are 6 ways to arrange the 3 seasons among the 3 people. That is a total of $4*6 = 24$ ways of having birthdays in different seasons. There are $4^3 = 64$ ways of choosing the seasons for the 3 people, if the seasons do not have to be different. Therefore, the probability that the three people have their birthdays in different seasons is $\frac{64-24}{64} = \frac{40}{64} = \frac{5}{8}$

The probability that there is NO match is:

The second is not the same as the first: 3/4

Multiplied by

The third is not the same as the first two: 1/2

So the probability of a match is 1 - 3/8 = 5/8

Let S be the event that at least two people share the same birth season, and denote by S’ the complement of that event. Then

$P(S’) = \dfrac{4\times 3 \times 2}{4 \times 4 \times 4} = \dfrac{3}{8}$ and hence $P(S) = 1 - \dfrac{3}{8} = \dfrac{5}{8} > \dfrac{1}{2}$

There are $4^3 = 64$ ways to assign birth seasons to the three people.

If they have no birth season in common, each season is assigned to one person or "nobody". There are 24 arrangements of $\{\text{person 1}, \text{person 2}, \text{person 3}, \text{nobody}\}$ . This is less than half of the 64 assignments.

Therefore, the majority of assignments has overlapping birth seasons; the probability is $P = \frac{64 - 24}{64} = \frac 5 8.$

It seems that anybody else solved this problem in a straightforward computational way. So, I will write an intelligible Python 3 algorithm to determine the probability (even if it's mathematically easy to calculate it):

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17

from random import randint
def choose():
    birthdays = []
    for person in range(0, 3):
        birthdays.extend([randint(0, 3)])
    return birthdays
def check(dates):
    for season in dates:
        if dates.count(season)!=1:
            return True
    return False
cnt = 0
t = int(input("Number of iterations: "))
for i in range(0, t):
    if check(choose()):
        cnt += 1
print(100*cnt/t)

Let's calculate the probability that there is no collision. When there's 1 person the probability is 1. If we add a second person, then ¾ of the possibilities of his birthday season lead to no collision. So for 2 people to have no collision the probability is 1 • ¾. If we add a third person, then ½ of the possibilities of his birthday lead to no collision. So for 3 people to have no collision the probability is 1 • ¾ • ½ = ⅜ The probability of having a collision = 1 - the probability to not having a collision. Thus the answer is 5/8.

We can calculate how many possibilities there are, so we have 4 seasons and 3 people, that means, according to combinatorial analysis, that there are 4³ possibilities, 64 possibilities. Now we can calculate the favourable cases over that 64 ones. For us, a favourable case would be a possibility which has a birthday repetition. So we can calculate how many cases do NOT have a reiteration, that's just a variation without repetition, according to combinatorial analysis. The unfavourable cases would be then 4•3•2. 24 cases are not favourable. It means that 40 cases are favourable. According to: favourable cases over possible cases, we end up with 40/64, which is the same as writing 5/8, and it's obviously greater than 1/2.

A sampling Python solution

import random

iterations = 0
matches = 0

while iterations < 10000:
    iterations += 1

    A = random.randint(0, 3)
    B = random.randint(0, 3)
    C = random.randint(0, 3)

    if A == B or B == C or C == A:
        matches += 1

print('ratio of matches to iterations', matches / i)

If you make a chart, you can see that there is $\frac{4}{12}$ or $\frac{1}{3}$ chances of them all having the same birthday season and $\frac{2}{3}$ chances of two of them having the same season.

A solution without the need of knowledge in probability:

The first one can be born on any season.

The second could either be born on the first ones season or on the other three

Now we have 2 cases:

1. the first two were born on the same season.

2. the third one have equal amount of seasons that someone was born in and not.

The second option give us 1/2 probability but since option 1 can occur the probability of 2 people to be born on the same season out of 3 people is greater than 1/2.

To calculate the probability of two people sharing birth seasons between 3 people, I will first calculate the probability of all 3 people having different birth seasons. For the first person, we have 4 options to choose from. For the second, we have 3 options to choose from, all four seasons except birth season of the first person. A similar thing can be argued for the third person, who has 2 options to choose from. So, we have $4 × 3 × 2 = 24$ options out of $4^3 = 64$ (four options for each person) this means the probability that the three persons have different birth seasons is $\frac{24} {64} =\frac{3}{8}$ which implies the probability that at least two of the three persons have the same birth season is $1 - \frac{3}{8} = \frac{5}{8}$ which is greater than $\frac{1}{2}$ . Hence, the answer is $\boxed {Yes}$

Maybe I'm missing something here, but is there any reason you can't solve this simply by examining the difference in proportions? It takes 23 people in order to get a greater than 1/2 chance so that's 23 people with a possible 365 days for each. With the seasons, there are 3 people and a possible 4 seasons for each. 23 out of 365 >1/2 so of course 3 out of 4 should be >1/2. It doesn't tell you how much more likely, but that's not what the question asks.

The total number of possible birth season combinations for the three persons in the room is $4*4*4 = 64$ . The number of combinations in which no two persons get the same birthday season is $4*3*2 = 24$ . Therefore, the probability that more than one person has the same birthday season is $1-\frac {24}{64} = 0.625$ , which is clearly greater than $\frac {1}{2}$ .

The probability of at least two people share the birth season can be expressed as:

P(n) = 4!/(4-n)!*4^n

Plugging in n=3, we get 62.5% of the chance.

Hence, the answer is true.