Birthday Paradox: 4 people What is the probability that two (or more) of them have the same birthday?

Question

Four people around a room. What is the probability that two (or more) of them have the same birthday?

However, I am not sure if my working out assume finds out about the 2 or more part. I am using the pigeon hole method the most suitable approach for answering this question

Here is my working out

$P(A)$ = 2 or more people having the same birthday

This is difficult to find. However, I can use the Pigeon Hole theory.

$P(A')$ = 2 people having the same birthday

Thus $P(A) = 1 - P(A')$

To calculate P(A')

$P(A') = (1/365)^4 * (365*364*363*362) $

$P(A') = 0.9836440875$

$P(A) = 1 - P(A')$

$P(A) = 1 - 0.9836440875$

$P(A) = 0.0163559125$

Thus it approx 1.635% that 2 or more people will have the same birthdays.

EDIT: For spelling errors and changing the value of P(A)

Answer 1

Your calculations are all correct except the percentage is wrong (your multiplication by $100$ is off). However your complement of the event $A$ should read

$A'=$ "None of the people in the room share the same birthday".

Fix these two small issues and it looks good.

Answer 2

We first find the probability that no two persons have the same birthday and then subtract the result from 1.Excluding leap years,there are 365 different birthdays possible.Any person might have any one of the 365 days of the year as a birthday.

A second person may likewise have any one of the 365 birthday: and so on.

Hence the total number of ways is given by $n$=$365^{N}$.

And the number of possible ways for none of the $N$ birthdays to coincide is

$m$=$365.364.....(365-(N-1))$ =$\frac{365!}{365-N!}$

therefore,The probability that no two birthdays coincide is $\frac{m}{n}$=$\frac{365!/(365-N)!}{365^{N}}$ hence the required probability=1-$\frac{m}{n}$=$1-\frac{365!/(365-N)!}{365^{N}}$.....(1)

here the number of people is 4.

So $N$=$4$,putting $N$=$4$ we get the required answer which is

$0.983$