TITLE: How Can the Birthday Problem be solved directly
QUESTION [5 upvotes]: How would one go about solving a variant of the birthday problem given below
"What is the probability of at least two people having the same birthday in a group of 23"
I know the answer is just above 0.5 but I am interested in how it is solved when one computes it by taking probabilities of two people having the same birthday, then of three people having the same birthday and so on rather solving it by finding the probability of people not having the same birthday.

REPLY [5 votes]: Solving it directly means enumerating and counting all the possibilities.  That's the reason the complement is easier -- it's easy to describe and compute.
Starting with three people, Alice, Bob, and Charlie, we have the following groupings of "same birthday":

$(A=B)$ with $P=\frac{1}{365}$, and $C \ne A$ with $P=\frac{364}{365}$ resulting in $\frac{1}{365}\frac{364}{365}$
$(B=C)$ similarly with $B \ne A$ for $\frac{1}{365}\frac{364}{365}$
$(A=B=C)$ with probability $\left( \frac{1}{365} \right)^2$

Probability of any of those is the sum of all probabilities, or 
$$\frac{1}{365}\frac{364}{365} + \frac{1}{365}\frac{364}{365} + \left( \frac{1}{365} \right)^2
= \frac{729}{365^2}$$
Add Donna to the mix and we have (for simplicity, say the common birthday is $x$):

$P(A=B; C \ne x; D \ne x,C) = \frac{1}{365} \cdot \frac{364}{365} \cdot \frac{363}{365}$
$P(A=C; B \ne x; D \ne x,B) = \frac{1}{365} \cdot \frac{364}{365} \cdot \frac{363}{365}$
$P(A=D; B \ne x; C \ne x,B) = \frac{1}{365} \cdot \frac{364}{365} \cdot \frac{363}{365}$
$P(B=C; A \ne x; D \ne x,A) = \frac{1}{365} \cdot \frac{364}{365} \cdot \frac{363}{365}$
$P(B=D; A \ne x; C \ne x,A) = \frac{1}{365} \cdot \frac{364}{365} \cdot \frac{363}{365}$
$P(C=D; A \ne x; B \ne x,A) = \frac{1}{365} \cdot \frac{364}{365} \cdot \frac{363}{365}$
$P(A=B=x; C=D=y, x \ne y) = \frac{1}{365} \cdot \frac{364}{365} \frac{1}{365}$
$P(A=C=x; B=D=y, x \ne y) = \frac{1}{365} \cdot \frac{364}{365} \frac{1}{365}$
$P(A=D=x; B=C=y, x \ne y) = \frac{1}{365} \cdot \frac{364}{365} \frac{1}{365}$
$P(A=B=C; D \ne x) = \left( \frac{1}{365} \right) ^2 \cdot \frac{364}{365}$
$P(A=B=D; C \ne x) = \left( \frac{1}{365} \right) ^2 \cdot \frac{364}{365}$
$P(A=C=D; B \ne x) = \left( \frac{1}{365} \right) ^2 \cdot \frac{364}{365}$
$P(B=C=D; A \ne x) = \left( \frac{1}{365} \right) ^2 \cdot \frac{364}{365}$
$P(A=B=C=D) = \left( \frac{1}{365} \right) ^3$

Probability of any of those is the sum of all:
$$ \frac{({_4C_2}+{_{364}P_2}) + ({_4C_3}+{_{364}P_1}) + ({_4C_4}+{_{364}P_0}) + 3 \cdot 364 }{365^3}  $$
The number of possibilities grows combinatorically.