There are 3 people in a room.What is the probability atleast 2 of them share same birthday

Question

I want to know how to solve by NAIVE APPROACH, NOT by taking complement. That is, not by calculating the probability that no one shares a birthday and subtracting from 1 to get the answer.

Answer 1

For the purposes of the calculations below, I assume that leap days do not exist.

Let's say that we have $3$ people, $A,B,C$. The probability that at least two people share a birthday is naively computed as $$ P = \frac{\text{possible ways for at least two people to share a birthday}}{\text{all possible birthdays}}. $$ To calculate the number on top, we note that we can uniquely construct every possibility where exactly two people share birthday as follows:

• Select a common birthday ($365$ choices)
• Select which people have that birthday ($3$ choices)
• Select another birthday for the remaining person ($364$ choices)

Thus, there are $365 \times 3 \times 364$ ways for exactly $2$ people to share a birthday.

We can uniquely construct every possibility where all three people share a birthday by selecting that common birthday, so there are $365$ choices.

So, the number on top is $3 \times 365 \times 364 + 365$.

To calculate the number on bottom, we note that we can uniquely construct every possibility by selecting $A$'s birthday, then $B$'s, then $C$'s. So, the number on bottom is $365^3$.

All together, we find $$ P = \frac{3 \times 365 \times 364 + 365}{365^3} = \frac{3 \times 364 + 1}{365^2} $$

Answer 2

Name the people A, B and C. Assume all birthdays are equally likely (not actually true, but good enough for an approximation).

What are the possibilities?

(1) A, B and C have all different birthdays.

(2) A and B have the same birthday, C's is different.

(3) A and C have the same birthday, B's is different.

(4) B and C have the same birthday, A's is different.

(5) A, B and C all have the same birthday.

We want P(2) + P(3) + P(4) + P(5), since these all satisfy your condition.

Note that there are $365^3$ different birthday permutations for A, B, C (if we're ignoring leap years).

So all you need to do is calculate the number of possible outcomes for each of (2), (3), (4) and (5), sum them, and divide that by $365^3$.

Answer 3

You must have: $$P(SSS)+P(SSF)+P(SFS)+P(FSS)=\\ \frac{365}{365}\cdot \frac{1}{365}\cdot \frac{1}{365}+\frac{365}{365}\cdot \frac{1}{365}\cdot \frac{364}{365}+\frac{365}{365}\cdot \frac{364}{365}\cdot \frac{1}{365}+\frac{365}{365}\cdot \frac{364}{365}\cdot \frac{1}{365}=\\ \frac{1}{365^2}\left(1+3\cdot 364\right)\approx 0.00820416588478138487521110...$$ Note: S - success, F - failure.

Answer 4

$$P=P_{A=B}+P_{A\ne B}(P_{A=C}+P_{B=C})=\frac1{365}+\frac{364}{365}\frac2{365}.$$