The Birthday Problem or Paradox

Let us assume that a person's birthday falls on any day of the year with equal
probability (chance). In other words, there is no inherent bias for one day
versus another. (In practice may not be strictly true, but close enough.)

Now suppose random 10 people assemble in a room.
How likely is it that two of them have the same birthday?

What is the likelihood if 25 people assemble?
How many people are needed to have a 50% chance of "two people with the same birthday"?

It's called the Birthday Problem, and sometimes Birthday Paradox because the answer 
is counter-intuitive.

One may think that for 50% probability, may be 365/2 = 182 people are needed, but
in fact only 23 suffice!!!

Let's start with the basics.
Suppose two random people in the room; what is the probability that they have the
same birthday?

Prob[2 people with same birthday] = 1/365
	
	(the second person has to have the same birthday as the first one, and
	since all birthdays are equally likely, this happens with prob 1 in 365]


Now suppose there are some k people in the room, and we want to calculate the prob that 
some two of them have the same birthday.  It will be easier for us to calculate the 
**complementary probability**: that is, what is the prob that no two have the same birthday.
This means all birthdays are distinct.
Write

P_i = Prob that the first i people have distinct birthdays.

Then,

P_1 = 1
P_2 = [1 - 1/365]
P_3 = 1.[1 - 1/365].[1 - 2/365]

For P_3, the first two have to have different birthdays, and then 3rd has to have
	birthday different from both.

In general,

P_k = 1.[1 - 1/365].[1 - 2/365] ... [1 - (k-1)/365]

We now use the approximation that

e^x <= 1 + x, for all x

Thus, [1 - i/365] >=  exp(- i/365)

Next, e^x * e^y = e^{x+y}

Thus, 

P_k <= exp[ 0 - 1/365 - 2/365 - 3/365 ... - (k-1)/365 ]
    <= exp[ - (1 + 2 + 3 + ... (k-1))/365 ]
    <= exp[ -k(k-1)/2*365]

If we choose k = 23, we get

P_23 = prob that a group of 23 people have distinct birthdays 
     <= exp[ -22*23/2*365]
     <= exp[ -506/2*365]
     <= e^{-0.69315}
     <= 0.499998


Thus, with probability > 1/2 the birthdays among 23 people are NOT distinct!!