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Did you know that in the room of 20 to 30 people you are very likely to find two people who share the same birthday? This is called birthday paradox. Try to calculate the probability of two students having the same birthday in the class of 30. This is counter-intuitive and you will be surprized by the result.
To follow...anyone has an idea how to solve it?
I've always found it easiest to calculate the complement of this problem rather then solve it directly, it makes it easier to understand to me. So in a room of 30 people find the probability that none of them have the same birthday.

First we need to find the entire sample space Ω or all the choices in which no condition occurs in the event. If we have 30 people, they can choose any birthday in ( 365 ) ^ 30 ways.

Now we can choose the number of people without the same birthday in 365 * 364 * 363 * ..... * 336 ways. ( First person can choose from 365 days and each subsequent has 1 less choice in order for none to have the same birthday )

Hence the desired probability is
( 365 * 364 * 363 * ..... * 336 ) / ( 365 ^ 30 ) = 29.4%

Thus the Probability 30 people have the same birthday in one room is

1 - .294 = 70.6%

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Jay your right when you start counting each possible pair of people, since each person compares their birthday to each other person. However in doing so each of these pairs has a probability of 1 / 365 of being a match meaning it must be taken to the power of ( 29 + 28 + ... ).

The probability 2 people have 2 different birthdays would then be

1 - ( 1 / 365 ) = 364 / 365

If we take this probability to the nth power for its n amount of pairs
( 420 in this case) we get :

( 364 / 365 ) ^ 420 = 31.6 %

// With the 2% error margin from my calculator not being able to accurately compute such large #s

This probability being the probability that 2 people had different birthdays we subtract it from one to get its complement.

1 - .316 = 68.4 %
Harris - excellent! And even better explanation!