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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 %