Expert: Paul Klarreich - 10/17/2010

Question
QUESTION: (a) Assume a 365-day year,find the probability that in a group of 3 people there will be atleast 1 birthday in common.
(b) if there are n people in the group,find the probability for an expression of atleast 1 common birthday.
(c) how many people need to be in the group before the probability exceeds 0.5 .
(d)how many people need to be in the group before the probability exceeds 90% .


ANSWER: Questioner:   athena
Country:  Australia
Category:  Advanced Math
Private:  No
 
Subject:  probability
Question:  QUESTION: (a) Assume a 365-day year,find the probability that in a group of 3 people there will be atleast 1 birthday in common.
(b) if there are n people in the group,find the probability for an expression of at least 1 common birthday.
(c) how many people need to be in the group before the probability exceeds 0.5 .
(d)how many people need to be in the group before the probability exceeds 90% .


The probability that the first person has the same birthday as himself (no, I have not neglected to take my anti-insanity medications today) is:

365
--- = 1
365


The probability that the second person has a different birthday from the first is:

365 364
--- --- =
365 365

364
---
365

The probability that, IN ADDITION, the third person has a different birthday from the first two is:

365 364 363
--- --- --- =
365 365 365

364 363
--- --- =
365 365

Now assume we have N people, not necessarily 3.  Then

The probability that, IN ADDITION, the N-th person has a different birthday from the first N-1, with all those being different is:

365 364 363 * ... * (365 - (N-1))
--- --- --- ---------------------=
365 365 365 * ... *  365

which can be written:
 365!
-------------------
365^N (365 - N)!

and computed using your calculator (or Windows's).  I suggest you use something like Excel -- very good for this kind of thing.

Of course, the probability that you get at least one 'matching' birthday is equal to

1.0 - <whatever you got above.>

And the classic one is (c), where, to most people's surprise, the answer is 23.  Yes, if you have 23 people selected at random, the probability is more than 50% that some two will have the same birthday.  (I used to pull this stuff on my classes the first day, AFTER determining that I had more than 23 students.)

and the answer for (d) is that as few as 41 makes it 90% that you get a match.


---------- FOLLOW-UP ----------

QUESTION: answers are
(a) 0.0082
(b) 1-365Pn/365^n
(c) 23
(d) 45

i still can't see how to get these answers from ur working.

Answer
The probability that the second person has a different birthday from the first is:

365 364
--- --- =
365 365

364
---
365

The probability that, IN ADDITION, the third person has a different birthday from the first two is:

365 364 363
--- --- --- =
365 365 365

364 363
--- --- =  (you can compute this)
365 365

Now assume we have N people, not necessarily 3.  Then

The probability that, IN ADDITION, the N-th person has a different birthday from the first N-1, with all those being different is:

365 364 363 * ... * (365 - (N-1))
--- --- --- ---------------------=
365 365 365 * ... *  365

which can be written:

365 364 363 * ... * (365 - (N-1)) *( 365 - N) *... 3  2  1
--- --- --- -----------------------------------------------  =
365 365 365 * ... *  365 *  ( 365 - N) *... 3  2  1  


which can be written, after you count the factors carefully:
    365!
----------------
365^N (365 - N)!


and computed using your calculator (or Windows's).  I suggest you use something like Excel -- very good for this kind of thing.

Program the following:

Cell 1:  1 - 365/365
Cell 2:  1 - Cell 1 * 364/365
Cell 3:  1 - Cell 2 * 363/365
....
Cell N:  1 - Cell (N-1) * (365-(N-1))/365

keep doing it until you get the answers you want:

0.50
0.90