Statistical probablity

Somewhere between 19 & 20

P(E)=0.5

Assumption a year is 365 hays (no leap year)

The propability of any two individuals having the same birthday is the product of the probability of any individual birthday (365/365), the probability of any second individual having the same birthday (1/365), and the number of combinations of n individuals taken 2 at a time { n!/[2!(n-2)!]}.

(365/365)*(1/365)* { n!/[2!(n-2)!]}=0.5

now n!/(n-2)!=n*(n-1) and 2!=2


1/365*1/2*n*(n-1)=.5

rearranging

n*(n-1)=2*.5*365=365

and for any number n>>1 then n*(n-1) is about equal to n squares

so n^2=365 & n=19.142 and change

so try 19

19*18 = 342

and

20*19 = 380

since people don't come in fractions, I'd say between 19 and 20.

BTW I've always heard it said it's a sure certainity that in a group of 30 individuals, two will share a birthday.

Rap c∫/

If the question asks for the probability of the events of exactly two of the party having the same birthday, then there is no way to have the probabilyt of 1/2 for such an event.

Satt,

Let me embelish my brief question. Since there are 366 days in a year, there would have to be 367 people gathered together in order to be absolutely certain that at least two people in the group had the same birthday.

Now, if you just wanted to be sure that given X number of people in a room, there is a 50% chance that two of them will have the same birthday. How many people would have to be in a room to make the probability?

Actually you need to have 367 people in a room for two to have the same birthday. Isn't that weird? You'd think that at least by 366 there would have to be two, but there can't be.

SCoates,

I wasn't clear. When I said 366 days in a year, that included a day for leap year. You're right that you need 367 to cover all years.

Raprap is close, but no cigar.

If your question is asking about the probability of the event of "at least" two persons having the same birthdays and a year contains 365 days then an answer would be as follows..

Think of a fixed person, say 0-th, and the birthday is Jan, 1, without loss of generality.
The probability of the event in which there are no pairs of the same birthdays in the X members is

Q = (364/365)*(363/365)*(362/365)*..((365-X-1)/365)

and hence the probability of the event of the existence of at least two persons of the same birthday is

P = 1-Q.

P=0.507 for n = 23.

This works only if you're matching the birthday of the first person in the room. Since the question is any two people in the room having the same birthday you have to include the possibility of permutations of any two birthdays. Consequently, if your going to use this method, once a third person enters the room you have to include the probability that the third person doesn't have the same birthday as the first and the second, the fourth person doesn't have the same birthday as the first, second, or third and so on.

Accordingly, the probability is the product of any two people having the same birthday is the product of the probability of any single individual having a birthday (1 or 365/365 that is unless you're John Smith) and the second person having the same birthday (1/365). The since there are n people in the room you have to consider that any two people in the room can have the same birthday, so you have to consider the permutations, That is, the common birthdays could be the first and the fourth person, or the fifth and the sixteenth. Fortunately Pascal developed a method to determine the number of permutations.

Your method would work, if you included the cumulative probabilities of any two common birthdays.

Or, P(E)=1-Q₀-Q₁-….-Q_n=0.5

BTW using the permutation method, then the probability [P(E)] of any two people sharing a birthday is 1 at n=28, which is a good agreement with the 30 people rule stated above.

Rap c∫/

It is a common method in probability calculation that the probability of the complementary event is obtained before getting the probability of a rather complicated event.

It's not really true that this would ensure anything.

But I didn't come here to pick on that but to throw out a complete (but for two seconds interesting) red herring.

How often will twins occur in a data set of 367 people?

You misunderstood me. I was being facetious... if I'd thought of leap years I would have made my answer 368. The plan was for the logic to be bad no matter what.