Thread: Probability birthday same day

There are two soccer teams (2x11 people) and a referee at a soccer stadium. What is the probability that at least two of the 23 people have birthday on the same day? And how do you work it out? Thanks in advance

Many solutions are available via google search: probability birthday same day

These "at least" kind of questions are worked out by calculating:
1 - P (none)
since the sum of probabilities of an event occuring and its not occuring is 1.

That is, P (event occurs) + P (event doesn't occur) = 1

So
P (at least two, i.e. two or more, people share their birthdate) = 1 - P (nobody has the same birthdate)

Now, if the numerator stands for the no. of favourable cases and the denominator represents the total no. of possibilities, then

The first person can have their birthday on any day of the year, i.e. 365 cases favourable out of 365.

For the second person to have a DIFFERENT birthdate, all days would be favourable EXCEPT the birthdate of the first person, i.e. 364 days out of 365.

For the third person to have a birthdate different from the first two, there would be (365-2) favourable days out of 365

and so on

Till the 23rd person, who needs to be born on the remaining (365-22) days [b'days of the previous 22 persons] to have a birthdate different from them all.

So the possibility that NOBODY shares the same b'day is:

365/365 * (365-1)/365 * (365-2)/365 * ...... * (365-22)/365

which, using permutation symbology, can be written as:

(365)_P_(23) / (365)^23

[Since the denominator gets repeated 23 times]

AND THE ANSWER IS:

1 - P (nobody shares the same birthday)

= 1 - (365)_P_(23) / (365)^23

so what does that translate to then?

Originally Posted by mrconfused

so what does that translate to then?

I really think you should understand the whole problem in order to be able to do a similar one on a test. Are you familiar with permutations and combinations?

Originally Posted by Apprentice

There are two soccer teams (2x11 people) and a referee at a soccer stadium. What is the probability that at least two of the 23 people have birthday on the same day? And how do you work it out? Thanks in advance

the solution is to find the chances of a birthday not occurring and then subtract from one. Like this. the first person can have a birthday of 365/365days. the next person must have a birthday on 364/365 days so as not to coincide with the first person's birthday. the third person must have a birthday on 363/365 days and so forth. the answer is found by multiplying the sequence 365/365 x 364/365 x 363/365 x .......x 343/365 for 23 people and then subtract this from one to get the chances of 2 people having the same birthday. the answer should be about .5

Originally Posted by Apprentice

There are two soccer teams (2x11 people) and a referee at a soccer stadium. What is the probability that at least two of the 23 people have birthday on the same day? And how do you work it out? Thanks in advance

50.73 % is exact answer

Originally Posted by mike

50.73 % is exact answer

No its an approximate answer for a number of reasons, one of which is
the assumption of 365 day years, and another is that it has been rounded to
four (hopefully) significant digits.

RonL