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

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- Oct 8th 2005, 12:32 AMApprenticeProbability (need it for Tuesday)
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

- Oct 8th 2005, 04:54 AMctelady
Many solutions are available via google search: probability birthday same day

- Oct 8th 2005, 11:06 AMAmiya
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** - Nov 7th 2005, 05:30 PMmrconfused
so what does that translate to then?

- Nov 7th 2005, 05:43 PMJamesonQuote:

Originally Posted by**mrconfused**

- Feb 14th 2008, 02:50 PMfurbish
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

- Feb 17th 2008, 10:11 PMmikebirthday
- Feb 17th 2008, 11:52 PMCaptainBlack