Birthday paradox: Difference between revisions
imported>David W Gillette (general formula and tweeks) |
imported>David W Gillette (more data) |
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The third person in the group, having to miss the birthdays of the first two people, has only 363 days, with a probability of <math>\frac {363}{365}</math> which is 0.9945. Multiplying this on to the previous result, gives 0.9918. | The third person in the group, having to miss the birthdays of the first two people, has only 363 days, with a probability of <math>\frac {363}{365}</math> which is 0.9945. Multiplying this on to the previous result, gives 0.9918. | ||
For four in the group the | For four in the group, the probability of a pair matching birthdays is 1 − 0.9836 = 0.0164, over 1 percent. | ||
The formula for the probability of having a matching birthday in a group of n people is | |||
:<math>p(n) = 1 - \frac {365!}{(365-n)! \cdot 365^n}</math> | |||
As more and more people come into the group, the number of days available | As more and more people come into the group, the number of nonmatching days available decreases; and the probability of everyone having a different birthday decreases. Thus the probability of there being two with the same birthday increases. Continuing to multiply more and more, smaller and smaller fractions, a point near 0.5 is exceeded as soon as the group consists of 23 people. | ||
The | The data below shows that as the size of the group grows, the probability of having a match increases very quickly. | ||
N p(n) | |||
5 0.0271 | |||
10 0.1170 | |||
15 0.2529 | |||
20 0.4114 | |||
22 0.4757 | |||
23 0.5073 | |||
25 0.5687 | |||
30 0.7063 | |||
40 0.8912 | |||
50 0.9704 | |||
60 0.9941 | |||
70 0.9992 | |||
80 0.9999 | |||
So in a group of 80 people, only one time in ten thousand, will everyone have a different birthday. | |||
[[Category:Mathematics Workgroup]] | [[Category:Mathematics Workgroup]] | ||
[[Category:CZ Live]] | [[Category:CZ Live]] |
Revision as of 13:53, 21 July 2007
The birthday coincidence, also known as the birthday paradox or the birthday problem, results when in a small group of people, there are two that celebrate their birthday on the same day of the year. It is surprising to most people, how small a group of people are needed to have a 50-50 probability of having a matching birthday. (In this article leap year is ignored.)
To show how to calculate the probability of a group finding such a match, it is simpler to first find the probability of all the birthdays being different. Consider a group of two people. The first person can have been born on any of the 365 days of the year, while the second must have been born on one of the other 364 days in order to not match. The first person has a probability of , which equals 1.0, and the second has a probability of which is 0.9973. Multiplying these probabilities together gives a net probability of 0.9973 for having different birthdays. Subtracting this number from 1.0 gives 0.0027 probability of having the same birthday.
The third person in the group, having to miss the birthdays of the first two people, has only 363 days, with a probability of which is 0.9945. Multiplying this on to the previous result, gives 0.9918.
For four in the group, the probability of a pair matching birthdays is 1 − 0.9836 = 0.0164, over 1 percent.
The formula for the probability of having a matching birthday in a group of n people is
As more and more people come into the group, the number of nonmatching days available decreases; and the probability of everyone having a different birthday decreases. Thus the probability of there being two with the same birthday increases. Continuing to multiply more and more, smaller and smaller fractions, a point near 0.5 is exceeded as soon as the group consists of 23 people.
The data below shows that as the size of the group grows, the probability of having a match increases very quickly.
N p(n) 5 0.0271 10 0.1170 15 0.2529 20 0.4114 22 0.4757 23 0.5073 25 0.5687 30 0.7063 40 0.8912 50 0.9704 60 0.9941 70 0.9992 80 0.9999
So in a group of 80 people, only one time in ten thousand, will everyone have a different birthday.