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We fix ‘N’ and determine the probability that no two people, in a room of N people, have the same birthday. Say that the people are P(1),P(2),….,P(N). Now person P(1) may have any of the 365 days of the year for his birthday without contradicting the condition that no two people in the room will have the same birthday. Once person P(1) has fixed a birthday, then person P(2) is not allowed to have that birthday if we are to maintain distinct birthdays. Thus person P(2) has 364 choices for his/her birthday, then person P(3) has 363 choices for his/her birthday, and so it goes,(if we are to maintain distinct birthdays).
In summary, the total number of birthday combinations for ‘N’ people, with no two being the same, is: (365)(364)(363)……..[{365 – (N – 1)}]
The total number of all possible birthday distributions among ‘N’ people, without regard for duplication or lack of duplication, is: (365)(365)(365)……..(365) (‘N’ times)
Therefore, the probability that N people in a room will all have distinct birthdays is:
P=[(365)(364)(363)……..[{365 – (N – 1)}] /[(365)^N]
or, P=(365/365)(364/365)(363/365)……..[{ 365 – (N – 1)}/365]
We now start multiplying the fractions together. When the product falls below ½, we are finished. The last fraction that we multiplied in will tell us what ‘N’ should be [because the last fraction you multiplied in was [{365 – (N – 1)}/365].
We multiply together 23 terms to obtain a probability of 0.4927027. Using only 22 terms yields a probability of 0.5243046. Clearly the least value of ‘N’ to yield a probability of less than 1/2 that each person in the room will have a distinct birthday is N = 23.
Therefore, we conclude that if there are 23 people in a room then the odds are better than even (in fact the odds are P=1-0.4927027=0.5072973) that two of them will have the same birthday. |
