If there are 60 people in a gathering, will any two persons have the same birth day? We are doubtful. we think if there are more than 365 persons, there may be a possibility of two persons sharing same birthday. But the statistics' answer is stunning. Here we go.
In a coin toss, chance of getting head or tail is 50 % or 1/2. Chance of head is 1/2 and the tail is 1/2. Hence the total probability is 1.
In a dice throw, chance of getting '1' is 1/6 and the chance of 'not' getting '1' is 5/6 = 1-1/6.
In a gathering, the chance for a person to have any date in a year as a birthday is 365/365=1=100%. For another person, the chance of not having a birthday same as first person is 364/365. For a third person, the chance of not having a birthday same as previous two persons is 363/365. Hence the probability for not having a match (two persons having same birth day ) among three persons is (365/365) * (364/365)*(363/365) (based on property of probability).
In another words, the probability of having a match among three persons is
1 - (365/365) * (364/365)*(363/365)
So the chance of finding a match of birthday among N persons is
= 1 - (365/365) * (364/365)*(363/365)* ........*(365-N+1/365)
= 1-[365*364*363*...........*(365-N+1)/365^N]
Using the above formula, let us calculate the probability of a match for different number of persons N = 1,2,3... and tabulate the results.
N P
no. of persons probability of a match
1 0
9 0.1 or 10%
23 0.5 or 50%
30 0.7 or 70%
58 1 or 100%
So the chance for two persons having same birthday (a match) is 50% even for 23 persons. A shocking result. If there are 58 persons in a room, there is a 100% chance for a match.Try to verify yourself.
This birthday problem(clash of dates, clash of data, clash of occupancy). Finds applications in spread of epidemics, photons occupancy and cryptography.
In a coin toss, chance of getting head or tail is 50 % or 1/2. Chance of head is 1/2 and the tail is 1/2. Hence the total probability is 1.
In a dice throw, chance of getting '1' is 1/6 and the chance of 'not' getting '1' is 5/6 = 1-1/6.
In a gathering, the chance for a person to have any date in a year as a birthday is 365/365=1=100%. For another person, the chance of not having a birthday same as first person is 364/365. For a third person, the chance of not having a birthday same as previous two persons is 363/365. Hence the probability for not having a match (two persons having same birth day ) among three persons is (365/365) * (364/365)*(363/365) (based on property of probability).
In another words, the probability of having a match among three persons is
1 - (365/365) * (364/365)*(363/365)
So the chance of finding a match of birthday among N persons is
= 1 - (365/365) * (364/365)*(363/365)* ........*(365-N+1/365)
= 1-[365*364*363*...........*(365-N+1)/365^N]
Using the above formula, let us calculate the probability of a match for different number of persons N = 1,2,3... and tabulate the results.
N P
no. of persons probability of a match
1 0
9 0.1 or 10%
23 0.5 or 50%
30 0.7 or 70%
58 1 or 100%
So the chance for two persons having same birthday (a match) is 50% even for 23 persons. A shocking result. If there are 58 persons in a room, there is a 100% chance for a match.Try to verify yourself.
This birthday problem(clash of dates, clash of data, clash of occupancy). Finds applications in spread of epidemics, photons occupancy and cryptography.