Q: What is the probability of a set of 28 people having two sets of matched birthdays?

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Birthdays are not distributed uniformly over a year but if, for the sake of probability games you assume that they are, then ignoring leap years, the probability is 0.5687. Including leap years, it is slightly lower.

Birthdays are not uniformly distributed over the year. Also, if you were born on 29 February, for example, the probability would be much smaller. Ignoring these two factors, the probability is 0.0082

In probability theory, the birthday problem, or birthday paradox[1] pertains to the probability that in a set of randomly chosen people some pair of them will have the same birthday. In a group of 10 randomly chosen people, there is an 11.7% chance. In a group of at least 23 randomly chosen people, there is more than 50% probability that some pair of them will both have been born on the same day. For 57 or more people, the probability is more than 99%, and it reaches 100% when the number of people reaches 367 (there are a maximum of 366 possible birthdays). The mathematics behind this problem leads to a well-known cryptographic attack called the birthday attack. See Wikipedia for more: http://en.wikipedia.org/wiki/Birthday_paradox

19.4%CALCULATION:The probability of at least 2 people having the same birthday in a group of 13people is equal to one minus the probability of non of the 13 people having thesame birthday.Now, lets estimate the probability of non of the 13 people having the same birthday.(We will not consider 'leap year' for simplicity, plus it's effect on result is minimum)1. We select the 1st person. Good!.2. We select the 2nd person. The probability that he doesn't share the samebirthday with the 1st person is: 364/365.3. We select the 3rd person. The probability that he doesn't share the samebirthday with 1st and 2nd persons given that the 1st and 2nd don't share the samebirthday is: 363/365.4. And so forth until we select the 13th person. The probability that he doesn'tshare birthday with the previous 12 persons given that they also don't sharebirthdays among them is: 353/365.5. Then the probability that non of the 13 people share birthdays is:P(non of 13 share bd) = (364/365)(363/365)(362/365)∙∙∙(354/365)(353/365)P(non of 13 share bd) ≈ 0.805589724...Finally, the probability that at least 2 people share a birthday in a group of 13people is ≈ 1 - 0.80558... ≈ 0.194 ≈ 19.4%The above expression can be generalized to give the probability of at least x =2people sharing a birthday in a group of n people as:P(x≥2,n) = 1 - (1/365)n [365!/(365-n)!]

It is to celebrate the day of your birth. and the reason why peoples 18th birthday is so important is because they are growing into an adult.

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Let us assume that there are exactly 365 days in a year and that birthdays are uniformly randomly distributed across those days. First, what is the probability that 2 randomly selected people have different birthdays? The second person's birthday can be any day except the first person's, so the probability is 364/365. What is the probability that 3 people will all have different birthdays? We already know that there is a 364/365 chance that the first two will have different birthdays. The third person must have a birthday that is different from the first two: the probability of this happening is 363/365. We need to multiply the probabilities since the events are independent; the answer for 3 people is thus 364/365 × 363/365. You should now be able to solve it for 4 people.

Birthdays are not uniformly distributed over the year. But, if you do assume that they are, then: ignoring leap years, it is approx 0.5982. If you include leap years it is 0.5687.

The probability with 30 people is 0.7063 approx.

Birthdays are not distributed uniformly over a year but if, for the sake of probability games you assume that they are, then ignoring leap years, the probability is 0.5687. Including leap years, it is slightly lower.

Birthdays are not uniformly distributed over the year. Also, if you were born on 29 February, for example, the probability would be much smaller. Ignoring these two factors, the probability is 0.0082

To determine the probability of 15 random people all having the same birthday, consider each person one at a time. (This is for the non leap-year case.)The probability of any person having any birthday is 365 in 365, or 1.The probability of any other person having that same birthday is 1 in 365, or 0.00274.The probability, then, of 15 random people having the same birthday is the product of these probabilities, or 0.0027414 times 1, or 1.34x10-36.Note: This answer assumes also that the distribution of birthdays for a large group of people in uniformly random over the 365 days of the year. That is probably not actually true. There are several non-random points of conception, some of which are spring, Valentine's day, and Christmas, depending of culture and religion. That makes the point of birth, nine months later, also be non-uniform, so that can skew the results.

People do celebrate their birthdays by hanging out with their friends or family. Maybe also by going out for a picnic or having fabulous food outside. I do celebrate my birthday by hanging out with my family in McDonald's or Pizza hut!

Millions around the world.

In probability theory, the birthday problem, or birthday paradox[1] pertains to the probability that in a set of randomly chosen people some pair of them will have the same birthday. In a group of 10 randomly chosen people, there is an 11.7% chance. In a group of at least 23 randomly chosen people, there is more than 50% probability that some pair of them will both have been born on the same day. For 57 or more people, the probability is more than 99%, and it reaches 100% when the number of people reaches 367 (there are a maximum of 366 possible birthdays). The mathematics behind this problem leads to a well-known cryptographic attack called the birthday attack. See Wikipedia for more: http://en.wikipedia.org/wiki/Birthday_paradox

obviosly, they celebrate birthdays but they do it with a lot of alchocol!

19.4%CALCULATION:The probability of at least 2 people having the same birthday in a group of 13people is equal to one minus the probability of non of the 13 people having thesame birthday.Now, lets estimate the probability of non of the 13 people having the same birthday.(We will not consider 'leap year' for simplicity, plus it's effect on result is minimum)1. We select the 1st person. Good!.2. We select the 2nd person. The probability that he doesn't share the samebirthday with the 1st person is: 364/365.3. We select the 3rd person. The probability that he doesn't share the samebirthday with 1st and 2nd persons given that the 1st and 2nd don't share the samebirthday is: 363/365.4. And so forth until we select the 13th person. The probability that he doesn'tshare birthday with the previous 12 persons given that they also don't sharebirthdays among them is: 353/365.5. Then the probability that non of the 13 people share birthdays is:P(non of 13 share bd) = (364/365)(363/365)(362/365)∙∙∙(354/365)(353/365)P(non of 13 share bd) ≈ 0.805589724...Finally, the probability that at least 2 people share a birthday in a group of 13people is ≈ 1 - 0.80558... ≈ 0.194 ≈ 19.4%The above expression can be generalized to give the probability of at least x =2people sharing a birthday in a group of n people as:P(x≥2,n) = 1 - (1/365)n [365!/(365-n)!]

They don't. They only celebrate 15th birthdays, which is a quinceanera.