if you 23 random people at least 2 of them will have the same birthday not including the year how come?

Question

need help please

Answer 1

The chance that 2 people *don't* have the same birthday is:
(365/365) x (364/365)

The chance that 3 people *don't* have the same birthday is:
(365/365) x (364/365) x (363/365)

The chance that 4 people *don't* have the same birthday is:
(365/365) x (364/365) x (363/365) x (362/365)

etc.

When you get to about 23 people if you figure out this probability if finally gets below 50%.

So it doesn't guarantee that some pair will have the same birthday, but it is more and more likely as you add people. 23 is the tipping point where it goes from unlikely (<50%) to likely (>50%).

The only time it actually gets to 100% is when you have 365 (or technically 366 people).

Answer 2

There are k people in a room. What is the probability that at least two of them have the SAME birthday, that is, month and day?

The common assumption is that every day is equally likely, that is, no twins, and no power blackouts so there was no unusual frequency of sex on a particular day, etc. We also ignore February 29 and assume a 365 day year.

For k=2, the probability they have DIFFERENT birthdays is (365 X 364) / (365 X 365). In the numerator, there are 365 possible birthdays for the first person, and once that birthday is chosen there are 364 birthdays for the second person. In the denominator, there are are 365 possible birthdays for the first person and 365 for the second person.

For k=2, the probability they have the SAME birthday is 1 - (365 X 364) / (365 X 365).

For arbitrary k, the probability they have DIFFERENT birthdays is (365 X 364 X … X (365 – k + 1)) / (365 ^ k). The carat "^" is the symbol for exponentiation., that is "365 raised to the k power." The numerator is denoted "365Pk," or "the number of permutations of 365 things, taken k at a time."

For arbitrary k, the probability that at least two of them have the SAME birthday is 1 - (365 X 364 X … X (365 – k + 1)) / (365 ^ k), or 1 – (365Pk) / (365 ^ k).

For k=23, the probability that at least two of them have the SAME birthday equals .507, which is better than a 50:50 chance. For k=30, the probability is .706.

Answer 3

If you select 23 ( 24 I think would be better ) people at random , at least two of them will celebrate their birthday on the same date is a probability , with odds in favor and not a certainty. Please note. The probability can be worked out and shown to be 27/50 i.e. better than 50 % . However in practice the conditions are such, birth dates tend to be much more in certain months than in others ,thus due to this concentration the probability is increased in favour .

Answer 4

The previous two answerers gave you the correct technical answer.
It sounds illogical, unless you look at it this way:
In order to verify that at least two people in the room share their birthday, you need to figure out how many pairs of people you can create, and then check the birthday of the two people in each pair.

With 23 people present you can create 22 +21 +20 +...1 pairs, or 253 different pairs of people!
No wonder there will, more often than not, be a pair whose members share their birthday.