### Brain Teasers

# Birthdays

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

How many people do you need to enter a room before the probability of any 2 or more people sharing a birthday (day and month only, not year) is greater than 50%?

Assume for the sake of the puzzle that birthdays in the population at large are equally spread over a 365 day year.

Assume for the sake of the puzzle that birthdays in the population at large are equally spread over a 365 day year.

### Hint

It is easier to calculate the probability that all birthdays in the group are unique.### Answer

We get the solution by calculating the solution to the opposite case: How many people are necessary before the chance of them all having unique birthdays is less than 50%.Let Pn be the probability of n people having unique birthdays.

Obviously, with 1 person in the room, P1 = 1.0

If there are n people in the room, the probability of the (n+1)th person also having a unique birthday is (365-n)/365.

Hence P(n+1) given Pn is (365-n)/365.

So P(n+1) = Pn * (365-n)/365.

If we start with n=1 and count upwards, calculating Pn, we find that P22 = 0.52 and P23 = 0.49.

Hence, with 23 people in the room, the probability that 2 or more people will share a birthday is 51%.

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## Comments

I have a question: does that mean that out of every random 46 people, 2 will share the same birthday?

probability is not my strong suit (many would ask what is?) but that seems awfully low to me. Are there any probablility experts out there then can vouch for the math in this one?

To einat16: No, it does not mean that with 46 people you are certain to get 2 people with the same birthday. That would be a bit (but not quite) like saying "If I toss 1 coin, I have 50% chance of tossing heads. If I toss 2 coins, do I have 100% chance of tossing a heads?"

I thought it was very logical. what is the mathematic explanation to this?

OK - To calculate the probability that 46 people have unique birthdays, we calculate the product series:

P(46) = ((365-0)/365) * ((365-1)/365) * ... * ((365-45)/365) = 0.05

I.E. We have a 95% chance that 2 or more will share a birthday. This is surprisingly high, but it is not 100%. In fact you do not get to 100% until you get 366 people in a room (assuming a 365 day year - sorry Feb 29 but you make the maths soooo much more complicated). There is still a chance, however small, that 365 randomly chosen people will have unique birthdays. However, you can't have 366 unique birthdays when there are only 365 days to choose from.

P(46) = ((365-0)/365) * ((365-1)/365) * ... * ((365-45)/365) = 0.05

I.E. We have a 95% chance that 2 or more will share a birthday. This is surprisingly high, but it is not 100%. In fact you do not get to 100% until you get 366 people in a room (assuming a 365 day year - sorry Feb 29 but you make the maths soooo much more complicated). There is still a chance, however small, that 365 randomly chosen people will have unique birthdays. However, you can't have 366 unique birthdays when there are only 365 days to choose from.

I got it wrong I thought the answer would be 730 people. Which is 2 normal years so the probability would be that two people are born on each day of the year. I don't understand the answer?!

The answer is 23 persons. The probability that two person donĀ“t share a birtday is 365/365 * 364/365, 3 persons 365/365 * 364/365 * 363/365. For more persons go on with this system. The probality that two persons share the same birthday is 1 - 365/365 * 364/365. With 23 persons you get a percent rate over 50. Its contra intuitiv

Jul 12, 2002

I have never lost a bet that out of the next 26 people to come into a room, 2 would share a birthday. 23 seems kind of low. Charlie6

Actually, a more interesting version of this puzzle would be, "Would you bet on two people in a football match(including the referee) having the same birthday?" The answer is highly counterintuitive.

I always show this problem to the probability and stats classes I teach. They are always amazed when we get two or more people sharing the same birthday in a relatively small class of 25 or 30 people.Intuitively, you would think it would take a considerably larger number of people before you would find a match of birthdays.

Quite a common problem in Australian Maths Classrooms.....Yes it actually is true for you skeptics out there

i don't know how to say this, but i found this teaser a lot like the one titled "birthday line", by something i think...and i'm not complaining of copyright or plagiarism or anything...it's just that i was rather amused that such similar problem have 2 distinct answers...you'd say 20 and 23 is not too big a difference but 3 out of 20 is 15%, and to me anything more than 2 is many, so it DOES make out much of something...

Answer is wrong, as pointed out. You only get a 95% interval confidence. Meaning that with the number of people you mention (was it 40?) you are reasonably certain that there will be a match; but reasonably does not mean it's guaranteed. If you want 100%guarantee you need 365 people in that room.

Addendum: you need 365 to guarantee 100% if you assume they all have unique birthdays, otherwise you need way more...

Please ignore my 2 precedent posts. I misread the pb (shame). Point is still valid on the 95% confidence level. 23 people only give you a better than 50% chance with a 95% confidence level. If you want a 100% guarantee that you'll get a better than 50% chance of having a match, you need considerably more than 23 people in that room.

dunt understand

no, you dont add probablitilies! they are percents, hence they are derived from division

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