[Red Badger]

. . . . . . . . . . . .

➡ The solution: This is the famous “Birthday Problem” that never ceases to amaze people. The easiest way to solve it is to consider its complement: For a party of a particular size, what is the probability that no one will share a birthday? If it’s not true that no one shares a birthday, then it must be the case that at least two people share a birthday!

Let’s consider a group of five people. The first person will have some random birthday between January 1 and December 31. The second person, then, will have a probability of (364/365) of having a birthday that is not the same as the first person’s birthday. The third person, then, has 363 “free” dates to choose from, so has a 363/365 probability of having a birthday that is not the same as either the first or the second person’s birthdays. The fourth person will have a 362/365 chance of not sharing anyone’s birthday, and so on.

To then get the probability that not one of the five people will share a birthday, we need to multiply these probabilities together:

P = (364/365)*(363/365)*(362/365)*(361/365) = 0.97

In other words, in a room of five people, no one will share a birthday 97 percent of the time. This seems pretty intuitive. This means there is only a 3 percent chance (100 percent - 97 percent) that at least two people will share a birthday in that group.

What may be less intuitive (at least for me!) is how quickly this number grows. The table below uses the same calculation to show the probability that at least two people will share the same birthday in parties of increasing sizes:

birthday probability chart

LAURA FEIVESON In a party of 30 people, there is a 71 percent chance that two people will share a birthday. Carol should not take that bet!

This problem always brings to my mind some family lore: Years ago, the legend has it that my father mentioned this birthday problem to a woman he was wooing as he was bringing her to a party of about 30 people. She did not believe that the probability could be so high. My dad suggested testing it out and asked her when her birthday is. “May 20!” she exclaimed. It turns out that’s the same birthday as my dad! They never went on another date—which worked out well for me!

[Bloody Sam Roberts]

Should Carol take that bet?

I'd say yes. Take the bet. Only if she KNOWS the guest of honor was born on February 29th.

[Born in 1950]

When two vowels go a-walking the first one usually does the talking.

[john drake]

I guess if you know in advance that a pair of twins were attending the party, your odds improve dramatically?

[kvanbrunt2]

the host has a twin brother.

[OldWarBaby]

Besides that a lot of stuff happens on Christmas eve and New Years. In a group as big as 30 people stuff probly happened. Kind of surprising it’s only 71%. Gotta refigger!

[Michael.SF.]

Assume that birthdays are distributed equally over a 365-day year

That is an incorrect assumption. Certain months have a higher birthrate than others. September usually tops the list.

[Michael.SF.]

Assume that birthdays are distributed equally over a 365-day year

That is an incorrect assumption. Certain months have a higher birthrate than others. September usually tops the list.

But 30 is a high number. Take the bet.

[Steve_Seattle]

There are probably 50 regulars in my local bar, and I think there are one or two cases of people with the same birthday.

[JimRed]

Easy, the Smith twins were invited and accepted.

[Dr. Sivana]

I intuitive thought it was a good bet, mainly because it isn’t the odds of anybody matching an individual borthday, but of any of them matching any of them. That gives you a nice multiplier.

I gamed it out in Excel, simulating 100 parties, each with 30 guests with an assigned random number between 1 and 365. Counted duplicates, and came up with 70%, which is close enough.

Formulae used:
=RANDBETWEEN(1,365)
=OR(COUNTIF(A1:AD1,A1:AD1)>1)
=COUNTIF(AE1:AE100,”True”)

[Verginius Rufus]

One of my high school teachers pulled this on our class. I already knew that one of the other students had the same birthday as mine. We were the only two with matching birthdays. I don’t remember how many were in the class—probably 30 to 40.

[ryderann]

Take the bet! $100 isn’t much to lose anymore.

[Strident]

table should say 0.27% for 2 people. C’mon man, no way it should say zero. I get rounding and precision errors, but zero means nil and obviously there is a non-zero chance for two strangers to have the same birthday. so 1-[364/365]

Nice job on this. Posters pointing out the non-even distribution of monthly births are entirely missing the point. I guess (but I’m not sure) if more than 3/12ths are clustered nine months after the fecund period, it may sway the odds slightly, very minutely, but not enough to spoil the point, which is how most people would have no idea of how fast the probability of at one least 1 coincident birthday pair mounts as n grows. Very cool illustration.

[pnut22]

My daughter had a son on her birthday and my son was born on his Great Grampa’s birthday. No family party is ever without 2 people sharing a birthday.

[alancarp]

That’s a lotta people in the same house. Certain locales frown upon that kind of thing these days.

[CarolinaReaganFan]

cool,thanks

[FamiliarFace]

I was recently at a (birthday) party, where it was discovered that I and another person had the same birthday, though 50 plus years apart. Also there was a set of twins, so 4 people out of about 30 people, maybe 35, but not quite that many. In my case, at that party, the probability held true.