My gr 12 math teacher told the class back in 94…
that the probability of 2 people having the same birthday is high when there are more than 20 people.
i forgot the exact number he gave but i think it was in the 50% range…
he asked the class(26 people) for their birthdays… and sure enough, 2 sets of people had the same birthdays(not the year)…
i forgot the mathematician that came up with this theory…
but how does that work?
Fisrt of all, it’s not a theory, it’s a statistical result.
The idea here is that the ones sharing a birthday can be ANY two in the group, in ANY day. This means that you can have a lot of possible pairings to check…
Check this site:
http://www.mste.uiuc.edu/reese/birthday/explanation.html
and this
http://www.studyworksonline.com/cda/content/explorations/0,,NAV2-76_SEP949,00.html
for easy explanations.
that second link was great!
thanks!
This is kind of interesting. I had a college orientation last week, and in a group of 7 people, there were 2 who were born on the exact same date. Not only that, but their birthdate was only 3 days later than mine. What are the odds?
Your scenerio is much more probable than if you selected any 7 people in the world at random. Your population was already narrowed by the fact that you were at the same point in your lives - entering college. That likely would place you within a year of each other, so there were only 365 or 366 days in which you all would have been born.
When I was in my late 20’s I went out with a guy who was born within 3 hours of me, but 1/2 a continent away. The romantic thing didn’t work, but we remained close friends & call ourselves “twin children of different mothers”.
Somebody born within 3 days of you (+ or -)? My calculator says about 1 in 52
There was a survey in The Economist a few years back that tried to measure the correlation between people’s ability to calculate rough probabilities and their believe in New Age junk, Astrology etc.
One of the main questions was something like “say you were at a party, how many people would there have to be in the room for you to have a 50:50 chance of having the same birthday as someone else there” (the answer is about 24 from memory).
My girlfriend of the time who regularly went for palm readings, tarot, read her stars everday etc said you would need about 600 other people. Me, the confirmed skeptic, said about 30.
That’s not to say my girlfriend was stupid (she had an undergrad degree in economics and a Masters in Economics at the time while I only had an undergrad degree back then) just that she, like many people, was too ready to accept that our lives are more ruled by the stars than by simple mathematics. Or, in the words of Corporal Nobby St John de Nobbs “Million to one chances come in nine times out of ten”.
Motog, that’s a whole 'nother ballgame.
The OP and its replies are about the probability of any two people in a crowd having the same birthday. The odds of that are 50% somewhere near 20 people.
The question you pose says “…you to have a 50:50 chance of having the same birthday…” which would require significantly more people. Just guessing, but I’d say that it’d be closer to 180.
Ben
To calculate the odds of someone sharing the same birthday as you in particular, you’d need to solve (364/365)^x=0.5.
Trial-and-error puts x at about 253 people.
Just because I’ve nothing better to do: with 30 people, there’s about an 8% chance someone will share your bday. With 600, there’s about an 81% chance.
Let’s do it the hard way.
If there are two people, A & B, what are the chances of them having the same birthday? 1 in 365 (forget about leap year, okay?).
Suppose there are 3 people, A, B & C. What are the odds that any two of them have the same birthday? Well, there’s a 1 in 365 chance A and B do, a 1 in 365 chance that A & C do, and a 1 in 365 chance that B & C do. So, the chances are 3 in 365. STILL pretty remote. With 4 people, A B C & D, there are six possible combinations that might match (AB, AC, AD, BC, BD, CD). STILL pretty remote. With 5 people, you have 10 possible combinations (AB AC AD AE BC BD BE CD CE DE). 10 out of 365 is still pretty slim.
With 20 people, you have 190 chances in 365. That’s the first point at which the odds say you’ll PROBABLY have two people with the same birthday.
With 21-27 people, again, the odds say you’d PROBABLY have two people with the same birthday.
However 28 is the number at which it becomes (almost) a mathematical certainty that you’ll have two people with the same birthday.
With 4 people, there are 6 chances in 365. With 5 people, 10 chances in 365.
astorian, your math is not quite right. The problem is that you can’t just add up the probabilities like you’re doing (unless the events are mutually exclusive, which they’re not in this case). In general, the probability of (A or B) happening is
P(A or B) = P(A) + P(B) - P(A and B).
The easiest way to figure out the probability from scratch is this:
Say, for simplicity, you have four people. First let’s find the probability that no two have the same birthday (again, disregarding leap days, and assuming birthdays are distributed uniformly).
Person 1 can have any birthday.
Person 2 can have any of the remaining 364 out of 365 days (probability 364/365).
Person 3 can be any of the remaining 363 (prob. 363/365).
Person 4 can be any of the remaining 362 (prob. 362/365).
Multiply them all together to get the probability that no two share a birthday:
365365365
Now subtract that from one to get the probability that at least two of them share a birthday.
23 is the break even point; with 23 people, the probability of (at least) two sharing a birthday is about 50.7%.
Damn, Cabbage! I was working on that as you posted. You’re right - what Astorian is correctly doing is tallying the number of comparisons you’d have to make, but he’s also ignoring the possibility of multiple matches (3 people having the same bday would count, for example).
Somebody please check my math on this, but one way you could do this is take the odds of getting NO matches (which is 364/365) and multiply that by the number of comparisons you make. Those shold be additive, since getting no matches is an exclusive outcome (the way that getting one match vs. getting two matches isn’t).
So like this: 20 people matched 2 at a time = 190 comparisons * 364/365 = 0.59, or a 41% chance of getting at least one match
21 people, 210 comparisons, 44% chance of a match
22 people, 231 comparisons, 47% chance of a match
23 people, 253 comparisons, 50% chance of a match
And at 42 people, you’re in the 90% range. Easy money.
Someone said it is a statistical fact that with 23 people, the odds are that two have the same birthday, but it is more accurate to call it a statment of probability. That might seem like a quibble, but it is based on a computation of probablilities, not by taking a large population and selecting sets of 23 and looking. For all I know, the statistics might actually be a little different since the computation is based on the assumption that birth dates are uniformly distributed through the year, which is not quite right.
I was once teaching a class of about 60 students a course in math for biologists that seemed to be mainly teaching how to count (various things). One day I asked them to calculate the chance that someone on the front row (there were 8) had the same birthday as someone else in the class. It was well over 50% and, sure enough, the seventh person had the same birthday as someone else.
Wikkit Point taken. Blame my bad memory. The survey actually referred to “any two people having the same birthday”, rather than you specifically having the same birthday as someone else.
Your girlfriend must have been pretty stupid then, If you think about it, the MOST the answer could have been is 367 since otherwise, you fill up all the days.
Shalmanese – You may have misread the question. Motog’s squeeziebunny was asking how many people are necessary to have a 50/50 chance of someone having the same birthday as YOU - a specified date – NOT how many to have a 50/50 chance of any two of them sharing a birthday. It’s quite possible to round up any number of people who don’t share YOUR birthday. (Say, at the Taurus’s Only Convention, if you’re born in December).
If Dynosaur’s math is right, la squeezie was about as correct in her guess as Motog was.
Quoth Hari Seldon:
It’s a perfectly valid fact. With evenly-distributed birthdays, it can be proven that the odds of two people out of 23 sharing a birthday is 50.7%. Those are, factually, the odds, and they’re over 50%. If one assumes that birthdays are not evenly distributed, then the probability will only go up, and the answer will still be over 50%.
Bit harsh there Shalmanese. She may have been a low level psychopath but I think I made an effort to establish that she wasn’t stupid. The point of the survey was to show that even reasonably intelligent people might have absolutely no understanding of basic probability concepts and that there was a high correlation between that attribute and beliefs in such things as astrology.
Quercus (what is your affinity with oak trees by the way?) my correction (4 posts up) made it clear that the question was about any two people having the same birthday rather than somebody having the same birthday as yourself.
From memory the article was called something like “sorting the sheep from the goats”
Just did a google looking for the article and came up with this:
http://www.psychologie.uni-bonn.de/sozial/staff/musch/propsi.pdf
An article in the British Journal of Psychology entitled “Probability misjudgement, cognitive ability and the belief in the paranormal”
It’s a study of 123 uni students that compared their belief in the paranormal with their ability to understand basic probability concepts (via practical tests like the one above) and cross-references the results to their uni marks.
It verifies the hypothesis about correlation between misunderstanding of probabilities and belief in the paranaormal but also says that its more prevalent in those with lower uni marks.
It’s a fascinating article if anyone is interested.
Maybe my girlfriend wasn’t as smart as I thought.
I can’t believe I am the only person who reports this event, but you have to realize that it must occur with a fair amount of frequency, and certainly leaves an impression.
My high school algebra teacher confidently predicts that at least two people in our class (31 people, counting him) share the same birthday. The mathematical explanation is promised, but first, we will check.
No one shares the same birthday. We have three people on consecutive days, and eleven people in one month. No one with the same day. Of course probability insists that such a class will exist. Math teachers have to explain that to you.
Tris