the smallest number of people in a room where the probability of 2 of them havn the same b-day is at least 50%

Answer 1

The probability in a room of 23 people is OVER 50%, are you sure that's not what you are looking for?

http://projects.felipc.com/birthday-paradox/
http://en.wikipedia.org/wiki/Birthday_paradox

Answer 2

Yes, the answer is 23 people, although I think you meant "at least two people" having the same birthday, rather than "...[just] two people having the same birthday." That's what all your "correct" responders have assumed.

I answered this question for my father at his request, in England, when I was 12 or so. Because of my answer, he then made bets about it quite confidently, with any takers present, whenever he was in a pub with about 30 or more people in it. Making sure to say "at least two people present ...", he would take bets on both current and future beer purchases. By betting in this way, he was largely kept in beer for almost 50 years !

Here is the logic of this answer. A calculator is almost indispensable. (I first did it by laboriously using look-up paper logarithm tables !)The argument proceeds by exclusion, and I'll ignore leap years:

Choose an arbitrary person #1. The chance that person #2 has a DIFFERENT birthday is 364 / 365. Then the chance that person #3 has a DIFFERENT birthday to either of these is 363 / 365, so the combined chance that both are different from #1 is 364*363 / (365)^2. Similarly, the chance that person #4's birthday is different to the preceding 3 is is 364*363*362 / (365)^3, etc. If you continue working out similar products, you'll find that once you're considering a total of 23 people (meaning there are 22 fractional tems separately multiplied), the resulting probability is less than one half, or < 50% .

Since the probability that ALL the birthdays are DIFFERENT is now less than 50% , the probability that AT LEAST TWO birthdays are the same is now greater than 50% . I hope that you really MEANT "at least two people." It would in fact be a different problem to work out the probability that JUST two people have the same birthday. Off hand, I'd think it would be harder. ALL the other probabilities: 2 separate PAIRS of people having the same birthday, ... 7, 5, 4 and 3 people separately sharing birthdays (!) etc. are ALL part of that fraction > 50% .

Live long and prosper.

Answer 3

Yeah, that's the birthday paradox, an interesting enough quesiton in its own right and also it has led to solving other applications in networking and information structure design. The proverbial chain is only as strong as its weakest link. However, the general solution to this question gives estimates to the number of weak nodes that a system can tolorate, before it fails. This in turn has led to the highly reliable yet very inexpensive data handling devices such as personal computers, calculators, phones, radios, etc., that we all rely on so much at this time.
Skim the wikipedia link (cited by Jerry P) for the gist if you're not that technical oriented. Or better yet, read Dr. Spock's explanation, above. Yet another example of the magic of 23!

Answer 4

You have stated the Birthday Paradox.
The accepted answer is 23.
http://en.wikipedia.org/wiki/Birthday_paradox

The problem is asking if *any* of the 23 people have a matching birthday with *any* of the others — not one in particular.

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Answer 5

I think it's like.....23, but that's a hip-shot. I know it's a small'ish number tho.

Answer 6

183 people.

The chance that somebody has a particular date as their birthday is 365.25 (Leap year accounts for the .25).

If there's 183 people, there's a little over 50% chance that two people have the same birthday.

Answer 7

366 just add one more person than the number of days.

Answer 8

2

if you'r talking of BAD-day