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114 Comments
- agnoster, on 10/12/2007, -0/+5Wow. I had no idea that the digg audience was so mathematically inept.
1) if you have 367 people (consider Feb 29th), I *dare* you to find a way in which no two people have the same birthday. Imagine you set up a square on the floor for each day of the year, and then progressively put people onto the squares. There are 366 squares and 367 people - if each person must be standing on a square then, yet, one of the squares must have 2 people standing on it. For God's sake, this is the freakin' pigeonhole principle here.
2) If you believe "probability is a long way from reality", I'd like to play some gambling games with you... the gambling industry makes billions and billions of dollars off of probability. If you don't believe in probability, you're really no better than the flat earthers. - SirBriggs, on 10/12/2007, -0/+4Heh. "5318008"
- copperhead, on 10/12/2007, -0/+3Rather a weak description of a birthday attack. As usual, Wikipedia has a pretty good writeup of how a birthday attack is a concern in cryptography.
http://en.wikipedia.org/wiki/Birthday_attack - geminitojanus, on 10/12/2007, -0/+2@magebomb:
The Birthday Paradox is mathematically proven, so well in fact that the Birthday Attack can be used to break certain cryptographic cyphers.
If you ever want to test it, walk into your doctor's office and ask everyone's birthday. You'll find that after a certain number of people, you're odds start improving drastically, for no reason at all except for this mathematic proof.
It's odd, it's interesting, but it's not a myth. (oh, and birthdays are pretty evenly distributed through the year; it's slightly a normal curve with more birthdays in the summer (after the parents have been shacked up all winter ;), but all in all, tis true. - Rhine23, on 10/12/2007, -0/+2The wikipedia article made more sense to me :/
- inactive, on 10/12/2007, -1/+3I think digg should have a minimum IQ threshold. Some of the people here are too dim to be allowed to continue living, let alone use this website.
- aznboi04k, on 10/12/2007, -0/+2this isn't a paradox! this is just statistic. at first it's not obvious but once someone explains it to you, it makes perfect sense.
- puck, on 10/12/2007, -0/+2@chicken101
You are misunderstanding the problem. The problem is not that any one individual in the room shares a birthdate with someone else. It is that ANY TWO people share the same birthdate. If you think along those lines it makes sense. - argonplatypus, on 10/12/2007, -0/+2@danthemanpr
Why don't the odds reach 100%? if there are 366 people, they are only 365 days on which they could have been born, so there has to be a repeating date, right? - smcgrath, on 10/12/2007, -0/+1Aye, good read.
- copperhead, on 10/12/2007, -0/+1chicken101 and magebomb both misunderstand the question, and why it's an issue. The question is not what are the chances that someone in a group of people will share my birthday... the question is whether *any* of those people will share a birthday.
It's important when dealing with hashes like SHA or MD5. Let's say I downloaded a RedHat ISO, and wanted to create an ISO with a trojan in it, but that created the same MD5 hash as the legitimate copy. Doing so is next to impossible.
However, let's say that I was responsible for creating the legitimate ISO, and I also wanted to create a copy with a trojan. In this case, I have a much better chance of finding a collision, since I control both files. I don't need to make a file with a particular hash... I just need to create a file that generates that same hash as the other.
Does that make sense? - Jugalator, on 10/12/2007, -0/+1"This is definitely not a paradox."
Agreed. :-s
Where the heck did that come from? - Jalexxi, on 10/12/2007, -0/+1How is this a paradox? I mean, the odds are about 11% if you are looking for someone who shares the same birthday with you, but that's not the case here. You can pick anyone from the room, and pair them up with anyone else. Hence, you haven't got a fixed birthdate to start with, you can choose from 40, and then compare it to 39 others. That greatly increases chances of finding a match.
- fadedecho, on 10/12/2007, -0/+1THIS IS DUMB!!!!!
Most math proofs are idealizations.
For instance "Assuming birthdays are evenly distributed throughout the year"
THEY'RE NOT!! People have more kids 9 months after christmas or valentines day or holidays or vacations etc. Ever wonder why so many people have birthdays in August?? This 'Assumption' is just not true.
It's not like we're matching a room of 40 people who each have a randomly distributed birthday, it's just not that simple. - tysonhy, on 10/12/2007, -0/+1Very interesting. +digg
- seanc28, on 10/12/2007, -0/+0@ waterdragon
what is total crap about the monty hall problem? The way the probability is distributed? - deepsub, on 10/12/2007, -0/+0"I stopped reading the article when I saw "Assuming that birthdays are evenly distributed".
@argonplatypus
That only works when the birthdays are evenly distributed."
Exactly.
If you increase the population, of course the likelihood of a duplicates increases.
Anyone who reads digg is intimately familiar with this phenomena. - WaterDragon, on 10/12/2007, -0/+0puck commented:
"Also, if you have 500 people in a room, are you suggesting it is possible that no two could share the same birthdate? Think about it!"
LMAO
Well, what if 136 of them were clones, and were never born? - Mousse, on 10/12/2007, -0/+0"Remember, common sense is that thing telling you that the earth is flat."
- yvovandoorn, on 10/12/2007, -0/+0This definitely seems right.
We have 21 people that work at our company. Three people share their birthday on the same day (incl. myself) on Feb 11.
Then there are two others born on other days in February (7th & 16th).
Thats approx 25% of our company born in the same month (and 18% of the entire company on the same day). - booc0mtaco, on 10/12/2007, -0/+0Oh, I see now the line of thought.
Let's rephrase... if you have 365+ people, and are looking for EXACTLY one pair with the same birthday, you get that 100% by selecting one more person.
If you are looking for AT LEAST one pair (2 pair, 3, 4, etc.) then use the above confusing prob. and stat.s.
No one is dumb... it is just a difference of = versus > - Wheeler, on 10/12/2007, -0/+0(N)
- WaterDragon, on 10/12/2007, -0/+0The link about the 'monty hall problem' is totally crap, and just an example of realy bad math, and pure stupidity pretending to be mathematics!
The ARROGANCE is astounding! - aggies11, on 10/12/2007, -0/+0Haha, this is awesome, and a perfect example of why it this "paradox" even exists.
Almost 50% of the comments, are using standard human intuition, which leads them to totally misunderstand / get the question wrong.
It's not about matching a specific individual's birthday. It's about finding a PAIR of two people, in the entire group, who have the same birthday.
If their is a group of 10 people, how many pairs of 2people are there? (Permutation theory says it's 10 choose 2 = 45). Thats right, out a group of 10 people there are 45 different ways to select a pair of them, 45 different pairs.
Now to see if ANY 2 people have the same birthday, you have to check each pair. So right there, that is 45 different chances of matching.
Now crank that number up to a room full of 40, and there is ALOT of pairs. Enough so that, chances are pretty good you are going to find a pair who have the same birthday.
Aggies - IKbot, on 10/12/2007, -0/+0"There are about 50 comments in here....Who has the birthday Oct. 25th like me?"
WEIRD. Mine is Oct. 26th!!!
It must be some sort of 8th dimension or a paradox. - seanc28, on 10/12/2007, -0/+0are you saying that if there are 366 people in a room that the odds arent 100% that two people share a birthday?
- seuss, on 10/12/2007, -0/+0I thought that the statistics were as simple as; every person in the room has a 1 in 365 chance of having a birthday on the same day as you. I wouldn't say that you could guarantee that two people had the same birth-date. I would say that is only highly probable.
- Urusai, on 10/12/2007, -0/+0Wow, I can't believe the stupidity.
Let us say there are 365 days in a year. If you have 365 people, but no two people with the same birthday, then how MUST the birthdays be distributed? Lessee: one for Jan 1, one for Jan 2, one for Jan 3, ...., one for Dec 31. That's 365 people, with one having a birthday on each distinct day of the year.
Add another person, for a total of 366. Guess what? His birthday MUST match one of the others, since every day of the year has a representative already. Ergo, chance of a match = 100% for 366 people. Thank you, statistics, for making people believe the dumbest *****. - seanc28, on 10/12/2007, -0/+0"Wow, I can't believe the stupidity.
Let us say there are 365 days in a year. If you have 365 people, but no two people with the same birthday, then how MUST the birthdays be distributed? Lessee: one for Jan 1, one for Jan 2, one for Jan 3, ...., one for Dec 31. That's 365 people, with one having a birthday on each distinct day of the year.
Add another person, for a total of 366. Guess what? His birthday MUST match one of the others, since every day of the year has a representative already. Ergo, chance of a match = 100% for 366 people. Thank you, statistics, for making people believe the dumbest *****."
Yep. The pigeon-hole principle says this has to be true. - Beacon, on 10/12/2007, -0/+0Statistics on the level you fools could understand always makes assumptions such that birthdays are evenly distributed. Stop trying to sound like geniuses by dissing a statistical principle because it makes an assumption that isn't true. If you do that, you'll dis basically every stat principle in the book, because dealing in unevenly distributed stats isn't easy.
Besides, the name of this process is only *called* "The Birthday Principle' -- it has absolutely nothing whatsoever to do with birthdays and has many applications in oth areas. - FreddyZ, on 10/12/2007, -0/+0"due to Math's convoluted reasoning..."
Kill me now - digitalunltd, on 10/12/2007, -0/+0Paradox? Seems really intuitive to me. If there were 40/360 numbers and none were the same that would be odd to me. if you look randomly pick 1/9 of 360 numbers and none are the same then wow, that seems odd! not the other way around.
- jccalhoun, on 10/12/2007, -0/+0Sigh, repeat after me: "This has nothing to do with the probability that anyone else has YOUR birthday." Yes, if you want to talk about how likely it is that someone shares my birthday, then you can NEVER guarantee that a group of random people will have YOUR birthday. You CAN guarantee that there will be two people who have the same birthday, but you can't predict whom nor can you predict what day the birthday will be. Just because you have never met anyone with your birthday doesn't mean that this is wrong. Just because you have met a million people with your birthday doesn't mean this is true. As Dr. Phil says, "It ain't about you!"
- spyres, on 10/12/2007, -1/+1It's cool and all, but this basic math problem has been old news for 40-50 years.
No Digg. - MOGua, on 10/12/2007, -0/+0One thing to remember is that the percentage chance DOES NOT mean whenever YOU are in a room with 40 people, the chances of someone having the same birthday as YOU is 90%. nope.
that chance is for SOMEONE, not necessarily you. (most likely not you) - Denamite, on 10/12/2007, -0/+0"There are about 50 comments in here....Who has the birthday Oct. 25th like me?"
Mine is Oct. 25th - gh3tto, on 10/12/2007, -0/+0With that matematical mind, you should actually put it to use in something that BENEFITS society.
- nonchallant0819, on 03/28/2008, -0/+0This is a great story... found this one through http://www.google.com
___________________________________
http://www.TopNotchCarpentry.com - qishi, on 10/12/2007, -0/+0"THIS IS DUMB!!!!!
Most math proofs are idealizations."
Ah. Of course they are. I'm sure you've seen a bunch of them. - jketch, on 10/12/2007, -0/+0Birthdays may not be perfectly distributed but its not like there is any particular day that is twice as likely to have people be born on it. This is one of those cases where a simpler model can be applied to a case and it is still close enough to have usuful results.
- Olle, on 10/12/2007, -0/+0@muffinking,
You misunderstand. Lets rephrase a bit. Consider it your task to fill a room with people and the requirement is that no two people can share the same birthday.
So if the first person you put in there has birthday 1st of January, then the next person can't. So you put the next person in there and you choose somebody with birthday 2nd of January.
It is still true that no two people in the room share the same birthday.
Now, you have to keep filling the room, and every person you put in there can not share a birtday with someone already in the room.
The reason that the probability will reach 100% eventually, is because you can put only 365 people, on a non leap-year, in a room where no two people share the same birthday. If you put one more person in there he will share birthday with at least one person in there. - inactive, on 10/12/2007, -0/+0muffinking - YOU'RE MISSING THE POINT. Yes no two people would have your birthday Feb 11th in your example. But lots of people would share the birthday Feb 12th. It's about at least two people having the same birthday, but not specifying what the birthday is.
Suppose I have a draw full of an infinite number of red socks and an infinite number of blue socks. How many socks do I need to have a matching pair? Not necessarily a blue pair or a red pair - just one that matches? - ChileanGoD, on 10/12/2007, -0/+0w00t!... today is my birthday!
- emiles, on 10/12/2007, -0/+0This is definitely not a paradox.
- ohsh1rt, on 10/12/2007, -0/+0I didn't realize the 5318008 untill just now
Ahh good times in 8th grade
-jeffrey - kidgenius, on 10/12/2007, -0/+0HappySwappy:
It's a very good book that I read over Christmas. I have a pretty good understanding of stats, but it was interesting to see things like the blood sampling, medical testing, etc., covered in there.
To anyone else that wants just a good basic understanding of stats (and not actually how to calculate everything, though parts of that are covered), I would recommned reading the book that HappySwappy recommended "Inummeracy" - DrEbola, on 10/12/2007, -0/+0This guy sounds like a Catholic priest. (Especially when he personifies math.)
- Olle, on 10/12/2007, -0/+0@Digger,
Hehehe... Don't you realise that Kris2pe is kidding. - inactive, on 10/12/2007, -0/+0I made a piece of software a while ago that's built to help people understand the Birthday Paradox. It's called the Birthday Grapher.
http://www.pushnshove.com/index.php?page=1 - seanc28, on 10/12/2007, -0/+0No its not a paradox but it definately goes against intuition. Its simple statistics using combinatorial probability but its still interesting. Nevertheless it can be easier to think about the problem as 1 - the probability that no two people sharing a birthday. @ fadececho, you're right birthdays arent distributed evenly throughout the year but assuming they are the "birthday problem" holds true. No its not that simple but it still is an interesting statistical problem.
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