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[…] have the same birthday. Seem counter-intuitive? Here’s a great description of the paradox: http://betterexplained.com/articles/understanding-the-birthday-paradox/ For Bamboletta drawings, let’s assume that there are 21 drawings, and 200 entrants each. In […]
…and then sometimes I put too much egg white into the batter…
oh dang it!! Wrong website!!
Hi KGW, not sure what you mean about samples. I.e., you tested 14 samples (of 23 people), and of those 14 samples, 12 had a match? Yep, in theory it should be about half, but with a relatively small population, it’s easy to skew. Also, in the real world, birthdays probably aren’t perfectly evenly distributed, and the “clumpiness” may make matches easier.
For the purposes of the paradox though, it’s still startling that such small groups have so many matches! (So I think the experiment still makes that point :)).
@Joe: Actually, in this case it means among your 130 friends it’s almost a 100% chance that two of them share a birthday (not necessarily with you though! Friend A and friend B could have a birthday in common).
@sonny: Great question – I don’t think my probability knowledge is strong enough :). The issue is you need to enumerate every possible type of collision: 1 with 3, 1 and 2 with 3, 1 and 3 and 14… all of which are “problem scenarios”. It’s a bit like writing a spellcheck where you keep track of the possible typos vs. having the correct word and seeing if what you wrote is different from that :).
Thanks Khalid. So is there a way to solve solve this without using the ‘negative’… that is not by calculating the probability of someone else in the group not having the same bday? Do it directly instead?
On average, Facebook members have befriended 130 other Facebook members, which means there is nearly 100% chance that every average member of Facebook shares a birthday with a person on his/her friend list. If any staff member from Facebook is listening, I’d like to know if the outcome matches the theory!
I think you discount the formula at the beginning of Appendix A (1-1*(1-1/365)(1-2/365)…) too much by jumping immediately into an approximate shortcut. Let’s see what happens if we simplify that equation first.
The denominator of the equation is simple to work out - it’s 365 multiplied by itself as many times as there are people. For x people, the denominator will be 365^x.
The numerator also has a familiar pattern. For x people, It will be 365364363*…*365-x.
So, we have a pattern something like a factorial, but that stops after x numbers. How do we handle that? Yep, it’s our old friend the permutation formula!
So, the short form of the formula would be P(365,x)/365^x. Writing the long form of the formula, we end up with: x!/(((365-x)!)365^x)
Prefer to think of it with combinations instead of permutations? Permutations are just combinations with redundancies taken into account to focus on particular orders of events, or mathematically: P(365,x)=(C(365,x))x!
This makes the full formula: ((C(365,x))x!)/365^x
Yes the formula you write out at the start of Appendix A looks bad, but it simplifies quickly to a clear and understandable form. Examining it several ways in terms of combinations and permutations helps make it clearer.
As I wrote the formula above, that is, of course, the formula for no 2 people sharing a birthday.
To find the probability of at least 2 people sharing a birthday, as mentioned, we still need to subtract all that from 1.
I used the Birthday Paradox concept in a math project of mine, and they told me some professor objected to it cause it’s his idea. It was my research and they were my results and raw data. What do you think i should do, submit it or change it?
Thank You for the awesome facts!!! I love it. Really helped with my algebra project xD. 
Why doesn’t the following work :
We start with the first person in the group. The probability of another person in the group with the same birthday is 22/365 (since the probability of any one person having the same birthday is 1/365 and these are independent probabilities). Then we go to the next person. There is 21/365 chance of finding another person in the group with the same birthday. The next one is 20/365 and so on. And since these are all independent if each other we can add the probabilities, which gives us 253/365. This is the probability of finding 2 people in the group with the same bday.
What am I missing here?
Thanks!
@sonny: You can’t add the probabilities :). By that reasoning, if we had 30 people in the group, the chance would be (1 + 2 + … + 30) / 365 = 435 / 365, which is greater than 1. It is the right idea to consider each pair, though.
Something doesn’t add up here. The first calculator shows that the birthday example with 365 persons would result in a 100% match, meaning at least 2 persons should have the same birthday. But it’s possible that all 365 persons have different birthdays (the first person born on January 1, the second on January 2 and the last on December 31).
Indeed “only consider the scenarios we’re involved in”. Thanks for the remanding.
[…] you’d like a further explanation of the birthday paradox, BetterExplained has step-by-step instructions, as well as a birthday paradox calculator. And if you have an […]
Hi Steve, great question. I might try this: put the kids in groups of 10 and have them guess (before they start) how many handshakes they need so everyone in the group shakes hands. They might guess 10 (or 9, since that’s how many THEY need to do), but you’ll see it’s quite a large number (10*9/2 = 45). In the same way, the number of “birthdays to check” is not you against everyone else (22), but everyone by everyone (a much larger number). In rough terms, that’s why the odds are much closer to 50-50 instead of 22/365.