Understanding the Birthday Paradox

why is 23 the number of people required for a probability of 50 % of two people having the same birthday? its my homework question^

Sir, I read it before the heading “Interactive Example” and hats off to you. You have explained it so nicely, that it can actually feel what is happening in this “paradox”!
Thanks a lot sir!

It is actually much easier than what is explained here.

The first person could have any birthday during the year.

This would be 365/365

The second could only have one birthday that matched (regardless of the number of people in the class.

This would be 1/365

Using the multiplication rule of probabilities that one event AND another will occur:

365/365 x 1/365
1 x 0.00274

0.00274

Sorry Josh, but it isn’t that simple and you are actually wrong. How do you use that equation to get 50.05% probability?
Certainly not having a go here, as this is what the website is about. However:

  1. the example you use doesn’t give an answer to the actual question involving 23 people.
  2. using your calculations is not the calculation of probably as you have the potential to get an answer greater than ‘1’, which in this circumstance you should never have an answer greater than ‘1’, that just isn’t probability… you need to look at the chances of someone NOT having the same birthday as someone else and work from there…

Straight from my Statistics class and WebAssign:

Same Birthday: Suppose two people are randomly selected from a class of 35 students. What is the probability that they have the same birthday? Round your answer to 3 significant digits*.
WebAssign will check your answer for the correct number of significant figures.0.00274
Correct: Your answer is correct.
seenKey 0.00274


*Significant Digits: Here are some probabilities expressed to 3 significant digits.
You start counting digits from left to right starting with the first non-zero digit.
0.123 0.0123 0.00123 0.102 0.350 0.300

Solution or Explanation
This is tricky because it doesn’t matter how large the class is. Also, it doesn’t matter what the first person’s birthday is. The probability that the second person has the same birthday is
1
365
= 0.0027397 ≈ 0.00274
to three significant digits.

Josh…it matter immensely how big the classroom is…

Under your example of 35 people you have 1190 pairs (35x 34)…factor in that is pair is twice (1:you & I, 2: myself and you) and you have 595 unique pairs.

lets go PROBABILITY calculations now…

Chances of someone NOT having the same birthday as the other
364/365 = 0.997260274. Ans ‘y’ 595 = 0.1954649. 1 - 0.1954649

Chances of two people having the same birthday is 80.4536%

Use the same maths as above for 25 people and indeed you get you 50.05%.

In your calculation you haven’t factored in the most important part of the calculation…the amount of people in the equation :slight_smile:

In two of my class (one a class of 20 and the other class of 21), there are two other people that share the same birthday as me, so 3 total. I find that really weird.

All depents on how frequently those births happen. Look at the below diagram, http://www.todaysparent.com/blogs/on-our-minds/birthdays-most-common/, the actual probability for someone to have a birth in a certain date is not equal for all 365 days. There are factors, we dont know (probably weather , moon etc), that pushes a lot of people to have birth in Jun, Jul, Aug, Sep, Oct. So that explains the probabilty to have 2 people in a class with same birthdate since most people are between those months, like 80% are in that range of 150 days.

That’s confusing.

Why you, for “x!”, do not use Stirling formula: x!=x^(x+0.5)exp(-x+1/(12x)-1/(360x^3)+1/(1260x^5)-1/(1680x^7)+1/(1188x^9) …), (“x” do not need to be whole number, but it have to be greater than, say, 10) so something like n!/((n-m)!*n^m) (n=365, m=23, for example), may look like (n/(n-m))^(n-m+0.5)*exp(-m+(1/n-1/(n-m))/12-(1/n^3-1/(n-m)^3)/360)+…), and this way it is not problematic to calculate “complementary probability”? After that you subtract that from one and … Of cause “IEEE float” format is not accurate enough for this formula, and it does not allow argument of exponential function to have absolute value greater than 88 or so, but that is a story for another day …

… so 1000!/(858! * 1000^142) is 2.69985824…e-5, and probability that among 142 “items” that share 1000 “unique item numbers” two or more have the same “item number” is 99.99730014176% (the last digit is “rounded up”), and it is not so close to 100%, even for “IEEE float” format, to be rounded to 100% …

… I am sory, but I forgot to mention that exp(-142) is too small for “IEEE float”. I swear: no more comments from me in a foreseeable future :slight_smile:

The probability of at least two people sharing a birthday in a group of 22 people is about 50.72972343239854072, not 50.05%. Kalid’s method faultily assumes that the probability any pair sharing a birthday is independent of the probability of another pair sharing a birthday, which is not the case because the pairs contain some of the same people. The exact probability of at least 2 people sharing a birthday out of a group of x people can be calculated by the formula 365Px/365^x, or 365!/((365-x)! 365^x).

Correction: the above is the probability of at least two people sharing a birthday out of a group of 23, not 22, people.

You really want to blow someone’s mind?

with a true random selection of 230 people, merely ten times the birthday paradox, there’s almost a 50% chance of not only having two people with the same birthday - but two people with the same birthDATE. (with a 100 year pool.)

Mathematically, it says that number is 191.11… (365.2425*100 = 36524.25 sqrt(36524.25) = 191.113186…)
However realistically, it doesn’t close in on 50% until you get above 220s
[365.2425 is the actual days per year to take leap years into consideration.]

And if the people are randomly selected from a certain pool - such as a college population - the chances increase greatly, obviously…

Isn’t math fun?

And if your population group was born born nine months after a natural disaster ?? ?? No electricity for the television, doesn’t everything get skewed ? ? ? ? ?

I understand the math part of this question but I have to answer this question in sort of algorithm way because we are using computers so can someone help me.

u suck

hi

Working on this problem for class and I found this website to not provide much depth and to be a little mathematically shaky. I stumbled upon this website which gives a more in-depth explanation: www.birthdayparadox.com .