
Ask HN: A random maths problem we started discussing at work - cleis
How many people would you need in a group to make it likely that at least one member had a birthday on each of 365 days of the year?<p>Let&#x27;s ignore leap years and define likely as &gt;50% probability.
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wnoise
This is the coupon collector's problem.

[http://en.wikipedia.org/wiki/Coupon_collector%27s_problem](http://en.wikipedia.org/wiki/Coupon_collector%27s_problem)

The formula in the article (1/2 + n*gamma + n log n) gives 2365 for the
expected number, but this is different than "at least 50% chance".

~~~
metaphyze
I don't think this is exactly the same as the proposed problem. What it's
giving you is the expected value. This includes in its calculation the lower
probability cases and the higher probability cases. For instance, there's a
slim change of this happening when the number of employees is 365. It's been a
while since I took probability, but I think the expected value is going to be
larger than the minimum value needed to achieve a .5 probability if the
probability graph is skewed, which in this case it is.

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DanielStraight
I have no answer for you. I mean... the solution seems to be 2121 or 2122 (no,
it isn't, see below) determined by brute force, but I don't know why.

But, it's worth noting that if this interests you, Project Euler almost
certainly will as well and is worth checking it out if you don't already know
about it:

[http://projecteuler.net/](http://projecteuler.net/)

EDIT: So if you're wondering why I got a different brute force result from
other users, it's because I had a stupid bug. I was looking at the ratio of
meeting the criteria to the ratio of not meeting it, not the ratio of meeting
it to total.

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metaphyze
Brute force gives me about 2287/2288\. Trying each number 1,000,000 times
(distribute X balls in 365 bins, count number times all bins are filled) so it
seems reliable, but maybe I have bug somewhere:

import java.util.Random;

public class Birthday {

private static Random rand = new Random(System.currentTimeMillis());

public static void main(String... args) {

    
    
    		final int n = 365;
    		boolean[] bins = new boolean[n];
    
    		final float maxNumberOfAttempts = 1000000;
    		for (int numberOfBalls = 2285; numberOfBalls <= 2400; ++numberOfBalls) {
    
    			int attempts = 0;
    			int numberOfTimesAllBinsFilled = 0;
    
    			while (attempts < maxNumberOfAttempts) {
    				if (distribute_X_Balls_in_N_bins_randomly_and_return_the_number_of_filled_bins(
    						numberOfBalls, bins) == n) {
    					numberOfTimesAllBinsFilled++;
    				}
    				++attempts;
    			}
    
    			float prob = numberOfTimesAllBinsFilled / maxNumberOfAttempts;
    			System.out.println((prob > .5 ? "PASSED:" : "FAILED:") + numberOfBalls + ":" + prob);
    
    		}
    	}
    
    	public static int distribute_X_Balls_in_N_bins_randomly_and_return_the_number_of_filled_bins(
    			int x, boolean[] bins) {
    		for (int inx = 0; inx < bins.length; ++inx) {
    			bins[inx] = false;
    		}
    
    		int whichBin = 0;
    		int total = 0;
    
    		while (--x >= 0) {
    			whichBin = rand.nextInt(bins.length);
    
    			if (!bins[whichBin]) {
    				bins[whichBin] = true;
    				++total;
    
                                    if (total == bins.length) {
    					return total;
    				}
    			}
    		}
    
    		return total;
    
    	}
    
    }

~~~
jackgolding
Oh God that function

------
dded
Here's my take:

Let n be the number of people needed. Probability that a person's birthday is
not today: 364/365

Prob. that no one's birthday is today: (364/365)^n

Prob. that someone's birthday is today: 1 - (364/365)^n

Prob. that someone's birthday covers every day: (1 - (364/365)^n)^365

Setting equal to 0.5 and solving, I get n==2284, which is close to metaphyze's
brute force number.

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darkxanthos
Here's a link to the wikipedia entry on the birthday problem.

[http://en.wikipedia.org/wiki/Birthday_problem](http://en.wikipedia.org/wiki/Birthday_problem)

~~~
DanielStraight
This isn't the birthday problem. OP is asking how many people to hit _every
possible birthday_ at least once. I thought Wikipedia might cover this on
their birthday problem page, but they don't.

------
minussohn
are you too stupid or too lazy to use a proper web search engine?

------
chaoticgeek
This solves the problem for you.
[http://en.wikipedia.org/wiki/Birthday_problem](http://en.wikipedia.org/wiki/Birthday_problem)

23 people is a 50% chance, 57 is a 99% chance, 367 is 100% chance (including
leap years).

~~~
apourchet
His question isn't related to the 'Birthday Problem'. Read the whole post...

