Ask HN: A random maths problem we started discussing at work

https://news.ycombinator.com/item?id=6895950

by cleis • 13 years ago

8 12 13 years ago

news.ycombinator.com

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?

Let's ignore leap years and define likely as >50% probability.

minussohn• 13 years ago

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

DanielStraight2• 13 years ago

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/

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.

chaoticgeek• 13 years ago

This solves the problem for you. 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• 13 years ago

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

darkxanthos• 13 years ago

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

http://en.wikipedia.org/wiki/Birthday_problem

DanielStraight2• 13 years ago

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.

wnoise• 13 years ago

This is the coupon collector's 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".

brd• 13 years ago

Well... you made my day complete. My brute force approach got me 2364.646, glad to know my math intuition isn't entirely broken.

Thanks for the link!

metaphyze2• 13 years ago

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.

metaphyze2• 13 years ago

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• 13 years ago

Oh God that function

dded• 13 years ago

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.

Ask HN: A random maths problem we started discussing at work

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