Skip to content


December 16, 2012

Three problems about birthdays this time. Each solution needs a different way of looking at the problem – very common when dealing with different ways of arranging people or objects, which an area of maths called permutations and combinations.

In my answers (in detail in the next post – just some hints here), I ignore both of the following in the general analysis, but comment on their influence on the problem outcome.
1. The existence of February 29
2. Some example data here shows that frequencies of birth dates are spread over a range of about 20% from highest to lowest, excluding a handful of extra low frequency dates (e.g. Jan 1, Jul 4, Dec 24, Dec 25). There also seems to be a weekly cycle, with about 2% less on weekends – presumably when doctors are playing golf.

1. Birthday pairings

How many people must be in a group before there is a better than 50% chance of having at least two people in that group with the same birthday?

With many problems like this, where the question has the form “At least one …”, it’s often easier to first work out the probability of the opposite problem. Here this means no pairings – no people with the same birthdays.

2. All dates covered

I had dinner with some old 🙂 university friends last night. Mike actually mentioned the above problem, but said he was wondering…

How many people does he need to add to a birthday book before he has a 50% chance of having at least one person for each of the 365 dates?

I thought about this for a while. Mike doesn’t care how many people in each date, just that there’s at least one person for each date. So each date is either live (awaiting its first entry) or not (already has at least one entry).

3. Get in line

Imagine a line of people, one behind the other. You get a prize if you are the the first person in the line (from the front person, working backwards) to have a person ahead of you with the same birthday.

Where should you stand (in which position in the line), to have the best chance of getting the prize?

Given a solution for problem 1 above, this is actually the easiest to solve.
In fact, it turns out that we can even ignore most of that analysis.

  1. Sean permalink

    I have a different problem. How many people do you need in a group on a given date, say the 1st of April, before a member of the group has their birthday on that date
    Say you go to a football match, how many spectators does there need to be before the probability approaches 1 for one of those spectators being there on their birthday?
    Seems simple but I just can’t get it.

    • This is actually quite similar to the first problem…
      You want the probability that at least one person in the crowd has a birthday on, say, 1st of April.

      Again, try to get the probability of the opposite problem – that of there being no person in a crowd with that birthday.
      If there are 365 days in a year, then the chances of each person in the crowd not having that birthday is 364/365.
      For n people, this becomes (364/365)^n.
      Then (364/365)^n<=50% when n>=253, and (364/365)^n<=10% when n>=840.
      So, if you want a better 50% chance that someone in a crowd has a birthday on 1st of April, this is when there are at least 253 people.
      If you want a better 90% (i.e. 10% – as above – less than 100%) chance that someone in a crowd has a birthday on 1st of April, this is when there are at least 840 people.
      If you want an even better 99% chance that someone in a crowd has a birthday on 1st of April, this is when there are at least 1679 people.
      The corresponding crowd sizes for 29th of February are 1013, 3363 and 6726.

Trackbacks & Pingbacks

  1. Birthdays – Get in line | mathsevangelist

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: