Birthday Question Answered - in full!

(July 03, 2020)

Birthday Problems – just how common are shared birthdays?

We all know the surprising result that in a group of 23 runners there is a 50% probability of a common birthday – assuming birthdays are evenly distributed throughout the year.  (Do consult Pete Francis as to why for romantic, biological and other reasons this may well not be the case.)

But what about this variant: how many runners do you need on the start line so there is a 50% probability that one of them has your birthday?

As Mark posted on the website yesterday: Meanwhile - a challenge for James Smyth's maths group and our treasurer:  of "62 random runners is it statistically anomalous that 2 pairs will share birthdays as well as one triplet of individuals"....Answers to the Treasurer.

So, we have a group of 62 runners:

  • 55 do not share a birthday within the group
  • 2 share a birthday
  • 2 more share another birthday
  • 3 share a birthday

Why has Mark passed this over to me?  Well, for those who attended our recent AGM, and were fortunate enough to hear me deliver another riveting treasurer’s report, you may recall that towards the end I lapsed into discussing the distribution of birthdays of all 271 members. Those of you who were still awake as I neared the end might recall that five members share a birthday, five sets of four share birthdays – and there were triplets and pairs.

Many problems in probability can be characterised by throwing balls into urns.  (I once used this very expression when giving a talk at Warden Park to their young maths boffins and I heard a lad in the second row turn to his mate (this was pre-Covid 19) and whisper “I wouldn’t want to put my balls in an urn.”)

The balls and urns may be distinguishable or indistinguishable; in the first birthday problem above the urns are treated as indistinguishable, whereas in the second problem they are distinguishable.

So, here we are throwing 271 balls at random into 365 urns.  How many balls will land in each urn? (Each urn represents one of the 365 (birth) days in the year)

Balls     A         E       

0          172      174

1          137      129

2            41        48

3             9         12

4             5           2     

5             1       0.3

So, for example, 172 urns remain empty – no Harrier has a birthday on these 172 dates; 41 urns contain 2 balls – there are 41 days of the year on which two Harriers share a birthday; one urn contains 5 balls (this urn represents 23rd January, which five Harriers share as their birthday.)

Column E contains the “expected” number and is derived using statistical theory.  As you can see the actual distribution of balls in urns is consistent with expectation, and a goodness-of-fit statistical test would reveal “no statistical anomaly”

Let’s now turn to the subset of Harriers who rocked up for the Blackcap Handicap.

Balls   A          E       

0          307      309

1          55         52

2            2           4

3             1       0.2

4             0           0

5             0           0

Again there is no statistical anomaly.

If anyone is still alive having got this far then lockdown has either increased your boredom threshold or, more likely, whetted your appetite for learning!

If you would like to learn about the theory that produces the expected values than I’m sure time can found at the next AGM – even if it means delaying the awards ceremony.





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