In the first Newsletter, Clark & John’s probability whizz-kid Rob Clark, informed us of the actual chances of winning a prize on the UK National Lottery. The favourable feedback generated by the Lotto article, has prompted Rob to turn his spotlight this time to birthdays.
To begin, let me pose a question to you…….
“How large should a group of people be, to make it more likely than not that at least two of them share the same birthday?” Remember your answer.
The answer, like many things in life is often a bit different to the one that most people would guess at initially.
Apparently, slightly more people are born in the summer than the winter and of course, February 29th only comes around every leap year but setting these two observations aside for the moment, we can assume that birthdays can arise on only one of 365 days and each one is equally likely.
The simple method to apply here (and elsewhere in the mathematical world) is, if a problem appears difficult to calculate, calculate the chance of it NOT happening and then subtract it from 100%. This method should help us here.
(For the more mathematically inclined reader, we have either two positions; either all birthdays are different in the group or, at least two of them coincide. We should calculate the chance (probability) that everybody in the group has a different birthday, then subtract this from 100%. This will then give us the probability, of at least one pair having the same birthday).
So, starting off with just two people in the group, the second person will have a different birthday 364 times out of 365. A group of three people, will have different birthdays when the first two are different and the third person has a birthday in one of the remaining 363 days. This means that the total chance of all three people having different birthdays is:
(364/365) x (363/365) which is about 99.18%
The chance that a group of four people have different birthdays is as above but with the fourth person then having their birthday on one of the remaining 362 days, hence:
(364/365) x (363/365) x (362/365) which is about 98.36%
I’m sure now, you will see a pattern emerging and it’s sensible to predict that the larger the group of random people we look at, the probabilities decline as the group gets larger, that they all will have different birthdays.
Using this method, we seek to find the number of people in the group required, where the chances of different birthdays crosses over from above 50% to below 50% chance. This point is where the odds change from favouring all different birthdays, to that of favouring at least two people having the same birthday.
What answer did you give for the starting question above?
The answer 183 is fairly popular, being just more than a half of 365. It’s rare to get initial guesses of below 40 and yet, the answer is……23!!!
For those that don’t want to do the longhand arithmetic, here’s the answer.
Using our method above, the chance that a group of 22 people have different birthdays
is 52.43%. The chance that a group of 23 people have different birthdays declines to 49.27%. So 23 is indeed the crossover point and is the point where the chances of two people having the same birthday, is more likely than not.
For conclusive proof of this, the science writer Robert Matthews, carried out an experiment and collected the birthdays of each 22 players and the referee, from the starting 10 Premier League football games at the opening day of the 1997 season. Six of these games had some matching pair, with the other four all with different birthdays. It would be difficult to find a more convincing real life result than this!
Lastly, I’ve had quite a few clients that gave their praise about my last Lottery probability article, which is much appreciated. For those interested, consider this…….
A pub syndicate in Kent, in 1997, won nearly £11 Million, when they won two of the five shares of a double-rollover jackpot. Their intention was to spend £28 to buy all combinations of six numbers from a set of eight. However, the man entrusted to filling-in the tickets made a mistake; instead of twenty-eight combinations he only entered twenty-seven and one combination was duplicated. By great fortune, that duplicated combination was the winning one!
The chances of that happening, well………….