The Birthday Bet

I am working on a post about risk and decision-making and I need your help.

I am trying to get a feel for the risk numeracy of the average doctor – I realise that none of my readers is “average” but please take the time to read the following scenario and give me an honest answer.  Please try to avoid Googling the “right” answer – I am after your gut feeling.  I want to know if this were you – and you had to take a guess what would you say?  It is not about the maths – it is about how doctors think…  yes I am data mining your brains.  Sorry.  That is the sort of thing I do for fun on nights here in Broome!

OK here is the scenario;

You are invited to the Hospital Xmas Party by a nice young nurse.  Unfortunately they tell you the wrong time – so you turn up in you fancy dress Chicken Suit an hour early.  Not sure what to do – you decide to engage in idle banter with the Hospital Tea Lady who is carefully arranging her sausage rolls on a long table.  After 5 minutes you discover that you both lived on the same street in Wagga Wagga – although she was there in the 1950s, 30 years before you were born.  Nice coincidence!.

After 20 minutes the conversation stalls a bit, and during and awkward silence she proffers the following bet:

I’ll bet you that there are two people coming tonight who share the same birth date – not necessarily the same year – but the same date eg. the 6th of May…

Of course there are 500 people coming to the party – so it is a pretty easy bet.

OK, to make it interesting.  Lets ask each person as they arrive their birthdate.  And see how long it takes before we have a match i.e.  a pair with the same Birthday.

OK – sure.  How many people do you think?  Tell me how many people need to arrive at the party before it is a better than 50:50 bet that there will be a shared birthday?  That is – when would it be in your favour to take the Tea Lady’s Bet

Thanks for putting your mathematical pride on the line.

The answer to the questions is….. *drum roll*….  23.

Yes, just 23 – this is much lower than most doctors answer when asked to give a gut feeling about the problem.  In my own offline informal torturing of friends and coworkers the usual answer is somewhere between 150 & 200 partygoers.

So how did the FOAMed crowd do on this version of the famous betting dilemma?

Here is the latest spread:

Screen Shot 2014-12-10 at 9.35.10 pm

Assuming that a heap of smart folk either knew the answer (and did not rely on intuition) or Googled the puzzle and cheated (once again – no intuition) – the correct answer was the most frequent answer.

What I was really after here was a measure of the average Docs “intuitive” risk / probability instinct for problems where the answer is calculable but there is uncertainty.

I think it is pretty likely that the average doctor went with 183 (i.e half of 365) as a best guess. There was a spread around this number if you exclude the folk who got it right.

If you want to see how the actual mathematic probability equations work – check out MATH IS FUN here

Of course if you are really into n! factorial equations then there is a 99% chance that you do not actually have 23 friends whom you might invite tot your Birthday – so its all a bit academic really!



  1. As a maths nerd in high school, I’m pretty sure I know the answer to this!

  2. 200. Just a random guess, what feels right -- ie more than half the days in a year.

  3. I think there’ll be a minimum of 135 people with same birthday out of 500 so I don’t think it will take long -- perhaps only 20-30 until a >50/50 chance as a guess..

  4. I think there’ll be a minimum of 135 out of 500 with same birthday so won’t take many. perhaps 20-30 as a guess?

  5. Jim Holland says:

    I seem to remember a problem like this in high school maths (30 years ago!). My sense is that it’s a low number. I picked 23

  6. thomas coombs says:

    back of the envelope calculations
    Chance of being born on any day assumed to be 1/365.25
    We have a remaining pool of 499 people
    It is by no means certain that another person will be in the group with the same birthday but the odds ratio favour this possibility -- 1.37
    If there is actually a person in the group the each person arriving has a 1/365.25 chance of matching the first persons birthday
    The cumulative chance that another person with the same birthday has arrived exceeds 50 % after the 182nd person but this does not mean that you can consider the next person arriving as a 50 50 chance of sharing the first persons birthday.
    Each person arriving still has just a 1/365.25 chance of sharing the first persons birthday.

    I don’t think your odds will ever be 50:50 unless you are tipped off that there is a person in the group which shares the birthday and even then will not be 50:50 until they haven’t arrived until after the 497th person.

    your question demonstrates the difficulty of conceptualizing low risk possibilities applied to actual people and groups, i couldnt pull any more complex math from distant stats and high school to help me

  7. Ewen McPhee says:


  8. Ok, I bit. More than 12, far less than 1/2 of 365, and then I’m grasping. Hate that I looked at this when I was on a stats kick years ago and simply forgot. Reading “How Not to Be Wrong” which surely has the answer. Except I think I got it wrong.

  9. When I taught computer science, this was part of my usual introductory spiel to Poisson distributions and hash tables.

    As a corollary quiz question, how many people would you need for two to have the same star sign? This is a much more tractable problem to do in ones’ head (rather than just knowing the answer, like Michael and me), but I wonder how many people would get it.

  10. If the 1st 365 DON’T share a birthday, then as soon as the 366th arrives (ignoring leap years), they have a 100% chance of sharing a birthday with someone already present.

    But you asked when it becomes 50% likely.

    For a Permutation:
    1st guest has a certain birthday.
    2nd has a 1/365 chance of matching the original birthday.
    A 3rd means there are now 2/365 and so on.
    We add guests until we get 182.5/365 -- 50%

    For a combination:
    2nd guest has 1/365 chance.
    3rd guest has only to match one of the first 2 guests’ days.
    4th has a 1/365 chance to match.
    Progressively, this looks like 183 total (original + 182 newbies) = 50% chance.

    BUT: Each guest must match only one other (of the growing new arrivals).

  11. It is easier to work out this problem by working out the converse -- what is the probability that NO ONE shares a birthday (let’s call this P). Then, the probability that someone DOES share a birthday is (1 -- P).

    First person -- the probability that NO ONE shares a birthday is obviously 1 (since there is only 1 person). That is, P(1) = 1

    Second person, the probability that NO ONE shares a birthday is 364/365, IN THE SETTING of the the prior probability of 1 person.
    That is, P(2) = 364/365 * P(1)

    Third person, the probability that this additional third person also SHARES their birthday with no one so far is 363/365, IN THE SETTING of the prior probability of 2 people.
    That is:
    P(3) = 363/365 * P(2)
    P(3) = 363/365 * 364/365 * 1

    You can probably see where this is headed. For the fourth person:
    P(4) = 362/365 * P(3)
    P(4) = 362/365 * 363/365 * 364/365 * 1

    Thus, for n, where n is a integer between 1 and 365:
    P(n) = (365) * (365 -- 1) * (365 -- 2) * … * (365 -- n + 1) / (365)^n

    I computed the values of P(n) up to n = 25:
    n P(n)
    1 1.000
    2 0.997
    3 0.992
    4 0.984
    5 0.973
    6 0.960
    7 0.944
    8 0.926
    9 0.905
    10 0.883
    11 0.859
    12 0.833
    13 0.806
    14 0.777
    15 0.747
    16 0.716
    17 0.685
    18 0.653
    19 0.621
    20 0.589
    21 0.556
    22 0.524
    23 0.493
    24 0.462
    25 0.431

    At 23 people, the probability that NO ONE shares a birthday drops to 0.493, just below half. Thus, the probability that someone DOES share a birthday (1 -- P), is just above half. Therefore, 23 is the correct answer to the puzzle in the post.

    Hildy suggested doing star signs. Using the same reasoning as above:
    Thus, for n, where n is a integer between 1 and 12, and P(n) is the probability that NO ONE shares a star sign:
    P(n) = (12) * (12 -- 1) * (12 -- 2) * … * (12 -- n + 1) / (12)^n

    Computing the values of from 1 to 12 people:
    n P(n)
    1 1.000
    2 0.917
    3 0.764
    4 0.573
    5 0.382
    6 0.223
    7 0.111
    8 0.046
    9 0.015
    10 0.004
    11 0.001
    12 0.000

    At the fifth person, there is a better than average chance that at least 2 people will share a star sign.

    • Right, it’s very tractable -- but if you gave it as a quiz, I wonder who would choose 4, 5, 6, or 7 -- the numbers are so close together that you would get a lot of smooshing of the distribution, I suspect.

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