I stumbled upon the following probability problem (translated from Swedish):

Calculate the probability that, out of 23 people, at least two of them have a birthday on the same day. Assume that a year has 365 days and that all birthdays are equally probable to be a birthday. Tip: Utilise the complementary event.

My Attempt

Based on my reading of the question, it seems to suggest that the classical definition of probability should be used to calculate the probability, e.g $P(A) = \frac{|A|}{|\Omega|}$, where A is the given event and $\Omega$ is the universe of outcomes. Furthermore, to solve this I tried to calculate $P(A) = 1 - P(A^*)$ instead. I defined the following in my solution:

  • $\Omega :=$ {"all bags of exactly 23 birthdays"}. I said bag because I believe that the order does not matter (e.g person 1 having birthday 51 or person 3 having birthday 51 does not matter), while at the same time multiple instances of the same value should be able to be represented in the bag. $|\Omega| =$ ${365 + 23 - 1}\choose{23}$ (I used the formula for choosing k elements from n where order does not matter and the element is returned after being chosen once).
  • $A^* := $ {"all bags of exactly 23 unique birthdays"}, $|A^*| = $ ${365}\choose{23}$. I chose combination and not permutation as I believe this reflects having a bag (i.e order does not matter) of exactly 23 unique birthdays.

However, this leads to an incorrect answer when the calculation is carried out.

Correct Solution

The correct answer uses $\Omega :=$ {"all tuples of exactly 23 birthdays"} and $A^* :=$ {"all tuples of exactly 23 unique birthdays"}, i.e tuples instead of bags. This means they care about who had which birthday.

They get the result $P(A^*) = \frac{\frac{365!}{(365-23)!}}{365^{23}}$, and carrying out the rest of the calculation (i.e 1 - Ans) yields the correct result.

My Question

I sort of understand why the provided solution works. What I don't understand (and what my question is) is: Why does my solution not work?

The only thing I could spot as being different is that I'm 'smashing together' the outcomes with similar but shuffled birthday values into a single outcome in my $\Omega$, compared to the $\Omega$ of the provided solution. But I don't understand why that does not work; a similar thing is done when calculating probabilities in, say, picking 3 cards from a shuffled deck. The outcomes can be seemingly grouped together into a single outcome "The cards X, Y and Z" and that works, and calculating probabilities on that still seems to work (i.e you can still use the above formula for $P(A)$ to calculate the probability of A happening).

I'm just all around confused on when to utilise combinations and when to utilise permutations in my solutions. I don't know if I've misinterpreted how to define $\Omega$, or if something else is misinterpreted, but I seem to run into this type of so many times and not a single book I've read throughout 3 courses involving combinatorics has ever touched on this in a way that I can understand.

I would be very grateful if anyone could help me clarify exactly what in my method is incorrect. Any help is appreciated! :)


The problem is that you treat every outcome in your 'bag of exactly 23 birthdays' as equally probable. But, getting all $23$ birthdays on, say, June $1$, is less probable than getting $22$ birthdays on June $1$ and one on June $2$, for that one person born on June $2$ can be any one of the $23$ people. And, of course, getting much of a 'spread' of birthdays will be a good bit more likely yet.

  • $\begingroup$ Thank you for the answer! It seems reasonable that that is the case. A followup; for probability problems involving choosing a hand of 3 cards from a deck of cards, these tend to use combinations rather than permutations. Why does it work to calculate 'good outcomes/total outcomes' using combinations in this case? Does it work because the outcome 'cards 1,2 and 3' and the outcome 'cards 6, 3, 8' (order does not matter, think of the deck as 52 'card-objects' with corresponding ids 1,2...) are (perhaps) equally probable? $\endgroup$ Mar 18 '18 at 6:35
  • $\begingroup$ In which case you may calculate the probability of an event A = 'getting three hearts' by calculating '# outcomes in A/# outcomes in universe', since each hand of 3 cards (disregarding any qualities of a particular card besides the fact that they are distinguishable) are eqully probable? $\endgroup$ Mar 18 '18 at 8:52
  • 1
    $\begingroup$ @SimonSirak Yes, that's exactly it: each particular hand of cards is equally likely, because each card is unique. But with the birthdays you can get repeat birthdays, and now certain combinations are more likely than others. $\endgroup$
    – Bram28
    Mar 18 '18 at 12:32

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