# Calculating Probability using Combinations as Alternative Solution

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.