I just watched the video on the Birthday paradox (The Birthday Paradox) and, though it all seemed indeed intuitive for me, there is just one issue that has been bothering me.
To find the number of cases in which n people do NOT share a birthday date, the formula used is 365!/(365-n)!. Again, all of this makes sense, since the “choice” method sounds like the most intuitive way to approach this problem.
However, when I stopped to think more about it, I realized I had a question: 365! / (365-n)! is essentially the permutation formula. However, here, the order does not matter, isn’t that so? Since, say, a group of 2 people whose birthdays are 08/12 and 15/08 is the same as if we had selected a group with birthdays on 15/08 and 08/12.
I understand that it does not work if we use the combinations formula, and laying out the cases before me makes it clear we should use the permutations formula. However, I still can’t wrap my head around the question: if we are using permutations, we are assuming that order matters, but isnt it true that order in fact does not matter in our case?
To sum up, I have no doubts about the fact that the permutation formula should be used; I just need some help understanding why this is the case even though the ordering does not matter (or does it, am I got something wrong?).
It is indeed a permutation: of 365 days, you have to chose n days (none of the birthdays repeated) where order does matter: since the people are clearly different, A and B having the birthdays a and b is not the same outcome as them having the birthdays b and a.
Thanks for the answer! But this is what I can’t wrap my head around: how is it not the same outcome? Let’s say X and Y have birthdays at 11/01 and 20/02. Wouldn’t a group in which I select X and Y be the same as a group in which I selected Y and X?
Like, I understand that if I were to randomly assign birthdays to people, then permutation should be used, since assigning 11/01 to X and 20/02 to Y is different from assigning 20/02 to X and 11/01 to Y. However, if I am selecting people (whose birthdays have already been defined), shouldn’t selecting first X and then Y be the same as selecting first Y and then X? I can’t wrap my head around that, combinations seem to be the way to go to find the number of possible birthday sets
I think my question is more related to a fundamental misundertanding of what “order matters” mean. For instance, take the question “the question what is the probability of getting three different numbers when rolling three dice?”
In my mind, simply doing (654)/(6^3) would be wrong, because this is assuming that order matters (so that a 1, 4, 5 result is different from a 4, 1, 5 result for instance). Like, how different it is from selecting people A and B to shake hands (which is a combinations problem)?
sorry for the lengthy question, but I am really confused it would be awesome if I could figure this out once and for all
Something just occurred to me: if the question was “how many distinct sets of birthday dates can we assemble?”, then the way to go would be using combinations, since {11/01, 22/02} and {22/02, 11/01} are the same set. The same for the handshake question.
However, when dealing with probabilities of independent events, it is best that we consider those as permutations. So, for instance, say you have 3 people in a room, and you want to know the probability of A and C shaking hands. One way would use combinations to notice that there are only 3 possible options: A-B, A-C, and B-C. 2 cases out of 3 satisfy our conditions, so the probability is 2/3.
Another way to go would be using the same rationale as “find probability of 1 case; find number of cases; multiply one by the other”, in which we consider those to be permutations (with repeats, if thats the case). Then, in the shaking hands example, We would say, considering that order does not matter: probability of 1 event (A being chosen, then C): 1/3 * 1/3 = 1/9. Number of ways in which this can happens: 3! = 6. Hence, p(A-C) = 1/9 * 6 = 2/3.
In the end, it is just a matter of how approaching / solving the problem, with the “permutation way” being much simpler and easier. Am I on the right track, or is there something else you’d like to add?
Instead of thinking contextually, you could simply think of a set containing all 2-tuples of birthday dates (in dd/mm format, if you like), where no tuple contains two identical elements (i.e., the same birthday).
This is correct. You have \frac{6}{6} \cdot \frac 56 \cdot \frac 46 because the “first” dice can assume any 6 numbers, the next dice would have one less option, and the pattern continues “recursively”. You can also just view this as \frac{6P_3}{6!} without a care about the permutation perspective (like evident from the above snippet).