I can chime in here. Think of the complementary situation: what is the probability within a group of n unrelated people of all n having different birthdays. The sample space size is 365^n. The number of possibilities for all birthdays different is 365 . 364 ... (365 - n + 1). So the probability that all birthdays are different is 365 . 364 ... (365 - n + 1)/365^n. Therefore, the probability that 1 or more birthdays are the same is 1 - 365 . 364 ... (365 - n + 1)/365^n or equivalently 1 - (1 - 1/365).(1 - 2/365) ... (1 - (n-1)/365). We want the smallest value of n such that the probability is greater than 1/2. There are various ways to find a good approximation of those expressions in terms of n but, if memory serves, n=23 will do it.
The idea of the Birthday Paradox is used in some mathematical theorems that give interesting results.
I think it overestimates (over-counting) the probability for a given n, in this case for n=20. You cannot simply add 1/365 190 times because they overlap. IOW, it ends up counting the cases where multiple pairs have the same birthday multiple times. So it is overestimating the probability for a shared birthday when it is 20 people. The probability is not 190/365 but less than this… less than 50% even.
You can try subtracting off the overlap, 2 pairs with the same birthday, then 3 pairs, etc… But this is cumbersome and that is why you take the shortcut outlined by Andy7 of doing the complementary problem instead.
P.S. Oh and you also have to subtract off the cases where 3 have the same birthday, because this got counted 3 times. and then 4 (counted 6 times - one above you already subtracted), 5, etc… then 2 and 3 at the same time, then 3 and 4 at the same time etc… all counted multiple times both in the original count and in some you already subtracted… quite a mess really…
Yes, I think you are on the right track on the error in your answer. Often in counting arguments like this, to count the positive outcome (eg. how many with one or more birthdays the same vs. how many with birthdays all different), you have to use the Inclusion–exclusion principle - Wikipedia, which as you note, can be messy. But, it is a valid way to arrive at a solution.