Here's a challenge. Find the probability that, in a group of 4 people, at least 2 have the same birthday. The catch is - try doing it the DIRECT way with cases and do NOT use the indirect method that has been flogged endlessly and everyone has seen. BTW this is NOT a paradox. So when you see this listed as "The Birthday Paradox" you can politely correct them.
What? No takers?
Hello there. I think we can make cases like 1) all 4 have same birthday
2)3 same 1 different
3)2 same 2 different
And then just apply the basic probability formula.
Or simply do 1-(probability of no common birthdays)
I couldn't think of some other out of box method but if you have any, please do share it here with the answer
Thanks,
That's exactly what I did. Except you have missed one case. 2same 2same. You can see why 1 - P(4 different) is much easier.
@Ian Fowler Yup Thank You