$11$ numbers between $1$ and $40$ are select. What is the probability there are some duplicated integers in those $11$?
I think the answer to this question is very straightforward: $1-P(\text{they are all different})$
Is this correct? Suppose I have $3$ equal numbers, does that still count as one duplicate? or $2$?
Thank you
Assuming the selections are independent and uniformly distributed, this is an example of the birthday problem, but with $d=40$ rather than $365$
You are correct to calculate $1-P(\text{they are all different})$ and that in this case this gives $1-\frac{40!}{(40-11)!40^{11}} \approx 0.78$
Indeed the probability exceeds $0.5$ with $8$ selected from $40$, corresponding to the $23$ from $365$ in the classic birthday problem
A slightly different question would be the expected number of distinct days selected, which would be $40-\left(1-\left(\frac{39}{40}\right)^{11}\right)\approx 9.72$, substantially less than $11$, so you should not be surprised that the probability of having duplicated dates is high