Probability for someone in room having same name as someone else in the room. This is follow-up question for post posted earlier here
There are 7 people in the group whom names are picked from 12 randomly picked names. What is probability for at least two people having same names.
This is my attempt to solve this--
Attempt one:
$$\sum_{n=2}^7 \frac{7!}{n!(7-n)!} (\frac{1}{12})^n(\frac{11}{12})^{7-n}$$
$$≈0.1100617083$$
To my understanding we are calculating sum of all probabilities that meet the condition. This should work similar to solution provided here
However you get correct answer with ( not exactly the correct answer but it's very close)
$$1-(\sum_{n=2}^7 \frac{7!}{n!(7-n)!} (\frac{1}{12})^n(\frac{11}{12})^{7-n})$$
$$≈0.8899382917$$
Attempt two:
$$\frac{12^7-{12\choose7}7!}{12^7}$$
$$≈0.888599537$$
Attempt three:
Python gives answer with:
import itertools
import math
names = ["name1","name2","name3","name4","name5","name6","name7","name8","name9","name10","name11","name12"]
count = 0   
product = itertools.product(names, repeat=7)
for array in product:
    if len(array) != len(set(array)):
        print(array)
        count += 1
print("final answer is", count/math.pow(12,7))

Output:
final answer is 0.8885995370370371
Process finished with exit code 0

If someone could explain why attempt one does not give correct answer, that would be much appreciated. I am certain that there is something i dont understand fully.
 A: There are two problems with your attempt one.
(1) There are twelve possible choices you can give to the people with the same name. So you need to multiply by ${12\choose 1}$.
(2) Assuming you fixed (1) by multiplying $12$. The remaining $7-n$ people can have same name as well and this duplication needs to be subtracted from your counting. For instance you double counted the case $AABBCDE$ when you choose $A$ as the name and when you choose $B$ as the name.
A: Although this doesn't answer your question, here is an alternative approach.
You can think of the solution as the probability that none of them have the same name, which is
$$
1 - \left(\frac{11}{12}\cdot \frac{10}{12} \cdot \ldots \cdot \frac{6}{12}\right)
$$
If Rumpelstiltskin is one of the names given to one of the $7$ people, then none of the other people can be named Rumpelstiltskin, which leaves $11$ out of $12$ names we can assign for our second choice. If Jerry is one of the names we can give, and we end up naming someone Jerry, then the next name we assign cannot be Jerry and it also cannot be Rumpelstiltskin. This leaves $10$ out of $12$ names for assignment and so on and so forth.
