Odds of getting pairs -- QM version

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SUMMARY

This discussion focuses on calculating the odds of obtaining pairs of electrons with the same energy level in a quantum system consisting of 16 energy levels filled with 16 electron pairs of opposite spin direction. The calculations show that the total number of cases for observing 2 pairs, 1 pair, and 0 pairs are 120, 1680, and 1820 respectively, resulting in probabilities of P(2)=0.033, P(1)=0.464, and P(0)=0.503. The user seeks clarification on whether the differences observed between classical and quantum cases are valid and if the calculations presented are reasonable.

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anuttarasammyak
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As a spin out of How to calculate these odds?
https://www.physicsforums.com/threads/how-to-calculate-these-odds.1012014/
I have a question for quantum objects.
Question : An atom system has 16 different energy levels filled with 16 electron pairs of opposite spin direction. We pick up or measuring energy of 4 of them. Calculate odds for numbers of pairs of same energy we get in the observation results.

I assume:
If we observe spin z component of electrons together, the result is same as the thread.
If we do not observe it the number of cases are
2 pairs : ##\ _{16}C_2 = 120##
1 pair : ##\ _{16}C_1 \ _{15}C_2 = 1680 ##
0 pair : ##\ _{16}C_4 = 1820##
Total case number is ##120+1680+1820=3620##
So the probabilities are
P(2)=0.033 > 0.003 which is the result in the thread
P(1)=0.464 > 0.187
P(0)=0.503 < 0.810

I would like to know whether such a difference between classical and quantum cases takes place or not and if yes also this calculation is reasonable or not.
 
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