Taking the n=10 case, because you think n=2 is confusing.
> the average winner will get prize_pool / (10/2) or prize_pool / 5.
No, the average winner will get expected_prize_pool / expected_number_of_winners.
If 5 is the number of winners averaged over all draws - including those without winners - the (average) pot they share has also to take into account draws without winners.
The average prize shared in this case is not prize_pool, it’s 1023/1024 times prize_pool.
> the average winner will get prize_pool / (10/2) or prize_pool / 5.
No, the average winner will get expected_prize_pool / expected_number_of_winners.
If 5 is the number of winners averaged over all draws - including those without winners - the (average) pot they share has also to take into account draws without winners.
The average prize shared in this case is not prize_pool, it’s 1023/1024 times prize_pool.