Since a+b = b+a shouldn’t the total number of ‘different sums’ be half of what you give? Fortunately the rest of the argument works completely analogously.
You can see this as sampling z(z+1)/2 times sorta-independently, or as sampling z2 times with less independence (ie most sums are sampled twice).
Either view works, and as you said, it doesn’t change the outcome.
Since a+b = b+a shouldn’t the total number of ‘different sums’ be half of what you give? Fortunately the rest of the argument works completely analogously.
You can see this as sampling z(z+1)/2 times sorta-independently, or as sampling z2 times with less independence (ie most sums are sampled twice).
Either view works, and as you said, it doesn’t change the outcome.