Earlier quoted context omitted.
The birthday paradox doesn't apply. All that matters it the value at the time the ack is applied, so it's always 1/2^64 (because the xor of any number of random numbers is still random).
No, the birthday paradox very much applies. The chance of success for your first ack is (2^64-1)/2^64. The chance of success for your first and second ack is the chance of success for your first ack times the chance of success for your second ack, ((2^64-1)/2^64)^2. And so on for your third ack, and your ack. By the time you reach your 2^32 ack, your chance of having all successes is ((2^64-1)/(2^64))^2^32, which is…
Suppose that the sequence of values you have after each ack is A, B, C, D, E (five acks total). So long as A, B, C, and D are all nonzero, we're OK. With the birthday paradox, we'd be looking at A=B, A=C, A=D, A=E, B=C, B=D, etc. -- many more combinations.