I hope you'll accept an 8-bit rand algorithm that demonstrates a greater than 256 period?
Sure. 
my @x = (0x00011011) x 24;

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But that's not 8-bits. You keep a state using 768 bits. You've no dispute from me that you can create long periods from that. Busy Beavers can go through an amazing number of steps with just very limited memory to keep state on. A trivial counter using a rollover can go through 2768 values before repeating itself.

However, considering that you are using 8-bits seeding, all you have are 256 different sequences. Regardless how long they are.

Assuming he allows it to self-seed -- no srand() -- even if perchance two of his runs picked adjacent seed-points in the sequence, on average, he'd have to generate 4e6001 / 4e9 = 1e5992 rands before the two sub-sequences would overlap.
That I do not understand. There are 232 seeds. Each of them starts a different sequence. You don't get to start at a random point in the sequence. You could of course keep track of where you are in the sequence, but that requires adding ⌈log2P⌉ bits to the seed, where P is the length of the period.
So, (ignoring the birthday paradox, imperfect PRNG etc. for a moment), for him to get a dup, he would have run his program 2**32 times and pick exactly 1 sequence each time. But if he generates 10 each time, that's 10 * 2**32 sequences before he gets a dup.
I read this as "the more he generates, the more it takes for a duplication to happen". That seems quite counter intuitive to me, and I'm not sure if that's what you mean.

In reply to Re^9: How likely is rand() to repeat? by JavaFan
in thread How likely is rand() to repeat? by desertrat

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