Now, if the next two spins were r, then b, that happened after 4 spins, not two, right? The odds of a red, followed by a black didn't change but the first two spins weren't red, then black, they were black, black. If that was black again, you would again have to start over. So, you would start over with the next spin.

What are the odds of the first spin being red, followed by a black spin? If you sat at the roulette table and the first spin was black, it would not be possible to count that when waiting for a red, then black. Now, why can the Wizard predict the future, but the past doesn't count? Let's look at 2 spins instead of 18. So if you placed your bets for 18 spins on the pattern rrbbrbrrbrbbbrbbbr you have a 1 in half million chance of winning all 18 spins, the same as winning 18 straight reds in a row. Sit at a roulette table and wait for either outcome, 18 straight reds or rrbbrbrrbrbbbrbbbr and you will find they both occur with the same regularity. Now, that may seem like a random set(it is, I just hit the r and b keys randomly till I got to 18) but IT IS NO LONGER RANDOM as the above set has been committed to paper(or computer screen in this case) and if the Wizard tried to calculate the odds of that EXACT outcome it would be the same as 18 straight reds. Let us say that instead of 18 reds straight, the odds were calculated for rrbbrbrrbrbbbrbbbr.

Once you try to calculate the outcome of 2 or more random spins(a sequence) they are NO LONGER RANDOM. The problem is that each spin being an independent event, the only (random)set of spins that can be calculated is a single spin which at double zero roulette would be 18/38.

As a layman and not a math guy, I hope I can explain it better as I get his confusion. Hopefully he'll read this and hopefully it won't be closed. I wanted to explain this to the Australian guy but that thread was closed.

Odds of red or black in roulette