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So I'm bored watching a PC update. . .

According to Google, and math, the odds of red coming up on a roulette wheel 3 times in a row is about 10%. But if you get 2 reds in a row, you are incorrect in thinking that you have anything different than a 46% chance of getting red again on the 3rd spin. Because the wheel doesn't remember.

I get "the wheel doesn't remember". But also, there's a pretty low chance of getting three reds in a row. So why would betting black be no different than betting red?

That's a Catt question for sure.

As a math dummy though, let me say that math in gambling and sports is sometimes extremely detrimental if you're trying to win.

Because of the perspective of things through time, and because each spin is an independent event.

At time zero (t0), the odds of red or black are 46%. The odds of 3 in a row are about 10%. The odds of 2 in a row are 21% (.46 * .46)

You spin, we get red. Now, at time one (t1), the odds of red or black are still 46%. So the odds of the t0-2 in a row are now 46%. The odds of the t0-3 in a row are 21%. But, the odds of a NEW 3 in a row, are still 10%.

You spin again, we get red again. Again, at time two (t2), the odds of red or black are still 46%. The odds of the t0-2 in a row are 100%, it hit. The odds of the t0-3 in a row are 46%.

This only applies to independent results (fair coins flips, fair roulette wheels, etc). The math for dependent results is more complicated.

You can see these odd shifts in real-time if you think of things like parlays. Just like a parlay, you are betting on 3 independent results (red at t0, red at t1, and red at t2). The odds of 3 independent results happening are their odds multiplied together. If you are betting on, say, NFL games, you can drop $10 on 3 50/50 games, and you'll get roughly $80 (at fair odds, lower when the place takes some) if you win all 3 of your bets. Say you're betting on a 1pm game, 4:30pm game, and the Sunday night (8:20?) game. After the first game, if you're right, the value of your $10 will raise to $20, and you can cash out early. If you're right after 2 games, the value goes to $40 and you can cash out early. The odds of that last game are still 50/50, despite the 1:8 odds of the original bet.

I like the visualization provided by binomial trees for things like this (they are more useful when the odds of each of the 2 events are uneven, but work just fine for this as well). You could substitute heads/tails coin flips (though the odds are 50/50 instead of 46-46-8)

The wheel may not remember, but the universe does. I'd still like to see stats on: How many times does a 7th red come up after 6 reds in a row compared to black. You should find that half the time on the 7th spin, it's both colors, right?

I mean, I have to assume that's correct, otherwise, roulette wouldn't be a thing. Or, if they see that's what you are doing, they stop you. Or something. But still.

It's the same thing with coin flips, or dice, or any other random fair process. You can simulate this yourself with coins.