What is the probability of landing a different color on 2 spins of roulette?

Well the problem here is that real-world roulette wheels can have a 0 which is neither red nor black and where operators are especially greedy wheels also have a 00 which is also neither red nor black.

Without knowing which kind of wheel this question cannot be answered perfectly for real-world wheels.

But if assume there are and equal number of black numbers and red numbers (and no 0 or 00), then the probability of landing on a particular color (red or black) is 0.50 .

In this question it does not matter what color comes up first, on that second the opposite comes up.

The probability of black coming up on the second spin is 0.50 .

The probability of red coming up on the second spin is 0.50 .

So no matter what came up first, the probability that the second will be the opposite is 0.50.

Possbilities

Black then red color the same

Black then black color the opposite

Red then red color the same

Red then black color the opposite

So out of four possible sequences which are all equally probable, 2 out of 4 (1/2) are that opposite colors come up. 1/2 = 0.50

The above answers are correct, I will just combine them and make it complete. Since you only have 2 spins, the first spin does not matter, because it is just a 'base spin' to set the color. The second should just be opposite. Which means, you only have to calculate the 2nd spin, which is 18/38 = 47,36% for an (American) roulette table with 2 green spots, or 18/37 = 48,36% for a (European) table with only 1 green spot.

Since the house edge is higher with American roulette, I would not advise you to play this version. But then again, maybe you just wanted the odds, and used roulette as an example.

Assuming a standard roulette wheel (18 black, 18 red and two green), you have 18 chances out of 38 to hit the alternate color on successive turns -- about 47%

If there is only one green, you have 18 out of 37 chances -- nearly 49%.