probability of Reb/Black results in a row

(Disclaimer: forget the zero)

Here is a little “paradox”. Not a paradox really, but an interesting observation.

Theoretically, if you make at the same time three even chance bets (High-Low, Red-Black, Even-Odd) the probability of loosing ALL your bets is:

1/2 x 1/2 x 1/2 = 1/8

Well, it is not exactly like that.
Depending on which even chances you bet (Low-Red-Even or Low-Red-Odd or High-Black-Odd etc.) the probability of loosing ALL your bets is either 1/9 or 1/7,2

For exmple, if you bet High-Odd-Black, there are only 4 numbers (12,14,16,18) which will cause you to lose all your bets.
Probability of losing ALL your bets = 4/36 = 1/9
If you bet High-Odd-Red there are 5 numbers (2,4,8,6,10) which will cause you to lose all your bets.
Probability of losing ALL your bets = 5/36 = 1/7,2

Of course this DOES NOT mean that betting High-Odd-Black is “better” than High-Odd-Red. I just want to point out this little “unbalance” in an exceptionally well balanced game like roulette, as it may open a window to interesting ideas on how it could be used to our advantage.
Nothing less, nothing more.