Okay, first let me clarify my position. I have no intention of using this strategy for real money. I am not claiming I can beat roulette doing this. I understand that the ZERO will put this in the houses favor and I will lose long term. That is obvious, so there's no need to suggest I try a different strategy. I get it.

I am interested in the MATH part which disproves that Red/Odd does not have any more advantage over Red/Even. Thatâ€™s all.

I am simply confused about the Red/Even and Red/Odd distribution, because according to a couple of posts in this thread, betting on ONE has a higher probability of winning than betting on the OTHER.

**Example #1:**

JUICE, in reply #6, wrote that betting on **Red/Even** has a **HIGHER **chance of hitting than betting on Red/Odd.

**Example #2:**

Scepticus wrote out the probabilities within 37 spins. He wrote that **Red/Odd** has a **HIGHER **chance of winning (10 out of 37) compared to Red/Even (8 out of 37 chance of winning)

Juice is saying the OPPOSITE of Scepticus. So, I'm not the only one who is confused :-) But, for the sake of argument, I am assuming that Scepticus is correct, since there are 2 more numbers in the Red/Odd category.

In any case, my question is the following.

If Red/Odd has a HIGHER chance of WINNING within 37 spins (compared to Red/Even), does this not mean that, on average, Red/Odd will win MORE times than Red/Even, in a 37 spin of the roulette wheel? Specifically, 2 more times?

Yes or No? (I understand both Juice and Scepticus confirming the answer is YES. Correct?)

If the answer is No, why not?

I am so confused :-)

And Dane, thanks for your advice on leaving if I don't win BOTH bets at the same time. In other words, if the first bet WINS, then PARLAY the 2nd bet. If the 2nd bet is a PUSH then take the Parlay bet down, do not keep it up.