*Originally posted by shortfuse*

no because in the instance numbers which are in 1st or 3rd AND are black come up, then you get a chance for a big payout.

So? If you hit a red number in the 2nd column, you loose it all. :-)

If you bet on a single number, the payout is much higher (like 35 times your bet).

*Originally posted by shortfuse*

NOTE: the best must be of equal value on 1st, 3rd, and black

(I'm unsure why you say 6 bets are placed?)

Yeah, you are right. On wiki they say you bet 2 bets on 1st, 3rd and black, 6 bets in total. You can obviously bet a different amount as long as all bets are equal.

*Originally posted by shortfuse*

This is why it breaks even.

It doesn't.

Lets do the math one last time.

You bet $2 on 1st column, $2 on 3rd column, $2 on black (that's the strategy from wiki).

If you hit a black number from 1st or 3rd column (there are 10 of them), you get 10 bets (4 for black and 6 for column).

If you hit a red number from 1st or 3rd column (there are 14), you get 6 bets (for column).

If you hit a black number from 2nd column (there are 8), you get 4 bets (for color).

If you hit a red number from 2nd column (there are 4), you get 0 bets.

If you hit zero (there is 1), you get 0.

That gives you grand total of (10*10 + 14*6 + 8*4 + 4*0 + 1*0)/37 (there are 37numbers), which is 216/37 = 5.8378 bets on average.

Because you always bet 6 bets total, your EV is 5.8378 - 6 = - 0.1622 (which is - 0.1622/6 = - 0.027 or - 1/37 EV on average on a single bet

**like always in roulette**).