The 0 & 00 are the casino’s profit…
Without them all other bets would pay even in the long run.
As engineer said, your best bet is to put everything on a single bet, the longer you play, the more the law of averages works against you.
The best bet is probably shares in a casino…
No, that’s not right. If you place 100 $50 bets, each with a winning probability of 48%, then you have a 30.8% chance of winning overall (51 or more wins). This was engineer_comp_geek’s professor’s point - in a game with the odds against you, by playing lots of times, it becomes nearly inevitable that you will lose money. By betting it all at once, you have the greatest chance of coming out ahead.
Note that the expected (average) return of both methods is exactly the same. It’s the distribution that will be different. In the bet-it-all-at-once method, a histogram of possible outcomes will just have two points: $0 and $10000. In the 100 bets method, it will be a bell-shaped curve centered around $4800, with only 30.8% of the area above the 50 wins centerline.
Let’s look at the expected value for covering two “dozen” bets – that is, the 1-12, 13-24, 25-36 wagers, or the 1st, 2nd, 3rd column wagers.
Each of these bets covers 12 numbers out of 38, and pays 36 for 12 = 3 for 1, or 2 to 1.
The probability of a win, covering two of these wagers, is 24/38. The probability of a loss is 14/38. Assuming a pair of $10 bets, a win returns $30 ($20 plus the wagered $10), and a loss returns nothing. The expected payout is $30 * 24 / 38 + $0 * 14 / 38 = $18.95
However, it costs you $20 to play this game. The expected value of this strategy is the expected payout less the cost to play the game. EV = EP - C = $18.95 - $20 = -$1.05. Dividing this by the $20 wagered produces an EV percentage of -5.26%.
In conclusion, this strategy expects to lose 5.26% of the total amount wagered on each spin of the wheel, in the long run. This is the house advantage, and with two exceptions, applies to any betting pattern or strategy in American (double-zero) roulette.
For completeness, the exceptions are the even-money wagers where the en prison rule is in effect, with a house advantage of 2.70%, and the 5-number (0, 00, 1, 2, 3) wager, which has a house advantage of over 7%.
Back to the original post:
With two house numbers you can still win a majority of the time. You can bet an equal amount on the first twelve numbers and on the middle twelve numbers (or middle and last, first and last, first and second column of twelve, et cetera). This gives you a 63.15789% chance of winning one of your bets. You have a 100% chance of losing the other bet, but the casino will pay you two-to-one odds on your winning bet. So if you put $100 on, say, middle 12, and $100 on the last twelve, you have just over a 36+% chance of losing $200, and a 63+% chance of winning $100. The odds favor you winning $100 on a given spin, but when you lose, you pay twice that. You can’t tie on any given spin, you can’t lose only $100 on any given spin, and you can’t win $200 on any given spin. Betting this way, you should win $100 63% of the time, and lose $200 36% of the time. The casinos’ bank rolls will typically outlast this kind of betting until the players who bet this way put together enough strings of losses to go bust.
Also, it seems to me that the odds would be fairly even over time if the odds of winning $100 were two-thirds and the odds of losing one-third. You’d win twice as many times as you’d lose, but pay twice as much when you lost. Again, the house numbers push the overall odds in the casinos’ favor.
This is not what I said. I said you can cover all the bets on the table. And you can. I wasn’t talking about red-black bets.
I believe you can get almost 50-50 odds at Craps if you bet “behind the line” (e.g., place a bet on Pass/Don’t Pass, then after the point is made, put 5x or 10x of your bet behind your original)…?
Taking the odds has no house edge and does decrease the house edge, but it’s a quick way to lose a lot of money if you keep getting seven out. Here’s a good table.
Taking odds decreases the house edge only if you consider only the expected loss as a percentage of the total bet. In absolute dollar terms, it makes no difference.
In other words, placing a $10 bet on the pass line costs you an average of 15 cents, so there is a 1.5% house advantage. If you take 2X odds, the bet has still cost you an average of 15 cents. However, you now have $30 on the table. If you want to view it as a percentage, then it will be lower, but that doesn’t seem to me to be a useful thing to figure.
However, 0/00 pays off 35 to 1 (or 17 to 1 if you’re an idiot, and your one bet covers both); you’re better off if it hits, assuming you bet the same amount on 0/00 as you did on the 1st, 2nd and 3rd twelves. Or on red and black.
Either way you lose the bet that didn’t hit and you lose money. Furthermore, the roulette dealer should not allow you to cover the board. They should never, ever let a player bet on both red and black, or an both even and odd, or on both 1-18 and 19-36. Every time I have seen someone try that the bets have been disallowed.
FWIW, I’ve seen it several times depsite not playing roulette much. Each time the dealer (are they really called dealers?) has noticed, he/she has discouraged, but if the players insists, they let them. Ditto for playing pass and don’t pass simultaneously at Craps.
CurtC, thanks for posting the math in response to Don’t Ask.
When I was young(er) and naive(r) I thought I’d come up with a foolproof system. However before I could waste my money I sat down and worked it out, and much disappointment ensued. 
This was similar to the OP, but involved covering the third 12 and the bottom row:
1 4 7 10 | 13 16 19 22 | 25 28 31 34
0 2 5 8 11 | 14 17 20 23 | 26 29 32 35
3 6 9 12 | 15 18 21 24 | 27 30 33 36 <-bet
^
bet
Now, labelling each ninth of the table from top left A B C and so on, you get, with 1 unit bet on the 12 and 1 unit on the row, a return of 3 units on each of sectors C, F, G and H, and a whopping 6 units on sector I, as both bets win.
So 5 out of the 9 sectors are a winner… 20 out of 37 possible outcomes are a winner! Wow! Unfortunately, yes but no…
My expected return is:
0: 1/37 x 0 = 0
A: 4/37 x 0 = 0
B: 4/37 x 0 = 0
C: 4/37 x 3 = 0.324324
D: 4/37 x 0 = 0
E: 4/37 x 0 = 0
F: 4/37 x 3 = 0.324324
G: 4/37 x 3 = 0.324324
H: 4/37 x 3 = 0.324324
I: 4/37 x 6 = 0.648649
Total = 1.945946 units
Unfortunately I am paying 2 units on each turn, so I am losing abiut 2.7% each spin, as before (this is for a European table)