Same here! 
You’d like to think so, wouldn’t you!. I didn’t say the system doesn’t work, only that it isn’t foolproof! If someone is a great fool, then it is possible that the system will not work for them. But you know I am not a great fool…in fact you would have been counting on it…and so I can clearly not pick the Martingale in front of me.
Good point. If someone is a big enough fool, they wouldn’t be following the Caro System, and therefore the system wouldn’t work, and they’d lose money.
By the way, I want in on this Wednesday Martingale action. What’s the current bet?
The bet is currently at $2^624.
I’m out.
Can someone briefly explain the Caro system?
Click the link, read the website. Really. 
S basically the system is that certain numbers win more frequently than others? The same numbers on all wheels?
Methinks someone didn’t spend enough time reading.
Although this is amusing, in the interests of fighting ignorance:
The Caro system is DON’T BET ANYTHING. That’s the only way to eliminate the house advantage in roulette.
He-he
You know Garfield, that ruins the whole set up an djoke- whcih is actualy quite funny. Sigh. 
:wally
Yep, no doubt about it, your friend has ironed out all the wrinkles and developed a sure fire road to riches.
Why don’t you and he toodle on off to Vegas and give it a go. Don’t forget to take out a second and third mortgage on the house so you can come back really, really rich.
Yes, and as you head off to Vegas to make your fame and fortune, here’s something else to try.
It seems the game of Chuck-A-Luck (3 large dice in an hour-glass shaped ‘birdcage’) is a no-lose proposition too. You don’t have to develop a winning system either. Here’s how it works.
You bet on 1 of the 6 numbers showing up on the 3 dice. Let’s suppose you bet 1 dollar. If your number shows up on just 1 of the 3 dice, you keep your 1 dollar bet and you receive another dollar. If 2 numbers show up, you receive 2 extra dollars and if the number shows up on all 3 dice you win 3 extra dollars (along with keeping your original bet).
So, logic would tell us that since each of the 3 dice have the numbers 1 through 6 on them, the chances of having your number showing up just once is 50%. The house pays off even more if 2 or 3 of your numbers show up. Wow - how can you lose?
From what I’ve heard, Chhuck-A-Luck is declining in popularity. I think I know why. It finally dawned on “the house” that they were losing money on this and so they are gradually “phasing out” this casino game before people get wise to the gravy train this game provides.
For those folks that can’t figure out why the Martingale roulette “system” is a losing proposition, then this Chuck-A-Luck “secret” should be right up your alley.
3 dollar payoff on 3 die win (1/6 * 1/6 * 1/6)
2 dollar payoff on 2 die win (1/6 * 1/6 * 5/6 * 3 (which die doesn’t turn up))
1 dollar payoff on 1 die win (1/6 * 5/6 * 5/6 * 3 (whic die does turn up))
-1 dollar payoff on 0 die win (5/6 * 5/6 * 5/6)
(31 + 253 + 1553 - 155*5)/216
(3+30+75-125)/216
-17/216 expected win per roll.
The house never lost a dime on this. Next!
Mathocist
I thought you have seen enough of my postings by now to know that I do understand mathematics and probability just a tad.
To quote my own posting:
For those folks that can’t figure out why the Martingale roulette “system” is a losing proposition, then this Chuck-A-Luck “secret” should be right up your alley.
I was agreeing with SandyHook about how the OP found a “winning system” and should go try it out. I was just suggesting to him that another “sure-fire” system existed where he couldn’t possibly lose either.
Unless, Mathocist, you were thinking the OP finally learned the roulette system was a loser and my recent posting might have given him new incentive to go to Vegas to lose his money.
More generally, to explain in small words to anyone who doesn’t see clearly why your statement was fallacious.
Minor nitpick: there’s an ‘h’ in there, you know? As in someone who enjoys painfully (masochistically) grinding his way through abstruse mathematical reasoning…
Table maximums prevent you from placing your 6th bet. Therefore, you must win no later than the 5th bet. 5 non-red (or non-whatever) spins in a row is highly likely in a very short number of trials. Therefore, the tripling system is worse than the true martingale. (I’m guessing half again as bad.)
The advanced Martingale is a much better system. I believe it is called the Labouchere, but it may be the Delambert system.
Tangent: I think the Delambert is start with one unit, increase the bet by one unit after each loss, decrease by one unit after each win. I lost my shirt doing this in Casino War.[sup]1[/sup] (Yes, the old card game you played as a kid.)
The Labouchere requires a pencil and a piece of paper. I’ve done it many times at craps. They don’t care if you have a pencil and piece of paper. That alone should tell you that systems don’t work. If you still don’t get it, you will be told by a dealer at the table, after s/he eventually asks you why you have the pencil and paper, that betting a system makes you systematically lose. I have heard this witticism from dealers every single time I’ve played with pencil and paper.
Before I get to my personal favorite systems, let me explain about 50-50 wagers, which is where you play virtually all systems. Roulette is one of the worst 50-50 wagers in the house.[sup]2[/sup] Casino War is better, but still not good. (And it is way faster for each decision.) Craps is where to go for 50-50 bets. The field is a one-roll bet, and the fastest resolving “50-50” proposition there is. However, it is not even close to 50-50; it has roughly the same odds as roulette.[sup]3[/sup] (Over 5% house advantage.) But the field pays double on 2 or 12, so if you happen to be pretty high up in your sequence when one of these numbers comes in, it doubles your huge bet. Which is very nice. I’ve had a $5 system up to $120 when a 12 came in and payed out an unexpected extra $120 above the original $5 I was playing for. All that said, the field is bad bad bad bad bad bad bad bad bad. Sorry for all the bads, but that’s how many non-fields in a row you’ll see within 10 minutes of getting to a table. So don’t play it.
All 50-50 systems should be played on the Don’t Pass line, for many reasons. First and foremost, the Don’t Pass has the lowest house advantage (27 / 1925 = 1.40%) of any fixed-expectation 50-50 bet in the Casino. (Blackjack can be card-counted, but that’s a pain in the ass to say the least.) The second major reason the Don’t Pass is the smartest 50-50 bet is because it is one of the (if not the) slowest resolving 50-50 bets in the casino. That reduces variance. Anything that reduces variance is good when betting a system.
Okay, so all 50-50 systems should use the Don’t Pass. Now on to my personal favorite systems:
1) $10 Don’t. It’s for a $10 craps table, and requires a $200 bankroll. Bet $10 Don’t Pass or Don’t Come (which is a Don’t Pass bet after a point is established) every roll of the dice (except bar 12, which is a push). You get paid (every established bet on the table wins) by the Seven Out. You can “get hurt” by the initial come out rolls when the point is off. (8-3 against you when coming out.) All money coming back goes to a different rack from the original $200. In other words, you make 20 bets. (Dip into the other rack to finish off the final shooter.) This system is good in two ways: It is very fun, because you get to make a whole lot of bets. And it is very, very low variance. You are almost guaranteed to walk away from this system with between $175-$215. Sure, that isn’t great, but nothing is. But, if you are down to your last $200, and your friend wants to keep playing, this can keep you at the table, and very active, for an extremely long time. I play this system when I first get to the casino. This is an excellent way to ease into gambling.
2) The Labouchere. This system is a complicated version of the Martingale with lower variance. All bets are on Don’t Pass. (No Don’t Come bets.) The actual numbers in the “train” are arbitrarily set. My example uses the numbers I like. You need a $5 table, and $400 for a bankroll. Write down, on a piece of paper, your initial “train”. Mine is:
2 3
Now, each bet is the sum of the first and last numbers in the train. 2 + 3 = 5, so the first bet is $5. (If there is only one number in the train, that number is the current bet.) Write down the current bet (5) on the right of the train. If you win, circle the current bet and cross off the first and last numbers in the train. If you lose, recalculate and continue. Let’s say, for example, you start with a Martingale killer, followed by a decent amount of wins. Say, 7 losses, 2 wins, a loss, and finally 3 wins. Okay, so after your first 7 decisions, all losses, your Martingale friend is bankrupt (5 10 20 40 80 160 320 = 635 down, needing 640 more for the next bet) and your train looks like:
2 3 5 7 9 11 13 15 17
Now you write down (and bet) 19 on the end, which wins. So you circle (I will bold) the 19 and cross off (I will sub) the 2 and the 17:
[sub]2[/sub] 3 5 7 9 11 13 15 [sub]17[/sub] 19
The next bet should be the first and last numbers in the train, which are currently the 3 and 15, so your next bet (and next number in the train) is 18, which also wins:
[sub]2[/sub] [sub]3[/sub] 5 7 9 11 13 [sub]15[/sub] [sub]17[/sub] 19 18
Next bet is 5 + 13 = 18, and it loses:
[sub]2[/sub] [sub]3[/sub] 5 7 9 11 13 [sub]15[/sub] [sub]17[/sub] 19 18 18
Remember, 18 is the new last number, so the next bet is 5 + 18 = 23. It wins:
[sub]2[/sub] [sub]3[/sub] [sub]5[/sub] 7 9 11 13 [sub]15[/sub] [sub]17[/sub] 19 18 [sub]18[/sub] 23
Your next bet, 7 + 13 = 20, also wins:
[sub]2[/sub] [sub]3[/sub] [sub]5[/sub] [sub]7[/sub] 9 11 [sub]13[/sub] [sub]15[/sub] [sub]17[/sub] 19 18 [sub]18[/sub] 23 20
And the next bet (9 + 11 = 20), also a winner, closes the train:
[sub]2[/sub] [sub]3[/sub] [sub]5[/sub] [sub]7[/sub] [sub]9[/sub] [sub]11[/sub] [sub]13[/sub] [sub]15[/sub] [sub]17[/sub] 19 18 [sub]18[/sub] 23 20 20
Once the train closes, start a new one. This system requires (roughly) half as many wins as losses to resolve, which is not an unreasonable thing to expect. Also, the variance is much lower than the Martingale, so it has more staying power. However, it is a mathematical system, so you will be guaranteed to lose in the long run. But it is far superior to the Martingale. Notice the totals of the losing and winning bets:
Losing: 5 + 7 + 9 + 11 + 13 + 15 + 17 + 18 = 95
Winning: 19 + 18 + 23 + 20 + 20 = 100
It’s a mathematical system: all resolved trains pay out exactly equal to the sum of all numbers in the initial train. I play 2 3, so I play for $5 per train. In books, they recommend initial trains like 5 5 10, or 5 10 15 20, or some such. That’s way too high variance for me. I used to just play a single 5 as my initial train, but that is even too high variance for me. In my example above of 7 initial losses, that would look like:
5 5 10 15 20 25 30 35
Which is a net loss so far of $140, plus the $40 next bet, instead of $77 plus the $19 next bet using my initial train of 2 3. Note: in some cases you may be stuck with a $3 or $4 final bet. Consider that train a wash and start a new one.
3) Aggressive Let It Ride. I always play this (and only this) system on the Pass Line, because it feels better on the “light side” than the “dark side”. It requires a $10 table, and a $300 bankroll. Bet $10 on the Pass. When it wins, leave the winnings up and add in another $10, for a total of $30. (The initial $10, $10 winnings, plus an additional $10). If it wins again, leave it up again and put in another $10. If it loses at any point, start over with the basic $10. You are waiting for a 4th win in a row:
10 30 70 150
The 4th win is a $150 bet plus a $150 win, which is the $300 you started with. Put those $300 in chips in your pocket, and continue the system until you are out of your original $300. Either you lose $300, break even, or win $300. All you need is 4 wins in a row, which can happen immediately at a craps table. Say the first shooter rolls 7, 11, 7, 7. You’ve just won $300 (actually, $260) in the span of about 30 seconds. Quite fun. 
On the flip side, you cannot possibly lose quickly. It takes a couple hours to get through your $300, and though you don’t get much activity, you get to play for a very long time, and you get several heart-pounding big bets to keep you interested.
Those are my top 3 systems. All have caps on the maximum loss. All three will keep you at a table for a long time, unless you really get screwed by the Labouchere. Last time I played the Labouchere, btw, I won $150 or so in like 4 hours, and then a hot shooter hit 14 wins in a row. Ouch! Luckily, I bailed on win 9 and switched to the Aggressive Let It Ride, winning back $300 of the $400 those 9 losses cost me. I left the casino up $50, thankful that I was able to identify that the worm had turned.
My typical casino trip will be a $1000 bankroll. I start with the $10 Don’t Pass / Don’t Come system for a while, making sure I don’t lose more than $100. Then I split $900 into 3 $300 Aggressive Let It Ride sessions. Pocketed chips, btw, never see the light of day for the rest of that casino trip. The few dollars left over go to dinner and slots on the way out. You’d be amazed how often I come home having at least $900 in my pocket. Sure, sometimes it’s only $600, rarely it’s $300, and once it was $0, but I’ve also come out with $1500 a couple times.
Gambling is fun if you understand the ground rules. Have a fixed bankroll. Limit your variance. And above all, have fun.
One final system that is just as bad as the Martingale: (very bad)
Everything Pays Place the 6 and 8 for $18 each, Place the 5 for $15, and put $10 on the field every roll except when coming out. Basically, every roll of the dice (except 7) pays roughly $10. Any seven loses roughly $60. Can you get 6 non-7s before the next 7? I think we all know the answer to that one. High variance and high activity. Definitely not for the faint of heart.
[sup]1[/sup] When I lost my shirt in Casino War playing the Delambert system, my friend was playing your exact Martingale strategy in the chair next to me. He lost $1300 in a single 5 minute period. (We won him $700 back in a roulette system I made up on the spot, but the lesson was still never forgotten. Martingale = bad.)
[sup]2[/sup] Roulette’s “50-50” odds are exactly equal to 18 / 38, which is 47.37%.
The Pass (and Come) “50-50” odds are exactly equal to 976 / 1980, which is 49.29%.
The Don’t Pass (and Don’t Come) odds are exactly equal to 949 / 1925, which is 49.30%.
If anybody is interested in why these craps numbers are exact, multiply each of the 36 possible rolls by 55 and make a chart. (I can post a chart if anyone cares.)
[sup]3[/sup] The field in craps is virtually the same, mathematically, as a 50-50 wager on roulette. In roulette, you have 18 ways out of 38 to win. With the field, you have 16 ways to win out of 36 combinations. But the 2 and 12 pay double, so you can count those as extra ways to win. Therefore, you end up with the same 18 out of 38 ways to win as roulette. It’s not precisely the same, but close enough statistically as to be indistinguishable. To demonstrate, assume $10 bets, and every possibility happens once. In roulette, you won $10 18 times and lost 20 times, for a net $20 loss. In the field, you won $10 14 times, $20 twice, and lost 20 times, for a net $20 loss. Same damn bet. (Though roulette had $20 more action.) The main difference is that you will get 6 or more rolls of the dice in the same time as a single spin of the roulette wheel, so the brutality of a 5.26% house advantage is easier to see on the field. (Okay, technically, 5.56% on the field. But anything over 2% is terrible.)
Yes, but my point is, even if you figure out that there is some sort of bias such that a certain number comes up more than it should, that doesn’t mean you are going to win. You have to get not just any edge, but an edge better than the house edge (around 5% apparently). Otherwise, in the long term, you’re still going to lose.
And I’ll bet anything (except something that’s actually worth something) that there ain’t no way you are going to find a bias in the wheel or the guy running the game that will give you an edge better than 5%
I agree completely. The biased-wheel theory in roulette really requires a single-zero wheel, where the house advantage drops to 2.70%. And even then, good friggin luck.