posted
If Casinos dont want people to count cards, all they have to do is shuffle after every deal.
Posts: 60 | From: London | Registered: Oct 2004
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posted
It was my understanding that to prevent people from counting cards in blackjack (or at least make the matter insanely complicated), the casino simply used multiple decks shuffled together.

Wonko "Lost Wages, Nevada" the Sane

-------------------- "It seemed to me that any civilization that had so far lost its head as to need to include detailed instructions for use in a package of toothpicks, was no longer a civilzation in which I could live and stay sane." Posts: 1462 | From: Outside the Asylum (Massachusetts) | Registered: Jul 2003
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DemonWolf
Ding Dong! Merrily on High Definition TV

posted

quote:Originally posted by nicky: If Casinos dont want people to count cards, all they have to do is shuffle after every deal.

In order to prevent cheating, cards are never re-used.

-------------------- Friends are like skittles: they come in many colors, and some are fruity!

quote:Originally posted by trollface: I'm pretty sure that I once heard a system that was actually unbeatable for roulette.

First you pick a number (or a colour, I forget which) and you place the minimum bet on it. Then, when you most likely lose, you put double that on the same number. Then double again, and so on. Eventually, you'll win loads.

Of course, the problem is that you very quickly need an almost infinite amount of money.

That's d'Alembert's system.

You bet 1 on a colour. If you win, the house gives you 2, so you've won 1. If you lose, you bet 2 on the same colour. If you win this time, the house gives you 4, so you've won 4-(1+2)=1. If you lose again, you bet 4 and so on. Theoretically, you're sure to eventually win, but your benefits will only ever be 1 each time you use the system.

Ulkomaalainen correctly points out that there's a limit on the amount that can be bet (on a colour, it's 1000 times the minimum bet). This evens the chances, since you've got 1 chance out of 1024 that your colour fails to appear 10 times in a row and you lose 1023 (1+2+4+...+512), and 1023 chances out of 1024 that your colour eventually appears and you win 1.

But once again the reason why d'Alembert's system doesn't work is that there is a zero (two in the US) on the roulette. Since you (basically) lose half your bet on a colour when the zero appears, someone playing d'Alembert's system would statistically lose (1/2)*(1/37)=1.35% of their bets, or (1/2)*(2/38)=2.63% on an American roulette.

The Fourth "not sure about the math, but that's the general idea" Man

-------------------- If you keep trying, you'll eventually succeed. Therefore, the more you fail, the higher your chances of success. -- Jacques Rouxel, 1931-2004 RIP :( Posts: 406 | From: Paris, France | Registered: Jan 2004
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quote:Originally posted by trollface: I'm pretty sure that I once heard a system that was actually unbeatable for roulette.

First you pick a number (or a colour, I forget which) and you place the minimum bet on it. Then, when you most likely lose, you put double that on the same number. Then double again, and so on. Eventually, you'll win loads.

Of course, the problem is that you very quickly need an almost infinite amount of money.

That's d'Alembert's system.

You bet 1 on a colour. If you win, the house gives you 2, so you've won 1. If you lose, you bet 2 on the same colour. If you win this time, the house gives you 4, so you've won 4-(1+2)=1. If you lose again, you bet 4 and so on. Theoretically, you're sure to eventually win, but your benefits will only ever be 1 each time you use the system.

Ulkomaalainen correctly points out that there's a limit on the amount that can be bet (on a colour, it's 1000 times the minimum bet). This evens the chances, since you've got 1 chance out of 1024 that your colour fails to appear 10 times in a row and you lose 1023 (1+2+4+...+512), and 1023 chances out of 1024 that your colour eventually appears and you win 1.

But once again the reason why d'Alembert's system doesn't work is that there is a zero (two in the US) on the roulette. Since you (basically) lose half your bet on a colour when the zero appears, someone playing d'Alembert's system would statistically lose (1/2)*(1/37)=1.35% of their bets, or (1/2)*(2/38)=2.63% on an American roulette.

The Fourth "not sure about the math, but that's the general idea" Man

You're close on the math. Yes, it theoretically *does* work, but the potential winnings are so small as to make it not worth trying. And, you have to leave the table once you win.

The chance of winning one bet is 18/38 = 9/19 = .4737. So, you lose your first bet 52.63% of the time. The chance of getting to and winning your second bet is 10/19 X 9/19 = .2493, and you "win" $2-1 = $1. The chance of getting to and winning your third bet is 10/19 X 10/19 X 9/19 = .1312, and you "win" $4 - 2 - 1 = $1. etc. If you stop at the third bet, you have a 85.4% chance of walking out $1 ahead, and a 14.6% chance of walking out losing $7, meaning your expected result is -$0.17 (this means that if you did this 100 times, you would win about 85 times and lose 15 times, and you would walk out $17 poorer). If you go for a fourth round, your chance of winning $1 is now up to 92.3%, but you are now risking $15, and your expected return is -$0.23. If you go for a fifth round, your chance of winning $1 is now up to 96%, but you are risking $31, and your expected return is down -$0.29. Etc. etc. etc. If the table has a $10,000 limit, your 14th bet is the last one under the limit. You have a 99.9875% chance of winning $1.00. This is a very very safe bet, but also one with not much up-side - you win $1.00 while risking $8192 on that bet and $4096 on the previous and... and your expected return is -$1.05.

So, yes, if you can go on doubling your bet forever, you will win sooner or later. There's a 99% chance you will win by the 7th bet. However, your total return is small, and the chance of losing at the high levels negates the possible winnings. If you lose that 7th bet, you have lost $127. If you do this until you win, then leave the table, you will win $1.00. If you do not leave the table, and keep coming back, sooner or later you're going to hit that 1%, and lose $127.

-------------------- "The large print givith, and the small print taketh away" -- Tom Waits, Step Right Up

"The only difference between me and a madman is that I am not mad." -- Salvador Dali Posts: 2443 | From: Illinois | Registered: Feb 2000
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quote:And, you have to leave the table once you win.

As a person who is very interested in games of all kinds, but don't play for money and so has little casino experience, that line made me interested.

As far as I can see, it will not affect your chances at all, and the casino would probably want the player to feel "on a roll" and continue playing before they get a chance to pause and think.

Could you please explain?

-------------------- /Troberg Posts: 4360 | From: Borlänge, Sweden | Registered: Nov 2005
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quote:If Casinos dont want people to count cards, all they have to do is shuffle after every deal.

Some of them do, in effect, by dealing their blackjack games from "continuous shuffle" machines.

quote:It was my understanding that to prevent people from counting cards in blackjack (or at least make the matter insanely complicated), the casino simply used multiple decks shuffled together.

Using more decks doesn't prevent players from counting or make counting any more difficult; it just lessens the probability that counting will be advantangeous for the player.

quote:In the long run, all games where you play against the house will cost you money.

That isn't quite true. There are actually some games that are favorable for the player in the long run, provided the player has the requisite skill and bankroll to achieve the expected long-term results.

quote:Originally posted by nicky: If Casinos dont want people to count cards, all they have to do is shuffle after every deal.

In order to prevent cheating, cards are never re-used.

What? Of course cards are re-used. A blackjack table using a couple of decks in a shoe uses the same cards for hundreds of hands.

One problem with the "double the bet when you lose until you win scheme" is that most tables have an upper bet limit. Once you hit that limit there is no way for the "stategy" to recover. Posts: 629 | From: Greenwood, IN | Registered: Dec 2005
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quote:And, you have to leave the table once you win.

As a person who is very interested in games of all kinds, but don't play for money and so has little casino experience, that line made me interested.

....

Could you please explain?

Sure. Let's take a simple example, where the max bet is $10.

You bet $1 on red, and lose. You bet $2 on red, and win. You're up $1. Thinking you have a system that will win money every time, you start again.

You bet $1 on red, and win. You're now up $2.

You bet $1 on red, and win. You're now up $3.00.

You bet $1 on red, and lose. You're up $2. You bet $2 on red, and lose. You're even. You bet $4 on red, and lose. You're down $4. You bet $8 on red, and lose. You're down $12.

If you play once until you win, you will walk out $1 richer than you walked in with (as long as you have enough money to keep doubling your bet and can play forever). If you keep playing and keep playing and keep playing, the law of averages will catch up with you and you will overall lose money. With the $10 table max above, your chance of winning is 93%. If you lose, you lose $15. The law of averages says you will lose money if you keep playing.

Don't be embarassed if you still don't get it. It has to do with infinity, and most people (including d'Alembert) can't quite grasp it. If this system really worked, people would be using it, casinos would be losing money, and there would no longer be roulette.

-------------------- "The large print givith, and the small print taketh away" -- Tom Waits, Step Right Up

"The only difference between me and a madman is that I am not mad." -- Salvador Dali Posts: 2443 | From: Illinois | Registered: Feb 2000
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posted
Easier (I think) yet: at roulette the player loses in the long run, as he loses "on expectation". So, if you win and continue, you just play more games, where you lose. Thus, you eventually lose all you've won. The highest probabilty of any strategy to get out with a gain is to leave the instant you've won. Every further game theoretically only loses you money, and thus decreases the probability of getting out with a gain.

-------------------- Movie characters never make typing mistakes. Posts: 586 | From: Hamburg, Germany | Registered: Sep 2005
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posted
I should also add, that "leaving the table" is only meaningful if you mean never to play again. Otherwise, the next time you sit down at that table, you're just picking up where you left off.
Posts: 418 | From: New Port Richey, FL | Registered: Apr 2005
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posted
sorry for bumping an oldie... but here's my 2 cents.

The "double up" is better known as the martingale system. Employed properly, it can actually yield a little bit of profit. But, that's ultimately determined by the table's minimum and maximum bets. Say you have a minimum bet of $4 and a max of $2000... you can only double up 9 times after the original bet, 10 if you bet only to recoop your losses. (i.e., $4, $4, $8, $16 etc.) The real trick in using a martingale system to your own advantage is to turn the house odds in your own favor. For example, betting on any color or even/odd section yields a 9/19th chance of hitting. To turn that in your favor... bet against it. If you see black come up 5X in a row, bet on red. Granted, the odds of it coming up red are 47%, but the odds of hitting 6 blacks in a row are 9/19ths to the sixth. Basically, red, black, even and odd will all hit 6 in a row about every 88 spins. So with an average dispersion of 22 spins between each occurance, you're somewhat "safe" to make that bet. Think about it... what are the odds that red will NOT hit 30 times in a row? Not so high. This sounds like a gambler's fallacy, but there's roulette world records for a reason. Take single numbers for example. The longest "streak" of a single number hitting is 6 times. just 6. Why? 1/38th to the 6th is in the neighborhood of 1 in 3 BILLION. What I do, strategy wise, is bet on sections of 12. I bet only collumns or sections. This way, Im betting on 24/38 to hit instead of 20/38. "maximizing them odds!" Anywho, I wait until two of them have not hit for at least 3 times, then I bet equal amounts on the remaining two sections/collumns. Because they pay of 2-1, I make back my investment or 50% more if I hit on the first shot. if I dont, then I engage a modified martingale to maximize the amount of spins I will last with the ultimate goal to be only recooping what Ive lost. For example: $4 on two sections. If I win, I get my $8 back plus $4 more. If I lose, I double up. That's the damage when playing sections of 12. However, the odds of NOT hitting two sections of 12 umpty times in a row are far less than NOT hitting even/odds or black/reds.

Comparatively, not hitting black 2X in a row equates to 10/19² = 27.7% odds or a little more than 1 in four.

Odds of NOT hitting 1 through 14 2X in a row is 7/19² = 13.57% odds or 1 in 7. Since it's less common to occur, I bet against it.

My two cents.

/did well in Reno a few weeks back.
Posts: 65 | From: Vallejo, CA | Registered: May 2006
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quote:Granted, the odds of it coming up red are 47%, but the odds of hitting 6 blacks in a row are 9/19ths to the sixth. (snip) The longest "streak" of a single number hitting is 6 times. just 6. Why? 1/38th to the 6th is in the neighborhood of 1 in 3 BILLION.

You are making a classical mistake here. Each spin is completely disconnected from the previous. The wheel has no memory. Even if a single number has come up six times in a row, the probability to get it the next time is still 1/38 (if 0 and 00 is on the wheel). Your chance will not be smaller than if the previous spin was another number.

In fact, the sequence 1, 1, 1, 1, 1, 1 is exactly as probable as the sequence 24, 8, 21, 17, 10, 3 or any other specific sequence. Any mathematician could tell you that the sequence 1, 1, 1, 1, 1, 1 is no more special than another sequence, it just looks that way to our brains as it tends to stand out more.

That was the reason for my earlier question on why one should switch tables after a win.

The odds you get on any bet are on parity with the risk. They are set as to even it out. If you win one time out of ten, you get ten times your money (minus the casino's share, which me be hidden as less payouts or lesser chance of winning (0 & 00 in this case)). System or no system, if you play long enough, this is where you end up. You can't beat the odds in the long run.

There are no systems that allows you to beat the odds, at most you can decide how you want to risk your money. Systems can give you control over if you want to take a big risk with a potential big win, or if you want to have a small risk, but also a small win.

The "double on loss" system is a system that gives you a fairly large chance to win a tiny amount of money (the minimum bet), and this is offset by a tiny chance of losing a big amount of money (the maximum bet). Overall, the probabilities remain the same, you have just redistributed the when and where a little.

Most systems try to obscure this by meaningless rituals (betting alternatingly on black and red), but the bottom line is this:

There are no systems that will give you a better chance to win than the casino. The odds will always be in favour of the casino.

There are two ways of getting rich from gambling:

1. Play games of skill against other players. As long as you are better than the average player, you will win in the long run.

2. Cheat.

Honest gambling against the house will never put the odds on your side. Period.

My one advice to prospective gamblers is to place a single, large bet. That way, you can at least get lucky and win. The more bets you place, the more the law of averages sneaks up on you, and sooner or later, it will get your money.

-------------------- /Troberg Posts: 4360 | From: Borlänge, Sweden | Registered: Nov 2005
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posted
A biased wheel is extremely hard to detect, as it will at best (or worst) drop the ball in the same area most of the time.

The best way to detect it would probably be to plot the number of results as a circular bar graph, modelled after the wheel (hope that description makes sense). If you see that, after a large series, the bars on one side of the wheel are significantly longer, it's time to blow the whistle.

-------------------- /Troberg Posts: 4360 | From: Borlänge, Sweden | Registered: Nov 2005
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quote:Originally posted by Troberg: There are three ways of getting rich from gambling:

1. Play games of skill against other players. As long as you are better than the average player, you will win in the long run.

2. Cheat.

Honest gambling against the house will never put the odds on your side. Period.

3. Own the casino.

As for cheating (and other aspects of roulette playing) I can highly recommened a book called The Newtonian Casino (AKA The Eudaemonic Pie) by Thomas A. Bass. It's very interesting.

-------------------- Små hönor skall inte lägga stora ägg för då blir de slarviga i ändan Posts: 1334 | From: Sweden | Registered: Feb 2000
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quote:In fact, the sequence 1, 1, 1, 1, 1, 1 is exactly as probable as the sequence 24, 8, 21, 17, 10, 3 or any other specific sequence. Any mathematician could tell you that the sequence 1, 1, 1, 1, 1, 1 is no more special than another sequence, it just looks that way to our brains as it tends to stand out more.

That is very true. However, it does not preclude gauranteed loss. When betting, you are betting on a single spin and any color or section of 12, etc is just as likely to come up as another. That is on a SINGLE spin. What I do is bet against "occurrances", as I call them. while in the grand scale, the percentages will more closely resemble statistical probability in tewrms of percentage, they also tend to reflect probability in terms of "streaks." That is why there exists roulette world records. For example, let's examine Pi. http://3.141592653589793238462643383279502884197169399375105820974944592.com/index1.html

Any streak of 6 would be 1/10th to the 6th. aka, one in a million. 10 numbers popping up 6 times in a row in the first million places of pi is the predicted outcome. (one for each digit) 1=1 2=1 3-1 4=0 5=2 6=0 7=2 8=0 9=2 0=0 for a total of 9 "occurances".

The real trick here is to acknowledge that there is an average distribution of 111,105 numbers between each "occurance" even though the predicted outcome is 99,994. What I bet against is that I happened to sit down and play 200 (about 7 or 8 hours of gameplay) spins that fall right on that "occurance". 99,994/200 = about 1 in 500. The same principle applies to roulette.

What are the odds that I am sitting at a roulette table when it hits say... 17 hits in a row without touching a section of 24? Including avg diustribution between occurances, thats almost 1 in 4 million. divided by the 200 spins I hang around for, you're talkin one in 20,000 that I happened to sit down at the right time.

while the order of 6 blacks in a row may be just as likely as rrbrrb, it is NOT as likely as any combination that produces a single red.
Posts: 65 | From: Vallejo, CA | Registered: May 2006
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quote:This fallacy is commonly committed by gamblers who, for instance, bet on red at roulette when black has come up three times in a row. The odds of black coming up next are the same regardless of what colors have come up in previous turns.

-------------------- "If I didn't see it and didn't know it was a real news report, I wouldn't believe it. I mean, how nutty can you get?"-Pat Robertson Oct 26, 2006. Posts: 2936 | From: Mean Streets of West Virginia | Registered: Feb 2003
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See for yourself. The fact that the numbers reach a sort of critical mass that isn't increased very often attests to the fact that 100 hits/losses in a row just ISNT that likely unless you're really trying to buck the odds by playing single numbers. In fact, after 250,000 "spins", the highest number of losses on a color or even/odd scheme was 19. 10/19ths to the 19th = 1:197,863.

So yeah, just try out the sim for yourself. Run it for I dunno, 10,000,000 spins and then tell me you wouldnt risk $4 that the wheel would hit red in under 50 spins.
Posts: 65 | From: Vallejo, CA | Registered: May 2006
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posted
The sim is ultimately going to return the same results regardless of the "system" your employ. You could stay put until you hit, work your way around the wheel, add one to the number last hit or play the number of drinks consumed by fellow gambers and select the color by a coin flip or play the day of the month (when appropriate) and choose black after the sun sets or red while its still daylight.

It doesn't matter. The results will statistically come out exactly the same. Each and every spin is still an independant entity with the same probability as every one before it and every one that will come after it. "Strategy" has no part in a game of absolute chance like Roulette or the lottery.

-------------------- "If I didn't see it and didn't know it was a real news report, I wouldn't believe it. I mean, how nutty can you get?"-Pat Robertson Oct 26, 2006. Posts: 2936 | From: Mean Streets of West Virginia | Registered: Feb 2003
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quote: It doesn't matter. The results will statistically come out exactly the same. Each and every spin is still an independant entity with the same probability as every one before it and every one that will come after it. "Strategy" has no part in a game of absolute chance like Roulette or the lottery. [/QB]

Let me ask you this... if you sat down at an unbiased table, (as you are aware that the casinos do what they can to avoid biased wheels) and you saw 17 spins in a row not hit red, do you think you happened to sit down when that 17 is going to be 27 in a row or will it hit a red sometime in the next ten spins? Just tell me which you think is more likely.
Posts: 65 | From: Vallejo, CA | Registered: May 2006
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quote: It doesn't matter. The results will statistically come out exactly the same. Each and every spin is still an independant entity with the same probability as every one before it and every one that will come after it. "Strategy" has no part in a game of absolute chance like Roulette or the lottery.

Let me ask you this... if you sat down at an unbiased table, (as you are aware that the casinos do what they can to avoid biased wheels) and you saw 17 spins in a row not hit red, do you think you happened to sit down when that 17 is going to be 27 in a row or will it hit a red sometime in the next ten spins? Just tell me which you think is more likely. [/QB]

At 10 spins you would have be certain the next spin was probably going to be red. But you would have been wrong that time and the next 6 times after it. If by then you don't realize that either color is just as likely I don't know why spin nuber 18 should be magic.

posted
What do the previous 17 spins have to do with the next 10? This is the very essense of the gambler's falacy. Some are going to look at that and say that black is "hot" and favor it for the next 10. Others are going to say that red is "due," because it hasn't come up in a while. Neither makes any logical or statistial sense.

Accounting for the zeros, you're basically looking at a roughly 47% chance of black or red on each of the next 10 spins. The thing is, we're not dealing with a bet on red or black, but on one of any possible combinations of color and number. Again, you cannot improve your odds by staying on the same number or by any other "strategy." It just doesn't work that way.

-------------------- "If I didn't see it and didn't know it was a real news report, I wouldn't believe it. I mean, how nutty can you get?"-Pat Robertson Oct 26, 2006. Posts: 2936 | From: Mean Streets of West Virginia | Registered: Feb 2003
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quote: It doesn't matter. The results will statistically come out exactly the same. Each and every spin is still an independant entity with the same probability as every one before it and every one that will come after it. "Strategy" has no part in a game of absolute chance like Roulette or the lottery.

Let me ask you this... if you sat down at an unbiased table, (as you are aware that the casinos do what they can to avoid biased wheels) and you saw 17 spins in a row not hit red, do you think you happened to sit down when that 17 is going to be 27 in a row or will it hit a red sometime in the next ten spins? Just tell me which you think is more likely.

At 10 spins you would have be certain the next spin was probably going to be red. But you would have been wrong that time and the next 6 times after it. If by then you don't realize that either color is just as likely I don't know why spin nuber 18 should be magic. [/QB]

it's not magic, it's math. 10X in a row without hitting a color or section will happen more frequently than 11X and that will happen more frequently than 12X and so forth. When you sit down after 10X in a row, you're already seeing a rare occurance that only happens every so often. (10/19 to the 10th = 1 in 613 spins) If you start on that, what youre betting against is that you will hit an even rarer occurance, which is 11X or 12X in a row. The more times it misses your target, the rarer an occurance it is. What you're betting is, the 200 or so spins you stick around (about 8 hours of play) wont happen to represent that fraction/be at the same time that the wheel manages to produce some rediculous number like 50 spins in a row without hitting red. If you start at 10 misses and only bet on black, the average dispertion between events is 613 spins-10 = 603 spins. Even on that occurance, you waited until at least 5 misses before you started betting which means you'd be able to overcome that occurance even if it went up to 15 misses. Which again... did you happen to sit down during the (15,000 spins divided by 200) 1/75th fraction of the time when the table will ruin you? Odds say no.
Posts: 65 | From: Vallejo, CA | Registered: May 2006
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posted
You are mixing up two differnt things. Math says that the chance of hitting red or black on each spin is 48%. The previous spin or any spin before that does not change this math.

The odds you are calculating would only come into play if one was betting that x spins in a row would be black. Even then, the odds of each individual spin would still be 48%.

posted
The confusion is understandable, but it involves mixing two different ideas (as BeachLife is saying).

The chance that 17 spins will all come up red is much lower than the chance that 2 spins will come up red. However, that does *not* mean that each successive spin is more and more likely to produce a red number if one has not come up yet.

The distinction is difficult, which is why the gambler's fallacy is so popular.
Posts: 333 | From: Columbus, OH | Registered: Dec 2005
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quote:In the long run, all games where you play against the house will cost you money.

That isn't quite true. There are actually some games that are favorable for the player in the long run, provided the player has the requisite skill and bankroll to achieve the expected long-term results.

- snopes

As a craps player and someone familiar with a lot of casino games I would like to hear you elaborate. I am not doubting you, I am just curious.

I know of one way of gambling which guarantee that you will be successful on some occasions. It is based on what I call streak theory.

If you flip a coin a thousand times then it is mathematically likely you will have a certain number of streaks of different lengths. These streaks are fairly regular, in other words you might expect to have 300 streaks of 2, 200 streaks of 3, 100 streaks of 4, 50 streaks of 5, etc and a certain number of longer streaks, 8 or above. You want to bet in a way that minimizes your losses and raises your wins.

So you start off with your base bet of $5. If you lose you always go back to your base bet. If you win you raise your bet to $10. If you win that you raise it to $15.00. If you win that you raise it to $20.00 and don't raise it again. This means that when you hit a streak of 8 or more wins you will be winning $20.00 at a time but when you lose you only lose you always reduce your bet to $5.00 and start again.

This won't work on every trip to the casino and doesn't work very well on long sessions but it does conserve your money and gives you opportunities to leave a winner. When I stick to this at the craps table I stand a better chance of leaving a winner then when I just bet according to whim.

posted
How do you know when you're entering a streak? Every system for games of chance is based on the same mixing of overall probability with the probability of the upcoming individual throw.

quote:This won't work on every trip to the casino and doesn't work very well on long sessions but it does conserve your money and gives you opportunities to leave a winner.

It does not and cannot work. Even if it did, you are contradicting yourself. Your streak hypothesis assumes probabilities over a great number of bets, but you say it doesn't work very well over long sessions. The reality is that most games are designed to draw money away from you slowly and give the proverbial step forward for every two steps back. Otherwise you would give up to soon. If you've had a taste of winning a small amount, you're more likely to stick around believing that something even better might come your way.

Don't get me wrong, I think games are kind of fun, but to me one has to decide whether or not the entertainment value of playing is worth the money spent, as there is no reasonable expectation of beating the house.

-------------------- "If I didn't see it and didn't know it was a real news report, I wouldn't believe it. I mean, how nutty can you get?"-Pat Robertson Oct 26, 2006. Posts: 2936 | From: Mean Streets of West Virginia | Registered: Feb 2003
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posted
As far as I've read, Craps, Baccarat, and Blackjack have the lowest house percentages of any casino game. Blackjack is the only one that you can "beat" (i.e. win money in the long run) but this can only be done by counting cards, and may no longer be possible since casinos have been trying to prevent this.
Posts: 333 | From: Columbus, OH | Registered: Dec 2005
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quote:Originally posted by BeachLife: You are mixing up two differnt things. Math says that the chance of hitting red or black on each spin is 48%. The previous spin or any spin before that does not change this math.

The odds you are calculating would only come into play if one was betting that x spins in a row would be black. Even then, the odds of each individual spin would still be 48%.

right... but even though my money is active each spin after x amount of misses, what I am betting against is the so called "streak." Each INDIVIDUAL spin has the same probability as the previous, but each set of 10 does not. Take flipping a coin for example. 2 flips and the probability of two tails is 1 in 4. Are you going to bet that you will see that or would you be safer to bet that at least one of those flips comes up heads? Obviously, you'd take the 3 in 4 possibility if you had to choose. This is what I do with roulette. Would you bet on red coming up some time in the next 10 spins or would you bet that it comes up nothing but black and green for 10 times in a row? How about 20? How about 1,000?
Posts: 65 | From: Vallejo, CA | Registered: May 2006
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You can bet against the streak in your head, but it doesn't change the probability, or the math, or the reality of the situation. Your odds remain at 48%. One more thought though, the longer the streak goes on the higher the probability that it's a malfunction in the equipment. In which case you're chances are even less than 48%. Which means that betting against the streak is probably more likely to be a losing strategy.

quote:Originally posted by BeachLife: You can bet against the streak in your head, but it doesn't change the probability, or the math, or the reality of the situation. Your odds remain at 48%. One more thought though, the longer the streak goes on the higher the probability that it's a malfunction in the equipment. In which case you're chances are even less than 48%. Which means that betting against the streak is probably more likely to be a losing strategy.

think about it like this...

how many combinations of 2 spins will NOT produce a red number?

For example, the combination of events that will produce only green is 4. 0,00; 0,0; 00,0; 00,00.

Just multiply the total spaces than can produce that result by itself. Black (18 options) plus green (2 more) = 20. 20 x 20 = 400 combinations that will produce no black hits at all.

How many total combinations can the wheel produce? 38 x 38 = 1444 combinations. Subtract the number that produce no red (wins) and you have 1,044 possible combinations that will produce a red at least once because you have only 400 that will produce none.

400/1444 = 0.277008. This is the same as 20/38²

20/38 x 20/38 = 0.277008 or 27.7008% probability. Therefore, 2 misses in a row has a 27.7008% probability of happening. Inversely, this means that there is a 72.2992% probability that a red will hit over the course of 2 spins.

As we get further and further into the amount of spins, the spread between probabilities gets greater. 20 x 20 x 20 x 20 x 20 = 3,200,000 possible combinations to produce no reds out of 38 x 38 x 38 x 38 x 38 = 79,235,168.

3,200,000/79,235,168 = 0.0403861, or 4.03861% probability.

If you go 15 times without hitting red, you're witnessing some seriously low probabilities, and the next spin, for it to continue it's trend, requires even lower probabilities to show up. It is because of this that records exist for roulette hitting any color x amount of times in a row. Even blackjack has a record of the longest streak. Any game that employs probability must succumb to it. Frankly, probability states that I wont happen to sit down at the same time a 1 in 75 bazillion chance occurance happens.
Posts: 65 | From: Vallejo, CA | Registered: May 2006
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