Even in the statistically implausible scenario where he lost 100 flips in a row, he could comfortably make his next linear progression bet of $101. He would always get a win eventually. His biggest bet would always be a win. And every time he won, he would start over at $1.

If you were to take the other side of all these bets, you would expect to lose 50% of them. But knowing that your biggest wager would result in a loss to him, wouldn't your overall EV be negative and the progressive bettor's EV be positive?

If you don't believe the system gives any advantage to the bettor, then you have to believe that a collection of bets varying randomly in amount with the winner of each unknown is analytically the same as a series of bets with one important piece of additional information -- namely, that the largest single bet will be a win for the bettor.

Can those who insist that all betting systems are worthless explain why they believe the winner of the biggest single bet in a series of 50/50 bets has no greater probability of coming out on top overall than the loser of that bet?

If the bettor put out random bet sizes, then after the fact had his largest bet automatically marked as a win by the game's scorekeeper regardless of the actual outcome, would that confer an advantage to the bettor? Of course it would. If rigging one of the game's recorded outcomes confers an advantage, then why wouldn't using a system that generates that same outcome also confer an advantage?

No casino allows a situation such as this.

When it is said all systems are worthless it is inferred in a casino under casino rules and conditions.

I can make any system a winner if I set the rules.

But you do need to define "worthless" and "worthwhile".

Can it dent the EV? Some can

Can it break the house edge? Almost none can.

in blackjack, previous wins are loosely corelated with high cards coming out, which in turn is loosely correlated with more likely future losses. Previous losses are correlated with small cards coming out. It's an extremely raw version of card counting, but it has can dent the baseline EV nonethenless.

That baseline dent is enough for me to beat a 0.07% House edge blackjack game easily with no card counting, just using previous wins/losses as a betting system.

Single deck, Das, H17, Split4,DDany

It can turn that into a positive EV game.

But of course, it would not be worth your time to play a near 0 EV game. Also good luck finding any Single deck 3:2 DDany game.

All of this is just to prove a point. "no", betting systems can dent the EV, in extreme cases, can turn a tiny house edge game into a tiny player edge game. Whatever floats your boat for definition of "worthwhile"

Quote:JackSpade

Can those who insist that all betting systems are worthless explain why they believe the winner of the biggest single bet in a series of 50/50 bets has no greater probability of coming out on top overall than the loser of that bet?

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My limited understanding is that a progressive betting scheme favors the side with the larger bankroll, as their counterpart is more likely to bust out first.

Even the smallest casinos seem to have vastly larger bankrolls than I.

How much I bet doesn't seem to influence whether a tossed coin lands obverse or reverse.

But most often when the progression is run on an "even money" bet, a win will occur on either the first or the second bet made, securing a profit. If a third bet is required, a win will result in a break-even outcome. After three losses in a row, returns become negative.

If your goal is to never lose money at the casino, then you should never step foot in one! But if your goal is to increase the odds of a winning session and cut into the house edge, then you should consider using a linear progression.

To be clear, this system does NOT give players an edge over the casino in games like roulette and baccarat. It does put the casino in a slightly less advantageous position versus the player. And when considering the value of comps, some players may be able to glean net benefits from playing negative expectation games by making progressive bets that increase their average bet size.

Quote:JackSpadeIf two millionaires decided to bet on coin flips where there was exactly a 50% chance of winning each time, neither would gain any advantage by flat betting. But if one used a system where he would start with a $1 bet and increase his bet by $1 until he won a flip, he would be able to guarantee himself that his largest bet would always be a winning bet.

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His guarantee would comfortably cover the eventuality that he might go bust, but his winning wager would not recoup prior losses so as a system, it's worthless.

The rest of your post was nonsense.. Sorry.

Quote:JackSpadeDieter, there is no realistic risk of a properly bankrolled player going bust before completing a linear progression that starts with a table minimum bet and terminates with a single win. (He's not doubling up after losses; he's only increasing his bet size by a single unit.) There is a risk of losing money overall after completing a progression or re-running it multiple times. The only claim being made is that the risk of experiencing negative total returns from a series of bets is lowered if the largest bet in that series is a win.

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Flat betting, the player with a smaller bankroll is more likely to bust out first.

With progressive betting, they are still more likely to bust out first, it just decreases the number of rounds until it can happen.

This saves time, and hopefully increases however entertainment is quantified.

Best of luck!

Quote:JackSpadeDieter, there is no realistic risk of a properly bankrolled player going bust before completing a linear progression that starts with a table minimum bet and terminates with a single win. (He's not doubling up after losses; he's only increasing his bet size by a single unit.) There is a risk of losing money overall after completing a progression or re-running it multiple times. The only claim being made is that the risk of experiencing negative total returns from a series of bets is lowered if the largest bet in that series is a win.

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Time for some quick math: let B the initial bet, and N the number of consecutive losses before a win.

Your total losses are B + (B + 1) + ... + (B + (N - 1)) - (B + N) = (N - 1) B + N (N - 3) / 2.

This is negative when (N - 1) B + N (N - 3) / 2 < 0

(N - 1) B < - N (N - 3) / 2

B < N (3 - N) / (2 (N - 1))

Notice that, if N >= 3, 3 - N <= 0, so B has to be <= 0, which is impossible.

The "risk of experiencing negative total returns from a series of bets" is 100% if you lose the first three bets.

That's why Martingale is the way that it is - a win guarantees that the total return from that series of bets is positive.

Quote:JackSpadeIf two millionaires decided to bet on coin flips where there was exactly a 50% chance of winning each time, neither would gain any advantage by flat betting. But if one used a system where he would start with a $1 bet and increase his bet by $1 until he won a flip, he would be able to guarantee himself that his largest bet would always be a winning bet.

Even in the statistically implausible scenario where he lost 100 flips in a row, he could comfortably make his next linear progression bet of $101. He would always get a win eventually. His biggest bet would always be a win. And every time he won, he would start over at $1.

If you were to take the other side of all these bets, you would expect to lose 50% of them. But knowing that your biggest wager would result in a loss to him, wouldn't your overall EV be negative and the progressive bettor's EV be positive?

If you don't believe the system gives any advantage to the bettor, then you have to believe that a collection of bets varying randomly in amount with the winner of each unknown is analytically the same as a series of bets with one important piece of additional information -- namely, that the largest single bet will be a win for the bettor.

Can those who insist that all betting systems are worthless explain why they believe the winner of the biggest single bet in a series of 50/50 bets has no greater probability of coming out on top overall than the loser of that bet?

If the bettor put out random bet sizes, then after the fact had his largest bet automatically marked as a win by the game's scorekeeper regardless of the actual outcome, would that confer an advantage to the bettor? Of course it would. If rigging one of the game's recorded outcomes confers an advantage, then why wouldn't using a system that generates that same outcome also confer an advantage?

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I apologize for the fact that one of our forum members has called your post "nonsense." I think it was well-written and thoughtful. Much of it was posed as questions and thought exercises, and, IMO, questions are almost never "nonsense."

Let's look at your stated system: linear progression in the wager size until you win, given a 50/50 fair coin flip. Here are some of the possible flip sequences.

_W_ You win your first coin flip with a probability of 0.5, and stop. You win +1 unit.

_LW_ You lose your first flip and win your second flip, which has a wager size of +2, and stop. This has a probability of 0.25 and a net payout of +1 unit.

_LLW_ This has a probability of 0.125 and a net payout of 0 units.

_LLL W_ This has a probability of 0.00625 (1/16) and a net payout of -2 units.

_LLL LW_ This has a probability of 0.003125 (1/32) and a net payout of -5 units.

_LLL LLW_ This has a probability of 0.0015625 (1/64) and a net payout of -9 units.

...and so on.

With this sequence

1. you will never have a net win for a session greater than +1 unit.

2. You will have a winning session 75% of the time (winning one unit.)

3. You will have a losing session 0.125% of the time, losing at least 2 units, but sometimes, losing 5, 9 or 14 units or more! The bigger your final winning bet, the more you will have lost in the session. So winning the largest bet in an infinite (or very long) sequence of such bets (using your liner progression method) doesn't sound attractive to me.

4. Over many such sessions, it is expected that you will win 1/2 of your bets and break even with a net profit of 0. Just as you would if you had not used a linear progression method.

what are your thoughts?

Quote:gordonm888Quote:JackSpade

4. Over many such sessions, it is expected that you will win 1/2 of your bets and break even with a net profit of 0. Just as you would if you had not used a linear progression method.

what are your thoughts?

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Since it is also expected that the largest bet amount will be a win, these equal probability bets do not carry equal weight. A 50% win rate can generate a greater than 0% expected value if the wins have a larger average bet size than the losses.

Some in this forum have suggested that we cannot expect the player to be able to secure a highest-bet win. They apparently think that the risk of the bettor going bust is meaningful enough to render the system useless. They either didn't read, didn't understand, or for whatever reason deliberately avoided addressing the actual contents of the post to which they were replying. The probability of losing 50 coin flips in a row is 0.000000000000001. The probability of losing 100 flips in a row approaches 0, but I showed how even such a streak would not come close to causing the bettor to go bust.

The bettor would have a greater chance of causing a casino to go bust by hitting multiple back to back royal flushes. At some point down the statistical abyss from improbable to impossible, we have to recognize that certain things that could still happen in theory -- like winning every poker hand or losing every coin flip -- never will happen in reality.

Securing the highest-bet win is not the issue. Oh I expect it all right. I acknowledge that the linear progression system will be fantastically unlikely to bust your millionaiireQuote:JackSpade

Some in this forum have suggested that we cannot expect the player to be able to secure a highest-bet win.

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If that highest bet win does not return the bettor's previous losses in full, it's rather a pointless exercise.

His bankroll is taking small steps backwards and forwards. He'd get a lot of play. His expected value for the period of play would be 0.

Quote:JackSpadeQuote:gordonm888

4. Over many such sessions, it is expected that you will win 1/2 of your bets and break even with a net profit of 0. Just as you would if you had not used a linear progression method.

Since it is also expected that the largest bet amount will be a win, these equal probability bets do not carry equal weight. A 50% win rate can generate a greater than 0% expected value if the wins have a larger average bet size than the losses.

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"Can," but not necessarily "will."

Suppose your first four bets are losses, and your next four are wins. You won 1/2 of your bets, as expected. However, your results are: lost 1, lost 2, lost 3, lost 4, won 5, won 1, won 1, won 1, for a total of a loss of 2. If the first four bets are wins and the next four are losses, that's four wins of 1 followed by losses of 1, 2, 3, and 4, for a total loss of 6.

Quote:OnceDearQuote:JackSpade

If that highest bet win does not return the bettor's previous losses in full, it's rather a pointless exercise.

His bankroll is taking small steps backwards and forwards. He'd get a lot of play. His expected value for the period of play would be 0.

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I'd like to see the outcomes of simulations. Other systems get stretched to their breaking point in simulated play and cause massive bankroll declines. This one would surely be less volatile. I suspect as long as it never busts, it would exhibit some statistical tendency toward generating positive returns. But I'm willing to be proven wrong.

Quote:JackSpadeQuote:OnceDearQuote:JackSpade

If that highest bet win does not return the bettor's previous losses in full, it's rather a pointless exercise.

His bankroll is taking small steps backwards and forwards. He'd get a lot of play. His expected value for the period of play would be 0.

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I'd like to see the outcomes of simulations. Other systems get stretched to their breaking point in simulated play and cause massive bankroll declines. This one would surely be less volatile. I suspect as long as it never busts, it would exhibit some statistical tendency toward generating positive returns. But I'm willing to be proven wrong.

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Jack, welcome, but I think you are going about this the wrong way. You can pick any number of bets you want, but if the casino has an advantage on each bet do you really (even just intuitively) think summing ALL these bets that you are at a disadvantage you will end up with an advantage?

The math about such systems is crystal clear, proven repeatedly, that there is NO system that can somehow combine -EV bets into some magic positive expectation.

Now….. I don’t think betting systems are worthless necessarily. They certainly can be used to give you a greater chance to achieve a specific goal. If your goal is to win $5 then martingaling $5 don’t pass bets at craps is a great system to achieve your goal. Lots of betting systems add fun to the player. I play pass line with odds, and continuous come bets with odds. I in NO WAY think I have a system that has an advantage over the casino, but I do have a system that gives me a decent chance to double my initial stake.

If someone ran a simulation on your system, if big enough, your results would mirror the house edge.

Quote:JackSpadeI'd like to see the outcomes of simulations. Other systems get stretched to their breaking point in simulated play and cause massive bankroll declines. This one would surely be less volatile. I suspect as long as it never busts, it would exhibit some statistical tendency toward generating positive returns. But I'm willing to be proven wrong.

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The key phrase is, "as long as it never busts." Remember, Martingale "generates a positive return" 100% of the time "as long as it never busts."

Your statement as given probably will not be proven wrong, as it sounds correct, since, with an infinite bankroll (so there's no risk of busting), pretty much any system should reach a profit on a 50/50 game "eventually."

In order to simulate it properly, I need to know all of the conditions. If I understand you correctly, your system is:

(a) Start with a bet of 1

(b) When you lose, increase the bet by 1

(c) When you win, stop

Do I have this right? Because if I do, then there's what a simulation would show:

1/2 of the time, you win your first bet, so you are +1; the EV here is 1/2 x (+1) = 1/2.

1/4 of the time, you lose your first bet, then win your second, so you are -1+2 = +1; the EV here is 1/4 x (+1) = 1/4.

Notice that 3/4 of the time, you make money.

1/8 of the time, you lose your first two bets, then win your third, so you are -1-2+3 = 0; the EV here is zero

1/16 of the time, you lose your first three bets, then win your fourth, so you are -1-2-3+4 = -2; the EV here is 1/16 x (-2) = -1/8.

1/32 of the time, you lose your first four bets, then win your fifth, so you are -1-2-3-4+5 = -5; the EV here is 1/32 x (-5) = -5/32.

1/64 of the time, you lose your first five bets, then win your sixth, so you are -1-2-3-4-5+6 = -9; the EV here is 1/64 x (-9) = -9/64.

I can show you the math, but the combined result, as you would expect for a 50/50 game, is zero.

Why, yes, the combined EV never actually reaches zero unless you lose an infinite number of bets - How About That!

Gee, you don't think this is one of the reasons why no casinos actually offer any 50/50 games, do you?

Even that is questionable. There is no guarantee that a win will ever come. Given enough time, you will eventually see a string of, for instance, 100 trillion consecutive losses.Quote:ThatDonGuy

The key phrase is, "as long as it never busts." Remember, Martingale "generates a positive return" 100% of the time "as long as it never busts."

Your statement as given probably will not be proven wrong, as it sounds correct, since, with an infinite bankroll (so there's no risk of busting), pretty much any system should reach a profit on a 50/50 game "eventually."

As a side note, it has already been reported that casinos set min/max bet limits to segregate players by bet level, not to thwart martingale bettors. Martingale will still fail eventually since no one has an infinite bankroll.

No, I don't think applying progressive betting strategy to casino games will in itself give the player an advantage over the casino. It will give the player an advantage over the way most people play.

Why are actual casino holds on games like baccarat, which require no skill to play optimally, consistently much higher than the reported house edges for those games? The biggest reason is that most players place side bets that offer terrible odds.

But there's something else going on, because players are even managing to lose on roulette at a higher rate than the house edge. There are typically no side bets in roulette. Could it be that players are losing more because they rarely walk away after a win? They usually quit after a loss either busts their bankroll or breaches their pain threshold. Consistently giving their last bet (and often their largest bet) to the casino instead of cashing out immediately after a previous win boosts the casino's EV and hurts theirs.

Wonderful, you break even over the long run. Can you tell us where one can freely martingale where the table max is unlimited while using real money, and they are giving you a true 50/50?Quote:JackSpadeSOOPOO, I invite you to read my original post which references 50/50 coin flips that have no house edge.

Quote:JackSpadeSOOPOO, I invite you to read my original post which references 50/50 coin flips that have no house edge.

No, I don't think applying progressive betting strategy to casino games will in itself give the player an advantage over the casino. It will give the player an advantage over the way most people play.

Why are actual casino holds on games like baccarat, which require no skill to play optimally, consistently much higher than the reported house edges for those games? The biggest reason is that most players place side bets that offer terrible odds.

But there's something else going on, because players are even managing to lose on roulette at a higher rate than the house edge. There are typically no side bets in roulette. Could it be that players are losing more because they rarely walk away after a win? They usually quit after a loss either busts their bankroll or breaches their pain threshold. Consistently giving their last bet (and often their largest bet) to the casino instead of cashing out immediately after a previous win boosts the casino's EV and hurts theirs.

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I don’t believe what you are saying is true. At a non thinking game like roulette, casinos will win the house edge times the amount bet. It’s really that simple. Players rarely ‘walk away with a win’ not because of a system, but because (mostly) of the house edge.

As far as your OP saying a 50/50 game, then yes, whatever system you are proposing over a large enough number of simulations will approach even.

Edit. A casino ‘hold’ will be larger than the house edge, because a player coming with a bankroll of $XX will tend to bet some multiple of $XX during his session. it is not the initial bankroll that is only exposed to the house edge, it is the total bet that is.

That’s been proven false. Having preset stopping points has no effect on house edgeQuote:JackSpade

But there's something else going on, because players are even managing to lose on roulette at a higher rate than the house edge. There are typically no side bets in roulette. Could it be that players are losing more because they rarely walk away after a win? They usually quit after a loss either busts their bankroll or breaches their pain threshold. Consistently giving their last bet (and often their largest bet) to the casino instead of cashing out immediately after a previous win boosts the casino's EV and hurts theirs.

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Quote:JackSpade

But there's something else going on, because players are even managing to lose on roulette at a higher rate than the house edge.

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The House edge does not apply to what you bring to the table or what you buy in at the table, but what you play through the table.

So roulette* player brings $100: Buy's in for $100 and keeps staking $1 per spin.....

After 100 spins, he has put $100 at the mercy of the house edge, and he might have something like $97 left.....

Which he then subjects to the house edge again, losing another £3 or so.... Rinse Repeat.

After a while, the house will hold all his buy in amount. Hold 100% House edge: 2.7% Or maybe he walks while he still has $50, in which case the hold is 50%

No contradiction or paradox.

* European single zero.

A casino boss would expect to benefit from any betting system that caused players to wager more than they otherwise would. But doesn't the casino also benefit when players end their session with a losing bet far more often than with a win (irrespective of whether they are up or down overall)?

In a scenario where 100 gamblers make a total of 5,000 bets with an average bet size of $20, the casino would process $10 million in total wagers and theoretically retain 2.7% ($270,000).

But if each of those 100 gamblers ended their session with a winning bet, and the remaining 4,900 bets were won or lost at the statistically expected rate, the casino would likely retain less than 2.7% of the total wagered on that particular night.

Quote:Ace2Even that is questionable. There is no guarantee that a win will ever come. Given enough time, you will eventually see a string of, for instance, 100 trillion consecutive losses.Quote:ThatDonGuy

The key phrase is, "as long as it never busts." Remember, Martingale "generates a positive return" 100% of the time "as long as it never busts."

Your statement as given probably will not be proven wrong, as it sounds correct, since, with an infinite bankroll (so there's no risk of busting), pretty much any system should reach a profit on a 50/50 game "eventually."

As a side note, it has already been reported that casinos set min/max bet limits to segregate players by bet level, not to thwart martingale bettors. Martingale will still fail eventually since no one has an infinite bankroll.

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And you make fun of Alan’s 18 yos. Bit hypocritical.

I never said 18 consecutive yo’s was impossible. But I’d bet everything I have that no human has ever seen it. Same goes for 100 trillion consecutive headsQuote:unJon[]

And you make fun of Alan’s 18 yos. Bit hypocritical.

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Quote:JackSpadeWhy are actual casino holds on games like baccarat, which require no skill to play optimally, consistently much higher than the reported house edges for those games? The biggest reason is that most players place side bets that offer terrible odds.

But there's something else going on, because players are even managing to lose on roulette at a higher rate than the house edge. There are typically no side bets in roulette. Could it be that players are losing more because they rarely walk away after a win? They usually quit after a loss either busts their bankroll or breaches their pain threshold. Consistently giving their last bet (and often their largest bet) to the casino instead of cashing out immediately after a previous win boosts the casino's EV and hurts theirs.

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My observation is that these larger than expected losses happen when table games players keep having unrealistic “do or die” win goals relative to bankrolls - and keep blowing out.

What’s really bad is that often the goal in these situations rises as the bankroll decreases.

For example the player with a $10K bankroll wants to win 500 or 1000. Does it regularly but then one session gets -6000 down and now with that remaining 4000 wants to win back the entire 10K. Of course it’s easier to win 500 or 1000 with $10K than to win 6000 with $4K - the risk of ruin is definitely greater in the latter situation.

Then say the player loses the entire $10K comes back the next day with another $10K wants to win all of the lost $10K “do or die.” If he doesn’t make it and comes back the next day with $10K unwilling to stop until he has the entire lost now $20K back, the chances of success drop yet again.

I see this sort of thing all the time. People down to less than half of their bankrolls unwilling to get up until they’ve either lost it all or recouped it all. I see them win decent sums that in a vacuum they’d be happy with as session results but - because they want it all, end up being given back on the way to a total blowout.

Of course these people aren’t flat betting either.

I liken it to getting into a sports car with an increasingly higher drag coefficient - imagine 😃 the car is sticky and is driving through a leaf and twig storm where more and more crap keeps sticking to its body - and trying to achieve the same top speed. Just gets harder and harder but doesn’t stop people from trying.

In short - if your win goal is increasingly unrealistic relative to your bankroll you’ll be more likely to just blowout before the house edge even becomes a real issue.

I wrote about this at greater length here.

Different game, but same general concept of chasing an ever increasing sum with an ever decreasing bankroll:

I've often seen these people, these squares, at the table.

Short-stacked and long odds against, all their outs gone,

one last card in the deck that can help them.

I used to wonder how they could let themselves get into such bad shape...

and how the hell they thought they could turn it around.

You also mentioned “quitting while ahead.” In the context of stopping when a realistic goal relative to bankroll is achieved versus pressing on for an unrealistic win relative to bankroll, stopping at certain points is a strategy that means something, especially if on the next session a consequence of not stopping is a higher goal against a same or smaller bankroll.

That ploppy’s last “do or die” $10k bet is a drop in the bucket to a Vegas casino. It’s just another regular bet on which their average profit will be the house edge for that gameQuote:MDawgQuote:JackSpadeWhy are actual casino holds on games like baccarat, which require no skill to play optimally, consistently much higher than the reported house edges for those games? The biggest reason is that most players place side bets that offer terrible odds.

But there's something else going on, because players are even managing to lose on roulette at a higher rate than the house edge. There are typically no side bets in roulette. Could it be that players are losing more because they rarely walk away after a win? They usually quit after a loss either busts their bankroll or breaches their pain threshold. Consistently giving their last bet (and often their largest bet) to the casino instead of cashing out immediately after a previous win boosts the casino's EV and hurts theirs.

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Then say the player loses the entire $10K comes back the next day with another $10K wants to win all of the lost $10K “do or die.” If he doesn’t make it and comes back the next day with $10K unwilling to stop until he has the entire lost now $20K back, the chances of success drop yet again.

If all you do is flat bet, and just stand there playing at the craps table until you get tired, then come back to more of the exact same with no clear goal in mind, and always the same relative bankroll, then it might make sense why you would say some of what you say.

Quote:Ace23 point Molly with 3-4-5 odds has a standard deviation of 9.4, MUCH higher than any table game I know of (even a single number roulette bet) and nearly 10x the SD of baccarat. Objectively speaking, baccarat is ten times more boring/tiresome

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It's interesting to me that you quote numbers but don't tell us about actual results.

That’s the difference. I’m always playing for material amounts. You play for pennies 90% of the time

As long as your bet level is high enough, you could flat bet on a coin flip (pretty much the same as baccarat) and it will be exciting.

My actual results are as posted.