Quote:DJGeniusQuote:AxelWolfQuote:DJGeniusJust in case anyone is interested I wrote a quick program to test this for myself.

The program flips a rigged "coin" which will come up heads 101 times and tails 99 times out of 200. So if my math is right, a player always betting on heads has a 1% advantage.

I set the program to flat bet 1 unit and d'Alembert bet on the same results and ran it for 100 million tosses.

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Thanks for your comment. Running the program with a 1% disadvantage (house edge) results in the same. Both systems approach the expected result of -1%.

However it's worth noting that the poor d'Alembert system is forced to increase again and again (since it always increases with a negative result, and things will be trending negative here). The final result was a loss of over 501 billion dollars! So the average bet was somewhere around 500,000 units higher than flat betting. In other words it was a train wreck!

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The max bet is smaller than the max bet would be for Martingale. You could have kept the probabilities the same and reduced the payoff to go from +1% to -1%.

the percentage of edge does not change because you use a system

the important thing is personal preference

would you prefer an occasional large win and are willing to accept small losses along the way____?_________then maybe a reverse martingale trying for 4 or five wins in a row

would you prefer to win very often and can accept an occasional large loss_________?___________maybe a 4 or 5 step marty

if you make very, very many bets and you have wagered for example 100K - your expectation will be the same no matter if you use a system or not

as long as you have made a great many of your largest bets

the most effective method is to bet proportionately ro the size of your bankroll - always betting a certain percentage of your bankrolll as it relates to the size of your edge - larger bets when your bankroll increases and smaller bets if the size of your bankroll decreases and also as your edge increases or decreases

this style of play is often referred to as "Kelly" - I think because somebody named Kelly published it as an effective method of play in advantage blackjack

.

That having been said, systems would be potentially inadvisable to someone playing with an advantage, particularly so with negative progression systems.

Again, what we tend to assume is that the casinos have an unlimited bankroll...which is as good as true for the purposes of assumption, and even when untrue, they can still cap the maximum bet the player is allowed.

Most players, however, do NOT have an unlimited bankroll.

Therefore, in theory, (and I don't feel like coming up with a specific example), a player could go from an effective Risk-of-Ruin of zero (they are well-bankrolled enough to meet their bet size, advantage and variance for the particular way such that they will never go El Busto) to potentially going to a non-zero RoR with a negative progression system, such as a Martingale.

Quote:MentalThe important point is that EV adds up proportionally to the bet size and variance adds as the square of the bet size. So, you intuition is right. For a given sum of total wagers over a given number of bets, you always minimize variance by flat betting. If betting one unit gets you one unit of variance for a certain wager, and you bet 1, 2, and 3 units, then you will incur 1*1+2*2+3*3 = 14 units of variance. Flat betting 2 units 3 times gets you 3*(2*2) = 12 units of variance. The EV is the same for both scenarios.Quote:DJGeniusApologies I'm sure this issue has been addressed in the past. Searching "betting systems" with "advantage" brings up lots of stuff very different from what I'm asking here.

I'm asking about a hypothetical situation where the player has a small mathematical advantage, for example somewhere between 0.50% and 0.75%

Would a betting system like the d'Alembert (increase 1 unit after a loss, decrease 1 unit after a win) increase profits, since you know in the long run you will actually win more than you lose?

Or would it be basically exactly the same as if you had bet the average bet size from your d'Alembert progression and just bet flat the whole time?

Does it make any difference at all? Seems like the d'Alembert would be more volatile.

Any help appreciated!

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That makes sense. Thanks for explaining the math on the variance because I haven’t seen it expressed quite that simply before.

in AP blackjack iirc it was often recommended to size your bet accepting a 2%, or 1.5% or 1% risk or ruin - which meant that your max bet would be 2% of your bankroll if you were accepting a 2% RoR

an interesting question - to me anyway - maybe for one of our mathletes to answer - attn. Mission - is this:

if an AP bj player instead accepted a 4% risk or ruin compared to a 2% risk of ruin____________

how much would his profit increase in percentage terms over the course of ten thousand hands assuming he didn't tap out______?

the intuitive answer is that compared to a 2% RoR his profit would double

but I'm not at all sure that that is correct

.

If you bet $10, then $20, then $30, then start over, your average bet is $20. In blackjack you can get the EV of any hand with EV(NumHands) = (NumHands*AvgBet)*(HouseEdge).

So if you flat bet $10 all night at a .5% HE game and play 75 hands per hour, 4 hours (so 300 hands).... EV(300 hands) = (300*10)*(-.005) = -$15.

If you use the "betting system" above, then what's the EV? Well, your average bet is $20, so... EV = (300*20)*(-.005) = -$30

So I hope you also see here why betting systems are meaningless... because they don't change the underlying house edge. Thus, if you're playing a negative edge game then all you're doing is raising your average bet, which gets hit with the house edge.

Now, if you have a slight advantage (say .5%) instead of disadvantage, then you would up your expected profit (but only because it upped your average bet). So one might think, then just bet all I can and maximize my EV, right?!?! Well, what if you bust on your first hand and can't play anymore? So how much should you bet to maximize your EV whilst taking in to account your bankroll restrictions? As SOOPOO pointed out, Kelly Criteria tells us this.

Certainly not me——————-ha ha ha

.

Quote:lilredrooster______________

in AP blackjack iirc it was often recommended to size your bet accepting a 2%, or 1.5% or 1% risk or ruin - which meant that your max bet would be 2% of your bankroll if you were accepting a 2% RoR

an interesting question - to me anyway - maybe for one of our mathletes to answer - attn. Mission - is this:

if an AP bj player instead accepted a 4% risk or ruin compared to a 2% risk of ruin____________

how much would his profit increase in percentage terms over the course of ten thousand hands assuming he didn't tap out______?

the intuitive answer is that compared to a 2% RoR his profit would double

but I'm not at all sure that that is correct

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It's a rather impossible question to answer. At an absolute minimum, we would need to know the precise bankroll size.

Another thing to consider is the fact that, if you want to maintain a 2% (or 4%) RoR relative to the advantage and your bankroll size, then that may eventually entail either sizing up or sizing down your bets.

With that, even operating under the assumption that our 4% RoR player doesn't bust out, that doesn't mean that the 4% RoR player wouldn't have to bet size down, at some point, in the 10k hands.

On the other hand, as bankroll increases, both 2% RoR and 4% RoR become greater bet amounts.

Of course, if you wanted to set a bet amount for 10,000 hands based on a 2% RoR and one based on a 4% RoR with the stipulation that both have the same bankroll, both can do not actually bust out AND neither varies from the bet amount associated with the initial RoR, then your intuition is correct that one would have double the expected profit of the other as that person is betting double the first one.

I am not seeing how doubling the ROR percentage implies doubling the bet. Doesn't the bet size vary roughly as the square root of percentage ROR? I am basing this on a random walk situation.Quote:Mission146Quote:lilredrooster______________

in AP blackjack iirc it was often recommended to size your bet accepting a 2%, or 1.5% or 1% risk or ruin - which meant that your max bet would be 2% of your bankroll if you were accepting a 2% RoR

an interesting question - to me anyway - maybe for one of our mathletes to answer - attn. Mission - is this:

if an AP bj player instead accepted a 4% risk or ruin compared to a 2% risk of ruin____________

how much would his profit increase in percentage terms over the course of ten thousand hands assuming he didn't tap out______?

the intuitive answer is that compared to a 2% RoR his profit would double

but I'm not at all sure that that is correct

link to original post

It's a rather impossible question to answer. At an absolute minimum, we would need to know the precise bankroll size.

Another thing to consider is the fact that, if you want to maintain a 2% (or 4%) RoR relative to the advantage and your bankroll size, then that may eventually entail either sizing up or sizing down your bets.

With that, even operating under the assumption that our 4% RoR player doesn't bust out, that doesn't mean that the 4% RoR player wouldn't have to bet size down, at some point, in the 10k hands.

On the other hand, as bankroll increases, both 2% RoR and 4% RoR become greater bet amounts.

Of course, if you wanted to set a bet amount for 10,000 hands based on a 2% RoR and one based on a 4% RoR with the stipulation that both have the same bankroll, both can do not actually bust out AND neither varies from the bet amount associated with the initial RoR, then your intuition is correct that one would have double the expected profit of the other as that person is betting double the first one.

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Quote:MentalI am not seeing how doubling the ROR percentage implies doubling the bet. Doesn't the bet size vary roughly as the square root of percentage ROR? I am basing this on a random walk situation.Quote:Mission146Quote:lilredrooster______________

in AP blackjack iirc it was often recommended to size your bet accepting a 2%, or 1.5% or 1% risk or ruin - which meant that your max bet would be 2% of your bankroll if you were accepting a 2% RoR

an interesting question - to me anyway - maybe for one of our mathletes to answer - attn. Mission - is this:

if an AP bj player instead accepted a 4% risk or ruin compared to a 2% risk of ruin____________

how much would his profit increase in percentage terms over the course of ten thousand hands assuming he didn't tap out______?

the intuitive answer is that compared to a 2% RoR his profit would double

but I'm not at all sure that that is correct

link to original post

It's a rather impossible question to answer. At an absolute minimum, we would need to know the precise bankroll size.

Another thing to consider is the fact that, if you want to maintain a 2% (or 4%) RoR relative to the advantage and your bankroll size, then that may eventually entail either sizing up or sizing down your bets.

With that, even operating under the assumption that our 4% RoR player doesn't bust out, that doesn't mean that the 4% RoR player wouldn't have to bet size down, at some point, in the 10k hands.

On the other hand, as bankroll increases, both 2% RoR and 4% RoR become greater bet amounts.

Of course, if you wanted to set a bet amount for 10,000 hands based on a 2% RoR and one based on a 4% RoR with the stipulation that both have the same bankroll, both can do not actually bust out AND neither varies from the bet amount associated with the initial RoR, then your intuition is correct that one would have double the expected profit of the other as that person is betting double the first one.

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You're probably right, and if you are, then doubling the expected outcome wouldn't even be implied.

Let me try to do this a different way using the work of someone smarter than I am, namely, Wizard:

https://wizardofodds.com/gambling/risk-of-ruin-calculator/

Bankroll: 186

Game: Blackjack

House Edge: -.01 (I gave the player a 1% advantage)

HPH: 100

Hours Played: 100 (To make 10,000 outcomes)

SD/Unit: 1.15 (This will remain constant)

RESULTS:

Probability of Ruin: .02 (2%)

Expected Loss -100 (You are expected to win $100)

Total SD +/- 115

Probability Win: 0.80773

Okay, so doubling the bet would be the equivalent of halving the bankroll, so if that increases the RoR beyond 4%, then we will know that you're right.

Bankroll: 93

Game: Blackjack

House Edge: -.01

HPH: 100

Hours Played: 100

SD/UNIT: 1.15

Probability of Ruin: 17.6%

Expected Loss: -100 (This is by Units and represents an expected win of 100 units, so has, in fact, doubled)

Total SD: +/- 115

Probability Win: 0.80773

This is interesting because it gives the same probability of finishing with a profit. As it turns out, it's just broken...I created a RoR in excess of 20% and it still gives the same probability of winning...which isn't possible.

In any event, it doesn't change the conclusion that doubling the RoR requires less than double the bet amount. Assuming everything else about the calculator is working, it says that a bankroll of 159 units or 159/186 = 0.85483870967 is low enough to cause the RoR to go from 2% to 4% for that particular situation.

New Experiment

I'm going to do a new experiment with the simulator at Beating Bonuses found here:

https://www.beatingbonuses.com/simulator.htm

I'm going to do a custom coin-flipping game where wins pay 1.02 units and losses cause a loss of 1.00 unit with both being 50% to happen, thereby creating a 1% player advantage. The first thing I am going to do is trial-and-error my way to finding the bankroll with a 2% bust rate based on one unit bets.

It looks like that's going to be in the 155-156 range. The SD for each sample of 10,000 simulations of 10,000 trials is about +/-100 relative to the average result, so that makes sense. The average return is a little under 100 units, as it should be, because the people who busted out can't play anymore.

Dropping the bankroll to 130 units (83.333%) is sufficient to increase the RoR to something close to 4% in multiple trials. The average return in units actually decreases, obviously, as a result of the fact that more people busted out and were unable to complete wagering.

However, even if we made it theoretically impossible to bust out, the average return (in units) is about the same---and we are not cutting the number of units in half (doubling the bet), so the expected profit does not increase at double the rate as you double RoR. In fact, for this example, if you call the original (2% RoR) unit $5, now you are only betting $6. In terms of actual gain, therefore, you're gaining an average of not even $100 more for $60,000 in bets as opposed to $50,000 in bets.

If the story has a moral, then I think the moral is that the best risk of ruin is 0.0000000000001% Risk of Ruin, then you functionally don't even have to worry about any of this.