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!

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!

link to original post

Without answering your question directly I’ll try and help. If you find a situation where you consistently have a small advantage, like card counting in BJ, hole carding in a few games, certain promo opportunities, etc…. You should learn the Kelly Criterion. It will help you figure out how much to bet, way better than any system like Dalembert.

My answer assumes you actually have a bankroll that you can define.

Quote:SOOPOOWithout answering your question directly I’ll try and help. If you find a situation where you consistently have a small advantage, like card counting in BJ, hole carding in a few games, certain promo opportunities, etc…. You should learn the Kelly Criterion. It will help you figure out how much to bet, way better than any system like Dalembert.

My answer assumes you actually have a bankroll that you can define.

link to original post

Thanks!

Quote:SOOPOOQuote: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!

link to original post

Without answering your question directly I’ll try and help. If you find a situation where you consistently have a small advantage, like card counting in BJ, hole carding in a few games, certain promo opportunities, etc…. You should learn the Kelly Criterion. It will help you figure out how much to bet, way better than any system like Dalembert.

My answer assumes you actually have a bankroll that you can define.

link to original post

does Kelly Criterion apply to mhb slots?

i usually max bet even $200 progressives..

most of the time I lose $. occasionally, i have a hand pay.

since i dont keep track of individual plays, i have no idea if max betting is profitable.

(overall, my bank acct grows.)

would kelly help me decide on the amount to bet to be more profitable?

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.

The d'Alembert betting started at 1 unit and was set to increase by 1 unit after a loss and decrease by 1 unit after a win (standard stuff).

The bankroll was unlimited.

The results were as follows:

FLAT BETTING:

After 100,000,000 tosses the final balance was 1,021,998. The total amount wagered was 100,000,000 units (flat betting) so the actual player advantage was 1.021998%. The lowest balance reached by the player at any time was -170.

D'ALEMBERT BETTING:

After 100,000,000 tosses the final balance was 50,504,783. The total amount wagered was 5,020,198,669 units so the actual player advantage was 1.00603%. The lowest balance reached by the player at any time was -10879.

The player advantage in both cases approaches the theoretical advantage of 1%. I don't think the discrepency is significant. I noticed bigger discrepencies when trying it out for a smaller number of trials. I suppose I could run it for a billion trials, but there doesn't seem to be much point.

So even when you have a statistical advantage, over the long run d'Alembert betting seems pretty useless.

VOLATILITY:

The d'Alembert player had a much lower "lowest balance" but we need to take into account the difference in bet sizes. The average bet for the d'Alembert player was 50.2 units (which is why their final balance of around 50.5 million units is comparable to the 1 million units of the flat bettor).

So if we imagine that the flat bettor was also betting 50.2 units then we have some more useful data to compare.

FLAT (ADJUSTED) vs

D'ALEMBERT

Final balance:

51,304,299.6

50,504,783

Total amount bet:

5,020,000,000

5,020,198,669

Lowest balance:

-8534

-10879

So it looks like the d'Alembert was a little bit more risky, and since it doesn't produce better results, why bother?

Not sure why I decided to mess with this today, but I hope the results are of some use to someone somewhere!

Quote:100xOddsQuote:SOOPOOQuote: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!

link to original post

Without answering your question directly I’ll try and help. If you find a situation where you consistently have a small advantage, like card counting in BJ, hole carding in a few games, certain promo opportunities, etc…. You should learn the Kelly Criterion. It will help you figure out how much to bet, way better than any system like Dalembert.

My answer assumes you actually have a bankroll that you can define.

link to original post

does Kelly Criterion apply to mhb slots?

i usually max bet even $200 progressives..

most of the time I lose $. occasionally, i have a hand pay.

since i dont keep track of individual plays, i have no idea if max betting is profitable.

(overall, my bank acct grows.)

would kelly help me decide on the amount to bet to be more profitable?

link to original post

I’m not the expert…. But I think Kelly will work if you know your edge, your bankroll, and usually most difficult, the variance. Or standard deviation of? Anyway, if it’s a high variance game, like the lottery, you will need a ginormous bankroll to take advantage of a small edge. In your example, the slightly off coin, Kelly would work very well. Probably easier for to just read about it on WOO than for me to try and explain it.

Quote:DJGenius

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!

link to original post

I am going to use a simpler system to make a few points. I assume a truncated Martingale system: you bet one unit and double to two units after a loss and then go back to one unit win or lose.

The one third of your bets will be 2 units and the rest will be one unit (on average). You will have a higher variance and a higher profit per bet compared to flat betting one unit. If you instead just bet two units every third time, you will have the same expected profit, but your variance will be slightly lower. This is because you will be betting one unit exactly 2/3rds of the time. In the Martingale case, you will be betting one unit 2/3rds of the time on average. If you run good on your one-unit bets, you will be betting less. If you run bad, you will be making more max bets. This is an extra source of variance.

If you truncate the Martingale after a bigger bet, let us say 8 units, the same principle applies. You will be betting 8 units only after a sequence incurs 3 losses to start off. If you figure out what fraction of bets end up being 8 units, you could just bet 8 units exactly that many times. You would have the same expected profit, but lower variance. The total variance and EV does not depend on what order you roll the 1/2/4/8 unit bets -- just on the fraction of bets that are at each multiple.

This is good stuff. It shows that if you have an advantage, you have an advantage, it doesn't matter if you have a system or not. Now run the same simulation with a -1% disadvantage.Quote: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.

The d'Alembert betting started at 1 unit and was set to increase by 1 unit after a loss and decrease by 1 unit after a win (standard stuff).

The bankroll was unlimited.

The results were as follows:

FLAT BETTING:

After 100,000,000 tosses the final balance was 1,021,998. The total amount wagered was 100,000,000 units (flat betting) so the actual player advantage was 1.021998%. The lowest balance reached by the player at any time was -170.

D'ALEMBERT BETTING:

After 100,000,000 tosses the final balance was 50,504,783. The total amount wagered was 5,020,198,669 units so the actual player advantage was 1.00603%. The lowest balance reached by the player at any time was -10879.

The player advantage in both cases approaches the theoretical advantage of 1%. I don't think the discrepency is significant. I noticed bigger discrepencies when trying it out for a smaller number of trials. I suppose I could run it for a billion trials, but there doesn't seem to be much point.

So even when you have a statistical advantage, over the long run d'Alembert betting seems pretty useless.

VOLATILITY:

The d'Alembert player had a much lower "lowest balance" but we need to take into account the difference in bet sizes. The average bet for the d'Alembert player was 50.2 units (which is why their final balance of around 50.5 million units is comparable to the 1 million units of the flat bettor).

So if we imagine that the flat bettor was also betting 50.2 units then we have some more useful data to compare.

FLAT (ADJUSTED) vs

D'ALEMBERT

Final balance:

51,304,299.6

50,504,783

Total amount bet:

5,020,000,000

5,020,198,669

Lowest balance:

-8534

-10879

So it looks like the d'Alembert was a little bit more risky, and since it doesn't produce better results, why bother?

Not sure why I decided to mess with this today, but I hope the results are of some use to someone somewhere!

link to original post

The 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:DJGenius

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!

link to original post

Quote:AxelWolfThis is good stuff. It shows that if you have an advantage, you have an advantage, it doesn't matter if you have a system or not. Now run the same simulation with a -1% disadvantage.Quote: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.

The d'Alembert betting started at 1 unit and was set to increase by 1 unit after a loss and decrease by 1 unit after a win (standard stuff).

The bankroll was unlimited.

The results were as follows:

FLAT BETTING:

After 100,000,000 tosses the final balance was 1,021,998. The total amount wagered was 100,000,000 units (flat betting) so the actual player advantage was 1.021998%. The lowest balance reached by the player at any time was -170.

D'ALEMBERT BETTING:

After 100,000,000 tosses the final balance was 50,504,783. The total amount wagered was 5,020,198,669 units so the actual player advantage was 1.00603%. The lowest balance reached by the player at any time was -10879.

The player advantage in both cases approaches the theoretical advantage of 1%. I don't think the discrepency is significant. I noticed bigger discrepencies when trying it out for a smaller number of trials. I suppose I could run it for a billion trials, but there doesn't seem to be much point.

So even when you have a statistical advantage, over the long run d'Alembert betting seems pretty useless.

VOLATILITY:

The d'Alembert player had a much lower "lowest balance" but we need to take into account the difference in bet sizes. The average bet for the d'Alembert player was 50.2 units (which is why their final balance of around 50.5 million units is comparable to the 1 million units of the flat bettor).

So if we imagine that the flat bettor was also betting 50.2 units then we have some more useful data to compare.

FLAT (ADJUSTED) vs

D'ALEMBERT

Final balance:

51,304,299.6

50,504,783

Total amount bet:

5,020,000,000

5,020,198,669

Lowest balance:

-8534

-10879

So it looks like the d'Alembert was a little bit more risky, and since it doesn't produce better results, why bother?

Not sure why I decided to mess with this today, but I hope the results are of some use to someone somewhere!

link to original post

link to original post

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!

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.

link to original post

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!

link to original post

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:DJGenius

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!

link to original post

link to original post

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

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.

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.

link to original post

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.

link to original post

link to original post

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.

Quote:Mission146If 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.

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That's probably true, but not necessarily realistic. The big players often raise their expectations such that even their huge session bankrolls don't always equate to no risk of ruin when considered against the max bet allowed.

Quote:Mission146

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.

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Thank for working up the examples. I never do anything where my RoR is much above zero, so I seldom mess with these RoR calculations.

I often have to worry about having enough cash on hand for a particular play. Whether its earning hurdle bonuses, playing leaderboards, of taking down must hit progressives, the loss in EV for running out of cash can be very large. Most of the time, I do not have accurate information about the probability distribution for the game at hand, so I have to make a guess based on the likely variance.

Quote:MentalQuote:Mission146

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.

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Thank for working up the examples. I never do anything where my RoR is much above zero, so I seldom mess with these RoR calculations.

I often have to worry about having enough cash on hand for a particular play. Whether its earning hurdle bonuses, playing leaderboards, of taking down must hit progressives, the loss in EV for running out of cash can be very large. Most of the time, I do not have accurate information about the probability distribution for the game at hand, so I have to make a guess based on the likely variance.

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Slot progressives, I take it?

There are ways that you can customize the Beating Bonuses simulator to give you something fairly reasonable if you want to make worst-case scenario assumptions, imo. I like to play around with it sometimes; the main thing is that you have to sort of standardize Free Games, which isn't ideal, because Free Games can have a wide variety of ($$$) results all on their own within the probability of just hitting Free Games to begin with.

When I sometimes play around, I start with the assumption that the machine pays whatever the minimum is for that jurisdiction, or 80%, whichever higher. Without knowing anything else, I figure Free Games (in total) makes up about 20%-30% of the base RTP on the game and then I just goof around with probabilities for, "High tier," standardized Free Games results, "Middle Tier," and, "Garbage Tier," standardized.

What you end up with is obviously still not going to be a very accurate reflection, but it probably ends up being basically within reason and it's kind of fun to play around with anyway.

Another thing you can do is use any known paytable/probability distributions for similar (known) games as a baseline for other games and then tweak it from there.

I can't emphasize enough that it's far from perfect, but it's better than nothing.

As fas as the point Mental made about a higher ror not meaning a higher bet - the way I always understood it and played it is that if you accept a 2% ror then that means that your max bet should be 2% of your BR

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Quote:lilredroosterGreat answers and discussion guys - thx a lot

As fas as the point Mental made about a higher ror not meaning a higher bet - the way I always understood it and played it is that if you accept a 2% ror then that means that your max bet should be 2% of your BR

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I think there are many more factors involved, although that may be close for some games.

Quote:lilredroosterGreat answers and discussion guys - thx a lot

As fas as the point Mental made about a higher ror not meaning a higher bet - the way I always understood it and played it is that if you accept a 2% ror then that means that your max bet should be 2% of your BR

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There are several rules of thumb about RoR that gloss over the very difficult problem of calculating a RoR for an arbitrary PDF. The 2% rule might apply for low-variance games like BJ or coin flipping. I rarely play low-variance games. If I need to know a session RoR, I do Monte Carlo sims. The RoR formula is really only good for low-risk games that tend towards a normal distribution very quickly. I have never used it.

If you play DK Rocket at max payoff, you have a 1 in 1030 chance of hitting a 999:1 payoff. If you bet 2% of your bankroll on DK Rocket with enough cashback to make it slightly +EV, you have a 95% chance of going broke, not 2%.

I have heard of edge over odd as a rule of thumb for Kelly betting. Your rule of thumb doesn't even have edge built into it, so it cannot be generally applicable. If you edge is negative, your RoR is 100% no matter how small your bet is.

Quote:Ace2For low-edge games, your RoR will be very close to double the probability of finishing the session with a busted bankroll. This is true for any variance level according to the Ace2 Conjecture

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Risk of ruin is defined for an infinite series of bets, What do you mean by 'finishing the session'? If a session is short enough, you can never bust.

I know a lot of folks mean something different when they talk about RoR. Risk of ruin as defined by Wikipedia, is the probability of busting a single finite bankroll before making infinite money from a +EV game.

I believe the link from Don Schlesinger's book "Blackjack Attack" gives an excellent explanation or ror for blackjack - the guy is truly great on bj imho

obviously, ror will be very different - and can get - I would presume - extremely complicated - for other games such as machine games with high variance

https://www.blackjackincolor.com/blackjackrisk1.htm

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RoR can be precisely defined as “Risk of Ruin”. This risk could be measured over a fixed session (+ev or -ev) or infinity (+ev) . I’ve never seen an exact formula for the former, but there are several for the latter. However, those infinite formulas only work for even-money payoffs (you either lose one unit or gain one unit). So you end up doing approximations in both casesQuote:MentalQuote:Ace2For low-edge games, your RoR will be very close to double the probability of finishing the session with a busted bankroll. This is true for any variance level according to the Ace2 Conjecture

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Risk of ruin is defined for an infinite series of bets, What do you mean by 'finishing the session'? If a session is short enough, you can never bust.

I know a lot of folks mean something different when they talk about RoR. Risk of ruin as defined by Wikipedia, is the probability of busting a single finite bankroll before making infinite money from a +EV game.

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The only way I know to calculate RoR exactly is via Markov chain, assuming the scenario being analyzed can be feasibly formatted into a Markov chain.

The Ace2 conjecture is simple and works very well for me since I’m calculating a reasonable bankroll amount to bring for an estimated length of play of a very low-edge game (pass line with full odds). It will also work well for a low edge game like blackjack.

Quote:Mission146Quote: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

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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.

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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.

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There is also an "Eternal" calculator here:

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

Can I ask, for blackjack what would be the number to input for "Standard Deviation"? It says it is the square root of the variance of each bet. I've seen numbers around 1.3 listed as the variance for blackjack, though I think it changes depending on the rules. So it's the square root of that?

I put in -0.005, with a standard deviation of 1.15, and a bankroll of 25, and got a risk of ruin of 1.2.

I raised my bankroll to 250, and got a risk of ruin of 6.6

I raised my bankroll to 500, and got a risk of ruin of 43.8

I raised my bankroll to 750, and got a risk of ruin of 290.3

I raised my bankroll to 1000, and got a risk of ruin of 1,922.4

I raised my bankroll to 1250, and got a risk of ruin of 12,729.5

I raised my bankroll to 2500, and got a risk of ruin of 162 million.

Just multiply each 250 increment RoR by 6.636 to get to the next result.

For a typical BJ game with a flat-betting player, the Standard Deviation (SD) is 1.15 units per round. The actual value varies a bit with different sets of rules, but 1.15 is a pretty fair average value. The SD varies TREMENDOUSLY if the player varies his bet (as card counters do, for instance). - Sep 29, 2014

In fact, the wiki post gives a formula for a random walk where "at every iterative step, is moved by a normal distribution having mean μ and standard deviation σ and failure occurs if it reaches 0 or a negative value." So it works for more than just coin-flip PDFs.Quote:Ace2RoR can be precisely defined as “Risk of Ruin”. This risk could be measured over a fixed session (+ev or -ev) or infinity (+ev) . I’ve never seen an exact formula for the former, but there are several for the latter. However, those infinite formulas only work for even-money payoffs (you either lose one unit or gain one unit). So you end up doing approximations in both casesQuote:MentalQuote:Ace2For low-edge games, your RoR will be very close to double the probability of finishing the session with a busted bankroll. This is true for any variance level according to the Ace2 Conjecture

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Risk of ruin is defined for an infinite series of bets, What do you mean by 'finishing the session'? If a session is short enough, you can never bust.

I know a lot of folks mean something different when they talk about RoR. Risk of ruin as defined by Wikipedia, is the probability of busting a single finite bankroll before making infinite money from a +EV game.

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The only way I know to calculate RoR exactly is via Markov chain, assuming the scenario being analyzed can be feasibly formatted into a Markov chain.

The Ace2 conjecture is simple and works very well for me since I’m calculating a reasonable bankroll amount to bring for an estimated length of play of a very low-edge game (pass line with full odds). It will also work well for a low edge game like blackjack.

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Also, your conjecture cannot be applicable without specifying how many bets are intended to be made in session (barring busting out). The OP does not specify a session length and there is no canonical session length. What exactly is your conjecture? What do you mean by "works very well for me"? We cannot prove or disprove it without knowing what it is in detail.

Quote:Ace2If you don’t know what standard deviation is and how to calculate it, then you need to learn a lot more before tackling RoR calcs

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You could have just said yes. Asking questions on forums is one of the ways I end up learning.

Have you yourself ever done a calculation of the variance of BJ. The variance does depend on the exact rules, so I suppose you have done dozens of these calculations. I have never done a single one of these variance calculations for BJ, but I have done the calculations from scratch for multi-play video poker.Quote:Ace2If you don’t know what standard deviation is and how to calculate it, then you need to learn a lot more before tackling RoR calcs

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To the original question, using Variance = 1.3 is close enough.

Quote:MentalHave you yourself ever done a calculation of the variance of BJ. The variance does depend on the exact rules, so I suppose you have done dozens of these calculations. I have never done a single one of these variance calculations for BJ, but I have done the calculations from scratch for multi-play video poker.Quote:Ace2If you don’t know what standard deviation is and how to calculate it, then you need to learn a lot more before tackling RoR calcs

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To the original question, using Variance = 1.3 is close enough.

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Wow I'm sure those variance calculations are pretty complex. The best I have ever done from scratch is an infinite deck BJ house edge calculation. Clearly I had a NEWBIE sized hole in my knowledge about standard deviation and variance. I knew those forumlas once, but I haven't used them for years. That hole has now been re-filled.

Thank you kindly for answering my question.

I said that I’ve seen exact formulas only for even-money scenarios. The formula you see on Wikipedia is an estimate, not an exact calculation. Like all estimates, it will work very well for certain scenarios and be useless for others. Incidentally, you seem to think that one short Wikipedia article is the full bible on RoR. It isn’tQuote:Mental[]In fact, the wiki post gives a formula for a random walk where "at every iterative step, is moved by a normal distribution having mean μ and standard deviation σ and failure occurs if it reaches 0 or a negative value." So it works for more than just coin-flip PDFs.

Also, your conjecture cannot be applicable without specifying how many bets are intended to be made in session (barring busting out). The OP does not specify a session length and there is no canonical session length. What exactly is your conjecture? What do you mean by "works very well for me"? We cannot prove or disprove it without knowing what it is in detail.

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The description of the simple conjecture and the game I usually apply it to are listed in the last few pages of this threadQuote:Mental[

Also, your conjecture cannot be applicable without specifying how many bets are intended to be made in session (barring busting out). The OP does not specify a session length and there is no canonical session length. What exactly is your conjecture? What do you mean by "works very well for me"? We cannot prove or disprove it without knowing what it is in detail.

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Quote:Ace2I said that I’ve seen exact formulas only for even-money scenarios. The formula you see on Wikipedia is an estimate, not an exact calculation. Like all estimates, it will work very well for certain scenarios and be useless for others. Incidentally, you seem to think that one short Wikipedia article is the full bible on RoR. It isn’tQuote:Mental[]In fact, the wiki post gives a formula for a random walk where "at every iterative step, is moved by a normal distribution having mean μ and standard deviation σ and failure occurs if it reaches 0 or a negative value." So it works for more than just coin-flip PDFs.

Also, your conjecture cannot be applicable without specifying how many bets are intended to be made in session (barring busting out). The OP does not specify a session length and there is no canonical session length. What exactly is your conjecture? What do you mean by "works very well for me"? We cannot prove or disprove it without knowing what it is in detail.

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Unless I’m missing something (which is very possible). there’s nothing about even money bets that makes an RoR calculation much more difficult than longer or shorter odds.

The concept gets applied in stock market investing through optimal portfolio growth theory. Kelly’s original paper, if my memory still serves, was a problem of information theory that is easiest to apply to a horse race (and which horses you should bet on when you have imperfect information but better than random information).

Yes, but that ideal 'betting system' is to raise you bet and hold it there. Flat betting at a higher level gets you the biggest increase in throughput with the smallest increase in variance and RoR. Any betting system which has this same average bet, but changes the bet up and down, is inferior in terms of risk. It will also be slower in many cases.Quote:Talldude90In general, wouldn't betting systems be a defined boon in a +EV situation as long as you are not raising your RoR significantly? Wouldn't you be getting a lot more coin-in against the +EV (as long as the system doesn't lower your EV), so your end profit should be bigger for a set amount of time gambling (say 4 hour session)?

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Quote:MentalYes, but that ideal 'betting system' is to raise you bet and hold it there. Flat betting at a higher level gets you the biggest increase in throughput with the smallest increase in variance and RoR. Any betting system which has this same average bet, but changes the bet up and down, is inferior in terms of risk. It will also be slower in many cases.Quote:Talldude90In general, wouldn't betting systems be a defined boon in a +EV situation as long as you are not raising your RoR significantly? Wouldn't you be getting a lot more coin-in against the +EV (as long as the system doesn't lower your EV), so your end profit should be bigger for a set amount of time gambling (say 4 hour session)?

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Isn’t the “ideal” betting system to bet the Kelly fraction of your bankroll each bet, which would vary as your bankroll fluctuates.

How practical would that be to do on the fly while playing?Quote:unJonQuote:MentalYes, but that ideal 'betting system' is to raise you bet and hold it there. Flat betting at a higher level gets you the biggest increase in throughput with the smallest increase in variance and RoR. Any betting system which has this same average bet, but changes the bet up and down, is inferior in terms of risk. It will also be slower in many cases.Quote:Talldude90In general, wouldn't betting systems be a defined boon in a +EV situation as long as you are not raising your RoR significantly? Wouldn't you be getting a lot more coin-in against the +EV (as long as the system doesn't lower your EV), so your end profit should be bigger for a set amount of time gambling (say 4 hour session)?

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Isn’t the “ideal” betting system to bet the Kelly fraction of your bankroll each bet, which would vary as your bankroll fluctuates.

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Quote:AxelWolfHow practical would that be to do on the fly while playing?Quote:unJonQuote:MentalQuote:Talldude90

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Isn’t the “ideal” betting system to bet the Kelly fraction of your bankroll each bet, which would vary as your bankroll fluctuates.

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What I've always been told about this was that it's not really wise to adjust your Kelly bet too often (on every single hand for example). I imagine you would base your flat bet on the formula and then play a pre-determined number of hands (a session?) and then recalculate from there.

But I don't know if that is actually something that is mathematically beneficial or if it's just recommended due to the sheer impracticability of recalculating and changing the bet that often.

Quote:DJGenius

What I've always been told about this was that it's not really wise to adjust your Kelly bet too often (on every single hand for example). I imagine you would base your flat bet on the formula and then play a pre-determined number of hands (a session?) and then recalculate from there.

But I don't know if that is actually something that is mathematically beneficial or if it's just recommended due to the sheer impracticability of recalculating and changing the bet that often.

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If I recall correctly, for a low variance game like Blackjack, Kelly says to wager a percentage of your bankroll equal to your percentage edge. But of course that edge fluctuates and you spread your wagers a lot. You are not going to sit placing bets of $101, $102, $103 $:o)

Just step up or down when your bankroll has changed significantly enough. E.g increased by 25%, maybe what was a $100 wager might be a $125 one.

Even with exact kelly betting, your bankroll chart would be all over the place and would not look very optimal.

Quote:AxelWolfHow practical would that be to do on the fly while playing?Quote:unJonQuote:MentalQuote:Talldude90

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Isn’t the “ideal” betting system to bet the Kelly fraction of your bankroll each bet, which would vary as your bankroll fluctuates.

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Not practical at all. I thought I was responding to a post about what was mathematically ideal in the abstract. But in the real world there are a bunch of practical considerations.