Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 31
  • Posts: 2095
November 24th, 2022 at 8:13:46 AM permalink
Quote: Mental

Quote: Ace2

For 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
link to original post



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.
link to original post

RoR can be precisely defined as ďRisk of RuinĒ. This risk could be measured over a fixed session (+ev or -ev) or infinity (+ev) . Iíve 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 cases

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 Iím 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.
Itís all about making that GTA
ChumpChange
ChumpChange 
Joined: Jun 15, 2018
  • Threads: 81
  • Posts: 3653
November 24th, 2022 at 8:24:06 AM permalink
Lose 10, win 1, lose 10, win1, lose 10, win 1...see, losing 30 times doesn't have to be all in a row! Oh, that wasn't the hard 6 I bet on, it was the PL!
DJGenius
DJGenius
Joined: Mar 5, 2010
  • Threads: 8
  • Posts: 45
Thanks for this post from:
Mission146
November 24th, 2022 at 8:36:39 AM permalink
Quote: Mission146

Quote: Mental

Quote: Mission146

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.
link to original post

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.
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.
link to original post



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?
"The Quest stands upon the edge of a knife. Stray but a little, and it will fail, to the ruin of all." - Elf Queen Galadriel, teaching Frodo about the importance of blackjack basic strategy.
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 31
  • Posts: 2095
November 24th, 2022 at 8:43:48 AM permalink
If you donít know what standard deviation is and how to calculate it, then you need to learn a lot more before tackling RoR calcs
Itís all about making that GTA
ChumpChange
ChumpChange 
Joined: Jun 15, 2018
  • Threads: 81
  • Posts: 3653
Thanks for this post from:
DJGenius
November 24th, 2022 at 8:44:52 AM permalink
For Black Jack:
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
Last edited by: ChumpChange on Nov 24, 2022
Mental
Mental
Joined: Dec 10, 2018
  • Threads: 4
  • Posts: 239
November 24th, 2022 at 9:00:11 AM permalink
Quote: Ace2

Quote: Mental

Quote: Ace2

For 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
link to original post



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.
link to original post

RoR can be precisely defined as ďRisk of RuinĒ. This risk could be measured over a fixed session (+ev or -ev) or infinity (+ev) . Iíve 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 cases

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 Iím 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.
link to original post

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.
DJGenius
DJGenius
Joined: Mar 5, 2010
  • Threads: 8
  • Posts: 45
November 24th, 2022 at 9:10:53 AM permalink
Quote: Ace2

If you donít know what standard deviation is and how to calculate it, then you need to learn a lot more before tackling RoR calcs
link to original post



You could have just said yes. Asking questions on forums is one of the ways I end up learning.
"The Quest stands upon the edge of a knife. Stray but a little, and it will fail, to the ruin of all." - Elf Queen Galadriel, teaching Frodo about the importance of blackjack basic strategy.
Mental
Mental
Joined: Dec 10, 2018
  • Threads: 4
  • Posts: 239
Thanks for this post from:
DJGenius
November 24th, 2022 at 9:24:28 AM permalink
Quote: Ace2

If you donít know what standard deviation is and how to calculate it, then you need to learn a lot more before tackling RoR calcs
link to original post

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.

To the original question, using Variance = 1.3 is close enough.
DJGenius
DJGenius
Joined: Mar 5, 2010
  • Threads: 8
  • Posts: 45
November 24th, 2022 at 9:56:08 AM permalink
Quote: Mental

Quote: Ace2

If you donít know what standard deviation is and how to calculate it, then you need to learn a lot more before tackling RoR calcs
link to original post

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.

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



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.
"The Quest stands upon the edge of a knife. Stray but a little, and it will fail, to the ruin of all." - Elf Queen Galadriel, teaching Frodo about the importance of blackjack basic strategy.
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 31
  • Posts: 2095
November 24th, 2022 at 10:44:01 AM permalink
I doubt the BJ variance of about 1.3 changes materially no matter what rule set you use
Itís all about making that GTA

  • Jump to: