Acceptable R.O.R

[BJC]

OK, you got me. Gonna have to explain this one. Top bet = (BR x edge)/variance. Assuming a variance of about 1.30 at higher counts, that would imply that your edge, when you make your top bet, is of the order of 9.75%! Damn! I must be playing the wrong games!

Don

Hi Don,

Variance = Standard Deviation?

Another Question - On terminology - it says Full kelly = 13.5% risk of ruin and according to BJA, that means a 2/3 chance of doubling bank roll before halving it and 5% chance of being wiped out.

For terminology - does the 13.5% mean the chance that we at some point drop to 50% bankroll of our original bankroll? so 1/7 chance of doing that overall, and a 1/3 chance of doing that before first hitting 2x original bankroll?

13.5% risk of ruin
36.7% chance of losing half bank roll
5% chance of total wipeout. I get this


Thanks,
BJC

[DSchles]

Variance = Standard Deviation?

'

No, variance = the SQUARE of standard deviation.

Another Question - On terminology - it says Full kelly = 13.5% risk of ruin and according to BJA, that means a 2/3 chance of doubling bank roll before halving it and 5% chance of being wiped out.

13.5% ROR refers to probability of tapping out if you never resize your original bets; i.e., you don't reduce your bet sizes as you are losing. If you do constantly resize (which no one does), theoretically your ROR is zero. While the statement about doubling before halving is true, it isn't the essence of Kelly, because doubling is simply an artificial barrier that doesn't have a lot of meaning for the long term.

and 5% chance of being wiped out.

Don't know where you got this from. It isn't correct. ROR = chance of being wiped out. They're the same thing.

13.5% risk of ruin

If you never resize, yes.

36.7% chance of losing half bank roll

No. 1/3 = 33.3%.

5% chance of total wipeout. I get this

No, unfortunately, you don't.

Don

[ShipTheCookies]

...If you do constantly resize (which no one does), theoretically your ROR is zero...

This is exactly why many of them fail at this endeavor.

[BJC]

Hi Don,

Thanks! I read the last paragraph in chapter 8 before the tables and few times and now i get it.

I didn't realize the ROR calculation assumed never readjusting your bankroll.

Another question - the maximum bet for kelly, in terms of blackjack, is that assuming a situation like being at your top bet, and then splitting and doubling?

Example, betting full kelly count is TC 5+ giving me 2.5% edge roughly, and i have my max bet out, on a 10k bankroll of about $200, but it i have to split and then double i could very easily end up with 8% of my total bankroll on the table.

This seems like massive overbetting. Should the max bet be adjusted so that in a situation like this with a split and double that you have out max $200?

Terminology wise, would the second example be called quarter kelly or full kelly?

BJC

[DSchles]

Hi Don,

Thanks! I read the last paragraph in chapter 8 before the tables and few times and now i get it.

I didn't realize the ROR calculation assumed never readjusting your bankroll.

Another question - the maximum bet for kelly, in terms of blackjack, is that assuming a situation like being at your top bet, and then splitting and doubling?

Example, betting full kelly count is TC 5+ giving me 2.5% edge roughly, and i have my max bet out, on a 10k bankroll of about $200, but it i have to split and then double i could very easily end up with 8% of my total bankroll on the table.

This seems like massive overbetting. Should the max bet be adjusted so that in a situation like this with a split and double that you have out max $200?

Terminology wise, would the second example be called quarter kelly or full kelly?

BJC

The reason you divide the edge by the variance is exactly to take into account the situations you describe. It's already factored into the Kelly wager. Variance in blackjack is virtually the same thing as the average squared bet size, so, again, no need to make any further adjustments.

Don

[BJC]

Thanks Don and Optimus.

In terms of the math, are splits treated an seperate hands?

For double downs, do most DD's favor the player by 2x the initial advantage?

[DSchles]

In terms of the math, are splits treated an seperate hands?

Not sure what you're asking.

For double downs, do most DD's favor the player by 2x the initial advantage?

Never counted them all. Many doubles, where you would not draw more than one card to the hand, are obviously double the initial edge. But many others, including some soft doubles and some doubles of 9, 10, and 11, vs. dealer's 7-A (some using indices) sacrifice some of the initial edge for the privilege of doubling.

Don

[Freightman]

Not sure what you're asking.



Never counted them all. Many doubles, where you would not draw more than one card to the hand, are obviously double the initial edge. But many others, including some soft doubles and some doubles of 9, 10, and 11, vs. dealer's 7-A (some using indices) sacrifice some of the initial edge for the privilege of doubling.

Don

I think a distinction should be made between double the edge and fewer hands won.

Case in point - for every 100 hard doubles available, you will win fewer hands doubled than non doubled, simply because you have an opportunity to take another card or cards on non doubled hands.

Goal of course, is most money at the end of the day, achieved by doubling. Easy to enough to show.

[DSchles]

Case in point - for every 100 hard doubles available, you will win fewer hands doubled than non doubled, simply because you have an opportunity to take another card or cards on non doubled hands.

Yes, that's what I said. But I believe the question was, do most doubles favor the player by twice the initial advantage? I said I hadn't counted them. Do you happen to know?

Don

[Freightman]

Yes, that's what I said. But I believe the question was, do most doubles favor the player by twice the initial advantage? I said I hadn't counted them. Do you happen to know?

Don

No, I don’t - It’s easy enough to make a table outlining differentials in dollars between doubling and non doubling over any number of hands at various percentages of successful doubles. I would start with something conservative like 60% and 55% for hard and soft doubles respectively.

Could probably explain it in under a paragraph

Maybe a different way to look at it. If Player first card us a 10, with its corresponding 10% advantage, then what is player advantage for first 2 cards holding 82,73,64,55,92,83,74,65 opposite dealer up card of anything 2 thru 9.

With varying advantages depending on dealer upcard - has to be huge.

[DSchles]

No, I don’t - It’s easy enough to make a table outlining differentials in dollars between doubling and non doubling over any number of hands at various percentages of successful doubles. I would start with something conservative like 60% and 55% for hard and soft doubles respectively.

Could probably explain it in under a paragraph

Maybe a different way to look at it. If Player first card us a 10, with its corresponding 10% advantage, then what is player advantage for first 2 cards holding 82,73,64,55,92,83,74,65 opposite dealer up card of anything 2 thru 9.

With varying advantages depending on dealer upcard - has to be huge.

The ability to determine DD edges is beyond simple. All the values are there. There's no mystery to the process. I'm simply saying that some doubles are done because the doubled value is greater than the undoubled value, but not by twice as much, while others are fully double the initial-hand edge. I just have never counted each variety to see which is the more prevalent. It's child's play to do, but I have no inclination to want to do it.

Don

[BJC]

Hi Don,

Where in BJA or Theory of BJ would I find the correct tables to add up?

I'm also considering it from a cutoff/shuffle track perspective of betting 1-2 units higher for the first .5-1.5 decks with the assumption that there is a higher frequency of high cards available. It would help me understand the odds of busting the dealer, but also whether it is worth doubling on a hand, eating up a 10 that could bust the dealer, or potentially help make a 21 on a later hand. (assuming the tracked section has a higher prevalence of aces also) Also, if i spread to multiple hands to try to catch an ace, it might again help make an educated decision whether burning cards is valuable depending on what actually came out.

Example, If i think a quarter deck segment has +5 (total ten's = 10 and 3 other), and I've spread 4 hands, and between myself and the dealer have seen 4 ten cards.

Let's say the dealer has a 5 or 6 showing, he likely will have a 15/16. I have 4 hands, and say one is 20, the other 2 tens are in 2 stiffs other, and the other hand could be 11 or under.

4/10 ten's have been seen, One is inferred for the dealer. i have 4 hands with 5 likely 10/Aces left.

10 of the 13 card segment has been dealt, I have seen -5 (-4 counted, -1 inferred) The next three cards are likely 10's, and the 2 ten's, that won't make it into this 13 cards segment, are probably just a card, two or three outside the original 13 card section. This is a realistic scenario as tracking is estimating versus difficult shuffles.

I would likely double on the 11 and under, stand on the stiffs, split the ten's, but if i pull another 10, it would put me into the situation of splitting again further depleting the 10 richnesss, and risking not busting the dealer. Doing this would potentially use another 3 10's and if forced to split another pair of 20's, by the time the dealer hits, the deck might be negative giving him better odds.

I know this is highly situational, but if tracking a particular segment and only playing with a higher degree of certainty that it is 10's rich by say +3 or more, you can easily get into situations where you have to make judgement calls.

I want to try to create a shuffle trackers basic strategy that assumes a high count, and a shallow penetration (to simulate utilizing this strategy for .5-1.5 decks). It would obviously recommend standing more often on all hands, but particularly stiffs. This is where the double down comes into play, and only doubling down if the advantage is truly double, since you are taking a chance burning a card that could bust the dealer.

This is all theory as i've played around with this on intuition and it seems to work, but it would be superior to have some math behind it the deviations


I am aware of methods of adding to the estimated True Count and playing the situation like that for betting purposes, but i think this doesn't do justice to the deviations, as the count tells us if low cards have been played and thus more hi cards are likely at some point in the future, whereas shuffle tracking tell us that the high cards are coming in the next 2-30 hands if we cut them off. It would be equivalent to a much higher true count. It's similar to end play for the tracked segments.

[DSchles]

Where in BJA or Theory of BJ would I find the correct tables to add up?

In all of the Appendix A charts.

I'm not going to comment on any of the rest of your post, as Freightman said it best: it's mostly voodoo. You're thinking too much, and the reasoning is invalid.

Don