I read that if you know that one of your cards is a A, then you should bet 40% of your entire bankroll. Has anyone done Risk of Ruin analysis for this? That would clearly depend on the table limit.
I can hack up my code to get some results, if no one else has done it.
If you are playing a game that has RSA, you would have to restrict your max bet to 25% BR -- split 4 ways -- oops lost entire BR on one hand! 8-)
On Soft hand you don't have to double or split.
Have you ever played into this game for real money? Don�t you remember what happen when you split ACES and get garbage on top of each one? Wow! .. 80% of your bankroll is down the tube now. What you do when you get some small card on top of your ACE and the dealer�s up card is 4,5,6 ? You Double Down for 80% of your bankroll and as you probably don�t know, the soft double is a very volatile play and you could lose 80% of your bankroll here again.
Now, you probably don't have any bankroll left or if you are lucky you do have some 20% of your bank. How you bet and play from this point on?
Alex_D30
If you don't play your hand properly, then you 'advantage' is less.
Plus, it is not if you are going to get an Ace, it is if your FIRST card is an Ace.
What if you threw out splits and doubles and just played the hand out?
Just wondering if a basic strategy (basically a when to stand strategy) has ever been worked out IF knowing that your first card was an Ace?
Thanks.
The figure is basically correct. With such a large fraction of bankroll bet some deviations to basic strategy are desireable for maximizing bankroll growth (not EV). You don't double or split.
Mathprof posted a comprehensive analysis of this some time ago. I think it is possible that a slightly smaller fraction might be optimal in some games so that you can split aces without so much risk, and there may be some minor changes to exploit the variance-reduction properties of a push-but the advice is very close to correct for practical purposes.
>> Plus, it is not if you are going to get an Ace, it is if your FIRST card is an Ace.
For all practical purposes, I think the two cards are dealt at the same time. What difference does it make whether your FIRST card or your SECOND card is an ACE? Care to elaborate on your comment?
Thanks,
~Black Jack.
For all practical purposes, I think the two cards are dealt at the same time. What difference does it make whether your FIRST card or your SECOND card is an ACE? Care to elaborate on your comment?
You are correct. Foreknowledge of the nth card in a hand sequence gives you equivalent advantage when n<3, deteriorating rapidly with every increment of n beyond that. I can create a formula demonstrating equivalence if this is of interest.
Concerning the Kelly criterion: If E is the expectation of a win/lose bet and p is prob of win, 1-p=q is prob of loss and payoff for win is A, payoff for loss is -1 (per unit bet), then the appropriate Kelly fraction is F = (Ap-q)/A = E/A. For even money payoffs we can express this as F=E (this is often erroneously recommended in blackjack literature). For A ><1, it's different. Minimizing E through strategy variation to some extent may be desireable if this results in a large increase in the value of F.
Depends on the method that you know you are getting an Ace.
If you get an Ace as your first card, the effect of removal on the dealer's first card (it is now less likely he will get an Ace). Also, with the Ace removed first, it is now more likely you will get a Ten as your second card. This is from a purely probabilistic type calculation. If you "know" that one of your cards is an Ace by some method, then the order doesn't matter.
Dear Dr. Seele,
>> Minimizing E through strategy variation to some extent may be desireable if this results in a large increase in the value of F.
This is somewhat a revelation to me. I personally would not use full Kelly betting even if I knew I'd get an ace, but just out of theoretical curiosity, could you please elabrote on this remark?
BTW, the orginal post by twin jacks is NOT correct then based on your statement. The majority (~30%) of the advantage, ~40%, comes from getting a Blackjack, whose payoff is 3:2. Therefore the correct (full) Kelly will be somewhat smaller than 40% of your bankroll. Probably closer to what E.J. suggested in this thread or around 30%.
TIA,
~Black Jack in the Box.
If you get an Ace as your first card, the effect of removal on the dealer's first card (it is now less likely he will get an Ace). Also, with the Ace removed first, it is now more likely you will get a Ten as your second card. This is from a purely probabilistic type calculation. If you "know" that one of your cards is an Ace by some method, then the order doesn't matter.
The probability of the dealer getting an ace is identical whether the player receives the card as first or second card. This is true for ace-system predictions (of whatever origin) with less than 100%efficiency, though the probability of the dealer receiving an ace itself changes according to efficiency of the system. The probability of a ten appearing as first card, when an ace is dealt second, is the same as that of a ten appearing when an ace is dealt first.
Simple test of theory: run random values of probabilities between 0.00 and 1.00 of receiving an ace as first and second card alternately, through Monte Carlo simulation or direct calculation determine probability of receiving a ten with ace as alternate first and second cards. Compare.
You will see the values symmetrize.
This is somewhat a revelation to me. I personally would not use full Kelly betting even if I knew I'd get an ace, but just out of theoretical curiosity, could you please elabrote on this remark?
I will in a minute, but I wanted to address what you said about avoiding full kelly betting, because it represents the common, and I believe, mistaken, wisdom about betting large fractions of bankroll on highly advantageous oppportunities.
When you bet a fraction l of your capital each time on a coin toss and win W times and lose L times, then the exponential rate of growth from Kelly's original paper is:
G = Lim{(W/N)log(1+l)+(L/N)log(1-l)
N->oo
= log(1+l)q + log(1-l)p
where p is the probability of losing and q is the probability of winning. You will see when you test this that for very high values of P resulting in a large l fraction the performance loss, as I think Thorp originally noted, is very dramatic in financial terms. Coin-toss games are simplifications but the wider principle holds for more complex games with large advantages relative to the kelly bet divisor.
Put in Layman's terms: Betting small when you have a rare opportunity like this costs you dear to the point of irrationality.
BTW, the orginal post by twin jacks is NOT correct then based on your statement. The majority (~30%) of the advantage, ~40%, comes from getting a Blackjack, whose payoff is 3:2. Therefore the correct (full) Kelly will be somewhat smaller than 40% of your bankroll. Probably closer to what E.J. suggested in this thread or around 30%.
My interpretation of this is that 40% is approximately correct. It is simply advantage when the ace is the first or second card divided by 1.3, which is the approximate kelly divisor in this case if a no-split, no-double strategy is employed. I believe your 30% figure may have arisen because you are looking at the main component of advantage rather than the "global" bet to determine the correct Kelly bet divisor.
Minor additional comment: in the literature the assumption is that blackjack has zero kurtosis, meaning it is normally distributed. This is an intelligent theoretical standpoint but does not reflect the true situation for many players who may be partially or wholly dependent on blackjack as a source of income.
For them, the game has what might be described as positive Fisher kurtosis, the extremes of negative and positive fluctuation as reflected by the normal distribution do not represent the true picture because a) extreme losses represent a reduction in an ability to cover living expenses and b) extreme wins may make expenses negligible relative to future earnings. Similarly the rate of debt repayment, if any, will be increased by bad luck at the tables, while a large bankroll arising from good fortune accrues something in interest payments.
Perhaps this is why many favour a small fraction of Kelly.
I made some changes to my code to get the Risk Of Ruin numbers.
Rules: Single Deck, DAS, DOA, Pen:65%
I used Bet spread 1-4 (except when Ace is known to be the first card)
No Double or Splitting on a hand with Ace as first card.
Risk of Ruin was done for 100000 hands.
Bankroll is 100 times the smallest bet.
Here are the numbers.
I. When there is no Table limit.
--------------------------------
% of Bankroll when first card is Ace: Risk of Ruin (%)
5% :15.50%
7% :9.89%
10% :6.21%
15% :9.80%
17% :22.58%
20% :51.16%
40% :58.90%
50% :77.84%
Optimal bet size is around 10 %.
II. When you put a table limit of 100 times the smallest bet.
-------------------------------------------------------------
% of Bankroll when first card is Ace: Risk of Ruin (%)
10% :5.19%
15% :3.64%
20% :4.16%
25% :5.03%
30% :6.52%
35% :8.84%
40% :11.17%
50% :16.35%
60% :21.59%
70% :25.72%
80% :30.82%
90% :34.60%
Optimal bet size is around 15%. But 40% has acceptable risk.
Dear Dr. Seele,
> >> Minimizing E through strategy variation to some extent may be desireable if this results in a large increase in the value of F.
> could you please elaborate on this remark?
Thanks for your message, but I couldn't really understand your answer regarding my earlier question (I read your post more than a dozen times ;-). But, in any case, I did some homework of my own, and found a very interesting story in the Theory of Blackjack 6th ed., pp 236 - 239. If anybody is interested, please take a look.
Anyways, I was somewhat confused with this whole Kelly thing, but I did some calculations again (based on the formula by Stanford Wong), and 40% came out to be a reasonable number *under the assumption* that you will get a blackjack with 30% chance, and otherwise you will win 40% and lose 30%. (I couldn't find the exact numbers for this win/loss/push probabilities. But I think this is a reasonable estimate.) If you plan to split or double then you should bet *less than* this fraction because this increases the variance. But, I can't say how much, and I don't know if this splliting/doubling is going to be justified in the sense explained in TOB.
Thanks,
~Dr. Black Jack in the Closet.
That's a great question and responses... but isn't the real question how large of a bet can you get away with? I think to pull this off you would have to use teamwork and play against a shoe- otherwise you're likely to get shuffled up on. Also, you have to consider what happens if you're off by 1 card, are you going to play 3 hands or take your chances? I know the question is hypothetical but I'm curious if anyone out there has actually tried this "ace sequencing" in a real casino?
That's a great question and responses... but isn't the real question how large of a bet can you get away with? I think to pull this off you would have to use teamwork and play against a shoe- otherwise you're likely to get shuffled up on. Also, you have to consider what happens if you're off by 1 card, are you going to play 3 hands or take your chances? I know the question is hypothetical but I'm curious if anyone out there has actually tried this "ace sequencing" in a real casino?
The "you are likely to get shuffled up on" argument is often advanced when this topic comes up for discussion. However, I have not heard of this ever actually happening nor experienced it in actual play. As I understand, many jurisdictions (Britain and AC for example) require this legally so a casino can't get out of the situation that easily - disclaimer: I'm not a lawyer so I'm not certain of that. You can make in a few seconds with a known ace what a counter could in a month so you really need to be sure that you a huge bet raise will result in the opportunity being eliminated. Post-deal barrings don't count-few places offer such great games that maintaining a welcome justifies passing up such a great opportunity.
Regarding Twin Jacks sim data, I appreciate the effort he put into this, and I don't mean to be critical but a) its not clear under what circumstances the player is getting the known ace and b) the 40% figure assumes no other knowledge about the shoe composition. In a high count the opportunity becomes much more valuable, since the chance of pairing a ten with the ace is generally increased, and a higher optimal bet would be justified. Counting may also have some kind of "drag" factor which deflates the known ace-opportunity.
The realtionship between counting and knowledge of an ace as first card is symbiotic not independant.
Also, c) I have had the occassional disagreement with Mathprof over external matters relating to gambling, but he is almost never wrong about the basic mathematics of the game. In this case, several other high profile math-types checked the numbers, so the probability of him being significantly wrong to my mind is <0.01%. That suggests to me Twin Jacks and Mathprof were answering different questions.
