Yah it's an industry oriented book, but it's only $70 hardcover. It has a lot of great information, especially on statistics and probability as relating to gambling. They didn't mention your strategy but it seems to me to contradict the whole idea. With a 10% rebate you WANT volatility in order to increase win and increase the reabate when you do lose. Also you're giving 5% to the house every other hand instead of playing at a 1% edge.
My opinion is, the boy could use a good spanking!
ETF
If you win one $10 bet, you must then lose 2 $10 bets to get a $1 rebate. When you are even, you are essentially betting $9 to win $10, a huge edge. As soon as you win your first bet this advantage seems to disappear. And when you are losing, when you win a bet you give back 1/10 of a bet in rebate, also no edge. But the trick is, it doesn't entirely disappear in one bet.
Thanks for clearing that up. I learned something.
Yah it's an industry oriented book, but it's only $70 hardcover. It has a lot of great information, especially on statistics and probability as relating to gambling.
Thanks. I've ordered a copy already.
They didn't mention your strategy but it seems to me to contradict the whole idea. With a 10% rebate you WANT volatility in order to increase win and increase the reabate when you do lose. Also you're giving 5% to the house every other hand instead of playing at a 1% edge.
Well, if you bet $100 on player and $100 on banker then you will get $200 back if players or $190 if banker wins. You either make a $10 or $20 profit once you factor in the rebate.
It is true that you would have a slightly higher expectation by just betting on bank. But expectation isn't everything. You would have a higher expectation, but because of the relatively high possibility of a player win and a loss, you could only risk a comparatively small fraction of your bankroll. With the bank/player hedging strategy you can bet everything because a win is certain.
In short, you would make a larger absolute amount of money even though your expectation is lower. 1% of $100,000 is a larger amount than 10% of $1000.
the overall EV (EW, in your terminology) can't be greater than 10% x $100 = $10, with the $100 house limit in your hypo.
Here is the problem I have with your calculation. You made the original EVS calculation for one wager -- that was fine -- but then you made the same calculation for multiple wagers. Your calculation would have some basis if the EVS applied on every wager.
If your calculation: "EW = 0.0367 x $300 = $11.01" was accurate, the player should increase his action as high as possible. For a "Stop at +10 or -10" strategy, we have "EW = 0.0367 x $1,000 = $36.70"
The other problem I have with your calculation is that the total action isn't necessarily $300 with a +3/-3 strategy. (Nor is it $1,000 for +10/-10.)
I think you did do a sim, because I think your +3/-3 strategy is probably right, as DD' points out. If you can get your sim to report the average win per week, over several thousand weeks, I think you'll find $4 < EW < $10.
ETF
Table of Variations and Outcomes
Here is a table of potential variations on the strategy. This table is based on my approximate simulation and the assumption that you make each bet at the table maximum of $100:
% of % of $ Avg. Weekly
Weeks Weeks Risked Hands EW StdDev
you you per Per EVS $ per per
Weekly Strategy Win Lose Week Week % Week $Risked
Stop at +1 or -1 49.7 50.3 100 1 4.43 4.43 1.0
Stop at +2 or -2 49.5 50.5 200 4 4.05 8.10 1.0
Stop at +3 or -3 49.3 50.7 300 9 3.67 11.01 1.0
Stop at +4 or -4 48.7 51.3 400 16 2.34 9.36 1.0
Stop at +5 or -5 48.2 51.8 500 25 1.58 7.90 1.0
I thought I'd add this. I just calculated using the same method as above that you quit if you are down 2.5 units, and you may quit if you are up 2.5 units. I think the best strategy is probably that you quit when +/- 2.5 units.
Up 2.5 bets:
.48^2.5 * .042 -.006= 0.0007=+ev, bet if you want to . Its close.
A different way to calculate that one, since you don't make half bets, is that the loss of two bets puts you at +half a bet. At this point a loss essentially gives you a rebate of 5%. The ev of a bet when you are a half unit ahead would be about: .05 * .48 -.006=.018, or +1.8%. So, needing two losses to get to this point from the +2.5 point, you get:
.48^2 * .018- .006=-0.0018528, negative ev, which would indicate not betting.
I don't know which method is correct. It being close with the first calculation I would likely not bet when +2.5 bets.
Down 2.5 bets:
.44^2.5 * .042 -.006=-0.0054= negative ev, quit.
or
.44^2 * .018- .006=-0.0025152, still negative ev, quit. Either way says you stop when down 2.5 units.
the overall EV (EW, in your terminology) can't be greater than 10% x $100 = $10, with the $100 house limit in your hypo.
I'm using EV in the way it is described in BJ21 glossary - as a percentage. I am also using EW the way it is described in the glossary - as a dollar amount over a period of time. In this case the period of time is one week. You are correct that EV cannot be greater than 10%. In fact, in this game the limit is about 4.4%, and only if you are making a single bet each week. However EW (which is different that EV) can be greater than the limit you suggest, because we are making multiple $100 bets during a week.
Here is the problem I have with your calculation. You made the original EVS calculation for one wager -- that was fine -- but then ou made the same calculation for multiple wagers. Your
calculation would have some basis if the EVS applied on every wager.
My intention is that EVS (EV of Strategy) represent the overall EV for whatever number of wagers the strategy (weekly series of bets) calls for. It could be one wager or many. EVS for the "Stop at +1 or -1" strategy would incorporate only one wager/week (excluding pushes). EVS for the "Stop at +3 or -3" strategy would incorporate a minimum of 3 wagers, an average of 9 wagers, with no theoretical maximum (although it would be under 30 wagers about 98% of the time). Under this strategy we do not know in advance the number of wagers that we will make each week, therefore we cannot calculate an EV per wager.
If your calculation: "EW = 0.0367 x $300 = $11.01" was accurate, the player should increase his action as high as possible. For a "Stop at +10 or -10" strategy, we have "EW = 0.0367 x $1,000 = $36.70"
The value 0.0367 in your above equation "EW=0.0367 x $1,000 = 36.70" is not a constant, but depends on your stopping point. See my post above entitled "Table of Strategies and Outcomes" for a summary of what happens at different stopping points. This table shows that if you play more hands, your chance of being behind at the stopping point increases. When the stopping point is 10, you are playing an average of 100 hands per week, and your chance of being behind is significantly greater. Your equation would become (roughly) EW = -0.01 x $1,000 = -10 (in other words you lose $10/week).
The other problem I have with your calculation is that the total action isn't necessarily $300 with a +3/-3 strategy. (Nor is it $1,000 for +10/-10.)
You are correct that by the conventional meaning of "total action", it is not necessarily $300 The conventional meaning of "total action" is the total amount of money you put in the bet circle. In a conventional blackjack game, every time you put money in the circle you are risking it. So in a conventional blackjack game you are risking your "total action".
This strategy is different. For any given week, you are saying "I will quit when I am either up $300 or down $300". Essentially you are only making a single $300 bet in any given week, even though that "bet" is spread over a variable number of blackjack hands. So you are only risking $300 in any given week.
I think you did do a sim, because I think your +3/-3 strategy is probably right, as DD' points out. If you can get your sim to report the average win per week, over several thousand weeks, I think you'll find $4 < EW < $10.
The sim for the +3/-3 strategy is showing an EW of about $11/week over about 50,000 weeks (assuming a table limit of $100). The +3/-3 strategy works because it effectively allows the player to circumvent the $100 table limit by combining a number of table limit bets into a "strategy" which is effectively equivalent to a larger bet. The EV decreases by a small amount with this strategy (from 4.43% to 3.67%), but because the effective bet size increases a larger amount (in this case from $100 to $300) the overall EW increases.
One final point that deserves amplification: If a player has a limited bankroll (they cannot bet above the table maximum) they should be using the +1/-1 strategy, which has the highest EVS. However if they have a larger bankroll, and can comfortably exceed the table maximum, then they should consider using the +3/-3 strategy. The +3/-3 strategy has a slightly lower EVS than +1/-1, but it has a much greater EW, and will therefore enable the player to win more money each week.
So is the whales' ROL worthless as they are required to bet for longer sessions? MathProf mentions ROL as one way to stay ahead of the game with flat betting BS. Does it apply only to such short sessions? Shark
I have believed for over a year that the "solution" to the loss rebate problem was (1 + 1/2 + 1/4 + 1/8 ...) x .5 x REBATE = 100% REBATE for infinite play (for a coin flip). As my head hit the pillow this morning, I realized I was wrong. I raced back to the computer this afternoon hoping to delete my last post before you saw it. SHUCKS! ;-)
It's easy to see there's no dollar limit on the value of a loss rebate on a fair coin flip. For instance, if you flip trillions and trillions of times, for a +10/-10 strategy, you can make the probability arbitrarily small (infinitesimal) that the coin will wind up anywhere BETWEEN +10 and -10. Ie., just keep flipping long enough, and you are, for all intents and purposes, certain to get to +10 or -10. A fair coin will then have a 50/50 chance of winding up at either node. So the value of the rebate becomes .5x(-9) + .5x(+10) = .5. Half a bet. Much more than 10%. For a hundred bets, the value is 5 bets, etc. There is no limit for a fair coin toss.
The problem with my reasoning was that once you are two or more nodes away from the "sweet" zero spot, the probability of getting back to zero is more than 2-n. For example, if you've lost two bets, there's more than a 1/4 chance to get back to zero before hitting the -3 node. 1/4 would be correct if you had to win two in a row, but there's more than one path to the Holy Grail. You could 1)win, then 2)lose, then 3) repeat 1) and 2) however many times, and THEN finally win two in a row. Hmmm... there's surely a series for that ...
Anyhow, I see now your calculation is kind of a "hybrid" with sim and EV mixed. I have nothing to dispute it. The limit for +3/-3 would be .5x(-2.7) + .5x(+3) = .15 bets, so your figure of $11.01 for $100 wagers is right in there.
GREAT JOB FREDDIE! Please tell me you've looked at this problem before 24 May 02, 11:24 a.m. I can tell you're a smart guy, but you're not THAT much smarter than the rest of us, are ya? ;-)
ETF
If you are playing for a rebate you want big variance with short enough sessions that the rebate more than offsets the session win/loss differential. If required to put in long hours you can have bets of sufficient size in a few of the hours that the variance will obscure the many hours of smaller bets. The house is attempting to get in the long run and you are increasing their long run index by use of non-uniform bet sizes. You can either bet big initially and then coast, or pour it on at the end. Doing it at the end gives the appearance that you are either pressing your winnings or you are betting big to try to get even. If playing good basic strategy, a 10% rebate on losing sessions will generally be more than enough to produce and overall positive ev using this betting strategy.
The ev of a bet when you are a half unit ahead would be about: .05 * .48 -.006=.018, or +1.8%. So, needing two losses to get to this point from the +2.5 point, you get:
You don't actually need two losses in a row -- with nothing else intervening -- to get two losses before hitting a higher node. Thus, the probability of that happening is greater than .48^2. See my "Retraction" below.
I'm afraid Freddie has beaten us all to the punch, here. Where'd he come from, dya think? ;-}
Sorry if I misled you with my faulty reasoning.
ETF
And I'm glad to see it generally agreed with the simulation. You may be right about using 2.5 units, however my simulation is not detailed enough to tell. It may turn out that the optimum is asymmetrical...something like "quit if up by 2.7 or if down by 3.4". The table I posted further down in the thread suggests the (symmetrical) optimum is somewhere between 3 and 4. However those results were based on a simulation that assumed a win-or-lose outcome for each bet; it did not cover outcomes like blackjacks, surrender, doubling etc). I'm not sure a spreadsheet simulator would be appropriate for addressing these complex outcomes, and my Visual Basic skills are a little rusty. However, it might be an interesting exercise. If I were going to try it, what would you suggest as a distribution of outcomes?
So the question would be - Playing BS, and with an initial bet of 1, what would be the:
% of hands with outcome of -2.0 (loss of Double or both halves of a split pair)?
% of hands with outcome of -1.0 (Player loses normal hand or dealer has blackjack)?
% of hands with outcome of -0.5 (Surrender)?
% or hands with outcome of 0.0 (Push or 50% win on Splits)?
% of hands with outcome of +1.0 (Player wins normal hand)?
% of hands with outcome of +1.5 (Player Blackjack)?
% of hands with outcome of +2.0 (Win of Double or both halves of a split pair)?
To simplify the analysis, it is probably reasonable to assume no RSP and no DAS (i.e. no hands with an outcome greater than +2 or less than -2). It is probably also reasonable to assume no insurance, since it is not part of BS.
Thanks,
Freddie
but it doesn't change the approximation all that much. You must get two losses prior to a net gain of one. You might go LWLWLL. It is basically the probability of getting two losses in a row before two wins in a row, without an initial win (pushes excluded). He didn't actually put up figures for +/-2.5. If you do this thing every week you will have cases when you are up or down 2.5 bets.
other than half bets, it would be a very strange if the win and loss stop points happen to be exactly the same. There is no reason why they should be. But if you are making flat bets, how do you ever get up 2.7 units? You would have to double for less, or insure for less, or something else odd. It is only the win of a natural that will give us an odd half unit. He won't be making insurance bets.
This is probably what I should have used, although I doubt the conclusions would differ much since +/-3 is not long enough for differences to add up to much. But this would limit the number of permutations that could result in going backwards by 2 bets. If you sum the probabilities of: LL, LWLL, etc, I seriously doubt that all the permutations will be significantly different than LL.
Of course, we needn't really worry about all the permutations that may go: LWLWLWLWLWLWLWLWLWLWLWLWLWL, for a long period of time. They are a wash anyway. We can say, if we take out all permutations that lead to a net wash, and ignore pushes, what is the probability of being down two bets before being ahead one bet, I suspect it will be very close to .52^2 about 27%. The chance of being down one bet before being ahead one bet will likely be very close to 52%.
When counting how would you increase variance to extract the rebate on session loss? would you use bigger spread? would betting high or spreading to more hands in TC where EV=0 do the trick? could you recommand me other strategies to maximize the rebate? Thanks a million, Shark
I don't think the people who are getting such offers are card counters. Counting might mess up the chance at something much better.
I guess I would use an anti-risk averse strategy. I would look for ways to increase variance... pay one big hand rather than spreading to two or three hands. The more variance the better. You get to keep the big wins while getting 10% back on the big losses.
Once you get a big flux you want to quit as soon as possible. After a big positive flux you will quit and blow off the 10% as you won't be getting it anyway. If you have a big negative flux you want to cut back on the action also and quit as soon as possible as you are giving up too much. For every hundred dollars you win you are giving back $10 in rebate.
What I've said above are big generalizations and certainly not any optimal strategy. If I had a chance at at 10% rebate on losses I'd probably prefer to play craps. I'm sure a simulator could be made that would tell you just when to quit, sort of like Fat Freddie's analysis. You could keep figuring the probability of winning another $1000 before getting back to the lucrative pivot, figure the ev cost, etc.
Card counting is something to do when there is nothing else better. Anytime you get a chance for a big ev play then throw card counting out the window. All it can do is increase your chances of getting tossed from the casino and missing a much better opportunity. For example, if you have a chance to make 4% at a game don't use card counting strategies in an attempt to capture an additional 1%.
... I'm not 100% sure it won't make enough difference to change the answer by half a bet. And the same problem is bound to come up for different games, with different edges etc.
I really suspect somebody has a complete answer to the loss rebate problem. Probabilist would know, but I'd rather see if I can make a dent in it myself, if I can ever find the time.
Sorry I don't have time to look through this whole thread right now, so maybe you've moved on. But I want to briefly comment on
"Of course you should use Kelly bet sizing principles to maximize the return on your bankroll. My guess is that #2 is preferable over #3 from a Kelly standpoint..."
If, for whatever reason, you have decide to play your strategy (play until +/- 3 units), then you can back out the probabilities of ending at either "+3 units" or at "-3 units + rebate". Hence you can treat this like one big bet, with known win/loss amounts, and known probabilities. You can use that to determine your unit size, in regards to Kelly betting. This is equivalent to what you called #3, and would be the only way to figure out a correct Kelly bet for such a situation.
Lefty
