How to profitably Red-bet a BR of 10,000 units over a hundred even-money Red & Black outcomes of which thirty-three are Red, and sixty-seven are Black?
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How to profitably Red-bet a BR of 10,000 units over a hundred even-money Red & Black outcomes of which thirty-three are Red, and sixty-seven are Black?
You go in knowing there'll be a distribution of 33/67?
Profitably bet, despite it being even money?
That's not possible.
And what do you mean by profitable? Largely profitable, or attempting to grind out a small/moderate win?
If attempting to grind, you can Martingale it, but that will take you down to zero when you have the occasional awful run of luck.
Absolutely would, maybe with the whole farm. (I'd double check some numbers first. It's been a few years since I solved this problem on a different forum.)Quote:
Originally Posted by RS__
A guaranteed reasonable profit of so much per Red. (Not saying, because that would be additional information at this point.)Quote:
Originally Posted by Dan Druff
No, not a Martingale as in doubling a bet upon losses. That's too easy to work here.Quote:
Originally Posted by Dan Druff
Hint: This is a fairly difficult problem for the AP school of thought.
Damn these problems -- they're tough. But I like them.
Initially, I'd think bet remaining # of blacks minus # reds divided by # of remaining spins of your session bankroll (i.e.: your advantage). First bet (67-33)/100 = 44/100 = 3,400 units. Now you'll either have 13,400 units with 66 black and 33 red, or you have 6,600 units with 67 black and 32 red.
I can't imagine that being the proper solution, though.
Or perhaps you wager 1/(red+1) of your BR. That way if you hit all red in a row, you'll end up with X units, all black and no reds remaining, and you would be betting 100% of your remaining BR the rest of the way.
As far as what happens once you know no red are remaining -- you parlay the rest of the spins.
What about the event of when red = black or red > black? I can't imagine quitting at that point would be proper, but not sure at which poin you'd quit because knowing how many of each more (that knowledge) is overcome by such a large disadvantage.
Edit: Nevermind, since youd just bet red and get an advantage that way. I'm stupid.
Edit2: There also may be a way to maybe get a larger advantage by betting the columns or other outside bets, not just the even money R/B. But don't quote me on this, more of a possible solution.
If you wanted to maximize value, I'm thinking, you'd bet it all on black until there are 33 red and 33 black remaining. Now you either bet half on each side (or 0 units). Then bet everyone on whichever one was 33 remaining. Then you bet half on each side, then bet everyone on whichever has 32 remaining. Etc.
Note: This would maximize your value (EV). But would damn near certainly lose. I can't think of any other way to maximize value.
Edit: Then again, I'm not sure, since you lose out on future value if/when you lose a wager. But this would be my guess at maximizing value.
Edit2: Everything* not everyone.
Your math is off in a couple of spots, and you interchanged a couple of references to red and black. However, I do see where you're going with the two posts above. I don't know an analytical solution to this problem.
That stuff will become much clearer to you given the answer. And poses more a supplementary value for persons really interested in this stuff. (You'll have to weaponize your own stuff.)
When I have time.
Working on this now. Rather lengthy, and explaining. Stand by.
The original problem as I found it on a different site three years ago,
Though you can't really take most of it too seriously on the gambling forums, you can get a lot of good raw and historically scientific gambling ideas. Not much real gambling (theory) takes place at today's casinos. Addiction, commercialism and "entertainment" instead of sincerely satisfying the gambling "bug".Quote:
We can argue forever if roulette can be beaten or not. However here is a challenge for you to test your creativity, knowledge and intelligence. Beat roulette given specific hypothetical certainties. The challenge. You have 20,000 units bankroll. Your bet can be from 1 unit to 500 units (500 is the table limit). You are only allowed to bet on RED, as often as you like. In the next 100 spins the RED will appear exactly 30 times, but we do not know in which spins. Make a profit of at least 1 unit after the 100th spin. How would you do it? http://www.rouletteforum.cc/index.php?topic=13422.0
I will include perhaps too much preamble here to help the reader better understand. My answer there was condensed and hurried to move on to better things, more of a tidbit, and I doubt anyone understood or cared. Relatively simple stuff, once you see the underlying trick. The problem has resurfaced unsatisfied in one form or another, like most other matters on the forums. (In fact, I know of no gambling forums which strive to satisfy the gambling "bug".) System players have a thing for the 40:1 BR:#bets ratio. How about a BR of 10,000 units? Clearly, going from a BR of 10,000 units to 20,000, or even 100,000, in the "land of the Martingales" can't change anything about answering the type of problem at hand. A proper answer should allow much more flexibility of terms with respect to BR.
As for my answer to the problem at hand - essentially the same as when I came upon it - I will give only the most relevant parts to allow the reader to build on, or not, as he sees fit. I'm sure each here has his own interest and direction with respect to gambling, so it's probably best that I don't stray too far from my own basics. An approach to gambling theory which includes both system and advantage play, combined with a deeper understanding of the math, itself. The broader approaches to gambling tend to quickly outstrip common calculation, knowledge and experience, but what one puts into it one gets back equally ten-fold. What makes the solution at hand so-simple is it's natural fit into such a broader and generalized gambling theory. Far as I know, mine is the only publicized answer to this problem. Original continued work on a few old betting systems. Fun to find old long-abandoned theoretical stuff which can still be continued in the same vein. Becoming part of the past in a simple way.
A specially constructed betting progression is required to answer this problem. Observe that the cancellation or Labby system ( http://www.rouletteonline.net/roulet...-for-roulette/ ) is the most versatile betting system, and the system best-suited to theoretical generalization. The very versatility or leeway which has so obscured its being commonly mis -understood and -applied with respect to its raw logical form. A betting system's simplest form, beyond an arbitrary start, middle, and end. After all, what is the logical way to commence betting? In general, how do the negative games blend into the positive, and the strategies overlap and complement? Why bet at all? I will first rework the betting setup, and then get a handle on the relevant betting system's application to all of the different theoretical overall L:W ratios for a given set of even- or non-even -money outcomes. The problem at hand specifies a 2:1 L:W ratio of outcomes. By L:W ratio, I mean the ratio of losses to wins in a given set of guaranteed outcomes of Black and Red. A 2:1 L:W ratio of Black to Red translates to twice as many Black or losing outcomes as Red or winning outcomes over the set of outcomes. Eg, the set or "registry" of losses/wins {LLWLWLLLWLLL} contains 3X's as many L's, so has an L:W ratio of 3:1.
How the required betting progression works. "Logically" start the betting at 1 unit under the pretend assumption that we lost our last bet. Little assumptions like this can have a HUGE impact in further developments of the general theory. Eg, working the micro-world Kelly stuff into the macro-world betting units and increments. Suffice it say for now that gambling systems, for the most part, are based on independent past events which are nonetheless "due" in some way. (These assumptions make sense only in the context of each other.) Record this loss of 1 unit and the sequence of further losses to the right as instructed in the Labby link above. Upon a win, we stroke that recorded lost bet of 1 unit out, and start over with another pretend loss. There is not yet a matter of how many bets to stroke out, because we had only one. For an L:W ratio of L:1, where L ≥ 1, stoke out up to as close to L many as possible of the lost bets which helped form the won bet. Here, L = 2, so stroke off one of the lost bets on the left-hand of the recorded sequence of losses, and one on the right-hand end of the recorded sequence of losses. Upon a loss, the next bet is the sum of the last loss, which is recorded on the right, and the first remaining recorded loss on the left. The two recorded losses which make up the next bet are the ones stroked out upon a next win. Wins equal losses. How to arrive at a desired and possible profit per win, and to not merely have the wins balance out corresponding sets of losses? With an L:W ratio of 2:1, add an additional 1 unit onto each new next bet's total. As the wins even out the losses in their prescribed ratios, the additional units shall form the overall profit. Of course - and there are many other generalities and specifics which come into play - the constantly changing L-W ratio composition of a given set of outcomes should be APPROXIMATELY reassessed and adjusted to for ongoing optimal betting. Regardless, degrees of such generalities and specifics aren't important to the betting system, itself, ie, to its short-run workings for the purpose of illustration.
Running through the betting progression. Start the betting at 1 unit, continue until a loss, at which point record the 1, as in {1}. Add 1 unit, for profit upon the next win, to that 1-unit loss to form the next bet. There were no other recorded losses on the left before the 1-unit loss, so add only the 1 unit for profit. Were there already other losses, form the new bet from the last loss, as well as the earliest other loss not yet stroked off on the left of the sequence of recorded losses, and the obligatory 1 unit for profit for when the bet is won. Bet 2 units. Upon a win, stroke out the 1, and start over. Upon a loss, record the 2 to the right of the 1. The sequence or list of losses is now {1, 2}. The next bet becomes the 2 (on the right of the list of losses) + the 1 (on the left of the list of losses) + 1 (unit for the profit) = 4. Bet 4 units. Upon a win, stroke out the entire list of {1, 2}, and start over. Two losses from the list formed the bet of 4 units which was won, so stroke out those two. Upon another loss, this time of 4 units, add it to the right of the list, as in {1, 2, 4}. Form the next bet, 4 + 1 + 1 = 6 units, in the same way the previous lost bet of 4 units was formed. The 2 isn't used yet because the L:W ratio here is only 2:1. With an L:W ratio of 3:1, add two numbers from the left of the list of losses, the last loss bet, plus a unit for profit, to form the next bet. The Reds are guaranteed to come, so every one of the lost "next bets" eventually wins, and removes their corresponding two losses from which they are formed. How the Labby has been re-engineered. Bet the 6 units. Upon a win, stroke out the 1 and the 4 from the curly brackets, and carry on from the {2} with a bet of 2 + 1 (for profit) = 3 units. But upon a loss of the 6 units, add the 6 the list, as in {1, 2, 4, 6}. The next bet in this spot becomes 6 + 1 + 1 = 8 units. If win the 8 units, stroke out 1 and 6 to get {2, 4}, and make a next bet of 4 + 2 + 1 (for profit) = 7 units. If lose the 8 units, the next bet becomes 8 + 1 + 1 = 10 units, from a list of {1, 2, 4, 6, 8}. If win the 10 units, the list of losses goes down to {2, 4, 6}. Stroke out the last lost bet of 8 units, and the first remaining recorded lost bet of 1 unit on the left. The next bet is 6 + 2 + 1 (for profit) = 9 units. Upon a loss, we record the lost 9 units, as in {2, 4, 6, 9}. Next bet is 12 units. If win, stroke out 9 and 2, the remaining list is {4, 6}. Next bet, 6 + 4 + 1 = 11 units. If win again, stroke out the 4 and 6, and start over with another pretend loss of one unit, and with a profit of 3 units from the available 6 L's and 3 W's in the example's given set of outcomes.
Another example. Suppose that the L:W ratio is 3:1. Again start at 1 unit. Lose the first four bets to arrive at a the lost-bet recorded sequence of {1, 2, 4, 8}. The 8 comes from adding the previously lost bet of 4 units, the two left-hand remaining recorded lost bets in the list, and the 1 unit for profit for when the bet does win. The list was {1, 2, 4} before the 8 units was lost. At this ratio, add the last lost bet, the first two remaining recorded lost bets, and the 1 unit for profit. Were the ratio of L:W at 4:1, then add the first three in the list to the last lost bet, along with the 1 unit for profit. At an L:W ratio of 3:1, strike out the three corresponding lost bets which made up the won bet after a win. Were the ratio of L:W at 4:1, strike out the four corresponding lost bets which made up the won bet after a win. The next bet after the loss of the 8 units is 8 + 2 + 1 + 1 (for profit) = 12 units. If win that, stroke out the 8, 2 and 1 from the list to leave only {4}. Those bets, with the 1 unit for profit, made up the 12 units. Next bet is 4 + 1 (for profit) = 5 units. If lose the 5 units, record 5 to the list, as in {4, 5}. Next bet would then be 10 units. Lose that, then {4, 5, 10} lost bets are left to recoup with a profit. If win the next bet of 20 units, strike out the entire list, and start over with a profit of two units. One unit profit for each guaranteed win within the allowed-for number of outcomes at the given L:W ratio.
Note that the maximal guaranteed profit per win doesn't depend on the order of losses and wins. If the majority of losses or Blacks comes almost right away, then it may be best to run a bunch of short parlays on the end to catch the runs of Reds; and if the majority of wins or Reds comes right away, it may be best to stop betting altogether. Guaranteed maximal profit is if all the losses come first; maximal guaranteed profit is 1 unit per Red if all the wins come first or at least manage to even out the Blacks within the given set of outcomes and ratio of. And a likely-guaranteed maximal profit in the silent majority of sets of outcomes when the standard random distributions of Red and Black outcomes tend to somewhat even out to their respective ratios throughout play, and the bet is steadily increased to what the BR can increasingly tolerate.
The problematic scenarios are when the Red outcomes continue to lag slightly or closely for a while and then stops; and when Red falls somewhat behind, catches up part way, and then stops. These two cases share the same math, and can be exactly solved with a calculus derivative-calculation of the point at which the increasing size of bet against all outstanding possible losses waxes and wanes. I eye-balled it to be around a bet of 470 units with 47 more losses to come. Give or take a few smaller bets on the table leading up to this. For a required total BR of about 22,500 units. Not quite under the challenged 20,000 units, but well within range all but for "a rainy day".
How to complete the challenge with 10,000 units? Not sure. Perhaps, we have to anticipate and preclude the more pesky of the problematic runs by breaking off those off a few times, after a couple of increases in size of bet, in the hope the Reds are likely coming enough to even out those few small runs which were beginning to going nowhere. Might help to increase the number of units of profit by 1 unit, and later by 2, to further recoup such losses. Would have to keep the break-off runs short, as tinkering around with those, eg, by decreasing or not increasing the bet would weakly put things off. Moderately long runs of losses shouldn't be attacked separately later properly by breaking up the BR. A cardinal rule of Kelly criterion, don't break up the BR into smaller BR's. The best way would likely be to scrub the 1 unit for profit added to next bets, and record the sequence of losing bets 1, 1, 2, 3, 4, and so on. And set the profit-goal to at least zero units. Or have the set of outcomes split into two sets with the same ratio B:R. I have worked more-sophisticated gambling theory methods, most of which lay well beyond the scope of such problems and discussions.
It's noteworthy at this time to point out also a couple of other items, at least in principle. These ratios must be expressed in terms of integers L and W of which at least one of L and W must be taken as 1 because it's not possible to stroke out/record a fractional number of lost bets. Can't work this betting progression, eg, with an L:W ratio of 2 1/2 :1. Sure, that can be simplified to a ratio of 5:2, which comprises integers, but not with at lest one term of the ratio a 1. However, there are a few techniques around this integer limitation later on, particularly with respect to the betting progression's inaccuracies which might accrue from the the varying ratio of remaining outcomes of a given set of. As noted before, the bet-amount and other variables are subject to also many other factors. In addition, this logically regular Labby reduces to the already(?) logical D'Alembert betting system at the 1:1 L:W ratio; and expands to the full-blown Martingale as L becomes much larger and/or the number of trials becomes much smaller. About D'Alembert, "While he made great strides in mathematics and physics, d'Alembert is also famously known for incorrectly arguing in Croix ou Pile that the probability of a coin landing heads increased for every time that it came up tails. In gambling, the strategy of decreasing one's bet the more one wins and increasing one's bet the more one loses is therefore called the D'Alembert system, a type of martingale." https://en.wikipedia.org/wiki/Jean_le_Rond_d%27Alembert . Fascinating thought processes. The more intuitive reader may be left to wonder, how is this special, nay, logically regular Labby further generalized into the favorable L:W ratios as W increases from 1, and L stays at 1 thus beyond the D'Alembert ratio of 1:1? Say, one loss to every two wins? A trivial way may be to show, with particular other restrictions, how it is possible to lose with the much better ratios. A topic for another post. Another theoretically interesting avenue, the Reverse Labouchère or Labby ( http://en.wikipedia.org/wiki/Labouch%C3%A8re_system ). How to tame similarly tame that? And for which purpose? The forward Labby, as a negative betting progression, is a great way to bring the negative expectation theory into the positive, without using positive betting progressions, which, aside from Kelly criterion's negligible-though-costly approach to avoid ruin, can't work well outright. Such connections to flat betting are "another ball of wax" entirely.
P.S. In the "old days" I would have re-banged this sort of thing out right away. But I went out of my way to find a woman who knows the best ways to nag the heck out of me. To take me just far enough away from my work and play, whatever, to see stuff I don't think I would have seen otherwise. In hindsight, I wouldn't give her up for the world. (She doesn't like it when I'm on the computer at home.)
Good to know the joys and lows of the years of each of living single and then married. Sometimes in any event, other things keep coming up for a while. This stuff I type up from home when a quiet moment. With the real work, of course, she is less able to immediately interrupt things from time to time.
Lol. I wonder if Linda comments to Alan about his computer time on the forums.