# More thoughts on Roulette accumulation strategy

Discussion in 'Roulette Tournament Strategy' started by London Colin, Mar 6, 2014.

1. ### London ColinTop Member

I come across these types of tournament from time to time. It is usually possible to shoot for your goal in one spin, covering as many numbers as you can. As discussed in the previous thread, that ought to be the best strategy, compared to any alternative, multiple-spin approach.

But I've been wondering about some possible reasons for this not to always be the case.

Reason #1, I think I've disproved to myself -

Roulette here is single-zero, with the rule that you get half your stake back on an even-money bet if zero comes up. So I thought there might be a benefit to starting with a sequence of one or more even-money bets, before going for the goal. However, a simple example seems to indicate it is still better to get it done in one spin -
BR:\$20, Target:\$180
Cover 4 numbers with \$5 each: chance of success = 4/37 = 10.81%.
Attempt to double up first with \$20 on even-money, then either cover 8 numbers (when bet wins, returning \$40) or 2 numbers (when zero comes up, returning \$10): (18/37 * 8/37) + (1/37 * 2/37) = 10.66%.

Presumably, starting by attempting to double up twice would perform even worse.

Reason #2 might be if the minimum bet increment would force you to overshoot your goal if you cover n numbers evenly, or undershoot if you cover n+1 numbers.

E.g., BR \$15, Target: \$360, bet in increments of \$0.50.
This is an extreme case, as you would massively overshoot if you bet \$15 on just one number. So the alternatives would seem to be either -
a) Bet on two numbers, one with \$10 and one with \$5, and if the \$5 bet wins, attempt a double-up with an even-money bet.
b) Bet \$7.50 on each of two numbers, and if you win (\$270), cover 27 numbers with \$10 each.
c) Start with an attempt at a double up; if that succeeds then cover three numbers evenly with \$10 bets; if the return is \$7.50 do whatever is optimal to try and turn this into \$360 over multiple bets (not sure what that would be).

I'm not entirely sure what question I am trying to ask. Just looking for general thoughts, I suppose. Discovering which of a, b, and c is best in the above example would be instructive, but general rules of thumb, applicable to any BR and target, would be the holy grail.

The closest thing to a conclusion I have at the moment is that there must be some occasions when two or more spins would have a higher probability of reaching the goal than one spin, and that an n-spin route to a given goal that incorporates some even-money bets (and hence some additional routes to the goal, involving more than n spins) presumably has a higher chance of success than an n-spin route that does not.

Last edited: Mar 6, 2014
2. ### gronbogTop Member

In any game, if you are close enough to overshoot your goal, then your chances of reaching it increase. You get the probability of hitting your goal on the first trial plus the probability of still being able to reach it from your diminished position should you lose. The closer you are to your goal, the better the chances of your recovery. If you are in a position where you can reach your goal outright on successive trials, your chances get really good (e.g. within 1/3 or 1/7 of a max bet in blackjack, where you can use a progression). I think that your example a) within "reason 2" above fall into this category.

However, in a negative EV game, if you are not within reach of your goal, setting up for multiple chances at attacking the goal later with an initial bet is always (as far as my research goes) worse than simply attacking the goal. This seems intuitive to me but I'm having trouble coming up with a proof. I'll give it some thought, however the best logic that I can come with at the moment is that introducing more trials erodes at your chances of success due to the negative expectation of each trial. I can say that when I've simulated various games with an accumulation goal these strategies have always performed worse than simply attacking the goal. I think that your Reason #1 falls into this category as do the other examples in "reason 2".

3. ### gronbogTop Member

One case that we can reach a conclusion on is the case where you need to cover 18 or more numbers, since (putting aside the case of zero) an even money bet of b is equivalent to covering 18 numbers with b / 18 and conversely covering 18 numbers with a bet of b each is equivalent to an even money bet of 18 x b. We can then say that if you need to cover 18 + n numbers with a bet of b to reach your goal, you will definitely be better off betting 18 x b on an even money wager plus b on n additional numbers. The reason is that the basic bet is equivalent, but by using an even money bet, you get the benefit of half your (ed: total) even-money bet returned when zero comes up. By using the even money bet, you've given yourself a second chance.

Last edited: Mar 6, 2014
4. ### London ColinTop Member

For the sort of scenario I am considering, you would always want to go all-in on every trial, rather than hold anything back in case you should lose. Once you've worked out how to cover as many numbers as you can, such that if any one of them hits then you reach your goal, betting any spare chips on the same spin has better odds than saving them for the next spin. You can either bet the spare chips on one of the numbers you didn't cover with your main bet (which looks to be the best option), or you can choose to cover all numbers equally, but in a way that will fall short of your goal whichever of them comes up, always requiring a second spin.

Example a) is an example of what I was just describing; you can bet \$10 on one number, and rather than hold back your remaining \$5 for a 1/37 chance on the next spin, bet it now for a 1/36 chance.

In this particular instance we are almost always within reach of our goal, because in the extreme case we can put everything on a 35:1 single number, and we are not constrained by the max bet size. In general, we can always find a way to attempt to reach the goal on the first spin.

The general principle that fewer trials is better is definitely true if we can always make the ideal combination of bets on any given trial. What I was thinking about was whether the constraints imposed by not being able to make the ideal bet combination (through the BR not being divisible in a way that can cover n numbers evenly without either overshooting or undershooting the goal), and/or the fact that not every available bet has the same EV in this particular roulette variant, might lead to cases where 2 (or more ) trials manages to come out better.

After a few back of the envelope calculations, it looks like the approach of example a) might be best. - Aim to win in one spin, and bet any unused chips on one or more of the numbers you didn't cover with your main bets.

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5. ### gronbogTop Member

I see what you're saying about example a) and I do agree that it would be better to bet the remaining chips on an uncovered number in event that you have enough spins to attack the goal again than to bet them on the next spin.

In the accumulation roulette events I have played, I have also found that it has always been possible to reach my goal in one spin. From a strategic point of view, if there are others at your table who can see your bet (i.e. not an online event alone at the table), it's best not to use the strategy of covering as many numbers as possible with the correctly sized bet too early. On every occasion where I have used this strategy, others at the table (and also among those watching) have remarked to me that they could see what I was trying to do, that they thought it was a good strategy and that they never would have come up with it on their own but would do it themselves next time. By using it too early, you can end up teaching this strategy to your opponents with time for them to use it. Of course, there are also those who tell me, when I lose, what a complete idiot i was and thanks for playing, my wife among them .

Because of this, and since we don't have the 1/2-even-money-bet-returned-on-zero rule here, I usually wait until the last spin before going for the goal and, as there are usually only a few chips remaining, I generally just keep them as a small buffer on my goal estimate.

6. ### London ColinTop Member

Agreed. Again, there can be practical difficulties in splitting the remaining chips evenly across n numbers which might cause you to have to diverge to some degree from the ideal case. But that's an issue that is always getting in the way.

This example does bring to mind a possible edge case - suppose you need to cover 17 numbers to reach your goal. 17/37 = 45.95% chance of success. How would this compare with covering 18 numbers, to be followed by a progression if you win, plus with the benefit of getting half your wager back if zero comes up?

I wish I could figure out a couple of useful things -
1) The odds of success in an n-step, even-money progression. (Awkward, since one or more of the steps might involve a zero)
2) The odds of doubling up by going all-in on even-money. (Again, you can conceivably halve your BR multiple times due to zero being spun, on the way to eventually reaching your goal.)

#1 would be needed for a precise answer to the above question, but for an approximation -
If we have 17 units available, then our goal is to reach 36 units (when one of the 17 equal, straight-up bets wins). A winning even-money bet leaves us with 34 units, 2 short. That's enough for a 4-step progression - 2+4+8+16 = 30. Ignoring the use that could be made of the extra 4 units, and ignoring the possibility of more units from zero being spun, the chance of success in 4 steps would be 1-(19/37)^4 = 0.93.

So we have 18/37 * 0.93 = 45.24% chance of success, before we consider the possibility of a zero on the first spin.
If we do get a zero, and half our stake back, that is 8.5 units. Assume that we follow this by covering 8 numbers to try to reach the goal.
1/37 * 8/37 = 0.58% additional chance of success.
45.24+0.58 = 45.82%, still less than what you get by covering 17 numbers in one spin.

Whether a precise calculation - including all the little factors I missed out - could take this result above 17/37, or whether there is some fundamental law that means the result can only ever be less than this is something I am totally clueless about.

7. ### gronbogTop Member

At the risk of being annoying, this is the kind of thing simulators are really good at .

As it happens, I do have a roulette simulator. I would just need to add to it the collection of these particular statistics.

8. ### London ColinTop Member

Not possible.
I'm not 100% sure I'm asking the right questions, so answers are welcome, but I wouldn't put too much effort into finding them.

In a moment of confusion, it occurred to me that a progression might not always be the best way to gain a few extra chips, compared to the alternative tactic of covering enough numbers to achieve your goal in a single spin. But if my arithmetic is correct, is seems a progression is better (albeit only marginally), so long as you insure you put all the available chips to use -

To turn 34 units into 36, you can cover 34 numbers for a 34/37 = 91.89% chance of success.
You can also improve matters by doing the trick you mentioned of placing 18 of the 34 units on even-money, making it 34/37 + (1/37 *9/37) = 92.55%.

Let's assume we have exactly 4 spins remaining; for the fourth step we can just cover as many numbers as possible. So if we again ignore the small chance of extra chips from zeroes during the progression, we bet 2, 4, 8 on evens, and then 20 on single numbers. And the probability of success is 1 - ((19/37)^3 * 17/37) = 93.78%.

If we only have 3 spins remaining, then the progression is 2, 4, then 28 single numbers. That gives 1 - ((19/37)^2 * 9/37) = 93.59%.

If we only have 2 spins remaining, then the progression is 2, then 32 single numbers. That gives 1 - (19/37 * 5/37) = 93.06%.
(still better than covering 34 numbers at once.)

Last edited: Mar 6, 2014
9. ### gronbogTop Member

I found some time to dust the cobwebs off of my roulette simulator and added the rule of only losing 1/2 of an even money bet vs 0 and 00. I then ran some sims to get you the answers to these questions. My first observation is that I think question 2 is a special case of question 1 in that it represents a 1-step even-money progression. The next decision was about what to do when the progression is derailed by losing 1/2 of a bet. At that point it loses the property of decaying as a power of 2. It is no longer required to double your bet in order to reach the goal. I decided that in this case, we should continue attacking the goal by betting the precise amount needed to reach it, if possible, and all-in otherwise. In this way, you still get n shots at the goal with a residual amount left over should all n spins fail. This will improve the results slightly over an n-stage progression where you either win or lose the full bet.

I ran sims for n=1 (question 2) to n=5. Here are the results for the single 0 wheel, since that's what we've been discussing:
n=1: succeeds 49.31%, fails 50.69%
n=2: succeeds 74.29%, fails 25.71%
n=3: succeeds 86.99%, fails 13.01%
n=4: succeeds 93.39%, fails 6.61%
n=5: succeeds 96.65%, fails 3.35%

Here are the results for a double zero wheel.
n=1: succeeds 48.611%, fails 51.39%
n=2: succeeds 73.58%, fails 26.42
n=3: succeeds 86.42%, fails 13.58%
n=4: succeeds 93.03%, fails 6.97%
n=5: succeeds 96.41%, fails 3.59%

If you have other specific results that you need to help with your calculations, feel free to ask.

10. ### London ColinTop Member

Thanks, Gronbog.
That makes sense. Having said that, my questions weren't fully thought through and I think had a degree of ambiguity about them, which I'm probably about to make worse -

The limiting factor when considering an n-step progression could be that you run out of chips, having sized your sequence of bets to allow for n successive losses, or you could gain one or more extra steps and/or resized bets, due to extra chips from zero(es), and the ultimate limiting factor may become the total number of spins available. I suppose this is similar to when you get pushes in the middle of a bj progression, but with the added complication of the mathematics of the progression getting messed around with.

I focused on even-money bets for any progression, because of the 1/2-back rule. But that rule has no benefit on the very last spin, as there will be no opportunity to re-bet the chips it may yield. And since there is generally some amount left over after dividing your BR by 3, or 7, or 15, etc., it makes sense for the final bet in the sequence to essentially be the standard, one-spin, tournament bet: i.e. all-in, covering as many numbers as possible such that any one of them will get you to your goal.

And, as in the example I went through in the previous post, the above is also true if the last bet in the sequence comes by virtue of being the last spin in the tournament, long before the ability to keep doubling your bet comes to an end. However, that example now seems even more complicated to me, as it looks like the optimal progression might involve a sequence of bigger, shorter-odds bets (i.e. using more chips to cover more numbers), rather than sticking to even-money bets until the final spin.

This is where the ambiguity in my questions inevitably gets reflected in the answers. From your result for n=1, it looks like you are actually allowing for up to 2 spins. Is that correct? That is, if following a zero you cover nine numbers on the next spin (assuming you can find a way to spread half your original stake evenly over nine numbers), then the probability of success is 18/37 + (1/37 *9/37) = 49.31%.

And for larger n, are you similarly allowing up to n+1 spins, with the extra spin making best use of any accumulated 'spare' chips from one or more zeroes appearing in the previous n spins?

Thanks again. My butterfly mind had moved on to other things, but there is in fact a tournament coming up this week which I am hoping to play in. So who knows what new questions that might throw up!?

Last edited: Mar 17, 2014
11. ### gronbogTop Member

The notion of there being a final spin is, of course, a practical one since, in real tournament situations, there is generally a limit. With the possibility of losing 1/2 to a zero in roulette, the probability of success of an n-stage progression will change, depending on the number of spins available. i.e. the probability for an n-stage sequence succeeding within s spins will be lower than if more than s spins are available. The results I posted were for an unlimited number of available spins. i.e. each progression proceeded until the goal was reached or the bankroll was exhausted.
Yes, whether the last bet arises because s is less than n or because some number of zeroes were hit making n + z > s, it would make sense to simply go all-in on the final bet.
Yes. I think the result of previous discussions has been that the optimal betting sequence is the one which minimizes the number of spins while at the same time maximizing the coverage of numbers on the layout, to a maximum of 35 numbers.
Not quite. As mentioned above, there was no limit on the number of available spins. Furthermore, the sim was limited to even-money bets only as implied in the wording of the questions.. For the case of n=1, if a zero was hit on the first spin, then the next bet would be all-in on another even-money bet. The actual algorithm is to bet the exact amount required to reach the goal on even-money and to bet all-in if the goal is not within reach.

I could certainly re-run the sims with some limit on the number of spins and with optimized betting choices if you would find that helpful.
I know the feeling. I sometimes get pulled in different directions as events draw near or as I get requests from different sources. All of this on top of the general direction in which I want my software to develop. It's a never ending series of different problems to solve which makes for a lifelong and interesting pursuit of infomation and improvement.

12. ### London ColinTop Member

Sorry I left this hanging...
Curious that the n=1 case gave exactly the same result as my calculation for going all-in on the second spin. I wonder if there might be some kind of equivalence between attempting repeated double ups (with an infinite number of spins available) versus going for the same result in a single spin.

I'd definitely find any results interesting, but my thinking on this topic has become a little too confused to make any meaningful, specific requests. Feel free to experiment.

That being said, I've managed to come up with a surprising result that might almost be worthy of a teaser. Based on some of the ideas expressed in this thread, an alternative approach to doubling up occurred to me that I thought ought to do better than the n=1 method. I did a calculation, and not only was it better, it had a greater than 50% chance of success!

When I saw that result on my calculator I thought I must have made a mistake, as it seems to contradict the negative expectation of the game. (A believer in betting systems might suggest that you can just repeat this method ad infinitum in order to make your fortune at the tables. But, after a certain amount of head scratching, I think I can explain what is going on, without recourse to voodoo!)

If you aren't in the mood for a teaser, just say the word and I'll explain all. (And I shall feel very foolish if it turns out I have in fact made a mistake.)

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13. ### gronbogTop Member

My suspicion is that the additional benefit of possibly surviving some number of consecutive zeroes (after the first one) is beyond the second decimal point.
How about leaving it for a few days to see if anyone can come up with it, or already knows about it (S. Yama?).

14. ### London ColinTop Member

Could be.

Inevitably, despite having checked my working multiple times, it was only after posting that I found the mistake! I now get a figure of less than 48%. However, that's an approximate, minimum figure; I don't know whether the precise answer will be greater than the n=1 (49.31%) result, let alone greater than 50%. My guess would be 'no' for the latter and 'maybe' for the former.

The logic I concocted to explain the paradox of a > 50% chance of doubling up in a negative expectation game still seems to make sense. But I'm not sure about anything anymore!

15. ### S. YamaActive Member

Great mental exercise guys!

I spent some (awhile) time looking at various stats for roulette accumulation with priorities to first understand all (some) the underlying principles, then confirm them with the real numbers, make elegant mathematical formulas (that’s better be left for real mathematicians, lol, mine where many times longer than simple multiplication of chances on the way to the goal score), then see the practical applications, etc..
Of course once I founded the key to one of the doors there were three new appearing, needed to be open, wish I could extend my 28 hours days to last longer.

There were many great ideas thrown in the thread but I think you, and I, as well, were all over the map.
I think one approach (more academic) would be to look at the issues using artificial roulette game where we can bet infinite small amounts and/or use a game design with a desirable large possible numbers to be spun, thus allowing us use our bankrolls to obtain precise payouts.

The vast possibilities of using different “angles” at relatively simple set of data makes it hard to make it just a brain exercise without using computer programs, spreadsheet, or at least a very long pencil and paper.
Though- all we need is the bankroll, the goal, and the (negative) game edge.
I, personally, like to look at the results of various methods compared to a fair game (zero edge), and mostly use the negative edge as it does apply to all bets (except the rule of return half bet even money with zero outcome) but I caught myself sometimes using and mixing either the edge or the chances of success, or failure at different stages, which are all derived from the same set of rules, and sometimes creating specific “rounded” or otherwise fitting numbers – all of it adding to the mess that at the same time was fun to deal with.

I am not sure if, and how much, my post may be helpful (you guys made great observations) but here are a few simple points that hopefully contain the all-imperative facts.

I think that to some people it would be clarifying to think of negative game as a “TAX”.
Every bet made that eventually contributes to reaching the goal causes a penalty of lowering the overall chances.
As a consequence the optimal strategy is to make bets that require a minimum of total “action” to achieve the desired objective.
Also, if we can reach the goal with a progression style betting but the progression doesn’t require additional bets if the progression is successful, it will be outperform a larger (whole) bet because total action is lesser for cases when we win a progression bet before its last step.

Some examples:

Let’s assume a roulette game with the same negative edge of 20% for all bets. That means “even” money bets have 40% of winning and 60% losing. There is a huge wheel with lots of numbers and we can bet fractions of money.
Bankroll \$30, goal \$40. For fair game (100% return) the chances are 30/40th = 75%
If we use progression on even money bets, 10 and 20 if we lose the first spin our chances are 0.4 + (1-0.4)*0.4 =64%
This can be looked at as tax of 20% (the edge) on 1/3 of the brl (\$10) plus 20% tax on the second spin 60% times 2/3 of the bankroll (\$20).
0.2 * 1/3 + 0.2 * 0.6 * 2/3 = 0.14667%
14.667% is the penalty compared to the fair game of 75% chance.
75% minus (75% x 14.667%) = 64%

If we try to achieve the same result with one spin using the whole brl, we could, for example, cover 60 numbers for 0.5\$ on 100 numbers wheel for a game paying 80 for 1, which is the same 20% negative edge - our chances are 60%.
Our total action was 100% bankroll, times 20% negative edge = 20% penalty compared to the fair game of 75% chance = 60%.

Let’s say we need to double our bankroll of \$1,000 in a regular (single zero) wheel, fractions of units allowed.
We could bet on an “even” payout for 18/37 chance = 48.65%
Or we could bet \$28.57 on the first spin, then \$29.39, then \$30.23, and so on until we hit and double our bankroll. After about 30 spins our chances are about 49.02%
This is a small 7.6%, but nevertheless, improvement versus betting all-in.
If the goals are higher the improvements gets also smaller (about 3% for quadrupling, and maybe a fraction of a promille for twenty-fold goal). This is because higher goals force bigger bets proportionally to the brl and less steps that can be successful.

In case of smaller edge (half bet returned on even money payouts when zero hits) versus higher edge.
It could be looked at as increase of the original bankroll by 1.35% (1/37/2) for the purpose of using the slightly higher edge. Total of any following bets leading to achieving the goal would have to carry a lesser penalty inherent for the game and the used strategy to make it a better choice.

The optimal bets are interesting but offer relatively very small improvements and are almost completely useless in practice. In most cases the optimal bets are the ones that offer the highest payouts (thus smallest table action) that fit reaching the goal by using allowed units and not overshooting the goal.

More when I return from a trip in a middle of next week.
Colin and gronbog, thanks for fascinating topic,

S. Yama

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16. ### gronbogTop Member

Valuable insights, as always and a lot to absorb. Thanks S. Yama!