Roulette system -- which optimization is better?

In summary, the conversation discusses a betting strategy for playing roulette. The strategy involves waiting for a certain number of reds in a row before betting on the opposite color using the Martingale system. The question is whether it is more profitable to increase the bet amount or decrease the number of times waiting for reds in a row. The probability of getting a series of reds decreases when waiting for four instead of five, making it more likely to occur and potentially resulting in higher profits. It is suggested that this can be mathematically calculated rather than relying on simulations.
  • #1
bernd
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TL;DR Summary
For given roulette gambling system decide wether betting on shorter series or betting more per series is more profitable on average?
Hey, gotta do some explanation first:
I assume you know how roulette works. (if you dont: ball is thrown into a pit and it can either land on red, black or zero, each having a certain likeliness to land there. you can bet on where the ball will land)
let's assume unrealistically you have the ultimate knowledge where you know that a ball will only land on a certain color in a row a max of like 15 times.

so if for example the ball this round lands on red, you could go ahead and repeatedly bet on black until you win (cause as assumed there will come at max 15 times red in a row, so you will win sooner or later.)
obviously the amount you bet on black has to increase with eahc round or you will get a negative total in the end.
so you like bet 1 buck on black.
won? great, series over.
lost? bet 2*1 on black next round.
rinse and repeat, betting double on black what you bet before.
martingale.
now if the first bet in such a series is 1 bucks, you would need a sh it ton of money to be able to bet 16 times or so. (we always assume worst case)

so we play smart and for example say:
we "empty bet" 5 times first, meaning we wait until we have 5 times red in a row.
And only THEN start betting 1 buck, 2 bucks, etc.
so the balance needed so you can pay up till the certain win is much bigger (it namely is like 1+2*1+2^2*1+...+2^11, which is less than 1+2*1+...+2^16*1).

anyways, say we got the needed balance and all so we can empty bet 5 times and bet the remaining 11 times on opposite color.
now, due to repeatedly winning, lets assume our balance has doubled in comparison to what we started with.

this gives us 2 possible choices:
now we either initally bet 2 bucks isntead of 1 bucks, basically doubling our profit for each series we do.
or we go ahead and emptybet only 4 times instead of 5 times. this also improves the profit because obviously in a given time span it is more likely to have a color appear 4 times in a row than it is for a color to appear 5 times.now that I hopefully have gotten the gist of the whole thing over, my big question would be:
is it better to increase the bet amount? or emptybet less? what on average should yield better returns, what is the better choice when "one upping" our system?
I know that this whole thing is likely unrealistic as f but please go along with it.
say someone does play rouletter for like 1 hour and, randomly assumed, a total of like 60 rolls are done, so we get 60 symbols in that time ( something like (RZBZRRRZBZRRRRRRRRBBBZBRRBBRZBRBR or such, 60 roll results, R= red, Z=zero, B=black). (obviously if we are in the middle of a betting series we will play it till the win)
I feel like we should be able to ind the answer with probabilities.
cause you can calculate how likely it is for like 5 times red in a row to occur and 4 times red to occur, so it should be comparable via probabilities.
but I cant get m head around the details yet.
someone any idea on this hypothetical roulette gambling scenario?

I know, one could go ahead and just do 2 trillion simulations where you do 60 rolls, look how mony rows of same color appear, how long the are, etc. and built the average about it all.
but I would assume there should also be a more proper mathematical way to solve this than jsut... trying it out.

Edit:
Attached is a picture which hopefully better shows what I mean:
Let's assume you currently wait till you see 5 times red in a row and then martingale bet (base bet amount 1 buck) until you win in the 16th round.
now you suddenly got enough money to make this more profitable.
shpould you still wait for 5 times red and have a base bet of 2 bucks isntead of 1?
or keep the base bet the same but wait for 4 times red isntead of 5 (which obviously should appear more often than the other)?
whats more profitable on average, how to mathematically calculate that?
 

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  • #2
Thinking for a bit, maybe I have higly overestimated the problem.
let's say so far base bet was 1 buck and
we waited for 5 times red.given that probabilty for red is 18/37 (18 fields are red, 18 black and 1 zero), that probabilty is (18/37)^5.

Now, we would instead only wait for 4 times red. the probablity of that is (18/37)^4.
now compared to before, getting a fitting series should be this likely compared to before:
(18/37)^4/(18/37)^5
=1/(18/37)=37/18=2.055=205,5%
so slightly more than double as likely as before.
so profitwise we made 1 buck per seires before, now we make on average
2.05*1=2.05 bucks per series.

now if we instead increased the base bet, we would make 2 bucks.
so lowering the amount of reds in a row aka how often we empty bet should be more profitable.

going for the general case of n+1 towards n times empty betting, we also get an increase of
(18/37)^n/(18/37)^(n+1)
=37/18, jsut as above.
so same result.

only question remaining would be:
here I only looked for continous series of red.

when in fct "n times the same color" is fulfilled when either we have n times blackor n times red.
since the whole situation and all is symmetrical.

how can I reflect this in my calculation and does it effect the end result? :-/
I would assume that the probability for n times same color is just 2*(18/37)^n since we basically jsut add the odds of the "n times red" and " n times black" paths together.

and the end result would be an increase of
(2*(18/37)^n)/(2*(18/37)^(n+1))
=same old 37/18
So same end result (waiting for shorter series is slightly better than upping the base bet)

Am I right or am I completely wrong here?
 
  • #3
It seems that you want to find a winning system for roulette. As if the sum of a series of bets, each with negative expectation could possibly have a positive expectation.

The optimal system (optimizing for expected value) is rather simple. Do not play.
 
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  • #4
jbriggs444 said:
It seems that you want to find a winning system for roulette. As if the sum of a series of bets, each with negative expectation could possibly have a positive expectation.

The optimal system (optimizing for expected value) is rather simple. Do not play.
The OP has an assumption that the same colour can come up a maximum of 15 times consecutively. That would allow a winning strategy of waiting for 15 reds or blacks then betting all you have on the certainty of the next turn.

I think the OP knows this assumption is false, but wants to look at the options under this assumption in any case.
 
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  • #5
while you are right about the maximum 15 times red or black, waiting for
actually 15 times red in a row would take way too long since it basically almost never happens (but is still possible, so cant ignore it).

so as explained before one might wait for 7 times red and then start betting.
bet 1 dollar on black.
black rolled? you won.
you got an 8th time red?
bet on black again but this time with double the amount, namely 2*1=2 dollars.

got a 9th time red and therefor lost again?
bet 2*2*1=2^2=4 dollars on black this round.

rinse and repeat, soone ror later you will win.
worst case you really have 15 time sred, bet (and lost) a couple of times already, but then the next bet on black (again with double the money as the bet before) will be a surefire win.

martingale basically.

here we waited for 7 times red in a row (which I would call 7 times "empty betting" as you dont bet any money in those rounds) and obviously had a basebet of 1 dollar (which upon each lose you doubled until you won. )
so yeah, thing is, given that you can only do one or the other, is it better to empty bet less (saying: only wait for 6 times red instead of the more unlikely 7 times red in a row)
or to increase the base bet from 1 to 2 dollars?
the later option doubles the profit per win round while the former increases the odds of win rounds happening.

which is better to do is the big question.
and then gotta keep in mind that it aint matter if you wait for series of red and bet blöack repeatedly.
or do it the other way around.

you wait for a series of the same color in a row.
and then keep betting on the opposite color, with increasingly bigger bet amounts, until you sooner or later win (the max 15 times color rule ensures a win, if you just have the money to go along with the rounds)
 
  • #6
Google the gambler's fallacy aka the monte carlo fallacy, then dump the idea of a "system" to win at roulette.

Google the longest run of reds in a monte carlo casino and ask yourself how much money you would need to have at the start while betting on black to make a profit of $1 / £1, then dump the idea of a system to win at roulette.

Google the longest run of reds at some USA casino, then dump the idea of a "system" to win at roulette.

Feel free to substitute black for red after doing the above, but using red is the famous one.

HINT: Each spin of the wheel is a single independent event, the wheel has no memory / knowledge of the previous spin.
The example I use is suppose I start by saying I wonder what will happen if I toss a newly minted coin 20 times, and you say you expect ten heads and ten tails. But I know something that the coin doesn't.
So I thrown ten consecutive heads (I've done seven in real life, when I was betting on heads against a local aggressive person, who kept saying double or quits after I won the first throw. He was so much bigger than me I was forced to keep trying, until tails came up. I never tried that again!)
Anyway, I have just thrown my tenth heads in a row, and you decide to raise the stakes to ten times the starting stake, and hold that bet for the next ten throws on the coins. The coin doesn't know what I will do next. I don't accept your bet, I drop the coin into the molten metal from which it was made, and no more throws ever take place. So the coin had a 100% record of throwing heads. It didn't "even things out" during its sad, short live.

The fallacy assumes things will "even out in the long run" because the coin is fair. There is no such statistical rule that things will even out. For all you know, your roulette wheel could have recorded 100 blacks in a row last night before being put away for the night as the casino closed. Does the wheel "know" this? In your plan it does know this, and is trying to even things up today. But NO, each spin is an independant event with the same odds on each spin.

PREDICTING ten reds or 15 reds before the bet starts is a different bet to betting on each spin and changing your stake.
 
  • #7
First, the OP is almost impossible to read with the lack of capitalization, punctuation and the stream-of-consciousness sfyle. Please have mercy on your readers and write clearly!

The point seems to be obscured by the wall-O-text, but the point seems to be that the limit of a run to 15 of the same color changes the strategy and allows for a winning strategy.

As described, these events are not independent. So any calculation assuming independence is wrong.

The best you can do is wait for 15 reds. Then you know the next color is not red, so you put 18/19 of your money on black and 1/19 on zero. Doing better than this requires details of the dependence between spins, and indeed, those details may not make this strategy possible. (If the reason you never get 16 of one color is that you never get 15 of one color...)
 
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  • #8
If there was any winning strategy, the casino would have shut down long ago. The marble, gold all over is also a giveaway in this regard.
 
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  • #9
The limitation on length of runs makes this a different game than roulette. I'd call it "screw-lette", except that's more appropriate for the casino version.
 
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  • #10
Yeah, we got it that there is no strategy to beat the bank.
but nobody cares! -.-
aside from that, not sure how else to explain my point.

"The best you can do is wait for 15 reds. Then you know the next color is not red, so you put 18/19 of your money on black and 1/19 on zero. Doing better than this requires details of the dependence between spins, and indeed, those details may not make this strategy possible. (If the reason you never get 16 of one color is that you never get 15 of one color...)"

I sure can wait for 15 times red, cause then something non-red must be next.
But one could also not wait for 15 times red but rather 7 times red and then bet between 1 and 8 times (depending on your luck) on black, each turn betting an appropriat amount on black and maybe zero.

there aint no real dependency of turns or such, each turn can yield red, black or zero (where zero is super unlikely with odds of 1/37=2,7%
 
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  • #12
bernd said:
there aint no real dependency of turns or such
You just said there was - a limit on N runs is exactly this kind of dependency.

@WWGD may be right and you may have a Martingale in mind, but all we can go by is what you write, not what you have in mind.
 

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