All betting systems produce the same results?
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Just a quick question. Mathematically, if you were to play a D'Alembert to a 5 stop (stopping if you lose a 5 unit bet, and starting over at 1), would you lose the same amount as if you were to add 1 unit after each loss and start over again at 1 unit if you win one time, or if you lose 5 in a row (the same as a 5 step Martingale, except only going up 1 unit at a time).
So, if playing a game where you win 47% of the time, mathematically, will these two games lose the exact same amount or not.
If you can show the math saying they both lose the same amount or not, that would be great.
Edit: If the above was confusing, the 2 games would look like this. D'alembert - 1-2-3-4-3-2-3-4-3-2-1-2-3-2-1 The other game 1-2-3-4-1-1-2-3-1-1-2-3-4-1-2-3-1
Quote: JyBrd0403All this Marty and Monkey stuff, got me thinking about a question I've had for a while.
Just a quick question. Mathematically, if you were to play a D'Alembert to a 5 stop (stopping if you lose a 5 unit bet, and starting over at 1), would you lose the same amount as if you were to add 1 unit after each loss and start over again at 1 unit if you win one time, or if you lose 5 in a row (the same as a 5 step Martingale, except only going up 1 unit at a time).
So, if playing a game where you win 47% of the time, mathematically, will these two games lose the exact same amount or not.
If you can show the math saying they both lose the same amount or not, that would be great.
Edit: If the above was confusing, the 2 games would look like this. D'alembert - 1-2-3-4-3-2-3-4-3-2-1-2-3-2-1 The other game 1-2-3-4-1-1-2-3-1-1-2-3-4-1-2-3-1
Interesting question. However mathematically, both methods and every other method in the planet loses 2,7% of the money wagered. You see sometimes mathematics doesn't tell the whole story. Mathematically any way of betting, even thowing your chips on the table has the same average expectation in the long run.
End result is the sucker loses all his money if he plays it long enough.
Quote: rudeboyoiYoud lose less because your average bet would be less.
If that's true, then all these betting systems produce different results. If the different systems produce different results, maybe, we can get someone to "rank" the different betting systems. Since, some will produce better results than others, the math should be able to "rank" the different systems from Best to Worst. Right?
Quote: RSOK. Best betting system: flat betting table minimum. Worst is flat betting table max. A good system is increasing your bet by 1 unit after 10 straight losses. A bad system is when you parlay your winning bets every time until you hit the max.
Flat betting is not a method, I wonder why people present "flat-betting" as a betting system.
Quote: RSOK. Best betting system: flat betting table minimum. Worst is flat betting table max. A good system is increasing your bet by 1 unit after 10 straight losses. A bad system is when you parlay your winning bets every time until you hit the max.
Okay, and from my original post which system would "rank" higher? And why?
Quote: JyBrd0403Okay, and from my original post which system would "rank" higher? And why?
I don't know which one is better. But, it'd be the one with a lower average bet than the other. If you can calculate the average bet then you can calculate the expected loss.
There's no real difference. It's an individuals own personal preference.Quote: JyBrd0403Okay, and from my original post which system would "rank" higher? And why?
Someone would need to know your goals and risk tolerance.
If you want to have a higher percent chance you will make a small amount of money but risk going broke fast. There's always Marty with a high 90 percent chance you will make 1 unit.
Quote: AxelWolfThere's no real difference.
So, you're saying these 2 betting systems produce the same results. Do you have math that shows that? Or, simulator results?
Martingale - start at 1; when you lose 5 in a row, go back to 1
D'Alembert - start at 1; when the bet reaches 6, go back to 1
Note that the same win/lose results were used for both simultaneously
I got that the average D'Alembert bet was about 2.35, and the average Martingale bet was about 2.61; as a result, you lost more with Martingale than with D'Alembert.
If you change the game to an even-money bet on double-zero roulette, it becomes 2.41 for D'Alembert and 2.74 for Martingale.
If you are playing a 50% game, then the expected result is zero regardless of what system you use. However, if the game is less than 50%, the one where your average bet is lower will lose less over the same amount of time. In either case, D'Alembert loses less over time.
Of course, using a "flat bet 1" "system", the average bet is 1, which is better than either D'Alembert or Martingale.
Quote: JyBrd0403If that's true, then all these betting systems produce different results. If the different systems produce different results, maybe, we can get someone to "rank" the different betting systems.
Two men were selected to participate in an offer, a mathematician and an engineer.
The offer was this: you start on one end of a football field, and on the other end is a beautiful woman. When you get to her, you can do with her as you like. (Both men seemed pleased with this idea.)
The catch is that they could only cover half the remaining distance every 5 minutes.
The mathematician heard this, and walked away in disgust, wishing the engineer luck, if he chose to continue. "You can never get there!"
The engineer started walking toward the 50 yard line, realizing that you can get close enough for all practical purposes, and it wouldn't even take that long...
With great respect, this place is loaded with mathematicians. Yes, the end result of any betting system on a -EV game is to go broke, if carried out fully. There are other ways to look at it - like how many rounds it takes to go broke given an N unit bankroll (more is better).
I don't know that you're asking the question in the right way to get a useful response. A lot of people are likely to get hung up on the part where it's still a negative expectation game, and nothing you're doing will get around the house edge.
Quote: JyBrd0403Do you have math that shows that?
I have math that shows that my earlier numbers are accurate.
Let p be the probability of winning, a q be the probability of losing; p + q = 1.
For Martingale, both the mean number of bets per run and the mean total bet per run can be calculated using the same formula, but with different parameters:
Win the first bet: A x p
Lose the first bet, win the second: B x q x p
Lose the first two bets, win the third: C x q2 x p
Lose the first three bets, win the fourth: D x q3 x p
Lose the first four bets, win the fifth: E x q4 x p
Lose five bets in a row: F x q5
For the number of bets, A = 1, B = 2, C = 3, D = 4, and E = F = 5;
the mean number of bets = p + 2qp + 3q2 + 4q3 + 5(q4p + q5)
= p + 2qp + 3q2 + 4q3 + 5q4 (p + q)
= p + 2qp + 3q2 + 4q3 + 5q4 {remember, p + q = 1}
For the total bet, A = 1, B = 3, C = 7, D = 15, and E = F = 31;
the mean amount bet = p + 3qp + 7q2 + 15q3 + 31(q4p + q5)
= p + 3qp + 7q2 + 15q3 + 31q4
Calculating D'Alembert is a little harder, since the bets go back and forth.
Again, the same method is used for both the number of bets and the total amount bet.
Let S(n) be the number of bets remaining in the current run when the current bet is n.
Since S(6) is reached after losing a bet of 5, which is when the run ends, S(6) = 0.
S(5) = E + p S(4) + q S(6) = 1 + pS(4).
S(4) = D + p S(3) + q S(5)
S(3) = C + p S(2) + q S(4)
S(2) = B + p S(1) + q S(3)
S(1) = A + p S(1) + q S(2) {since winning a bet of 1 means the next bet is also 1}
The first bet is 1, so we are interested in calculating S(1) for both the number and total amount.
This is five equations in five unknowns;
S(1) = ((A + B + C + D + E) q4 + (A + B + C + D) pq3+ (A + B + C) p2q2 + (A + B) p3q + Ap4) / q5
For the number of bets, A = B = C = D = E = 1;
the mean number of bets = 5 q4 + 4 pq3 + 3 pq2q2 + 2 p3q + p4
For the amount bet, E = 5, D = 4, C = 3, B = 2, and A = 1;
the mean amount bet = 15 q4 + 14 pq3 + 12 pq2q2 + 9 p3q + 5 p4
For p = 0.495 (the first example I used above):
Martingale:
Mean total bet = 5.10100501
Mean # of bets = 1.95385037562
Mean Bet = 2.6107449545
D'Alembert:
Mean Total Bet = 67.9507148123
Mean # Of Bets = 28.9302885898
Mean Bet = 2.3487741784
Quote: JyBrd0403Okay, and from my original post which system would "rank" higher? And why?
I'll tell you.
I'd rank higher the 1-2-3-4-5 system. With a little important modification:
The next progression will start 2 steps below the end of the previous progression.
That is:
If you lost the full 5 steps, next I'll begin with 3 units.
If you got a hit at step 4, next I'll begin with 2 units.
If I got a hit in the first 3 spins, next I'll begin with 1 unit.
This I consider a very good progression.
The problem in your initial progression is that you get too many 1 unit wins.
Quote: KavourasI'd rank higher the 1-2-3-4-5 system. With a little important modification:
The next progression will start 2 steps below the end of the previous progression.
That is:
If you lost the full 5 steps, next I'll begin with 3 units.
If you got a hit at step 4, next I'll begin with 2 units.
If I got a hit in the first 3 spins, next I'll begin with 1 unit.
This I consider a very good progression.
Just to clarify - is this what you are suggesting:
Current Bet 1: after a win, bet 1; after a loss, bet 2
Current Bet 2: after a win, bet 1; after a loss, bet 3
Current Bet 3: after a win, bet 1; after a loss, bet 4
Current Bet 4: after a win, bet 2; after a loss, bet 5
Current Bet 5: after a win, bet 3; after a loss, bet 3
Quote: KavourasThe problem in your initial progression is that you get too many 1 unit wins.
Wouldn't your change increase the number of 1-unit bets? Instead of a 3-unit win being followed by a 2, you follow it with a 1; a 4-unit win is followed by a 2, which would be followed by a 1 after a win (instead of the bets being 3 and 2); a 5-unit win is followed by a 3, which would be followed by two 1s after wins (instead of 4, 3, 2).
to be even simplerQuote: ThatDonGuyFor Martingale, both the mean number of bets per run and the mean total bet per run can be calculated using the same formula <snip>
For p = 0.495 (the first example I used above):
Martingale:
Mean total bet = 5.10100501
Mean # of bets = 1.95385037562
Mean Bet = 2.6107449545
for Mr. Marty
the avg total bet for the system = (1 -((2*q)^S)) / (1 - (2*q))
q = 1-p and S = 5 (steps max)
the expected number of bets = (1-q^S) / (1-q)
for extra credit for that Don
the D = ?
Mully
really?Quote: KavourasI'd rank higher the 1-2-3-4-5 system. With a little important modification: <SNIP>
This I consider a very good progression.
in my opinion
a progression that can lose the most (and win the least)
real $$$$ in an average amount of time is NOT a very good progression
i guess many will just think that when they lose, they actually win
this is the law of the fifth, btw
thank you for sharing your opinion
and not using any math
Sally
Quote: ThatDonGuyJust to clarify - is this what you are suggesting:
Current Bet 1: after a win, bet 1; after a loss, bet 2
Current Bet 2: after a win, bet 1; after a loss, bet 3
Current Bet 3: after a win, bet 1; after a loss, bet 4
Current Bet 4: after a win, bet 2; after a loss, bet 5
Current Bet 5: after a win, bet 3; after a loss, bet 3
Wouldn't your change increase the number of 1-unit bets? Instead of a 3-unit win being followed by a 2, you follow it with a 1; a 4-unit win is followed by a 2, which would be followed by a 1 after a win (instead of the bets being 3 and 2); a 5-unit win is followed by a 3, which would be followed by two 1s after wins (instead of 4, 3, 2).
Hi,
Yes, you got the system right.
The problem with the second system is that 5 steps are not enough. Meaning we would see many:
bet 1 lose, bet 2 lose, bet 3 hit, bet2 lose, bet 3 lose, bet 4 hit, bet 3 lose, bet 4 lose, bet 5 lose kind of sequences.
A comparison:
Version 1 would do much better in such a sequence IMO
1-2-3-1-2-3-1-2-3
Quote: mustangsallyreally?
in my opinion
a progression that can lose the most (and win the least)
real $$$$ in an average amount of time is NOT a very good progression
this is the law of the fifth, btw
thank you for sharing your opinion
and not using any math
Sally
Do you have a specific suggestion, which version is better and why?
And please when you post equations be sure to include a legend explaining every Letter you use in the equation. Your posts would be much more readable this way.
Quote: mustangsally
for extra credit for that Don
the D = ?
"The D" is a hotel/casino downtown (it used to be Fitzgerald's) - that's where I rented my car last year.
To what other D were you referring?
Quote: ThatDonGuyI simulated 2 billion bets in a game with a 49.5% win probability (house edge 1%) under the following conditions:
Martingale - start at 1; when you lose 5 in a row, go back to 1
D'Alembert - start at 1; when the bet reaches 6, go back to 1
Note that the same win/lose results were used for both simultaneously
I got that the average D'Alembert bet was about 2.35, and the average Martingale bet was about 2.61; as a result, you lost more with Martingale than with D'Alembert.
If you change the game to an even-money bet on double-zero roulette, it becomes 2.41 for D'Alembert and 2.74 for Martingale.
If you are playing a 50% game, then the expected result is zero regardless of what system you use. However, if the game is less than 50%, the one where your average bet is lower will lose less over the same amount of time. In either case, D'Alembert loses less over time.
Of course, using a "flat bet 1" "system", the average bet is 1, which is better than either D'Alembert or Martingale.
Don, was the Marty 5 step you did a regular Martingale double after each loss (1-2-4-8-16), or the 5 step one in my OP (1-2-3-4-5). It sounded like you did the 5 step double after a loss, which is also interesting. But, I'm still wondering about the comparison of the 2 systems in my OP.
But, those results you gave are kind of the point. Since all these different systems will give different results, the math should be able to distinguish between which systems are Better and which are worse. Maybe, we could have a reference here on Wov for which systems work best or worst. Just a thought.
Quote: JyBrd0403Since all these different systems will give different results, the math should be able to distinguish between which systems are Better and which are worse. Maybe, we could have a reference here on Wov for which systems work best or worst. Just a thought.
Flat betting one bankroll sized bet has the mathematical advantage over Marty and any other system. See https://wizardofvegas.com/forum/gambling/betting-systems/21359-debunking-roulette-marty-with-pictures/3/#post441243
Quote: ThatDonGuyIf you are playing a 50% game, then the expected result is zero regardless of what system you use. However, if the game is less than 50%, the one where your average bet is lower will lose less over the same amount of time. In either case, D'Alembert loses less over time.
Thanks for showing the math on the other post (and Sally too). But, I am curious as to how you would get an avg. betting amount for the D'Alembert. For the marty it's pretty simple, but for the D'Alembert how would you determine if the 4 unit bet would occur more often then the 2 unit bet? Ex. 1-2-3-4-3-4-5-4-3-2-3-4-5-4-5-4-3-2-1 There's 3 2 unit bets and 6 4 unit bets there.
Quote: OnceDearFlat betting one bankroll sized bet has the mathematical advantage over Marty and any other system. See https://wizardofvegas.com/forum/gambling/betting-systems/21359-debunking-roulette-marty-with-pictures/3/#post441243
Okay, but for the other systems, which be the 2nd best choice, 3rd, 4th, 5th, 6th etc. Since, they all change the EV of the game.
Quote: JyBrd0403Don, was the Marty 5 step you did a regular Martingale double after each loss (1-2-4-8-16), or the 5 step one in my OP (1-2-3-4-5). It sounded like you did the 5 step double after a loss, which is also interesting. But, I'm still wondering about the comparison of the 2 systems in my OP.
Yes, I did 1-2-4-8-16 Martingale. That's what I thought you meant.
A 1-2-3-4-5 "Martingale" is probably going to do better than a standard D'Alembert as the average bet should be smaller.
Both systems are the same at any particular betting point with a loss (increase by 1, except if the bet is 5, in which case reset to 1); on a loss, Martingale always goes to 1, while D'Alembert goes to something higher than 1 after a bet of 3 or more, and never goes to something less than 1.
Quote: JyBrd0403Thanks for showing the math on the other post (and Sally too). But, I am curious as to how you would get an avg. betting amount for the D'Alembert. For the marty it's pretty simple, but for the D'Alembert how would you determine if the 4 unit bet would occur more often then the 2 unit bet? Ex. 1-2-3-4-3-4-5-4-3-2-3-4-5-4-5-4-3-2-1 There's 3 2 unit bets and 6 4 unit bets there.
The way I did it is, I calculated both the expected number of bets needed, and the total of those bets, in a particular "run". A run starts with the first bet of 1, and ends when you lose a bet of 5 (and your next bet would be 6).
Let B(N) be the number of bets remaining before reaching a bet of 6.
The number of bets you need at level N to reach level 6 = 1 + (the probability of winning) x (the number of bets you need at level N-1, or at level 1 if you are currently at Level 1) + (the probability of losing) x (the number of bets you need at level N+1).
B(6) = 0
B(5) = 1 + p B(4) + q B(6) = 1 + p B(4)
B(4) = 1 + p B(3) + q B(5)
B(3) = 1 + p B(2) + q B(4)
B(2) = 1 + p B(1) + q B(3)
B(1) = 1 + p B(0) + q B(2)
This is five equations in five variables; you can get values for each of B(1) through B(5) in terms of p and q (and remember that q = 1 - p). The number you are looking for is B(1), since you start at level 1.
Calculating the average total amount bet is similar, except that you replace the 1s in the equations with N (e.g. B(5) = 5 + p B(4) + q B(6), since you are betting 5 at level 5).
Once you have the average total bet and the average number of bets, divide the total by the number of bets to get the average bet size.
Since the expected loss amount per bet = the average bet size x the house advantage, and the house advantage is the same regardless of bet size or method used, the better system is the one with the smaller average bet.
However is there substance in your opinion to the title "Better System"? Haven't they all the same average expectation if risking the same amount of money?
I think the last thread about Martingales was very thought provoking in a sense that it showed that depending on your play-style/strategy, bankroll and win target there ARE "better" and "worse" systems. Or not?
Quote: JyBrd0403Don, was the Marty 5 step you did a regular Martingale double after each loss (1-2-4-8-16), or the 5 step one in my OP (1-2-3-4-5). It sounded like you did the 5 step double after a loss, which is also interesting. But, I'm still wondering about the comparison of the 2 systems in my OP.
The 1-2-3-4-5 Martingale has an average bet of 1.85040487 in a 1% game, and 1.90068217 in double-zero roulette.
Yes, there's a generic solution for this as well: (1 - (N+1) qN + N qN+1) / (p (1 - qN))
I have a feeling this can be reduced further.
...and it can: N + 1 / p - N / (1 - qN)
Quote: JyBrd0403Okay, but for the other systems, which be the 2nd best choice, 3rd, 4th, 5th, 6th etc. Since, they all change the EV of the game.
None of them change the EV of the game as a proportion of amount bet which for gambling is all anyone ever talks about. They change the amount bet which can change the expected loss per hour but again none of them to any better then just flat betting table minimum so what does it really matter.
Betting systems do vary. The other important factor is variance.
If you compare over infinite plays, yes all systems lose your bankroll. Not everyone plays until their bankroll is empty...
Quote: TwirdmanNone of them change the EV of the game as a proportion of amount bet which for gambling is all anyone ever talks about. They change the amount bet which can change the expected loss per hour but again none of them to any better then just flat betting table minimum so what does it really matter.
Because the EV is a fraction of the bet. The more money you bet, the more you are expected to lose.
If your stop point is, "I'll stop after betting a total amount X", then you're right; all systems are the same in that all of them are expected to lose the same fraction of X.
Example: even-money betting on a double-zero roulette wheel (probability of winning = 9/19).
You start with a bankroll of 2, and you want to double it before losing everything.
If you make a single bet of 2, you always end after betting a total of 2, your probability of success = 9/19.
On the other hand, if you stick to bets of 1:
81/361 of the time, your first two bets win, and you have reached your target.
100/361 of the time, your first two bets lose, and you have lost your bankroll.
The other 180/361 of the time, you win one and lose one, and are back to your starting point; repeat the process.
You might end up betting a total of 2, or 4, or 6, or 8, or any positive even number.
Your probability of success is 81 / (81 + 100) = 81/181 < 9/19.
Quote: thecesspit'Better' depends on you definition to compare two systems and what the goal is. Better for one goal can be worsevfor another.
Betting systems do vary. The other important factor is variance.
If you compare over infinite plays, yes all systems lose your bankroll. Not everyone plays until their bankroll is empty...
Yes, but if you were to compare the different systems based simply on 1 million trials. You should be able to "rank" the systems based on how much money each system lost after 1 million trials. Should be pretty straight forward, since they all produce different results.
Quote: ThatDonGuyThe 1-2-3-4-5 Martingale has an average bet of 1.85040487 in a 1% game, and 1.90068217 in double-zero roulette.
Yes, there's a generic solution for this as well: (1 - (N+1) qN + N qN+1) / (p (1 - qN))
I have a feeling this can be reduced further.
...and it can: N + 1 / p - N / (1 - qN)
Thanks for this Don.
Quote: JyBrd0403Yes, but if you were to compare the different systems based simply on 1 million trials. You should be able to "rank" the systems based on how much money each system lost after 1 million trials. Should be pretty straight forward, since they all produce different results.
It's pretty straightforward in the sense that in the "long run" (1 million spins can be considered long run I guess) all systems are expected to return to you 97,3% of what you bet.
So, without a specific (win) target, bankroll requirements and some sort of plan/strategy all systems are the same as throwing your chips on the table, if you plan to mindlessly play a progression over 1 million spins, without bankroll and table limitations.
1 million spin tests are useless. Because we already know the result.
Quote: KavourasIt's pretty straightforward in the sense that in the "long run" (1 million spins can be considered long run I guess) all systems are expected to return to you 97,3% of what you bet.
I think that's the point - what systems have what average bets? The lower the average bet, the longer your bankroll is expected to last.
i say your guess is wrongQuote: KavourasIt's pretty straightforward in the sense that in the "long run" (1 million spins can be considered long run I guess)
just wrong
exactlyQuote: Kavourasall systems are expected to return to you 97,3% of what you bet.
key word = expected
in other words
expected = average
or in your opinion expected = ??
hahahaQuote: KavourasSo, without a specific (win) target, bankroll requirements and some sort of plan/strategy all systems are the same as throwing your chips on the table, if you plan to mindlessly play a progression over 1 million spins, without bankroll and table limitations.
thank you for your opinionQuote: Kavouras1 million spin tests are useless. Because we already know the result.
a wrong one in me opinion
take 18/37 = p (p=probability of success)
after 1,000,00 spins and equal average bets (1 unit okay)
1 in 15,787 (on average) would be outside the range of 4SD or four standard deviations
so with 1 million players playing 1 million spins (happens every second i say)
i would expect more than a few would be outside of these ranges and NOT exactly or close to your 97,3%
0.976967429 to 0.968978517
and as a percentage
97.6967429% to 96.8978517%
(return to player)
and in real $$$$ lost
because gamblers do lose real money
that range = -23,032.57 to -31,021.48
not close to exactly -27,027.03
(all values are in units)
you may check my math as Excel did it for me
Sally says in Oh so many words
all opinions
except Excel wanted to do the math real bad!
Quote: ThatDonGuyI think that's the point - what systems have what average bets? The lower the average bet, the longer your bankroll is expected to last.
Yes. And as we all know the lowest average bet is zero. Do not play at all :-)
Quote: ThatDonGuyI think that's the point - what systems have what average bets? The lower the average bet, the longer your bankroll is expected to last.
Yes, exactly my point, since they all have different average bets, which produce different results, it should be able to clearly "rank" which ones are better or worse, mathematically.
Just thinking about it. A Reverse D'Alembert or Reverse Martingale would both, I would assume, have lower average bet sizes than playing the normal way. So, there's two games that would rank higher than a regular D'Alembert or Marty. That's surprising, to me at least. I wouldn't have guessed that. Anyway, I think it would be interesting to see the different Systems ranked. That's all I'm saying.
Quote: JyBrd0403Just thinking about it. A Reverse D'Alembert or Reverse Martingale would both, I would assume, have lower average bet sizes than playing the normal way. So, there's two games that would rank higher than a regular D'Alembert or Marty. That's surprising, to me at least. I wouldn't have guessed that. Anyway, I think it would be interesting to see the different Systems ranked. That's all I'm saying.
What's "the normal way"? The obvious answer to "what's the best system" is, flat bet the minimum. By definition, a reverse anything system starts out with a higher bet than the minimum.
"The best system is not to play" reminds me of WarGames - speaking of which, ever hear the story of "WarGames: The Next Five Minutes"?
"Sir, our launch system has been deactivated."
"Incoming missiles - confirmed visually!"
"WOPR, what is happening?"
"You left one scenario out of Global Thermonuclear War - Soviets launch, USA doesn't; I win. I disabled your launching system. If Kirk is allowed to 'alter the conditions' of Kobayashi Maru, why can't I?"
Quote: KavourasFlat betting is not a method, I wonder why people present "flat-betting" as a betting system.
