martingale math - what say ye?

So I get that martingale system is mathematically sound, but in real world it doesn't work because casinos set table limits, etc.

But out of curiosity for you mathematical minded folk, at what point would it work?

For example, it doesn't work in roulette because most tables have low maximums, and the chances of black hitting (for example) in a large series (although unlikely statistically speaking) is going to happen frequently.

That is in a system where probability of success/failure is 50/50 (or very close to it).

So the question is, what ratio would you say that system could possibly work practically speaking? For example, lets say the probability of success of single trial is 2/3, and the chance of of loosing is 1/3. Would you think you could use martingale in that setting? If not, how about .75 chance of probability of success, vs .25 of losing?


Even more real world, starting with $7000, if you lay a $20 bet on 4 (0.083333333% chance of hitting a 4 hitting each roll), a 4 would need to hit 6 times before losing your bank role (20+40+120+360+1080+3240). Using the binomial formula, the chances of a 4 hitting 6 times lets say in 12 roles is 0.000198744170760046. However, the chance of a 7 coming up in 12 roles is 0.887843345215385.

That seems pretty good. However, I could never lay down that much money (don't have the balls), and I also know that casino's know math better than I do.

What if I could lay a 2, would the martingale system work then? At what odds in your favor would real mathematicians take in order to say they had a fighting chance with the martingale?

Quote: slackyhacky

So I get that martingale system is mathematically sound, but in real world it doesn't work because casinos set table limits, etc.

No, it's not mathematically sound. It's only neutral (not advantageous) for a fair bet.

In casino games, martingale increases the effective house edge. I'll quote one of my older posts on this.

...
For n-deep martingale in roulette, the possible outcomes of a martingale progression are:
1-(20/38)^n: Win 1 unit
(20/38)^n: Lose 2^n-1 units
For a total EV of 1-(20/38)^n-(2^n-1)(20/38)^n = 1-(20/38)^n*2^n = 1-(40/38)^n ~= 1-1.0526^n

Or, for other games, 1-(1+HA)^n.
It is a very simple function. For any HA>0, it exponentially approaches -infinity as possible martingale progression depth (n) increases.
If you just martingale 2 bets deep, you are playing a 10.8% house edge game instead of 5.26%. If you do it 3 bets deep, 16.7%. Playing $10-$1,000 game will let you martingale 7 bets deep, for a house edge of 43%.

In other words, you expect to encounter almost twice as quickly a row of 7 losing bets opposed to a fair 50/50 game, because of that seemingly "little" house advantage....

Quote: P90

No, it's not mathematically sound. It's only neutral

Ditto.

To be able to have even a small shot at it "working", you'd need both an unlimited bankroll, and a table with no maximum bet.

It doesn't matter what the game is.

Quote: DJTeddyBear

To be able to have even a small shot at it "working", you'd need both an unlimited bankroll, and a table with no maximum bet.

Which of course is a situation to be found only in mathematical textbooks and debating societies. In the real world this "Double Up to Catch Up" would always encounter rather limited bankrolls and would most definitely encounter table limits.

Sure, some "short run" statistics can be generated wherein it "works" adequately but in the real world there is always that "nth" spin wherein red appears in succession simply too many times for the system to work.

If you have ANY advantage, as the OP is asking, then you should Kelly bet, not Martingale. Kelly bet grinds out the advantage using bets proportional to the size of your bankroll.

Martingaling an advantage will not win as much in the long term as Kelly betting.

The leftover question is "under what circumstances would Martingaling still have an advantage?" I don't care to find out.

Quote: dwheatley

The leftover question is "under what circumstances would Martingaling still have an advantage?"

Doubling your bet after a loss in any succession of losses would work only under the circumstances of your knowing when the losses would end and if you knew that, you wouldn't have any losses to begin with.

Quote: P90

No, it's not mathematically sound. It's only neutral (not advantageous) for a fair bet.

How is that true. If A or B are the outcomes, and everytime B happens, it clears the debt. Everytime B happens consequetively, I gain. Over time, that gain persists.

Quote: P90

...
For n-deep martingale in roulette, the possible outcomes of a martingale progression are:
1-(20/38)^n: Win 1 unit
(20/38)^n: Lose 2^n-1 units
For a total EV of 1-(20/38)^n-(2^n-1)(20/38)^n = 1-(20/38)^n*2^n = 1-(40/38)^n ~= 1-1.0526^n

Or, for other games, 1-(1+HA)^n.
It is a very simple function. For any HA>0, it exponentially approaches -infinity as possible martingale progression depth (n) increases.
If you just martingale 2 bets deep, you are playing a 10.8% house edge game instead of 5.26%. If you do it 3 bets deep, 16.7%. Playing $10-$1,000 game will let you martingale 7 bets deep, for a house edge of 43%.

I don't follow this at all. First of all, where are you getting a house edge of 5.26%? I get lost in the math quickly so you tell me if this is wrong. Assuming a 5% vig on 20 dollars, I bet $21 to win $10. 10/21= .47619. The odds of a 7 coming up before a 10 is exactly 2/1 (or 0.5). So the house advantage is about 2.38%.

Also, using my example, after lossing the series I mentioned (20+40+120+360+1080+3240) - the 5% vig on each of those comes out to be 243. How are you coming up with 43%?

So the question was, in what odds (and when I say odds, I mean in your favor for probability of the event happening, not on the house edge, because you can easily add that into your losses since you know as sure as the sun comes up that your event is coming that will clear debt) in your favor would you feel comfortable playing the martingale system, and the resounding answer is NONE?


Interesting......

Quote: slackyhacky

How is that true. If A or B are the outcomes, and everytime B happens, it clears the debt.

B doesn't have to happen in your play session at all, even once. Since your bankroll is limiting it.

Quote: slackyhacky

I don't follow this at all. First of all, where are you getting a house edge of 5.26%?

It's for roulette. For craps, use whatever it is.

Quote: slackyhacky

How is that true. If A or B are the outcomes, and everytime B happens, it clears the debt. Everytime B happens consequetively, I gain. Over time, that gain persists.

It doesn't persist. You lose it back the next time A happens two times in a row. Then B comes and clears the debt. Your long term expectation is 0 (cleared "debt", but no profit).

Quote: slackyhacky

So I get that martingale system is mathematically sound, but in real world it doesn't work because casinos set table limits, etc.
...
What if I could lay a 2, would the martingale system work then? At what odds in your favor would real mathematicians take in order to say they had a fighting chance with the martingale?

Step back for a moment and consider the following:
1) Let E be the expected return (or expectation) on a casino game. In all the casino games you're thinking about, E < 1. That is, the house has the edge because for every dollar you put in, you expect to get back E which is less than that dollar. Obviously there is a wide variation in actual results, but the average result is less than 1. For the American roulette Red bet, for example, E = 1 - 2/38 = 0.94737. For the craps pass bet E = 1 - 7/495 = 0.98586.
2) When you make a bet of any size, B, the expected return is B*E. When you stake $1, you expect to get back $0.95; when you stake $2, you expect to get back $1.89; when you stake $1000, you expect to get back $947.37.
3) In the relevant casino games (e.g. roulette or craps), subsequent games or plays are independent, such that the probability and expectation of one play is not affected by any prior play. That means if you make a first bet of B1 and a second bet of B2 on the same game, the expected return is B1*E + B2*E. This expectation is the same as if you had made one larger bet of (B1+B2): (B1+B2)*E = B1*E + B2*E. In other words, your expectation is the same if you make a single $1000 bet or 1000 $1 bets.
4) From #3, it follows that the expected return of any gambling system is equal to the expectation of a single bet worth the sum of the wagers in that system. If you're playing a 6-bet progression on the pass bet of 20, 40, 120, 360, 1080, 3240 then your expectation is the sum of those wagers times E, in this case .98586, or 4791.27. In other words, you expect to lose $68.73 every time you play that sequence. You probably won't, of course, but that's the average result. If you don't like that average result, bet less. Don't hang your hopes on the wrong-headed idea that making bets which individually are long-term losers can somehow make the combined lot of them a long-term winner.

Your sum of the bets can't be quite right, as you don't always get to the end of the sequence.... (your average sequence bet is not the sum of all terms in the sequence)

You either win $20 or lose the total amount. You can work out the chance of losing all 6, and thus the inverse is the chance of winning $20, and thus the overall loss for each time you run the sequence.

I don't have time to do a worked example.

That's true, but the edge for any given sequence is equal to the edge on a single bet of the summed wagers of that sequence. That's what I was attempting to convey. If you're going to bet a total of $X, it doesn't matter whether you do that $1 at a time, in some convoluted pattern, or all at once. The edge is the same each way. What changes is the distribution of likely outcomes.

Quote: MathExtremist

If you're going to bet a total of $X, it doesn't matter whether you do that $1 at a time, in some convoluted pattern, or all at once. The edge is the same each way. What changes is the distribution of likely outcomes.

The catch here is "If you're going to bet a total of $X", however. In this condition, $X is the sum total of bets placed, not bankroll or typical bet size or anything else significantly relevant to a gambler.

Thus, betting systems are not all equal. Different patterns result in different $X for equally desirable distribution of outcome probabilities. As such, patterns that result in minimum $X for desired outcome distribution will be more efficient, others will be less efficient. Martingale is a very inefficient one. This efficiency difference manifests in different EV (and different house edge and vigorish) if you treat the entire sequence as one bet.

What's common however is that no betting pattern can produce a positive EV from negative-EV bets, or result in lower vigorish than individual bets it is composed of.

Assuming a straight martingale on a Evens bet.

p = probability of winning a bet
x = number of doubling levels

P(losing after x levels) = (1 - p)^x

Units bet after losing x levels = (2^x) - 1

Therefore if we look at each Martingale sequence as one bet, where you win 1 unit or lose (2^x)-1 :

P(win) = 1 - (1-p)^x
P(loss) = (1-p)^x

Expected Profit/loss = P(win) * 1 - P(loss) * (2^x) -1

Return = 1+Expected Loss

So, for a 6 level Martingale on a American Roulette wheel :

p = 18/38 = 9/19
x = 6

P(win) = 97.87%
P(loss) = 2.13%

Expected Profit/Loss per martingale sequence = -0.36
Expected Return = 64%

Every sequence you expect to loss .36 units on a six level martingale.

For a straight 50/50 game : 100%
For a game with 55/45 edge : 146%

Lets look at this as a percentage growth/loss of entire 63 unit bank roll :

50/50 Game = 0% growth
55/45 Game = 0.7% growth
Roulette Game = -0.6% reductions

Now the Kelly Criterion says to not bet without an edge.

But what does it say to bet on a game with a 55/45 on a evens game : 10% of your roll. 6.3 Units.

On 55/45 game, you expect to grow 0.5% per game (not per Martingale sequence).

The average length of a 6-level Martingale sequence is 1.9 games.

So to compare per game :

Kelly 10% stake on a 55/45 game = 0.5% growth in bankroll
Martingale stake on a 55.45 game = 0.7%/1.9 = 0.36%

Thus the Kelly is a better bet. Plus you can never go bust betting the Kelly stake (however once you are down to around 10 units, you are pretty much done, but the chances of going that low are much less than the chances of losing 6 in a row).

Quote: rdw4potus

It doesn't persist. You lose it back the next time A happens two times in a row. Then B comes and clears the debt. Your long term expectation is 0 (cleared "debt", but no profit).

I must be retarded cuz I think it does. Each time the debt it clears, it clears it to the starting point of where you were at that moment - not the starting point of when you started betting. Here is a link to an excel spread sheet I just made to show this. If you are betting on A (wins) and B (looses) and each time B hits you bet whatever amount you need to cover your losses, it will bring you back to whatever point you were at that round. Every time A hits in a row, it brings you up some.

https://docs.google.com/open?id=0B3eruTU5Xu_jM2IyYWViYzgtMTdlYS00ZjE3LTg1ZmItOTY0NTRkZWI5NDYz

Quote: rdw4potus

It doesn't persist. You lose it back the next time A happens two times in a row. Then B comes and clears the debt. Your long term expectation is 0 (cleared "debt", but no profit).

So I was curious....

Not only does it persist, but I added in the roulette percentage of expected wins on black or red to see the results of the martingale system - and it didn't make a difference, it always gains with time.

Again that is because, every time you hit a "win" it brings you back to where you were when you "lost", not back to zero.

Here is the spreadsheet for that.

https://docs.google.com/open?id=0B3eruTU5Xu_jZGNlZDM2NGEtN2ZjNy00MDU4LWFlZTEtZDEyMmE0MjAwZDYw

If you would have fair odds (eg 2.00 for a 50% chance), with martingale system you must also martingale the wins (but must have a bigger bankroll divided into several smaller allowing for a loss of them), otherwise you always gain 1 dollar. Then if you have a 6-step martingale, you would need 63 wins if always gain 1 dollar. But on average the 6 losing row will always come sooner then your required 63 winnings.

Quote: edward

But on average the 6 losing row will always come sooner then your required 63 winnings.

Although I have not followed the math in this thread at all, I do indeed understand that whether you choose Red or choose Black: six in a row of one of them is far more likely to take place than 63 in a row!

Quote: MathExtremist

For the craps pass bet E = 1 - 7/495 = 0.98586.

I worked out the expectation for the pass line bet:

(8/36) - (4/36) + 2(1/36) - 2(2/36) + 2(8/180) – 2(12/180) + 2(25/396) – 2(30/396) = - .0141

I see 7/495 = .0141, but how did you get 7/495?

By hand, without using Excel. It's the same value.

Here's how Wolfram Alpha solves "(8/36) - (4/36) + 2(1/36) - 2(2/36) + 2(8/180) – 2(12/180) + 2(25/396) – 2(30/396)"

Thanks.

Quote: P90

No, it's not mathematically sound. It's only neutral (not advantageous) for a fair bet.

In casino games, martingale increases the effective house edge. I'll quote one of my older posts on this.

...
For n-deep martingale in roulette, the possible outcomes of a martingale progression are:
1-(20/38)^n: Win 1 unit
(20/38)^n: Lose 2^n-1 units
For a total EV of 1-(20/38)^n-(2^n-1)(20/38)^n = 1-(20/38)^n*2^n = 1-(40/38)^n ~= 1-1.0526^n

Or, for other games, 1-(1+HA)^n.
It is a very simple function. For any HA>0, it exponentially approaches -infinity as possible martingale progression depth (n) increases.
If you just martingale 2 bets deep, you are playing a 10.8% house edge game instead of 5.26%. If you do it 3 bets deep, 16.7%. Playing $10-$1,000 game will let you martingale 7 bets deep, for a house edge of 43%.

Okay,

So again, I don't agree with this. If I bet (call it whatever system you want) so that my next win covers ALL my loses, including the VIG, then each bet only has a house advantage of 5.26%...it doesn't compound as you say because you cover that loss with our next bet.

Here is another excel spreadsheet that shows this nicely using a lay bet and 5% VIG. Both the roulette and the lay bets show a profit over a million rolls. The lay bet showed a $597,000 profit. I'm not sure where you guys are getting that a martingale system isn't mathematically sound. It isn't practical, but that wasn't the question.

https://docs.google.com/open?id=0B3eruTU5Xu_jOTg0YjgxMGMtMTk5Yi00NzllLWEzZGUtMGZiNzE0OTE2YTNm

Each individual bet has indeed the house advantage of 5.26% for roulette.

But each string of bets in a x-level Martingale is only ever going to win you 1 unit. So what P90 (and I) did was look at the expected value of that string. Either you win 1 unit of lose X^2 units. Whats the expected value? And it's not 5.26% for the complete string.

Your sheet for roulette is 40,134 spins long with one 15 long losing streak.... something you expect about once in 40,000 spins. As you used RAND, I've seen streaks up to 18 in there. In short, for the bank roll you've given yourself (1 million) and the max number of spins you wish to risk, I'm not surprised you've never gone bust. I don't see a million results here, but I suspect that might the number of items I seen.

This doesn't make a Martingale "Mathematically sound". You've shown only that it works in some experiments. Try the experiments with a "cap" on your bankroll (or amount willing to bet on a spin). I think you'll see more failures to show a profit.

Quote: thecesspit

Each individual bet has indeed the house advantage of 5.26% for roulette.

But each string of bets in a x-level Martingale is only ever going to win you 1 unit. So what P90 (and I) did was look at the expected value of that string. Either you win 1 unit of lose X^2 units. Whats the expected value? And it's not 5.26% for the complete string.

Your sheet for roulette is 40,134 spins long with one 15 long losing streak.... something you expect about once in 40,000 spins. As you used RAND, I've seen streaks up to 18 in there. In short, for the bank roll you've given yourself (1 million) and the max number of spins you wish to risk, I'm not surprised you've never gone bust. I don't see a million results here, but I suspect that might the number of items I seen.

This doesn't make a Martingale "Mathematically sound". You've shown only that it works in some experiments. Try the experiments with a "cap" on your bankroll (or amount willing to bet on a spin). I think you'll see more failures to show a profit.

ONly 40,000 - you are right, but I refreshed 25 times and kept track of the results. I'm not excel savy, but I am sure there is a way to tell the thing to run the sheet x number of times, and sum the results. But whatever.....

Like I said, it isn't practical.

But....

Perhaps I am missing something. For something to be mathematically sound in your mind, what do you think that means?

Proof by induction or deduction, rather than an example or simulation.

I've seen two different proofs that the Martingale does or does not work with infinite bank rolls and doubling. The biggest problem is the word infinity... things go wibbly at infinities.

I have no problem with simulation, as it uncovers interesting (*) behaviour's as you have shown. I did similar things with various progressions to see what they looked like. As I've said, have a play with your sums with a practical values of streaks.

(*) Interesting in the sense of discovering new stuff for yourself.