martingale math - what say ye?

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!