Topic: Does martingale really works? - page 79. (Read 123304 times)

No, you win 6.65 times to every 1 loss.
So you win 6.65 units and lose 7.21.

if it is some one have already bankrup all the dice site ...

Here's the formula for expected value:

EV = Possible outcome#1 * probability of outcome 1 + Possible outcome2 * probability of outcome2

Let's do the math with Doog's provided numbers:

Possible outcome#1:  +1 unit
Probability of that outcome:  99.16%

Possible outcome#2:  -127 units
Probability:  0.84%

Expected value = -.0752

There's no way to have that EV come out positive if the game has a house edge.

You can never have positive expectancy unless you're not in an independent series of bets... and then its up in the air because it depends on how you define high reward low risk opportunities.

Yep, I could.  If you went all in every time you bet to infinity, you are going to lose eventually.  Thus, negative expected value.


Yes, even with a house edge you can have a positive expected value(math someone?)  Like in my example.  20 losses, or if you had 200 losses to cover, and you only wanted to win 1 base bet.  Your expected value should be darn close to the base bet no?  Any math people out there, Doog you have been good at this, I could be wrong as well just figured

Sorry I was just going back to read your math. You made a valid optimization on using martingale to effectively increase expectancy if one were hell-bent to use it as their strategy in a dice game, maybe winning one jackpot and leaving in the black?. Bottom line is yes you can tweak an algorithm to offer a better expectancy but it will always be negative on a dice game. To effectively you martingale you would need to do it in a non-independent environment where the expected outcome depends on a previous state, and even that I would take my money and run because I would not want randomness to catch up to me which it will sooner or later in the long run. End of the day in my humble oppinion, experience mastering swings and support/resistance, or having inside info is really the only way to get an edge in trading/gambling.

Hmm doesn't that mean that #2 is actually better? 1 in every 7.65 bets is a loss, yet your risk is only 7.21 btc?

If you're playing a game with a house edge then you always have a negative expected value. Anything else is silly.  Consider a game where the site gave the player the edge.  Could you think of any way of betting that would give you a negative expected value?

That seems a bit rude.

Why don't you work it out and see for yourself?


That's your mistake. You lose one every 111 spins, so you recoup 111 BTC and lose 127 BTC. That's a negative expectation.


I was rounding to a single decimal place.

It's actually a little better:

>>> (1 - 0.505**7) * 100
99.1623942561664

>>> 1/0.505**7
119.38791100251332

So that's a 99.1624% chance of success, and a 1-in-119 chance of losing.

A 1-in-119 chance of losing 127 units is a negative expectation, not positive.

To break even, you need a 99.21875% chance of turning 127 into 128:

>>> 127/1.28
99.21875

This is where you are wrong.  What we are saying is you have a 99.1% chance of winning 1BTC.  Even when you try and win 2BTC, your odds get worse QUICK.  This is why Martingale fails eventually.  Because you might be up 50BTC-100BTC.  Well 1 streak of losses take you down 127BTC.  And if you double it again(which you would because you are playing martingale, you would be bankrupt.  Martingale works, but its just hard to hit the right amount of victory before you bust!

Your 2nd part, Martingale doesn't always have negative expectancy(I don't think).  If you were to play up to 20 hands, and you had enough to cover 20 losses.  The odds of you winning just 1 roll, you would have a positive expectancy...

I suspect your numbers are wrong. According to your math you can simply use the formula in #3 and bankrun any of these dice games.

#2 7.21 risk 1 reward & 86.94% outcome (You would need 7.65 spins to recoup one loss but it is less than the reward, so negative expectancy)
#3 127 risk reward 1 & 99.1% outcome (111 spins to recoup one loss, positive expectancy)

According to your change of success numbers looks like #3 is a good option. Are you sure its 99.1% chance of success? Mathematically and independent random walk with 50% chance or less martingale'd will always have negative expectancy. Again in the markets its different but arguably may be still considered random on a long enough timescale.

Why? Because even at 50% chance to win each roll you would need infinite funds to cover for long sequence of losses... Cutting it to a 7 series loss cutoff will affect your chance to win each roll and thus reduce your chance of success overall.

On 999dice I've used a martingale bot, it does work great. Much of the btc that I own is from it, I didn't bet my btc so it was all profit because they were gifts to me to play with. Martingale is best used for short terms or the house edge catches up to you and you lose it all.

Dooglus is correct.  Martingale used in very small doses is by far the best method to use.  I have been saying it since the beginning but people kept calling me martinfale....anyway its nice to have the numbers to back it up..

Thanks Doog!

Did you read my recent posts in this thread?

Suppose you have 3 choices:

1) bet 127 BTC at 98.2265625% to win 1 BTC with 98.2% chance of success
2) bet 7.21 BTC at 86.94% to win 1 BTC with 86.94% chance of success
3) bet a 1,2,4,...,64 BTC martingale sequence at 49.5% to win 1 BTC with 99.1% chance of success.

Which of these 3 is the best?

2 and 3 both risk an expected 7.21 BTC, but 3 wins more often
1 and 3 both risk a maximum of 127 BTC, but 3 wins more often

Isn't 3 clearly the best of the options?

No, there's no way to change the house edge by changing the size of your bet.

Books like Fortune's Formula have some good stories about this.

Martingale does not change expectancy (in fact you can try any MM/staking system, and you’ll reach the same conclusion). All martingale does is increase the percentage of positive outcomes, but the losses that do occur are commensurately bigger.

This holds true if trials are independent like in Casino games... however in the stock market trials are not independent.

However, in a market setting, it’s perfectly possible that some trade setups have a higher probability of success than others, and this justifies a higher bet size. (Analogy: That’s how blackjack pros beat casinos: higher bets when there are more high cards remaining in the deck, and the probability is in their favor). Another example might be where markets deliver a period that favors your system, causing wins to cluster, or losses to cluster when conditions are unfavorable (e.g. a trend following system prospers in a trending market, and loses money in a sideways market). Statisticians call this clustering phenomenon ‘serial correlation’.

In summary, then, the idea is to have higher position sizes for higher probability setups; but of course you need to know which setups are superior, before placing the trades. Where many traders go wrong with martingale is that they use it to try to recover recent losses. There is no mathematical basis for this, because the market takes no cognizance of the trader's personal P/L. It is simply gambling. Moreover, recent losses have no greater effect on eventual overall P/L than any historical losses.

However, all trading systems ultimately are a balance between maximizing income and minimizing risk of ruin (intolerable or irretrievable drawdown). Increasing position sizes causes risk of ruin to increase exponentially. Hence, while it’s justifiable to increase position size in favorable situations, the martingale progression (1,2,4,8,16,….) is probably too steep for most situations where serial correlation exists. The betting progression should remain approximately commensurate with the increase in probability. In other words, to justify a bet that’s 16x the original bet size, the probability of a successful trade should be ~ 16x as great as it was originally, which (for most systems) is very unlikely to be the case.

The problem becomes identifying higher probability setups, where the block sholes led us to conclude that the market follows a random walk. In that case even market setups become independent trails and we fall into the aformentioned category of a static expectancy over aggregrate trading over the life of a system(s). In other words, it's only a matter of time before you lose your shirt.. however the time may vary and may depend on your luck (another random walk).

doog, if everyone played dice like this, couldn't this have a disastrous effect on the house.

Let's say max bet is at 0.5% of the house bankroll.  If everyone turns into a martingale dicer.  They're effectively turning your house edge into a 0.0568% game.

We've determined (at least anecdotally) that a max bet % larger than the house edge is disastrous.  Wouldn't this hold true in this situation?

i'm thoroughly confused now.

Exactly as illustrated in #523, the bettor is expected to risk 7.21 btc on average with the martingale.
While on the other hand, if you put 127 bitcoin in a 98.2265625% bet, you are risking 127 btc.

No, because you're not betting 127 units every time.

Passionate calculations

What would be interesting to know is the percentage of the amount of BTC risked where doing the martingale has the same expected loss than playing the whole amount in one shot