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Topic: Martingale revisited - page 2. (Read 2536 times)

legendary
Activity: 3514
Merit: 1280
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May 25, 2020, 06:42:56 AM
with which you can catch a nice outlier before getting busted

You are thinking in correct direction

However, you only made so many rolls. As I suspect, to reveal a particular pattern of variance at so low win chances, you would have to literally make billions of bets and not bust in the process. If you bust, all your effort will be spent in vain. With win chances close to 50% we can be damn sure that the variance of variance (let's call it second-order variance) is less random that the first-order variance (this also confronts us with the question as to how truly random is random)

To cut to the chase, with lower multipliers you can safely assume that you will see fewer longer losing streaks than shorter ones. However, with small chances second-order variance may get out of hand, and you will see an entirely different picture. In practice, it basically means that you can't use the approach described here with very small odds as you are set to hit a very long losing streak kind of "all of a sudden" that would likely wipe you out
legendary
Activity: 3374
Merit: 2198
I stand with Ukraine.
May 25, 2020, 05:23:06 AM
With such strategy martingale is useless. Any martingale, with any "improvements"

Then you are on your own

Without any additional info, proper adjustment assumes linear change. This is not what I observed. If variance changes abruptly (like spikes exponentially), and you don't know by how much exactly, there is no way to lower the increase on loss that would match this spike. Finding it out empirically would probably mean busting in the process

I think, you can bust in the process anyway

Okay, let's now ignore traffic lights cuz we are still going to kick the bucket somewhere down the road, right?

For example, if you were betting with 40% win chance and increasing 160% on loss, you'd have to switch to 80% increase on loss when betting with 20% win chance; to 40% increase on loss when betting with 10% win chance, and so on

You are now suggesting what I already went through in my previous post

You are implicitly assuming a linear change. But if there is a linear change, then increasing 160% on loss with a 40% win chance should be the same in the long run as increasing 80% with a 20% win chance, which is otherwise known as six of one and half a dozen of the other. However, the change is non-linear. It basically means you can bust sooner with an 80% increase instead of 160% with a win chance only a few percentage points less. But if you want to check it for yourself, go for it since who am I to stop you?

In fact I tried it quite some times already, and the result was always a win. Below are some latest examples:







But I wouldn't try this strategy with BTC though. Smiley

I understand that 10% increase on loss is a bit too much for betting with 1% win chance, and that I was just lucky that I didn't bust. With 2% increase on loss it looks more like your strategy



with which you can catch a nice outlier before getting busted.
full member
Activity: 1750
Merit: 118
May 24, 2020, 09:13:28 PM
The good news is that it is the same as with any other complicated strategy out there that requires knowledge, expertise, and genuine understanding of how things hang in practice
That must be true, it all depends on the knowledge and experience practiced in the game. Maybe for you this strategy is useful, but not for me because of lack of knowledge when applying it.

Meanwhile, dice is my favorite game besides sports gambling. I can only win a few bets or more if the winning percentage is set at 60% or more. That is the best opportunity for me so far on dice. At least I can produce something even though its not much.

60 percent is or more is more winnable compare to lower percents but you can only earn less with it and if you want to earn big you need to adjust and up your bets but be careful because higher percent chance are also deadly  . you can loose imdiately upon upping your bets   .  so many newbies i see play that strat with bigger bets and sadly they end up easily  .   this is why i play only with low chance win rate but with martingale strat because this makes me last long and still be able to hit my targeted multi with a nice profit
legendary
Activity: 1974
Merit: 1150
May 24, 2020, 05:26:55 PM
The good news is that it is the same as with any other complicated strategy out there that requires knowledge, expertise, and genuine understanding of how things hang in practice
That must be true, it all depends on the knowledge and experience practiced in the game. Maybe for you this strategy is useful, but not for me because of lack of knowledge when applying it.

Meanwhile, dice is my favorite game besides sports gambling. I can only win a few bets or more if the winning percentage is set at 60% or more. That is the best opportunity for me so far on dice. At least I can produce something even though its not much.
legendary
Activity: 3514
Merit: 1280
English ⬄ Russian Translation Services
May 24, 2020, 02:33:34 PM
5 times in a row applying this technique, but I failed and lost. No matter how good the technique is, we also have to have a lucky day in gambling. Technique will only be a medium that helps us get luck

You should probably stay away from using martingale

By and large, it all comes down to properly balancing your chances of winning versus losing so that on a specific timeframe (a month, a year, a decade) your chances of coming through and making it would be higher than busting and losing your balance (actually, way higher, for a mighty safety margin). If you can't do that, martingale will be a losing strategy for you. Long story short, blindly applying it and hoping for the best will likely end in a disaster. The good news is that it is the same as with any other complicated strategy out there that requires knowledge, expertise, and genuine understanding of how things hang in practice

I believe that there is no perfect gambling technique, even though someone can prove that it works but that doesn't apply to everyone because we have to have luck

With a thoroughly thought-out martingale strategy you must be not so much lucky to win as unlucky to bust
legendary
Activity: 1974
Merit: 1150
May 24, 2020, 01:44:39 PM
Yes, but as i wrote above, if you for example spinning dice with 50 chance to win, you will not get in average "one win/one lose/ one win / one lose"
but rather you can get 10 loses in a row and 3 wins after and then again 2 loses. With such strategy martingale is useless. Any martingale, with any "improvements".
5 times in a row applying this technique, but I failed and lost. No matter how good the technique is, we also have to have a lucky day in gambling. Technique will only be a medium that helps us get luck.

I believe that there is no perfect gambling technique, even though someone can prove that it works but that doesn't apply to everyone because we have to have luck.
legendary
Activity: 3514
Merit: 1280
English ⬄ Russian Translation Services
May 24, 2020, 10:57:25 AM
With such strategy martingale is useless. Any martingale, with any "improvements"

Then you are on your own

Without any additional info, proper adjustment assumes linear change. This is not what I observed. If variance changes abruptly (like spikes exponentially), and you don't know by how much exactly, there is no way to lower the increase on loss that would match this spike. Finding it out empirically would probably mean busting in the process

I think, you can bust in the process anyway

Okay, let's now ignore traffic lights cuz we are still going to kick the bucket somewhere down the road, right?

For example, if you were betting with 40% win chance and increasing 160% on loss, you'd have to switch to 80% increase on loss when betting with 20% win chance; to 40% increase on loss when betting with 10% win chance, and so on

You are now suggesting what I already went through in my previous post

You are implicitly assuming a linear change. But if there is a linear change, then increasing 160% on loss with a 40% win chance should be the same in the long run as increasing 80% with a 20% win chance, which is otherwise known as six of one and half a dozen of the other. However, the change is non-linear. It basically means you can bust sooner with an 80% increase instead of 160% with a win chance only a few percentage points less. But if you want to check it for yourself, go for it since who am I to stop you?
legendary
Activity: 3374
Merit: 2198
I stand with Ukraine.
May 24, 2020, 09:57:49 AM
~
I have no explanation how it is ever possible, but winning chances below 37% bring about an exponential surge in variance, and martingale no longer remains safe. I can't come up with any plausible reason for this phenomenon but I've read about it before and can confirm it (call it broscience or a gambling street wisdom of sorts, or whatever)

I think it's simply a question of proper adjustment. If you lowered your "increase on loss" exactly in accordance with how the win chance was lowered, your strategy would work just fine

Without any additional info, proper adjustment assumes linear change. This is not what I observed. If variance changes abruptly (like spikes exponentially), and you don't know by how much exactly, there is no way to lower the increase on loss that would match this spike. Finding it out empirically would probably mean busting in the process

I think, you can bust in the process anyway Smiley, but what I meant was that with a proper adjustment you could create a betting strategy that would work exactly like your previous one in regards to the probability of busting.

For example, if you were betting with 40% win chance and increasing 160% on loss, you'd have to switch to 80% increase on loss when betting with 20% win chance; to 40% increase on loss when betting with 10% win chance, and so on.
legendary
Activity: 2436
Merit: 1849
Crypto for the Crypto Throne!
May 21, 2020, 05:01:32 PM
You can lose 100 bets in a row on a 50% win chance, but the probability of such an outcome would be infinitesimal (read, you won't see it during your lifetime), and thus it is discarded for entirely practical reasons as a purely hypothetical construct

Yes, but as i wrote above, if you for example spinning dice with 50 chance to win, you will not get in average "one win/one lose/ one win / one lose"
but rather you can get 10 loses in a row and 3 wins after and then again 2 loses. With such strategy martingale is useless. Any martingale, with any "improvements".
legendary
Activity: 3514
Merit: 1280
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May 21, 2020, 04:57:18 PM
My wise point was that martingale isn't loosing strategy if you have unlimited resources. Mathematically unlimited, i mean. Endless

Your "wise" point would hold only on an infinite timeframe (and then it would remain to be seen)

Yep, i meant exactly this. It's being seen from my quote

Then it is irrelevant to the point in question

We are not talking about infinite bankrolls on infinite timeframes. We are talking about a very determined range of concrete parameters, e.g. base bet, increase on loss, multiplier, and, yes, timeframes too. This puts us in a very probabilistically determined environment where statements like "or you will lose 100 times in a row. Or 10, or 2" simply make no sense. You can lose 100 bets in a row with a 50% win chance, but the probability of such an outcome would be infinitesimal (read, you won't see it during your lifetime), and thus it is discarded for entirely practical reasons as a purely hypothetical construct
legendary
Activity: 2436
Merit: 1849
Crypto for the Crypto Throne!
May 21, 2020, 04:12:24 PM
My wise point was that martingale isn't loosing strategy if you have unlimited resources. Mathematically unlimited, i mean. Endless

Your "wise" point would hold only on an infinite timeframe (and then it would remain to be seen)

Yep, i meant exactly this. It's being seen from my quote  Smiley

If your odds of losing are 1 to 100, then you'll be losing only once every 100 rolls or series of rolls on average (what average actually amounts to in a real gambling environment, you can learn here)

or you will lose 100 times in a row. Or 10, or 2. You can't be sure. of course, with 1/100 chances it's east to say something pathos, but in gambling we mostly faces with 40/50, 50/50, 22/78 or something like that. Noone plays on dice with winning chance around 1/100 (or loosing, whatever). And with such probabilities you can;t be sure about "average"  Smiley


Whether you are talking about a probability of a single roll or a series of rolls, you can't get around this. With this in mind, if your chances of hitting a losing streak of 20 rolls in a hundred years are not high, it essentially means that you are not likely to see such a series in a hundred years (you still can, but that would be called ill luck). As simple as it gets

As i said above, it;s mostly works with great difference in probabilities, when one event is significant more to occur than another.
legendary
Activity: 3514
Merit: 1280
English ⬄ Russian Translation Services
May 21, 2020, 05:50:31 AM
Since I was redirected (by you Smiley ) to this thread, I'll repeat my question here.

Is it possible to know exactly how big your highest bet during the whole experiment was?

With that information

1. I would estimate the minimum bankroll needed for your strategy to work

Bankroll as such is irrelevant

What you need is the capacity to endure a long enough losing streak for a given multiplier.

Which means, you need a big enough bankroll. So, it's not irrelevant actually

It is irrelevant as long as your base bet can be any. In other words, "a big enough bankroll" is only as big as you base bet is. Put shortly, "big" is relative here, and depends on very specific parameters

For example, at ~40% win chance it is pretty much a safe bet if you can survive 35 losing rolls in a row. If I remember correctly, at 37% win chance I've seen a losing streak of 30 rolls only once.

That would result in a pretty big final bet

Your "final" bet will always be half of your balance (or something to that tune), when you stake all of what is left. Indeed, I calculated the rolls in such a way that I would be able to win back everything with this last all-or-nothing roll and earn as much (actually, more as my win chance was below 50%)

I have no explanation how it is ever possible, but winning chances below 37% bring about an exponential surge in variance, and martingale no longer remains safe. I can't come up with any plausible reason for this phenomenon but I've read about it before and can confirm it (call it broscience or a gambling street wisdom of sorts, or whatever)

I think it's simply a question of proper adjustment. If you lowered your "increase on loss" exactly in accordance with how the win chance was lowered, your strategy would work just fine

Without any additional info, proper adjustment assumes linear change. This is not what I observed. If variance changes abruptly (like spikes exponentially), and you don't know by how much exactly, there is no way to lower the increase on loss that would match this spike. Finding it out empirically would probably mean busting in the process
legendary
Activity: 3374
Merit: 2198
I stand with Ukraine.
May 21, 2020, 05:28:55 AM
Since I was redirected (by you Smiley ) to this thread, I'll repeat my question here.

Is it possible to know exactly how big your highest bet during the whole experiment was?

With that information

1. I would estimate the minimum bankroll needed for your strategy to work

Bankroll as such is irrelevant

What you need is the capacity to endure a long enough losing streak for a given multiplier.

Which means, you need a big enough bankroll. So, it's not irrelevant actually. Smiley


For example, at ~40% win chance it is pretty much a safe bet if you can survive 35 losing rolls in a row. If I remember correctly, at 37% win chance I've seen a losing streak of 30 rolls only once.

That would result in a pretty big final bet. Even with 100% increase on loss, that bet would be over 10 DOGE, given that the base bet was 0.00000001 DOGE. But, as far as know, your increase on loss was higher than 100%, right?


I have no explanation how it is ever possible, but winning chances below 37% bring about an exponential surge in variance, and martingale no longer remains safe. I can't come up with any plausible reason for this phenomenon but I've read about it before and can confirm it (call it broscience or a gambling street wisdom of sorts, or whatever)

I think it's simply a question of proper adjustment. If you lowered your "increase on loss" exactly in accordance with how the win chance was lowered, your strategy would work just fine.
legendary
Activity: 3514
Merit: 1280
English ⬄ Russian Translation Services
May 21, 2020, 05:01:42 AM
My wise point was that martingale isn't loosing strategy if you have unlimited resources. Mathematically unlimited, i mean. Endless

Your "wise" point would hold only on an infinite timeframe (and then it would remain to be seen)

However, as soon as we are talking about limited amounts of time and time spans like a month, a year, etc, it all comes down to evaluating probabilities. If your odds of losing are 1 to 100, then you'll be losing only once every 100 rolls or series of rolls on average (what average actually amounts to in a real gambling environment, you can learn here)

Whether you are talking about a probability of a single roll or a series of rolls, you can't get around this. With this in mind, if your chances of hitting a losing streak of 20 rolls in a hundred years are not high, it essentially means that you are not likely to see such a series in a hundred years (you still can, but that would be called ill luck). As simple as it gets
legendary
Activity: 2436
Merit: 1849
Crypto for the Crypto Throne!
May 20, 2020, 05:12:27 PM
As much as bets are independent of each other, exactly the same can be said about the overall probability of losing. No matter for how long you have been rolling, the odds of busting remain as they were before you started rolling, and they don't increase despite what so many people erroneously assume (I call it the Gambler's Fallacy in reverse). Further, even if the outcomes are independent of each other, this probability does nevertheless depend on the number of rolls, surprise!

Ehhhhm, you just take my idea in other words.  Sad
Maybe i was not so clear but in fact, you're telling the same. My wise point was that martingale isn't loosing strategy if you have unlimited resources. Mathematically unlimited, i mean. Endless.

You just betting and betting, and at one moment of time you will win (it's like Infinite monkey theorem). But in real world your resources will end much faster. Because of the reason that probability for each event/roll still the same due to any time, you can't be sure when you will win. Something like that.
member
Activity: 773
Merit: 17
May 20, 2020, 02:59:52 PM
To make things clear right at the start, I know that martingale is a losing strategy in the long term as there is no way to beat the house and its edge if only by chance alone (or by exploiting a bug in the system). And since martingale effectively removes the chance part from the equation, it is set to fail in the end

With that said, though, it is an "old-school" martingale which is a sure way to lose all but what about using martingale when you constantly lower your chances to lose at each red streak by extending the number of losing rolls till you go bust? I don't know if it can actually help but it is certainly worth discussing here

Obviously, it can be done by "reinvesting" everything we earned at previous rolls without changing any other setting (like odds, initial bet amount, increase, etc) but we are not necessarily limited to only that. For example, we could continually add to our balance at each roll, thereby postponing our final moment until it gets lost in the vague future

Does it change anything even if it doesn't make a lot of sense as a strategy on its own?
Here you can read about the pitfalls of Martingale strategy: http://sportstatist.com/the-pitfalls-of-martingale-money-management-strategy/
legendary
Activity: 3514
Merit: 1280
English ⬄ Russian Translation Services
May 20, 2020, 01:27:06 PM
To make things clear right at the start, I know that martingale is a losing strategy in the long term as there is no way to beat the house and its edge if only by chance alone (or by exploiting a bug in the system). And since martingale effectively removes the chance part from the equation, it is set to fail in the end

Does it change anything even if it doesn't make a lot of sense as a strategy on its own?

Martingale is losing strategy because few reasons: probability of event doesn't depend on result from previous one

Has it never occurred to you that it works both ways?

As much as bets are independent of each other, exactly the same can be said about the overall probability of losing. No matter for how long you have been rolling, the odds of busting remain as they were before you started rolling, and they don't increase despite what so many people erroneously assume (I call it the Gambler's Fallacy in reverse). Further, even if the outcomes are independent of each other, this probability does nevertheless depend on the number of rolls, surprise!
legendary
Activity: 2436
Merit: 1849
Crypto for the Crypto Throne!
May 20, 2020, 10:33:44 AM
To make things clear right at the start, I know that martingale is a losing strategy in the long term as there is no way to beat the house and its edge if only by chance alone (or by exploiting a bug in the system). And since martingale effectively removes the chance part from the equation, it is set to fail in the end

Does it change anything even if it doesn't make a lot of sense as a strategy on its own?

Martingale is losing strategy because few reasons: probability of event doesn't depend on result from previous one. You can lose 100 times in a row, or win. So, you can't be sure on what iteration you will win. But our money is limit resource, so you can't lose for example 1000 time in meaning of available funds. But in meaning of probability - you can  Smiley
So, you with bigger chance lose all your limit, than win with this strategy. But from mathematical point, it's normal strategy.

And yeah, your ideas to "update" martingale is even worse than original strategy, my tho.
legendary
Activity: 3514
Merit: 1280
English ⬄ Russian Translation Services
May 20, 2020, 10:02:21 AM
Since I was redirected (by you Smiley ) to this thread, I'll repeat my question here.

Is it possible to know exactly how big your highest bet during the whole experiment was?

With that information

1. I would estimate the minimum bankroll needed for your strategy to work

Bankroll as such is irrelevant

What you need is the capacity to endure a long enough losing streak for a given multiplier. For example, at ~40% win chance it is pretty much a safe bet if you can survive 35 losing rolls in a row. If I remember correctly, at 37% win chance I've seen a losing streak of 30 rolls only once. I have no explanation how it is ever possible, but winning chances below 37% bring about an exponential surge in variance, and martingale no longer remains safe. I can't come up with any plausible reason for this phenomenon but I've read about it before and can confirm it (call it broscience or a gambling street wisdom of sorts, or whatever)

2. It would be clear whether we could use your strategy for betting with BTC(taking into account the restrictions on the max bet/profit on various gambling sites)

I can tell you straightaway that it is not possible. I tried and I failed miserably, losing one hell of a lot of money
legendary
Activity: 3374
Merit: 2198
I stand with Ukraine.
May 20, 2020, 09:02:40 AM
Since I was redirected (by you Smiley ) to this thread, I'll repeat my question here.

Is it possible to know exactly how big your highest bet during the whole experiment was?

With that information

1. I would estimate the minimum bankroll needed for your strategy to work;

2. It would be clear whether we could use your strategy for betting with BTC(taking into account the restrictions on the max bet/profit on various gambling sites).
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