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Am I right that when I want to make 1 btc profit from eg. 7 btc it is better to start doing this with a martingale strategy from 1 btc instead of going all in on a lower odd to earn this. Since with this strategy you wager on average less (1, 3 or 7 btc) than with the all-in strategy (7btc) I should on average lose less so I should have a higher chance making 8 btc from 7 btc using martingale than from an all-in bet?

felt bored and calculated it
so imagine you have 7 btc and want to make it 8 btc and there is a house edge of 2%
chance you'll be able to do this with martingale is 86.375%
chance you'll be able to do this with a bet of 7 btc is 85.75%
just a difference for exactly the same outcome

if you have 1000BTC bankroll and you have only 1satoshi as base bet then you can win with martingale :)

I started with a .1BTC bank roll and with the power of martingale

https://i.imgur.com/eKQUJ59.jpg

TBH, you could've also lost all of that in a series of losses. Martingale always make you feel good because you think you have a winning strategy when in fact all it does is convince you to keep playing until you lose it all. That's why it's so profitable to run a large dice site (and why Stunna can lose 200 BTC due to exploits without blinking an eye).

I started with a .1BTC bank roll and with the power of martingale

https://i.imgur.com/eKQUJ59.jpg

Am I right that when I want to make 1 btc profit from eg. 7 btc it is better to start doing this with a martingale strategy from 1 btc instead of going all in on a lower odd to earn this. Since with this strategy you wager on average less (1, 3 or 7 btc) than with the all-in strategy (7btc) I should on average lose less so I should have a higher chance making 8 btc from 7 btc using martingale than from an all-in bet?

felt bored and calculated it
so imagine you have 7 btc and want to make it 8 btc and there is a house edge of 2%
chance you'll be able to do this with martingale is 86.375%
chance you'll be able to do this with a bet of 7 btc is 85.75%
just a difference for exactly the same outcome

Don't misunderstand me, martingale is a losing strategy, just like every strategy, the house wins in the long run due to the edge.
but it's a better (or less worse) strategy than going all in, that's all i said

I'm not misunderstanding you, you're misunderstanding me. Your calculations are wrong, because martingale doesn't give the same probabilities for different starting amounts.

i think he/she just mean that with 7BTC, if you want to make it to 8BTC it would be better to make a martingale until you reach 1BTC profit instead of yolo bet

Martingale works great until it doesn't.  I agree with waterpile that the key is to set your win target and stick to it.  That seems to be what trips up most gamblers.

I like the idea of comparing it with one all in bet.  It seems that dooglus did something along these lines awhile back.  IIRC he found that you were actually better off betting smaller amounts at high odds rather than one all in bet.  This makes sense if you think about it, your expected value is the same but your amount at risk is lower.  But I'm not sure if that extends all the way to martingale.  Your calculations suggest it might.

Regardless, good luck!

felt bored and calculated it
so imagine you have 7 btc and want to make it 8 btc and there is a house edge of 2%
chance you'll be able to do this with martingale is 86.375%
chance you'll be able to do this with a bet of 7 btc is 85.75%
just a difference for exactly the same outcome

Don't misunderstand me, martingale is a losing strategy, just like every strategy, the house wins in the long run due to the edge.
but it's a better (or less worse) strategy than going all in, that's all i said

I would also contribute something to this thread. We have a player "snmp" on www.bitdice.me (http://www.bitdice.me) who is playing martingale for more than 2 weeks now and he keeps on winning. He has taken 2 loosing series but he has a stop in place at around 30 btc. But that being said, he is almost at 100BTC in profit.

NOt really a fact that martingale is a winning strategy, atleast he know when to stop, Martingale is a strategy but it is not always a winning strategy, all we need to do is the tendency to stop whenever we are in profit, otherwisethe losing streak will strike and bust our bankroll

No no im not saying it is a winning strategy but it can work very well :)

Your calculations are wrong, with a 0% house edge doing it with martingale is a 87.5% and doing it all in with 7 btc would be also 87.5% so it doesnt matter wheter you use martingale or all in, the only thing you win is time

Your calculations are wrong, with a 0% house edge doing it with martingale is a 87.5% and doing it all in with 7 btc would be also 87.5% so it doesnt matter wheter you use martingale or all in, the only thing you win is time

How can you prove that his calculation is wrong if your calculation shown to us is from 0% house edge?

No idea why XinXan talks about 0% house edge, but the 86.375% numbers is slightly off under the 2% house edge assumption.

The chance to win a x2 bet under 2% house edge is 49%.
The chance to lose it thrice is 51% ^ 3 = 13.2651%.
So the chance to make it 8 btc from 7 btc successfully is 1 - 13.2651% = 86.7349%.

The number 85.75% is correct by the way.

Your calculations are wrong, with a 0% house edge doing it with martingale is a 87.5% and doing it all in with 7 btc would be also 87.5% so it doesnt matter wheter you use martingale or all in, the only thing you win is time

Am I right that when I want to make 1 btc profit from eg. 7 btc it is better to start doing this with a martingale strategy from 1 btc instead of going all in on a lower odd to earn this. Since with this strategy you wager on average less (1, 3 or 7 btc) than with the all-in strategy (7btc) I should on average lose less so I should have a higher chance making 8 btc from 7 btc using martingale than from an all-in bet?

Martingale makes it easier to win a small amount, but it also makes it easier to lose a big amount than a flat betting strategy, and that is where the problem lies.

Here is a math problem to find out why martingale isn't the best idea. Whatever game you are playing, what are the odds of winning 10 games in a row, and what are the odds of losing 10 games in a row? Very close, if not the same. How much do you win when you win 10 in a row and how much do you lose when you lose 10 in a row? Big difference, and the problem.

The reality is that it's just not worth it in the long run, that big losing streak will come, and it will wipe out of all of your previous winnings and then some.

please read the OP. I'm comparing an all in method with a martingale one, with the same goal (eg 8 btc from 7) and then the all in method is worse than the martingale...

A major difference between OP's strategy and normal martingale players' strategy is that OP will only do 1 round of bets. When he wants to make 1 btc profit, he simply set his base bet to 1 btc and stop right after achieving his goal.
This way he effectively lower his wagered amount from 7 btc to about 3 btc (50%*1btc+25%*3btc+25%*7btc) and as a result has a higher probability of making 1 btc than the all-in strategy.

But if he set his base bet to, say 0.0001 btc, he will need to make 10000 rounds of bets to make the 1 btc, which will have a very high wagered amount and much lower success probability than the all-in strategy.

with martingale you end on 0 eventually.
that tactic works only with infinity amount of money

no, the total average amount wagered would be exactly the same. Trying to earn 1 btc in 1 martingale bet loses on average x, trying to earn 0.0001 btc loses on average 0.0001*x, trying this for 10.000 times (to have 1 btc profit) will give an average loss of 1*x too. 1 martingale or 10000 small martingales, it's all the same. Average wagered amount will be the same too taking into mind that the average loss is a % of the total amount wagered. Please think before you say something

Let's say you set the base bet to 0.0001 btc.
For each round of martingale, you will wager 0.0001 for 50% chance, 0.0003 for 25% chance, 0.0007 for 12.5% chance, ... , 3.2767 for 0.00305% chance, 6.5535 for 0.00305% chance, right?
So you will on average wager 0.0016 btc per round, and a total of 16 btc assuming you didn't hit a 16-loss streak, right?

with martingale you end on 0 eventually.
that tactic works only with infinity amount of money

My record was lost in 15 consecutive times , at this moment I realized we never could win dealer with any method

Yes.

And you're the first person who responded in such an open manner.

Everyone else just tells me I'm wrong.  :)

The trick is to split up your bet (the amount you were going to risk in a single bet) into a series of amounts which sum to a the same, and which form a sequence such that you can bet the smallest amount, and if it wins, you make the same as if you bet the whole amount at 49.5% (so you'll be betting with a smaller chance, and higher payout multiplier).  And if it loses, you want betting the 2nd amount to cover the first loss and make the same net profit.  Etc.

If you can find such a sequence (and you always can, though it can involve some hairy math depending on the length of the sequence you're looking for) then the amount you expect to risk is less than your whole amount (since there's a non-zero chance that you will win before the last bet, and stop at that point), and so the amount you expect to lose, being 1% of the amount you risk, is less than when you make the single bet.

Here's a very simple example:

you have 1 BTC and want to double it.

* you could bet it all at 49.5%, and succeed in doubling up with probability 0.495

* or you could bet 0.41421356 BTC at 28.99642866% with payout multiplier 3.41421356x, and if you lose, bet the rest at the same chance.  If you win either bet, you double up, else you lose.  Your chance of doubling up is 0.4958492857 - a little higher than the 0.495 you have with the single bet.

Cool, huh?

That's breaking the single bet up into a sequence of length 2.

If you break it up into more, smaller bets, then the probability of success increases further.

The more steps, the closer to 0.5 your probability of success gets.

You'll be limited by real-life barriers, like the invisibility indivisibility of the satoshi, and the limit of 4 decimal places on the chance at JD.  But in theory you can get arbitrarily close to 0.5.  I think.  :)

I cannot believe I'm saying this but I think you might be right.

I went ahead and played around with a case where the player starts at 1 wants to move to 2 by making two bets (of any size between 0-1 inclusive). Hence I went to go graph the function to see if it was true and I got this:

https://www.desmos.com/calculator/obkfifgbnl

Where y = probability of succeeding
and d = the value of the first initial bet

Notice how at both d = 0 and d= 1 the probability is 0.495 as expected (as you are either betting nothing then 1 or 1 then nothing and both are equivalent cases). And in between you get a probability higher than the 0.495 offered for the single bet.

I've tried a few set values for cases where you split your value up to more than two and you do get a better result. I can only theorise that this is because as your bet size approaches 0 with the number of bets approaching infinity your expectation approaches 1.

However, what I do not understand at the present is why this is so. I almost fell out of my chair when the numbers came out (I checked like 6 times), as it's inferring that you can get better than what the house 'technically' offers. The problem with this is that your expectation is better than just flat betting and logically that doesn't make sense. Both should have the same expectation.

I'm going to mull it over.

When you bet your whole bankroll in a single bet, you expect to lose 1% of it.

When you split it up and bet the pieces in order from smallest to biggest, and stop when any bet wins you often don't end up betting the whole bankroll, and so you expect to lose 1% of less than the whole bankroll.

By splitting it up you reduce the amount you expect to bet, and so you reduce the amount you expect to lose.

does playing using martingale system not profitable in the long run?

because from my experience, my bet always end in loss if i use it for more than 3 hours..>.<

This is also false. The distributions are not the same, even though the end effect is the same.

Consider one family with a 1.80m tall father, 1.60m tall mother, and a 1.70m tall son. The "expected height" of the family is 1.70m.

Consider another family with a 2.00m tall father, a 1.60m tall mother, and a 1.50m tall child. The Expected height is also 1.70m, but obviously the distribution is different.

It's the same thing with gambling strategies. The EV is the same, but distribution is different. In the case that you want to win 1BTC and EXACTLY 1BTC, no more, no less, with EXACTLY a 7BTC bankroll*, there are two optimal routes**:

1. Bet all-in at high odds
2. Martingale with a BASE BET of 1BTC.

These result in the same odds for earning exactly 1 BTC. Any other variation of the martingale would result in a lower chance.

*Any difference in ANY of the variables would change this conclusion.

**There are possibly other routes that would arrive at the same % chance, but these are not within the topic of martingale vs all-in.

Exactly. The OP can't do math, yet keeps telling others they're wrong. eh.

dig up an old conversation me a dooglus had about losing 'less' by betting less (ie. rather than all in betting, splitting your bets into n portions where you aim to get to x as a value then stop)

does playing using martingale system not profitable in the long run?

because from my experience, my bet always end in loss if i use it for more than 3 hours..>.<