Does martingale really works?

[Berau]

Martingale=Martingay. Famous bitcoiner saying.

[deisik]

I've had varied success with martingale, sometimes it works, sometimes it doesnt. You just got to know when to stop it, and it should work out just fine.

What you just said is just like saying "I've had varied success with 1% bets. Sometimes I win and make profit and others I don't." A system either works (meaning has a 100% success rate) or it doesn't. There is no in between. Otherwise it's no different than rolling the dice (no pun intended) and hoping you won on that roll.

The system works if it consistently brings profit, this is the only viable criterion. Whether it has a 100% success rate or just 1% (yes, one percent) is ultimately irrelevant...

martingale will bring you a disaster. how many your balance will empty if you meet with long streak. i think it will save if you bet 100 btc with base bet 1 satoshi. we must use other strategies if want to save margin. if you using martingale surely your balance runs out instantly some day.

I wasn't referring to martingale in particular. In fact, I had been thinking about trading when I wrote my reply. 100% success rate means that all your trades are profitable there, which is obviously not the case in real life. And with just one trade you can cover losses from a train of losing ones, e.g. if you average down. This makes it look like martingale to a degree (and as dangerous)...

[steven.G999]

I've had varied success with martingale, sometimes it works, sometimes it doesnt. You just got to know when to stop it, and it should work out just fine.

What you just said is just like saying "I've had varied success with 1% bets. Sometimes I win and make profit and others I don't." A system either works (meaning has a 100% success rate) or it doesn't. There is no in between. Otherwise it's no different than rolling the dice (no pun intended) and hoping you won on that roll.

The system works if it consistently brings profit, this is the only viable criterion. Whether it has a 100% success rate or just 1% (yes, one percent) is ultimately irrelevant...

martingale will bring you a disaster. how many your balance will empty if you meet with long streak. i think it will save if you bet 100 btc with base bet 1 satoshi. we must use other strategies if want to save margin. if you using martingale surely your balance runs out instantly some day.

lol who wants to bet 1 satoshi if someone has 10btc, too long to make a profit

not a few days to balance runs out, but maybe only a few hours

[picolo]

I have tried some times and I can say I'm in loss with martingale, so it's not for me

its better to do all in on 50 % chance as doing martingale makes the house edge bigger when you roll bunch of times. i never use this strategy

Apparently this is not true. You can actually "lower" the house edge if you bet with a certain strategy other than 50% bet. There was a long discussing on previous pages.

It's not really lowering the house edge, per se, but rather exploiting the rounding of numbers. Some sites have already fixed that issue, though, and it relies on very specific sets of numbers to work even on those that haven't fixed it. With that said, it's less of a "beating the house" and more of an exploit than anything.

Oh I see. I thought it had something to do with the fact that you on average reach the goal earlier before wagering all of the 1 BTC and thus incurring less house edge.

That post I made with all the math I think has to be off. I spent a lot of time thinking about it and I don't see how raising bet amounts (without changing the % chance) could possibly affect the house edge. I just don't know where I went wrong with it...

I guess it's possible that it's correct. It just doesn't make logical sense.

Yes, it boggles my mind too. My explanation made the most sense in my head, but could be dead wrong.

I'll probably need to get in touch with Dooglus and see his opinion on this. He's pretty solid when it comes to figuring these things out and if he has time, should be able to see what's wrong.

If you bet your whole bankroll on a single bet in a 1% house edge casino, your expected loss is 1% of your bankroll.

If instead you split your bankroll up into a series of bets, starting small and increasing on loss, you can reach the same goal with a slightly higher chance of success. You didn't change the house edge, you just expect to risk less, and so expect to lose less

For example, the guy earlier talking about how much you need to have a 99% chance of winning $50 in a casino.

Let's assume the house edge is (100/37)% like it is in roulette.

That means that the probability of winning a bet times the payout for that bet is 36/37.
"chance * payout = 36/37"

For example, when you bet on red, the payout is 2x. The chance of winning is 18/37 (there are 18 reds and 37 numbers). 2 times 18/37 is 36/37.
When you bet on a single number, the payout is 36x. The chance of winning is 1/37. 36 times 1/37 is 36/37.
And it's the same for all the possible bets.

So if we want a 99% chance of winning $50 from a single bet on roulette:

  chance = 0.99
  chance * payout = 36/37
  payout = 36/37/0.99 = 0.9828x

So if we want a 99% chance of winning a single bet, the house will pay us back less than our stake if we win. You can't have a 99% chance of making a profit on a single bet when the house edge is bigger than 1%.

So we're going to have to split the bet into a two-step martingale sequence.

It turns out that you only need $8839 (*) to have a 99% chance of winning $50, and here's how:

We'll bet at 90% (we have to imagine they offer any percentage you like, like dice sites do).

The 90% chance bet has a payout of 36/37/0.9 = 1.081081x

First bet $616.67 at 90%.
If it hits, we get 1.081081 times 616.67 back.
1.081081 * 616.67 = 666.67 and we've made our $50 already, so stop.

If we lose that first bet, bet the remaining 8839 - 616.67 = $8222.33 at 90%.
If it hits, we get 1.081081 times 8222.33 back.
1.081081 * 8222.33 = 8889.00, and if we take off the 616.67 we lost on the first bet we get 8272.33, and we've made our $50 again.

So if we hit either the first or 2nd bet, we're $50 up, and if we lose both we're $8839 down.

There's a 10% chance of losing each bet, so a 1% chance of losing them both, so a 99% chance of winning one of them.

ie. we only need $8839 to have a 99% chance of winning $50 in a casino game with the same house edge as single-zero roulette.

Note that if we put all $8839 on a single bet to win $50, the payout would need to be (8839+50)/8839 = 1.00566x
and so the chance of winning would be (36/37)/((8839+50)/8839) = 96.75%

So again, we see that trying to win $50 starting with $8839 has a lower chance (96.75%) if you make a single bet and a higher chance (99%) if you split it into two bets, and do a mini 2-step martingale progression. And that's because 90% of the time you only have to make the small first bet, and so you're betting less, and so on average you lose less.


(*) set p = 36.0/37/0.9 - 1, then the amount you need is 50*(2*p + 1)/(p*p) = $8838.888888

(The two bet sizes are really 616.66666 recurring and 8,222.22222 recurring)

Actually an interesting read ! You pushed the calculations pretty far.
we only need $8839 to have a 99% chance of winning $50 in a casino game with the same house edge as single-zero roulette.

[dooglus]

we only need $8839 to have a 99% chance of winning $50 in a casino game with the same house edge as single-zero roulette.

The 'only' was in reference to the point I was replying to:

in casino i hear a report from scientist that you need 170.000 $ to have a 99% chance to win 50 $
which is insane

He claimed you need $170k, but you 'only' need $9k...

If there was no house edge, you would 'only' need $5k I presume.

payout = (stake+50)/stake
chance = 99

payout*chance = 100
(stake+50)/stake * 99 = 100
99.(stake+50) = 100.stake
stake = 99*50 = 4950

OK, so I was close. You 'only' need $4950.

[ranlo]

we only need $8839 to have a 99% chance of winning $50 in a casino game with the same house edge as single-zero roulette.

The 'only' was in reference to the point I was replying to:

in casino i hear a report from scientist that you need 170.000 $ to have a 99% chance to win 50 $
which is insane

He claimed you need $170k, but you 'only' need $9k...

If there was no house edge, you would 'only' need $5k I presume.

payout = (stake+50)/stake
chance = 99

payout*chance = 100
(stake+50)/stake * 99 = 100
99.(stake+50) = 100.stake
stake = 99*50 = 4950

OK, so I was close. You 'only' need $4950.

Wouldn't all of this theoretically mean, though, that with $9k, you can make infinite money? If that's enough to make $50, then now you have $9050. Then you go up by 50 and have $9100, and continue for infinity.

[deisik]

we only need $8839 to have a 99% chance of winning $50 in a casino game with the same house edge as single-zero roulette.

The 'only' was in reference to the point I was replying to:

in casino i hear a report from scientist that you need 170.000 $ to have a 99% chance to win 50 $
which is insane

He claimed you need $170k, but you 'only' need $9k...

If there was no house edge, you would 'only' need $5k I presume.

payout = (stake+50)/stake
chance = 99

payout*chance = 100
(stake+50)/stake * 99 = 100
99.(stake+50) = 100.stake
stake = 99*50 = 4950

OK, so I was close. You 'only' need $4950.

Wouldn't all of this theoretically mean, though, that with $9k, you can make infinite money? If that's enough to make $50, then now you have $9050. Then you go up by 50 and have $9100, and continue for infinity.

Your chances to win are still not 100%, and if you lose (you will eventually), you will lose much more than 50$, namely, your whole stake (as I got it)...

[dooglus]

Wouldn't all of this theoretically mean, though, that with $9k, you can make infinite money? If that's enough to make $50, then now you have $9050. Then you go up by 50 and have $9100, and continue for infinity.

Well, no.

You have a 99% chance of making $50 and a 1% chance of losing $9k.

On average you're going to win 99 times and lose once every 100 times you try.

The 99 wins will make you 99*50 = $5k.
The 1 loss will lose you $9k.
So every 100 times you try you'll make a net loss of $4k - on average.

Sometimes you'll win 100 out of 100 tries. Other times you'll lose 2 or 3 times out of 100 tries. But on average you'll lose 1 in 100 tries.

I'm by no means trying to say "use martingale, it works and lets you beat the house".

All I'm saying is that martingale, used carefully, can give you a slight advantage over just going all-in. All it means is you'll probably lose a bit slower, or a bit less. It can't change you from having a negative expectation to having a positive one.

Edit: for a real-world demonstration... Just-Dice allows you to martingale from a single satoshi all the way up to around 100,000 CLAMs. ie. the biggest bet is 10,000,000,000,000 (ten trillion) times bigger than the smallest bet. Even with such a wide range of bet values, and with no rules against Martingale betting, the players still lose:

[GannickusX]

Wouldn't all of this theoretically mean, though, that with $9k, you can make infinite money? If that's enough to make $50, then now you have $9050. Then you go up by 50 and have $9100, and continue for infinity.

Well, no.

You have a 99% chance of making $50 and a 1% chance of losing $9k.

On average you're going to win 99 times and lose once every 100 times you try.

The 99 wins will make you 99*50 = $5k.
The 1 loss will lose you $9k.
So every 100 times you try you'll make a net loss of $4k - on average.

Sometimes you'll win 100 out of 100 tries. Other times you'll lose 2 or 3 times out of 100 tries. But on average you'll lose 1 in 100 tries.

I'm by no means trying to say "use martingale, it works and lets you beat the house".

All I'm saying is that martingale, used carefully, can give you a slight advantage over just going all-in. All it means is you'll probably lose a bit slower, or a bit less. It can't change you from having a negative expectation to having a positive one.

But you said that with half of 9k you had 99%, with 9k you should have more, am i right?

[dooglus]

But you said that with half of 9k you had 99%, with 9k you should have more, am i right?

How much you need to have a 99% chance of winning $50 depends on the house edge.

I worked out how much you'd need if the house edge was around 2.7% like in single-zero roulette ($9k) and how much you'd need if the house edge was zero like in no-zero roulette ($5k).

[ranlo]

Wouldn't all of this theoretically mean, though, that with $9k, you can make infinite money? If that's enough to make $50, then now you have $9050. Then you go up by 50 and have $9100, and continue for infinity.

Well, no.

You have a 99% chance of making $50 and a 1% chance of losing $9k.

On average you're going to win 99 times and lose once every 100 times you try.

The 99 wins will make you 99*50 = $5k.
The 1 loss will lose you $9k.
So every 100 times you try you'll make a net loss of $4k - on average.

Sometimes you'll win 100 out of 100 tries. Other times you'll lose 2 or 3 times out of 100 tries. But on average you'll lose 1 in 100 tries.

I'm by no means trying to say "use martingale, it works and lets you beat the house".

All I'm saying is that martingale, used carefully, can give you a slight advantage over just going all-in. All it means is you'll probably lose a bit slower, or a bit less. It can't change you from having a negative expectation to having a positive one.

Edit: for a real-world demonstration... Just-Dice allows you to martingale from a single satoshi all the way up to around 100,000 CLAMs. ie. the biggest bet is 10,000,000,000,000 (ten trillion) times bigger than the smallest bet. Even with such a wide range of bet values, and with no rules against Martingale betting, the players still lose:

That math actually shows that you have a 2x greater chance to bust than to double up, as well (which does coincide with my own findings). Amazing how big that is! Basically, you'll bust 2x for each time you double. I think that really helps put it into perspective, rather than just looking at the 1%, 0.5%, etc. house edge.

[dooglus]

That math actually shows that you have a 2x greater chance to bust than to double up, as well (which does coincide with my own findings). Amazing how big that is! Basically, you'll bust 2x for each time you double. I think that really helps put it into perspective, rather than just looking at the 1%, 0.5%, etc. house edge.

It's even worse than that.

To double up you need to win $50 177 times.

The chance of winning 177 times, when each individual win is a 99% chance is:

0.99 ** 177 = 16.88%

So your chance of busting is 83.12%

ie. the chance of busting is 4.9 times bigger than the chance of doubling up (if you do it this way, trying to win $50 each time)

If your goal is to double up, there are more effective ways of doing it. But the chance of doubling up will always be smaller than the chance of busting if the house edge is positive.

[GannickusX]

But the biggest problem with betting 1 satoshi as base bet is that you would need to play insane amount of bets to even get 0.01 it would take like a year to make 1 btc if you dont bust

[deisik]

But the biggest problem with betting 1 satoshi as base bet is that you would need to play insane amount of bets to even get 0.01 it would take like a year to make 1 btc if you dont bust

You would reach 0.01 btc starting at 1 satoshi with just 20 steps if doubling at each step (classic martingale). And many bitcoin casinos allow automatic betting which makes it even easier (to lose all your coins)...

[GannickusX]

But the biggest problem with betting 1 satoshi as base bet is that you would need to play insane amount of bets to even get 0.01 it would take like a year to make 1 btc if you dont bust

You would reach 0.01 btc starting at 1 satoshi with just 20 steps if doubling at each step (classic martingale). And many bitcoin casinos allow automatic betting which makes it even easier (to lose all your coins)...

What do you mean 20 steps? You would need to play around 2 million bets to grt 0.01

[dooglus]

But the biggest problem with betting 1 satoshi as base bet is that you would need to play insane amount of bets to even get 0.01 it would take like a year to make 1 btc if you dont bust

You would reach 0.01 btc starting at 1 satoshi with just 20 steps if doubling at each step (classic martingale). And many bitcoin casinos allow automatic betting which makes it even easier (to lose all your coins)...

What do you mean 20 steps? You would need to play around 2 million bets to grt 0.01

You two are talking at cross-purposes.

One of you is talking about reaching a profit of 0.01, and the other is talking about reaching a bet size of 0.01.

[deisik]

But the biggest problem with betting 1 satoshi as base bet is that you would need to play insane amount of bets to even get 0.01 it would take like a year to make 1 btc if you dont bust

You would reach 0.01 btc starting at 1 satoshi with just 20 steps if doubling at each step (classic martingale). And many bitcoin casinos allow automatic betting which makes it even easier (to lose all your coins)...

What do you mean 20 steps? You would need to play around 2 million bets to grt 0.01

You two are talking at cross-purposes.

One of you is talking about reaching a profit of 0.01, and the other is talking about reaching a bet size of 0.01.

He said about using 1 satoshi as base bet, so I was thinking that he meant martingale (as the headline suggests). With a negative expectation he would never get 0.01 btc in profits without successively increasing his bets starting at 1 satoshi (doubling or however else)...

[Hfleer]

To be able to see the server seed means that it's an inside job or some intruder compromised the website and took advantage of the system?

cant tell for sure if it is an inside job, but it is an intruder compromised, which enable him to few the next rolls, making him able to determine the outcome of the game

if you know the unhashed seed, you can just change your seed so that you will win every bet. Or you can just alter hi/lo and then win every bet.

yes I've done that in 999dice, I always change my seed every play. when a great game I always persist in the seed, but when I lose, I tried to change the seed and the results are pretty good.

I did not know it was really influential or not, or is it just my luck

although sometimes I finally lost, because too much change seed 

Actually you are talking about something completely unreleated. I was talking about basically cheating the house, you were just using selective memory and luck

[dooglus]

He said about using 1 satoshi as base bet, so I was thinking that he meant martingale (as the headline suggests). With a negative expectation he would never get 0.01 btc in profits without successively increasing his bets starting at 1 satoshi (doubling or however else)...

I understand, but he is talking about having to bet 0.01 BTC in a single bet, which you would have to do on around your 21st bet in a losing streak if you were doubling each time. (2^20 ~= 1e6)

[deisik]

He said about using 1 satoshi as base bet, so I was thinking that he meant martingale (as the headline suggests). With a negative expectation he would never get 0.01 btc in profits without successively increasing his bets starting at 1 satoshi (doubling or however else)...

I understand, but he is talking about having to bet 0.01 BTC in a single bet, which you would have to do on around your 21st bet in a losing streak if you were doubling each time. (2^20 ~= 1e6)

That is in 20 steps as I said in my post. I just don't know how one would have to play around 2 million bets to get 0.01 btc if he started at 1 satoshi...