Does martingale really works?

[cp1]

In very simple terms:

The house has an edge.  If you're making bets that the house has an edge at then you'd expect the house to make a profit.  If you expect the house to make a profit then you have to expect to lose.  You can't expect both you and the house to profit.

[emu512]

from my experience martinagle has some limitations so to say.Its like when a point starts there is a chance it could keep on going towards that very point.It could be profit or loss.

[leex1528]

You are confused about what "expected profit" means.

I think you are slightly confused about what "expected profit means.

Here's what everyone except you is talking about when they say 'expected profit':

    http://en.wikipedia.org/wiki/Expected_value

What are you talking about?

I am talking about a 1 time run where your odds are 99% chance of winning.  I have been saying that the entire time.  I have already said countless times that over the course of time, you are going to lose money.

I'm asking what you mean by the term "expected profit". If you're not talking about mathematical expectation then what is it you're referring to whe you say "expected profit"?

But if you say, had a 20 bet limit, and you rolled 20 times, you have a 99.99% chance of gaining a very small amount of coins/money...etc

If you roll 20 times with a 99% chance of winning a tiny amount and a 1% chance of losing a lot, then you only end up with a profit if you win all 20 rolls. The chance of winning 20 out of 20 99% rolls is 0.99^20 = 81.79%.

So you have a 81.79% chance of winning a small amount and a ~20% chance of losing a lot.

Your mathematical expectation in such a case is negative.

I am confused why you think you only end up in profit if you win all 20 rolls?  I mean, technically if you lost 19 in a row, then won the 20th roll, you'd be in profit.  Basically you would be promised to be in profit if you won the 20th roll.  I didn't mean that as a fixed number though, so you rolled 18th and won the 18th, you could stop and you would probably be in profit around 9 base units(give or take a few)

I am using the expected profit term in the short period of a limited trial, if you did  this once and whatever the odds may be, they will be so in your favor more than anything else offers.  I am not saying you will win, but the odds are definitely in your favor to win a small amount of coin.

[cp1]

He's talking about condensing your martingale series into one roll with the same chance of winning as the entire series had and doing that over again 20 times.

[sed]

You are confused about what "expected profit" means.

I think you are slightly confused about what "expected profit means.

Here's what everyone except you is talking about when they say 'expected profit':

    http://en.wikipedia.org/wiki/Expected_value

What are you talking about?

I am talking about a 1 time run where your odds are 99% chance of winning.  I have been saying that the entire time.  I have already said countless times that over the course of time, you are going to lose money.  But if you say, had a 20 bet limit, and you rolled 20 times, you have a 99.99% chance of gaining a very small amount of coins/money...etc

It's a very weird casino game where you have a 99% chance of winning.  I guess the win is <1% of the bet?  What game is this?

[leex1528]

You are confused about what "expected profit" means.

I think you are slightly confused about what "expected profit means.

Here's what everyone except you is talking about when they say 'expected profit':

    http://en.wikipedia.org/wiki/Expected_value

What are you talking about?

I am talking about a 1 time run where your odds are 99% chance of winning.  I have been saying that the entire time.  I have already said countless times that over the course of time, you are going to lose money.  But if you say, had a 20 bet limit, and you rolled 20 times, you have a 99.99% chance of gaining a very small amount of coins/money...etc

It's a very weird casino game where you have a 99% chance of winning.  I guess the win is <1% of the bet?  What game is this?

No, you would have a 50% chance of winning any given bet, or 49.5 for most house edge places. 

But Martingale works in the very short run.  So I am saying you do a small amount of bets...and then quit...

[Phildo]

You are confused about what "expected profit" means.

I think you are slightly confused about what "expected profit means.

Here's what everyone except you is talking about when they say 'expected profit':

    http://en.wikipedia.org/wiki/Expected_value

What are you talking about?

I am talking about a 1 time run where your odds are 99% chance of winning.  I have been saying that the entire time.  I have already said countless times that over the course of time, you are going to lose money.  But if you say, had a 20 bet limit, and you rolled 20 times, you have a 99.99% chance of gaining a very small amount of coins/money...etc

It's a very weird casino game where you have a 99% chance of winning.  I guess the win is <1% of the bet?  What game is this?

No, you would have a 50% chance of winning any given bet, or 49.5 for most house edge places. 

But Martingale works in the very short run.  So I am saying you do a small amount of bets...and then quit...

It works until it doesn't, which is the problem, because when it doesn't work the loss is devastating.

[sed]

You are confused about what "expected profit" means.

I think you are slightly confused about what "expected profit means.

Here's what everyone except you is talking about when they say 'expected profit':

    http://en.wikipedia.org/wiki/Expected_value

What are you talking about?

I am talking about a 1 time run where your odds are 99% chance of winning.  I have been saying that the entire time.  I have already said countless times that over the course of time, you are going to lose money.  But if you say, had a 20 bet limit, and you rolled 20 times, you have a 99.99% chance of gaining a very small amount of coins/money...etc

It's a very weird casino game where you have a 99% chance of winning.  I guess the win is <1% of the bet?  What game is this?

No, you would have a 50% chance of winning any given bet, or 49.5 for most house edge places. 

But Martingale works in the very short run.  So I am saying you do a small amount of bets...and then quit...

Ok, well if the game is a 50/50 game (or 49.5/50.5, similarly) then you're completely wrong about having a 99% chance of winning.  I guess as typical in martingale you bet 1 unit to start, so you are going to bet 20 times, is that right?

I think maybe your mistake is that you tried to consider that the probablility of losing all 20 bets is very low, is that right?  But consider that you don't have to lose all 20 bets to come out negative.  For fun, I generated a random series of 20 plays using python's bin(random.getrandombits(20)).  Assume you start with 1000 units and that you start betting at 1 unit, first column is W/L, second column is the amount won/lost, third column is your balance.

W +1    1001
W +1    1002
L   -1    1001
L   -2      999
L   -4      995
L   -8      987
L   -16     971
L   -32     939
L   -64     875
L   -128   747
L   -256   491
L   -512    -21  !!! you just went broke!!!
W  +1024 ...
L   -1
L   -2
L   -4
L   -8
W  +16
W  +1
L   -1

Ok, for fun assume you had 10,000 starting units instead of only 1000, then in that case, the simulation can continue:

W +1    10001
W +1    10002
L   -1    10001
L   -2      9999
L   -4      9995
L   -8      9987
L   -16     9971
L   -32     9939
L   -64     9875
L   -128    9747
L   -256    9491
L   -512    8979
W  +1024  10003
L   -1       10002
L   -2       10000
L   -4       9996
L   -8       9988
W  +16    10004
W  +1      10005
L   -1       10004

So, in this single instance simulation, if you had 10000 units and you play martingale starting with 1 unit then you came out +4 units after 20 bets.  However, I hope you see that that long string of negatives can just as easily happen towards the end of your 20 bets and then you come out negative.  Also, I hope you see that you need a huge bankroll in order to support losses that double each time and not go broke.

[leex1528]

You are confused about what "expected profit" means.

I think you are slightly confused about what "expected profit means.

Here's what everyone except you is talking about when they say 'expected profit':

    http://en.wikipedia.org/wiki/Expected_value

What are you talking about?

I am talking about a 1 time run where your odds are 99% chance of winning.  I have been saying that the entire time.  I have already said countless times that over the course of time, you are going to lose money.  But if you say, had a 20 bet limit, and you rolled 20 times, you have a 99.99% chance of gaining a very small amount of coins/money...etc

It's a very weird casino game where you have a 99% chance of winning.  I guess the win is <1% of the bet?  What game is this?

No, you would have a 50% chance of winning any given bet, or 49.5 for most house edge places. 

But Martingale works in the very short run.  So I am saying you do a small amount of bets...and then quit...

Ok, well if the game is a 50/50 game (or 49.5/50.5, similarly) then you're completely wrong about having a 99% chance of winning.  I guess as typical in martingale you bet 1 unit to start, so you are going to bet 20 times, is that right?

I think maybe your mistake is that you tried to consider that the probablility of losing all 20 bets is very low, is that right?  But consider that you don't have to lose all 20 bets to come out negative.  For fun, I generated a random series of 20 plays using python's bin(random.getrandombits(20)).  Assume you start with 1000 units and that you start betting at 1 unit, first column is W/L, second column is the amount won/lost, third column is your balance.

W +1    1001
W +1    1002
L   -1    1001
L   -2      999
L   -4      995
L   -8      987
L   -16     971
L   -32     939
L   -64     875
L   -128   747
L   -256   491
L   -512    -21  !!! you just went broke!!!
W  +1024 ...
L   -1
L   -2
L   -4
L   -8
W  +16
W  +1
L   -1

Ok, for fun assume you had 10,000 starting units instead of only 1000, then in that case, the simulation can continue:

W +1    10001
W +1    10002
L   -1    10001
L   -2      9999
L   -4      9995
L   -8      9987
L   -16     9971
L   -32     9939
L   -64     9875
L   -128    9747
L   -256    9491
L   -512    8979
W  +1024  10003
L   -1       10002
L   -2       10000
L   -4       9996
L   -8       9988
W  +16    10004
W  +1      10005
L   -1       10004

So, in this single instance simulation, if you had 10000 units and you play martingale starting with 1 unit then you came out +4 units after 20 bets.  However, I hope you see that that long string of negatives can just as easily happen towards the end of your 20 bets and then you come out negative.  Also, I hope you see that you need a huge bankroll in order to support losses that double each time and not go broke.

So I am guessing you missed tons of this conversation we had?  Its okay I will review it for you.  I don't mean an exact amount of hands.  For example, if you wanted to just win 1 base unit, and you were rolling 5-20 times.  Your odds are EXTREMELY high that you are going to win.  It doesn't mean yeah I am just rolling 20 times...or x amount of times.  It just means Martingale works in small doses, which GREATLY increase your odds.  But the problem with Martingale is that you keep doing it, and odds are you are going to bust eventually.  All I was saying is it works in extra small to small doses, and only once. 

[Wardrick]

The people who actually have the money to make a decent profit off of the martingale system wouldn't be the type of people who'd even use it. They already have a lot of money why would they risk so much just for a few bucks? A regular person using martingale will end up broke or with a small amount of profit before they go broke.

[Zebra]

It depends on how you define the word "work".
You could make bets with extremely high win chance with martingale, but you won't beat the house at all in the long term.

[dooglus]

It's a very weird casino game where you have a 99% chance of winning.  I guess the win is <1% of the bet?  What game is this?

I think he's considering a single Martingale progression as a "game".

Suppose you have 127 BTC, play a dice game with a 49.5% chance of winning each roll, and double your stake when you win.

You start by betting 1 BTC. If you win, you made 1 BTC profit. If you lose, you double your bet and repeat.
You will bet 1, 2, 4, 8, 16, 32, 64 in turn until you win one. If you win any one of those bets you stop, and have a net profit of 1 BTC.

Your chance of losing all 7 bets is 0.505^7 = 0.8376%

So your chance of winning any one of them (and stopping with a profit) is 99.1624%.

I think that's the 99% he's talking about.

Notice the expected value of the game:

99.1624% of the time you win 1 unit
0.8376% of the time you lose 127 units (that's about 1 in 120 times that you play)
expected profit = 1 * (1 - 0.505^7) - 127 * 0.505^7 = -0.0721

So you expect to lose 0.0721 BTC each time you play (even though you will win 99.1624% of the time).

Intuitively, if you win one unit almost every time you play, but every 120 times or so you lose 127 units, that's not a winning game. If you lost 127 units every 130 times you played on average, that would be a winning game, just about. But losing every 120 times on average is a losing game.

By an amazing coincidence the expected loss per game is exactly 1% of the amount you expect to risk, which is 7.21 BTC per "game":

>>> w = 0.495
>>> l = 1 - w
>>> w * (1 + 3*l + 7*l^2 + 15*l^3 + 31*l^4 + 63*l^5) + 127*l^6
7.21353521

[leex1528]

It's a very weird casino game where you have a 99% chance of winning.  I guess the win is <1% of the bet?  What game is this?

I think he's considering a single Martingale progression as a "game".

Suppose you have 127 BTC, play a dice game with a 49.5% chance of winning each roll, and double your stake when you win.

You start by betting 1 BTC. If you win, you made 1 BTC profit. If you lose, you double your bet and repeat.
You will bet 1, 2, 4, 8, 16, 32, 64 in turn until you win one. If you win any one of those bets you stop, and have a net profit of 1 BTC.

Your chance of losing all 7 bets is 0.505^7 = 0.8376%

So your chance of winning any one of them (and stopping with a profit) is 99.1624%.

I think that's the 99% he's talking about.

Notice the expected value of the game:

99.1624% of the time you win 1 unit
0.8376% of the time you lose 127 units (that's about 1 in 120 times that you play)
expected profit = 1 * (1 - 0.505^7) - 127 * 0.505^7 = -0.0721

So you expect to lose 0.0721 BTC each time you play (even though you will win 99.1624% of the time).

Intuitively, if you win one unit almost every time you play, but every 120 times or so you lose 127 units, that's not a winning game. If you lost 127 units every 130 times you played on average, that would be a winning game, just about. But losing every 120 times on average is a losing game.

By an amazing coincidence the expected loss per game is exactly 1% of the amount you expect to risk, which is 7.21 BTC per "game":

>>> w = 0.495
>>> l = 1 - w
>>> w * (1 + 3*l + 7*l^2 + 15*l^3 + 31*l^4 + 63*l^5) + 127*l^6
7.21353521

Exactly what I was talking about Doog.  Thanks for backing that up.  like so many people have said, there are no sure ways to gain money at a casino.  But if you just wanted to win once, Maybe twice(pushing your luck) You have some great odds at doing so, but you can also lose a lot...

[zimmah]

Not coincidence dooglus, that's just math.

whichever system you use, you'll ALWAYS expect to lose exactly the house edge over the total amount of money wagered.

(in practice obviously it's more likely to deviate from it, but that has to do with probability distributions).

[dooglus]

Not coincidence dooglus, that's just math.

whichever system you use, you'll ALWAYS expect to lose exactly the house edge over the total amount of money wagered.

True, but this is a neat result:

99.1624% of the time you win 1 unit
0.8376% of the time you lose 127 units (that's about 1 in 120 times that you play)

Betting 127 units to win 1 unit is a 1.007874x payout.
On a site with a 1% edge, you would get a 98.2265625% chance of winning such a bet.

But we have a 99.1624% chance of winning it by using this martingale progression.
That is significantly better.

Looks to me like we've reduced the 1% house edge to an effective edge of 0.0568%!

>>> (0.505^7 * 127 - (1 - 0.505^7))  /  127.0 * 100
0.05679948984803941

[leex1528]

Not coincidence dooglus, that's just math.

whichever system you use, you'll ALWAYS expect to lose exactly the house edge over the total amount of money wagered.

True, but this is a neat result:

99.1624% of the time you win 1 unit
0.8376% of the time you lose 127 units (that's about 1 in 120 times that you play)

Betting 127 units to win 1 unit is a 1.007874x payout.
On a site with a 1% edge, you would get a 98.2265625% chance of winning such a bet.

But we have a 99.1624% chance of winning it by using this martingale progression.
That is significantly better.

Looks to me like we've reduced the 1% house edge to an effective edge of 0.0568%!

>>> (0.505^7 * 127 - (1 - 0.505^7))  /  127.0 * 100
0.05679948984803941

So in small case sets, Martingale is actually better and helps you win more?

Would be interesting to see as the # of test runs go up and up....

[dooglus]

So in small case sets, Martingale is actually better and helps you win more?

Would be interesting to see as the # of test runs go up and up....

It helps you lose less, not win more...

What I showed is that if you want to turn 127 BTC into 128 BTC then it's better (as in better chance of success) to do a 49.5% 2x martingale from 1 BTC up to 64 BTC than it is to risk the whole 127 BTC on a 98.2265625% bet to make 1 BTC profit.

That's because the martingale allows you to risk less, in most cases, and so your expected loss is smaller.

[leex1528]

So in small case sets, Martingale is actually better and helps you win more?

Would be interesting to see as the # of test runs go up and up....

It helps you lose less, not win more...

What I showed is that if you want to turn 127 BTC into 128 BTC then it's better (as in better chance of success) to do a 49.5% 2x martingale from 1 BTC up to 64 BTC than it is to risk the whole 127 BTC on a 98.2265625% bet to make 1 BTC profit.

That's because the martingale allows you to risk less, in most cases, and so your expected loss is smaller.

True I should pick my words more carefully....

Still, so say you wanted to turn 127 BTC into 135 BTC...something like that....would it be better then trying to bet once at say 95% or something?  Not sure...

[dooglus]

Still, so say you wanted to turn 127 BTC into 135 BTC...something like that....would it be better then trying to bet once at say 95% or something?  Not sure...

To turn 127 into 135 with a single bet, you would have to use a 93.133333% bet:

>>> 99 / (135/127.0)
93.13333333333333

That obviously has a 93.133333% chance of success.

To do the same thing with the 1,2,4,8,16,32,64 martingale, we need to play (and win) 8 times.

We know that the chance of winning any single time is:

>>> 1 - 0.505**7
0.991623942561664

so the chance of succeeding 8 times in a row is:

>>> (1 - 0.505**7) ** 8
0.934923407802361

or 93.49234%

So yes, you're still slightly better off (93.49% vs 93.13%) to run the 1,2,4,...,64 martingale 8 times in a row.

In fact it's better to use the 1 BTC martingale over and over for all profits up to 11 BTC, but for 12 BTC a single large bet is better:

>>> p=8; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(93.13333333333333, 93.4923407802361)

>>> p=9; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(92.4485294117647, 92.70924356381637)

>>> p=10; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(91.77372262773721, 91.93270561466116)

>>> p=11; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(91.1086956521739, 91.16267199197111)

>>> p=12; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(90.45323741007194, 90.39908821513419)

Edit:

By an amazing coincidence the expected loss per game is exactly 1% of the amount you expect to risk, which is 7.21 BTC per "game":

So that's weird. The martingale risks an average 7.21 per game, whereas the large bet risks 127.  127 / 7.21 = 17.6. So the martingale risks 17 times less, on average. So why does running it just 12 times give a lower chance of success? That seems wrong to me. Maybe it's because failing the 12'th martingale leaves you with the 11 you won from the first 11 successes, whereas failing the single large bet leaves you with nothing.

[boumalo]

Still, so say you wanted to turn 127 BTC into 135 BTC...something like that....would it be better then trying to bet once at say 95% or something?  Not sure...

To turn 127 into 135 with a single bet, you would have to use a 93.133333% bet:

>>> 99 / (135/127.0)
93.13333333333333

That obviously has a 93.133333% chance of success.

To do the same thing with the 1,2,4,8,16,32,64 martingale, we need to play (and win) 8 times.

We know that the chance of winning any single time is:

>>> 1 - 0.505**7
0.991623942561664

so the chance of succeeding 8 times in a row is:

>>> (1 - 0.505**7) ** 8
0.934923407802361

or 93.49234%

So yes, you're still slightly better off (93.49% vs 93.13%) to run the 1,2,4,...,64 martingale 8 times in a row.

In fact it's better to use the 1 BTC martingale over and over for all profits up to 11 BTC, but for 12 BTC a single large bet is better:

>>> p=8; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(93.13333333333333, 93.4923407802361)

>>> p=9; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(92.4485294117647, 92.70924356381637)

>>> p=10; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(91.77372262773721, 91.93270561466116)

>>> p=11; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(91.1086956521739, 91.16267199197111)

>>> p=12; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(90.45323741007194, 90.39908821513419)

Edit:

By an amazing coincidence the expected loss per game is exactly 1% of the amount you expect to risk, which is 7.21 BTC per "game":

So that's weird. The martingale risks an average 7.21 per game, whereas the large bet risks 127.  127 / 7.21 = 17.6. So the martingale risks 17 times less, on average. So why does running it just 12 times give a lower chance of success? That seems wrong to me. Maybe it's because failing the 12'th martingale leaves you with the 11 you won from the first 11 successes, whereas failing the single large bet leaves you with nothing.

Still, so say you wanted to turn 127 BTC into 135 BTC...something like that....would it be better then trying to bet once at say 95% or something?  Not sure...

To turn 127 into 135 with a single bet, you would have to use a 93.133333% bet:

>>> 99 / (135/127.0)
93.13333333333333

That obviously has a 93.133333% chance of success.

To do the same thing with the 1,2,4,8,16,32,64 martingale, we need to play (and win) 8 times.

We know that the chance of winning any single time is:

>>> 1 - 0.505**7
0.991623942561664

so the chance of succeeding 8 times in a row is:

>>> (1 - 0.505**7) ** 8
0.934923407802361

or 93.49234%

So yes, you're still slightly better off (93.49% vs 93.13%) to run the 1,2,4,...,64 martingale 8 times in a row.

In fact it's better to use the 1 BTC martingale over and over for all profits up to 11 BTC, but for 12 BTC a single large bet is better:

>>> p=8; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(93.13333333333333, 93.4923407802361)

>>> p=9; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(92.4485294117647, 92.70924356381637)

>>> p=10; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(91.77372262773721, 91.93270561466116)

>>> p=11; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(91.1086956521739, 91.16267199197111)

>>> p=12; (99 / ((127+p)/127.0), (1 - 0.505**7) ** p * 100)
(90.45323741007194, 90.39908821513419)

Edit:

By an amazing coincidence the expected loss per game is exactly 1% of the amount you expect to risk, which is 7.21 BTC per "game":

So that's weird. The martingale risks an average 7.21 per game, whereas the large bet risks 127.  127 / 7.21 = 17.6. So the martingale risks 17 times less, on average. So why does running it just 12 times give a lower chance of success? That seems wrong to me. Maybe it's because failing the 12'th martingale leaves you with the 11 you won from the first 11 successes, whereas failing the single large bet leaves you with nothing.

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