Does martingale really works?

[futureofbitcoin]

a lot of what was said on this page can be summed up in two words: Selective Memory.

That's all it is. I admit I'm guilty of it as much as the next person. You just gotta understand that your intuition sucks.

A personal anecdote: in 2005-2007 I played this game called Kingdom of Loathing. It's a pretty interesting stickman semi texted based mmorpg with lots of references and jokes and stuff. The currency in that game is "meat".

Anyway, there's a gambling mini-game within the game, which is essentially 50/50 double or nothing, with a small house cut.

Long story short, I managed to make 135,000,000+ meat starting from 70,000 meat, using a modified martingale system.

What happened after that, you might ask. Well, I lost it all . If only I could repeat this with real life money... but I never gamble with real money, except "gambling" by investing in bitcoins.

[freeyourmind]

To your point, perhaps it has involved luck, but I don't think it's very odd for people to double their money before losing it all when following martingale.

No, it's not unusual. It just has to happen less than half the time on average, due to the house edge. Otherwise this simple strategy would be profitable:

* start with X; martingale until doubling; quit

If you successfully double Y% of the time, your expected profit would be:

    (Y*X - (100-Y)*X) / 100
    = X(2Y - 100) / 100

which is positive exactly when 2Y > 100, ie. when Y > 50%

]

Yeah, good point, so the question is what is Y?

If I look at a high/low game with 47.5% odds of doubling your bet, after the house takes its cut, then let's say I'm willing to go 11 consecutive bets before I lose it all.

Bet   Total Loss
1             1
2             3
4             7
8            15
16           31
32           63
64          127
128         255
256         511
512         1023
1024       2047

So assuming an initial bet of 1 satoshi, using martingale, my 11th bet would be 1024 and at that point I lose 2047 satoshi.  The odds of losing 11 bets in a row is (0.525^11) = 0.0835% or 1/1197 bets.  So for every 1197 bets, there will be one losing streak, on average, which will lose 2047+ on making 1196...not the best strategy in the long run.

Hopefully the math makes sense - feel free to correct if I made a mistake.

[birdcat90]

well in my view, doing martingale is risking more for the small profit/ constant..

i will never use it,i love bet using small but has good chance to gain more..

[Hfleer]

To your point, perhaps it has involved luck, but I don't think it's very odd for people to double their money before losing it all when following martingale.

No, it's not unusual. It just has to happen less than half the time on average, due to the house edge. Otherwise this simple strategy would be profitable:

* start with X; martingale until doubling; quit

If you successfully double Y% of the time, your expected profit would be:

    (Y*X - (100-Y)*X) / 100
    = X(2Y - 100) / 100

which is positive exactly when 2Y > 100, ie. when Y > 50%

]

Yeah, good point, so the question is what is Y?

If I look at a high/low game with 47.5% odds of doubling your bet, after the house takes its cut, then let's say I'm willing to go 11 consecutive bets before I lose it all.

Bet   Total Loss
1             1
2             3
4             7
8            15
16           31
32           63
64          127
128         255
256         511
512         1023
1024       2047

So assuming an initial bet of 1 satoshi, using martingale, my 11th bet would be 1024 and at that point I lose 2047 satoshi.  The odds of losing 11 bets in a row is (0.525^11) = 0.0835% or 1/1197 bets.  So for every 1197 bets, there will be one losing streak, on average, which will lose 2047+ on making 1196...not the best strategy in the long run.

Hopefully the math makes sense - feel free to correct if I made a mistake.

I think Y will be close to 1-(1196/2047) = 41.573%

[dooglus]

Yeah, good point, so the question is what is Y?

If I look at a high/low game with 47.5% odds of doubling your bet, after the house takes its cut, then let's say I'm willing to go 11 consecutive bets before I lose it all.

Bet   Total Loss
1             1
2             3
4             7
8            15
16           31
32           63
64          127
128         255
256         511
512         1023
1024       2047

So assuming an initial bet of 1 satoshi, using martingale, my 11th bet would be 1024 and at that point I lose 2047 satoshi.  The odds of losing 11 bets in a row is (0.525^11) = 0.0835% or 1/1197 bets.  So for every 1197 bets, there will be one losing streak, on average, which will lose 2047+ on making 1196...not the best strategy in the long run.

Hopefully the math makes sense - feel free to correct if I made a mistake.

The math looks good, though I didn't check your calculations. There's a problem with it however: it doesn't answer your question: "what is Y?"

We want to know your chance of doubling up using this system. You seem to have 2047 satoshi, since you're able to lose 11 times before busting. So in order to double up, you need to win 2047 times. The probability of winning 2047 sequences in a row is (the probability of winning one sequence) to the power of 2047. The probability of winning one sequence is (1 minus the probability of losing 11 rolls in a row), and the probability of losing 11 rolls in a row is, as you said, 0.525^11.

So Y as a probability (not a percentage) is:

(1 - 0.525^11)^2047 = 0.1808

So Y is 18.08%. Considerably less than 50%, and so it's a losing strategy.

Most Bitcoin games (at least the ones I'm familiar with have a house edge of 1%, and so give you a 49.5% chance of doubling your stake. In that case, Y is:

(1 - 0.505^11)^2047 = 0.3277, or 32.77%

Much better, but still less than 50%.

Interestingly, even with a 0% house edge the chance of doubling up before busting is:

(1 - 0.5**11)**2047 = 0.3679 or 36.79%

This came up recently and surprised me (*). How can it be less than 50% with a 0% house edge?

It turns out the reason is that by the time you lose your 2047 satoshis, you've probably already won a bunch of times, and you get to keep those winnings. So it turns out you're not choosing between doubling up or busting completely, you're choosing between doubling up, and getting left with only however much you were able to win before you hit your big losing streak.

So this makes me think I was wrong here:

No, it's not unusual. It just has to happen less than half the time on average, due to the house edge. Otherwise this simple strategy would be profitable:

* start with X; martingale until doubling; quit

If you successfully double Y% of the time, your expected profit would be:

    (Y*X - (100-Y)*X) / 100
    = X(2Y - 100) / 100

which is positive exactly when 2Y > 100, ie. when Y > 50%

because I was thinking that there were only two possible outcomes:

a) you double up
b) you lose everything

but by far the most common outcome is:

c) you hit a big losing streak and can't afford to make that last bet, even though you're not bust yet

What this does tell me is that in theory you should find that you are able to double up less than 37% of the time if you stop and call it a loss when you can't afford to double to make your next bet.


(*) Here's where it came up before. I'm talking about a bankroll of 15 instead of 2047, and a house edge of 2.702%, but other than that it's the same issue. I found I doubled up just 38% of the time, busted only 6% of the time, and the rest of the time ended up with from 1 to 14 satoshis:

Code:

3041169 wins out of 8000000 rounds (38.015%)
0:6.24%, 1:5.86%, 2:5.49%, 3:5.13%, 4:4.82%, 5:4.54%, 6:4.24%, 7:3.98%, 8:3.73%, 9:3.50%, 10:3.27%, 11:3.06%, 12:2.87%, 13:2.71%, 14:2.54%, 30:38.01%
expectation: 15.008940625

So only 6.24% of the time do you actually bust.
5.86% of the time you end up with 1 unit,
5.49% of the time, 2 units, etc.

[futureofbitcoin]

Y changes with your bankroll (or the % of your bankroll you bet per bet).

For example, if you bet all or nothing, assuming a 1% house edge, Y = 49%, quite obviously.

[dooglus]

For example, if you bet all or nothing, assuming a 1% house edge, Y = 49%, quite obviously.

Don't you mean 49.5%?

[futureofbitcoin]

I guess I misused the term house edge. I just meant if you had 49% chance to double, then you have 49% chance to double, lol.

In the case, that for example, the chances of winning/losing are exactly 50/50, but you get a 99% return if you win, it's slightly more difficult to calculate, because even when you win, you haven't doubled your amount, so you'll need a

WW or WLWW or WLWLWW, in essence, you need to win twice before losing twice, while winning the first game* which is basically

1/2 * (1/4 + 1/4(1/2) + 1/4(1/2^2)...) = 1/2 * (1/4 / (1 - 1/2)) = 1/2 * 1/2 = 1/4.

*The actual chances are smaller, due to the fact that the amount lost for every L is greater than the amount won for every W, so after a certain number of iterations, you'll need 1 extra W to make up for the lost amount. In particular, after the first W, and after 99 LWs in a row, you'll end up with 100% of your starting bankroll, effectively repeating the process from the beginning.

I have to say that back in highschool I used to be a prodigy at probability, and did very well at math competitions because of it. Probability came naturally to me, and everything just made sense. I didn't need to learn it. But for some reason, my brain power is drastically weaker than it used to be, and I no longer intuitively get these concepts.

I bet there's a much easier way to calculate the above, with more accuracy. Oh well, at least my point is made.

[ranlo]

I guess I misused the term house edge. I just meant if you had 49% chance to double, then you have 49% chance to double, lol.

In the case, that for example, the chances of winning/losing are exactly 50/50, but you get a 99% return if you win, it's slightly more difficult to calculate, because even when you win, you haven't doubled your amount, so you'll need a

WW or WLWW or WLWLWW, in essence, you need to win twice before losing twice, while winning the first game* which is basically

1/2 * (1/4 + 1/4(1/2) + 1/4(1/2^2)...) = 1/2 * (1/4 / (1 - 1/2)) = 1/2 * 1/2 = 1/4.

*The actual chances are smaller, due to the fact that the amount lost for every L is greater than the amount won for every W, so after a certain number of iterations, you'll need 1 extra W to make up for the lost amount. In particular, after the first W, and after 99 LWs in a row, you'll end up with 100% of your starting bankroll, effectively repeating the process from the beginning.

I have to say that back in highschool I used to be a prodigy at probability, and did very well at math competitions because of it. Probability came naturally to me, and everything just made sense. I didn't need to learn it. But for some reason, my brain power is drastically weaker than it used to be, and I no longer intuitively get these concepts.

I bet there's a much easier way to calculate the above, with more accuracy. Oh well, at least my point is made.

It's also worth noting that you have almost the same chance to win 13x in a row as you do losing 13x in a row. At the end of the day, it's all really based on luck.

Also, with your math, basically every win puts you at 1% below where you should be (I believe), so it would need to be 49 losses:50 wins (as long as after 99 repetitions you had one more win than loss you'd be doubled).

[Fiiasco]

The martingale strategy should not be used as a dependant strategy, you'll find your balance on 0 in a matter of minutes.

[freeyourmind]

The math looks good, though I didn't check your calculations. There's a problem with it however: it doesn't answer your question: "what is Y?"

We want to know your chance of doubling up using this system. You seem to have 2047 satoshi, since you're able to lose 11 times before busting. So in order to double up, you need to win 2047 times. The probability of winning 2047 sequences in a row is (the probability of winning one sequence) to the power of 2047. The probability of winning one sequence is (1 minus the probability of losing 11 rolls in a row), and the probability of losing 11 rolls in a row is, as you said, 0.525^11.

So Y as a probability (not a percentage) is:

(1 - 0.525^11)^2047 = 0.1808

So Y is 18.08%. Considerably less than 50%, and so it's a losing strategy.

Most Bitcoin games (at least the ones I'm familiar with have a house edge of 1%, and so give you a 49.5% chance of doubling your stake. In that case, Y is:

(1 - 0.505^11)^2047 = 0.3277, or 32.77%

Much better, but still less than 50%.

Interestingly, even with a 0% house edge the chance of doubling up before busting is:

(1 - 0.5**11)**2047 = 0.3679 or 36.79%

This came up recently and surprised me (*). How can it be less than 50% with a 0% house edge?

It turns out the reason is that by the time you lose your 2047 satoshis, you've probably already won a bunch of times, and you get to keep those winnings. So it turns out you're not choosing between doubling up or busting completely, you're choosing between doubling up, and getting left with only however much you were able to win before you hit your big losing streak.

So this makes me think I was wrong here:

No, it's not unusual. It just has to happen less than half the time on average, due to the house edge. Otherwise this simple strategy would be profitable:

* start with X; martingale until doubling; quit

If you successfully double Y% of the time, your expected profit would be:

    (Y*X - (100-Y)*X) / 100
    = X(2Y - 100) / 100

which is positive exactly when 2Y > 100, ie. when Y > 50%

because I was thinking that there were only two possible outcomes:

a) you double up
b) you lose everything

but by far the most common outcome is:

c) you hit a big losing streak and can't afford to make that last bet, even though you're not bust yet

What this does tell me is that in theory you should find that you are able to double up less than 37% of the time if you stop and call it a loss when you can't afford to double to make your next bet.


(*) Here's where it came up before. I'm talking about a bankroll of 15 instead of 2047, and a house edge of 2.702%, but other than that it's the same issue. I found I doubled up just 38% of the time, busted only 6% of the time, and the rest of the time ended up with from 1 to 14 satoshis:

Lol yeah I totally missed calculating Y - appreciate the response.  

Your math makes sense...I didn't realize the long-term probability would be affected so greatly by a slight change in the house edge.  It's also surprising to see that the odds are still against you.

I was gambling on a site with 47.5% odds to double your bet...and from 80,000 satoshi I made it up to 2,200,000 before hitting that losing streak.  Given our calculations, there seems to be a lot of luck involved there.

[Landplop]

No margintale not always work!

[jorjito25]

Martingale dosnt work if there is a maximum bet.
All u can do is win 2 pennys or lose it all.
May be if 7 or 8 ppl play together against the house can work.

[WhatTheGox]

Martingale dosnt work if there is a maximum bet.
All u can do is win 2 pennys or lose it all.
May be if 7 or 8 ppl play together against the house can work.

Not if you've not got infinite money to bet any infinite size amount.

[GannickusX]

Martingale dosnt work if there is a maximum bet.
All u can do is win 2 pennys or lose it all.
May be if 7 or 8 ppl play together against the house can work.

It doesnt work even if there is no maximum bet since you wont have enough money and you start to bet high pretty fast, if you bet with 0.1 as base bet

on your 20th lost bet you would have to bet 52.428,8 BTC so yeah..

[Coef]

Martingale dosnt work if there is a maximum bet.
All u can do is win 2 pennys or lose it all.
May be if 7 or 8 ppl play together against the house can work.

Even if there is no max bet, martingale won't work.
You may be able to sustain a longer loss streak assuming you have a huge bankroll, but still you are going to get a long enough streak sooner or later.
Don't forget every bet are always in favour of the house because of the house edge, and they are all independent. No matter how you play, the house still has the upper hand.

[deisik]

Martingale dosnt work if there is a maximum bet.
All u can do is win 2 pennys or lose it all.
May be if 7 or 8 ppl play together against the house can work.

Even if there is no max bet, martingale won't work.
You may be able to sustain a longer loss streak assuming you have a huge bankroll, but still you are going to get a long enough streak sooner or later.
Don't forget every bet are always in favour of the house because of the house edge, and they are all independent. No matter how you play, the house still has the upper hand.

It could work but just one time only (well, a few times really). And, as with any one-off strategies (otherwise known as fool's luck), once you win, you'd better never come back and try again. But since greed and adrenaline rush have nothing to do with reason and common sense, being mutually exclusive, this is rarely the case...

[jorjito25]

Without maximum and minimum bets it works.
Thats why casinos puts the limits bets.
I wouldnt waste my time explaining this... I think is very clear.

[GannickusX]

Without maximum and minimum bets it works.
Thats why casinos puts the limits bets.
I wouldnt waste my time explaining this... I think is very clear.

Nope it doesnt, if you are going to bet 1 satoshi as your main bet it would take you years to even win 1 btc since you have to play 200.000.000 bets to get 1 btc

[Coef]

Martingale dosnt work if there is a maximum bet.
All u can do is win 2 pennys or lose it all.
May be if 7 or 8 ppl play together against the house can work.

Even if there is no max bet, martingale won't work.
You may be able to sustain a longer loss streak assuming you have a huge bankroll, but still you are going to get a long enough streak sooner or later.
Don't forget every bet are always in favour of the house because of the house edge, and they are all independent. No matter how you play, the house still has the upper hand.

It could work but just one time only (well, a few times really). And, as with any one-off strategies (otherwise known as fool's luck), once you win, you'd better never come back and try again. But since greed and adrenaline rush have nothing to do with reason and common sense, being mutually exclusive, this is rarely the case...

Well then it will be just like risking all your bitcoin and playing a few 98% bets, hoping that you are not unlucky enough to get a loss. For that, I wouldn't call it "works".