Does martingale really works?

[PacoMartin]

Martingale can work if sites didnt play dirty like max bet limit and you have lots of money.

It's not dirty to set a maximum bet limit, it's absolutely necessary.

If there's no limit then you can place a bet that the house is unable to pay out on if you win.

That would be dirty.

JD has published daily results. For the first five weeks they were winning and losing on different days, but overall the average was positive. Then a big whale started betting big, and the house lost 2500 BTC in one day. The house was negative for 10 whole days.

Can you imagine what would happen without max bet limits?

Date (2013)	Average BTC/day	Profit
Jun.20	69	69
Jun.21	52	35
Jun.22	15	-58
Jun.23	29	71
Jun.24	39	77
Jun.25	25	-46
Jun.26	14	-51
Jun.27	27	117
Jun.28	16	-68
Jun.29	19	48
Jun.30	15	-27
Jul.01	18	54
Jul.02	19	22
Jul.03	16	-25
Jul.04	40	380
Jul.05	40	35
Jul.06	41	68
Jul.07	81	750
Jul.08	76	-2
Jul.09	85	259
Jul.10	85	68
Jul.11	72	-201
Jul.12	80	260
Jul.13	-27	-2,485
Jul.14	-41	-368
Jul.15	-63	-610
Jul.16	-57	96
Jul.17	-91	-1,005
Jul.18	-81	177
Jul.19	-48	907
Jul.20	-12	1,092
Jul.21	-5	188
Jul.22	-8	-102
Jul.23	9	573
Jul.24	45	1,284

 The house was ahead 1836 BTC on July 12th

[leex1528]

i'm not sure what you mean by roll counter.

I have been trying to make a version 2 where it adds the option to adjust the win chance, but it proves to be rather difficult to get it working properly.

I think the percentages are pretty accurate, however the maximum lose streak might not be accurate, and also the amount of bitcoin lost is sometimes several orders of magnitude greater than the bankroll (or even all bitcoins in existence for that matter).

so there's still some flaws in the calculations but i can't be arsed to find them at the moment.

If you're interested in seeing it here is the link: https://docs.google.com/spreadsheets/d/1urhC3Su1ByvstzfKu5SOwUKYQm2xO0cXOisK8hawNn8/


/edit i may or may not have found the cause, it may or may not be fixed

you mean like the amount of attempts (for example stop after 'x' amount of losses and than start again with the same conditions)?

I could probably do that.

Yes that is what I mean, I mean it would be nice to say, I want to have enough cash to cover 5 losses, if I roll 50 times, what are the odds of me winning/losing?  Expected Profit, etc

[CryptoKilla]

Never bet against Doog... Everything he touches turns to gold

[zimmah]

i'm not sure what you mean by roll counter.

I have been trying to make a version 2 where it adds the option to adjust the win chance, but it proves to be rather difficult to get it working properly.

I think the percentages are pretty accurate, however the maximum lose streak might not be accurate, and also the amount of bitcoin lost is sometimes several orders of magnitude greater than the bankroll (or even all bitcoins in existence for that matter).

so there's still some flaws in the calculations but i can't be arsed to find them at the moment.

If you're interested in seeing it here is the link: https://docs.google.com/spreadsheets/d/1urhC3Su1ByvstzfKu5SOwUKYQm2xO0cXOisK8hawNn8/


/edit i may or may not have found the cause, it may or may not be fixed

you mean like the amount of attempts (for example stop after 'x' amount of losses and than start again with the same conditions)?

I could probably do that.

Yes that is what I mean, I mean it would be nice to say, I want to have enough cash to cover 5 losses, if I roll 50 times, what are the odds of me winning/losing?  Expected Profit, etc

I can do that, but i have no idea when i have time/are in the mood to make such an update, it's will require pretty much effort to add these statistics.

[ondratra]

24 pages? Realy guys?

Mb we should make some page about basics of gamble, probability, etc.

[Cluster2k]

Even without a betting limit and the ability to start a martingale at 1 satoshi, the laws of mathematics dictate eventually the player will hit a long enough run of losses to lose the lot if they play long enough.  The other factor is that after 10 or so losses the bet amount required is ridiculous compared to what the player is trying to win.  Betting $100k to win back the initial $1 bet and all the losses.  It's insane.

But having said that... gambling is built around stupid people trying to beat the laws of mathematics.  Play on as long as I'm an investor 

[PacoMartin]

Yes that is what I mean, I mean it would be nice to say, I want to have enough cash to cover 5 losses, if I roll 50 times, what are the odds of me winning/losing?  Expected Profit, etc

To clarify, "I want to have enough cash to cover 5 losses" would mean you have 31 units, or 1+2+4+8+16=31 . Or if you lose the first 5 times you play, then your bankroll will be exactly zero.

If you were playing toss the fair coin, on average you would see 5 losses in a row with 62plays
If you were playing Just Dice, on average you would see 5 losses in a row with 59.5 plays
If you were playing European Roulette, on average you would see 5 losses in a row with 55.5 plays
If you were playing American Roulette, on average you would see 5 losses in a row with 50 plays

The zero house edge probability is easily calculated as 2^(k+1)-2 where k=5 . It's easy to do that calculation and then subtract a little for different house edges.

I am ignoring the latter part of your question because you might be in the middle of a losing streak at exactly 50 rolls.

It is more exact to pick your game and designate what the average number of rolls before you see your designated losing streak.

If you are having trouble envisioning this, then consider a streak of only 2 losses. On average this will happen every 2^(2+1)-2=6 tosses of a fair coin.

I've simulated it in a spreadsheet if you are having trouble doing it yourself. Download the file so you can re-caculate the spreadsheet with different random numbers. You can add more than 100 runs. But the average before getting 2 losses in a row will be around 6.

https://docs.google.com/spreadsheet/ccc?key=0Aor7l039-pqHdDR1SXNONV9KQW1Id3F6cGQ0bUdPblE&usp=drive_web#gid=0

[cp1]

Yes that is what I mean, I mean it would be nice to say, I want to have enough cash to cover 5 losses, if I roll 50 times, what are the odds of me winning/losing?  Expected Profit, etc

Don't you mean expected loss?

[leex1528]

Yes that is what I mean, I mean it would be nice to say, I want to have enough cash to cover 5 losses, if I roll 50 times, what are the odds of me winning/losing?  Expected Profit, etc

Don't you mean expected loss?

No, I mean expected profit.  Just because you don't understand Martingale doesn't mean it fails 100% of the time:) 

Yes, if you had 5 losses and you rolled 500 times, your loss % is probably close to 100.  I would like to know the expected profit though

[leex1528]

Yes that is what I mean, I mean it would be nice to say, I want to have enough cash to cover 5 losses, if I roll 50 times, what are the odds of me winning/losing?  Expected Profit, etc

To clarify, "I want to have enough cash to cover 5 losses" would mean you have 31 units, or 1+2+4+8+16=31 . Or if you lose the first 5 times you play, then your bankroll will be exactly zero.

If you were playing toss the fair coin, on average you would see 5 losses in a row with 62plays
If you were playing Just Dice, on average you would see 5 losses in a row with 59.5 plays
If you were playing European Roulette, on average you would see 5 losses in a row with 55.5 plays
If you were playing American Roulette, on average you would see 5 losses in a row with 50 plays

The zero house edge probability is easily calculated as 2^(k+1)-2 where k=5 . It's easy to do that calculation and then subtract a little for different house edges.

I am ignoring the latter part of your question because you might be in the middle of a losing streak at exactly 50 rolls.

It is more exact to pick your game and designate what the average number of rolls before you see your designated losing streak.

If you are having trouble envisioning this, then consider a streak of only 2 losses. On average this will happen every 2^(2+1)-2=6 tosses of a fair coin.

I've simulated it in a spreadsheet if you are having trouble doing it yourself. Download the file so you can re-caculate the spreadsheet with different random numbers. You can add more than 100 runs. But the average before getting 2 losses in a row will be around 6.

https://docs.google.com/spreadsheet/ccc?key=0Aor7l039-pqHdDR1SXNONV9KQW1Id3F6cGQ0bUdPblE&usp=drive_web#gid=0

What would be cool I think is this: 
Initial bet: 1 unit
Max bet: 16 units
# of bets till you quit:  5
# of bets you want to make:  40
% chance of busting:  30
Average Profit:  16 units
Where you can fill/edit those numbers and it would calculate how well you would do.

[cp1]

Your expected profit is negative.

[leex1528]

Your expected profit is negative.

Actually that just isn't true.  If you have a buffer of 10 rolls, and you roll 100 times.  Your expect PROFIT is going to probably be around 40 base units...

[PacoMartin]

What would be cool I think is this:  
Initial bet: 1 unit
Max bet: 16 units
# of bets till you quit:  5
# of bets you want to make:  40
% chance of busting:  30
Average Profit:  16 units
Where you can fill/edit those numbers and it would calculate how well you would do.

A table like this would say that after 40 plays you have a 46.792% of losing 5 in a row. Now you may or may not be out of chips at this point, because if you started with 31, then you will have won some in the previous throws. So it partly depends on your definition of "bust".

The odds of losing 5 in a row after 5 plays is 1/2^5=3.125% . Now the longer you play, the odds keep increasing until you play a lot of times, then they get close to 100%.

But on average you will play 62 times before getting 5 losses in a row.

A specified average profit will be difficult to say. If you play 40 times you should win half of them on average.

Now these numbers have to be adjusted for the house edge of a real gambling game.

Plays	Probability of losing 5 in a row
5	3.125%
6	4.688%
7	6.250%
8	7.813%
9	9.375%
10	10.938%
11	12.451%
12	13.940%
13	15.405%
14	16.846%
15	18.262%
16	19.653%
17	21.021%
18	22.366%
19	23.688%
20	24.987%
21	26.264%
22	27.520%
23	28.754%
24	29.967%
25	31.159%
26	32.331%
27	33.483%
28	34.616%
29	35.729%
30	36.823%
31	37.899%
32	38.956%
33	39.996%
34	41.017%
35	42.021%
36	43.009%
37	43.979%
38	44.933%
39	45.870%
40	46.792%
41	47.698%
42	48.588%
43	49.464%
44	50.324%
45	51.170%
46	52.001%
47	52.818%
48	53.622%
49	54.411%
50	55.188%
51	55.951%
52	56.700%
53	57.438%
54	58.162%
55	58.875%
56	59.575%
57	60.263%
58	60.940%
59	61.605%
60	62.258%
61	62.901%
62	63.533%

            

[leex1528]

What would be cool I think is this: 
Initial bet: 1 unit
Max bet: 16 units
# of bets till you quit:  5
# of bets you want to make:  40
% chance of busting:  30
Average Profit:  16 units
Where you can fill/edit those numbers and it would calculate how well you would do.

A table like this would say that after 40 plays you have a 46.792% of losing 5 in a row. Now you may or may not be out of chips at this point, because if you started with 31, then you will have won some in the previous throws. So it partly depends on your definition of "bust".

The odds of losing 5 in a row after 5 plays is 1/2^5=3.125% . Now the longer you play, the odds keep increasing until you play a lot of times, then they get close to 100%.

But on average you will play 62 times before getting 5 losses in a row.

A specified average profit will be difficult to say. If you play 40 times you should win half of them on average.

Now these numbers have to be adjusted for the house edge of a real gambling game.

Number of plays	Probability of losing 5 in a row
5	3.125%
6	4.688%
7	6.250%
8	7.813%
9	9.375%
10	10.938%
11	12.451%
12	13.940%
13	15.405%
14	16.846%
15	18.262%
16	19.653%
17	21.021%
18	22.366%
19	23.688%
20	24.987%
21	26.264%
22	27.520%
23	28.754%
24	29.967%
25	31.159%
26	32.331%
27	33.483%
28	34.616%
29	35.729%
30	36.823%
31	37.899%
32	38.956%
33	39.996%
34	41.017%
35	42.021%
36	43.009%
37	43.979%
38	44.933%
39	45.870%
40	46.792%
41	47.698%
42	48.588%
43	49.464%
44	50.324%
45	51.170%
46	52.001%
47	52.818%
48	53.622%
49	54.411%
50	55.188%
51	55.951%
52	56.700%
53	57.438%
54	58.162%
55	58.875%
56	59.575%
57	60.263%
58	60.940%
59	61.605%
60	62.258%
61	62.901%
62	63.533%

            

I agree it would be hard to calculate expected profit.  Maybe if you just used your base earnings...it would be easier?  hmm who knows..

[cp1]

Your expected profit is negative.

Actually that just isn't true.  If you have a buffer of 10 rolls, and you roll 100 times.  Your expect PROFIT is going to probably be around 40 base units...

If you do that your expected "profit" is -5 base units.

[leex1528]

Your expected profit is negative.

Actually that just isn't true.  If you have a buffer of 10 rolls, and you roll 100 times.  Your expect PROFIT is going to probably be around 40 base units...

If you do that your expected "profit" is -5 base units.

I don't think that is quite accurate...

If I remember correctly, losing 10 rolls in a row roughly should happen 1 sequence out of 32,000.   

So my Guess, saying you either win around 50 hands with a couple hands before 100 you lose....I am guessing you will be up...not negative...

[sed]

Your expected profit is negative.

Actually that just isn't true.  If you have a buffer of 10 rolls, and you roll 100 times.  Your expect PROFIT is going to probably be around 40 base units...

If you do that your expected "profit" is -5 base units.

I think some of the folks here don't know the definition of expectation or the definition of profit

Folks, expectation is the value you get when you calculate the long term average for your situation (it factors out all local variance).

Profit is the amount you have *above* what you started out with.  Now if you can forsee the result of all of your bets in such a way that at a real casino your expectation is positive profit, then you are a prophet.

[leex1528]

Your expected profit is negative.

Actually that just isn't true.  If you have a buffer of 10 rolls, and you roll 100 times.  Your expect PROFIT is going to probably be around 40 base units...

If you do that your expected "profit" is -5 base units.

I think some of the folks here don't know the definition of expectation or the definition of profit

Folks, expectation is the value you get when you calculate the long term average for your situation (it factors out all local variance).

Profit is the amount you have *above* what you started out with.  Now if you can forsee the result of all of your bets in such a way that at a real casino your expectation is positive profit, then you are a prophet.

Well that isn't true.....

I could easily go walk into a casino today, lay down 1000 dollars and probably end up + 5 dollars so I would walk away at $1005 dollars.

It hardly makes me a prophet, playing Martingale at a casino and winning 1 hand the odds are HIGHLY in your favor.  But, if you keep doing that, eventually you are going to lose the $1000 dollars...

[PacoMartin]

If I remember correctly, losing 10 rolls in a row roughly should happen 1 sequence out of 32,000.   

No that is too big of a number. I assume by roll you mean coin toss. The odds of losing the next 10 tosses in a row are 1024 to 1.

The expected (or average) number of rolls until you lose 10 in a row is 2046 (2*1024-2).

If you are playing a game with a house edge (like American roulette and betting black or red) then the expected number of times is 1292

[cp1]

Here's a histogram of your results (100 rolls, 1% house edge; 100,000 simulations; 1024 bet BR).  Expected loss = 4.9 base bets

Code:

-1024.000000    3493
-1013.020000    22
-1002.040000    40
-991.060000     40
-980.080000     124
-969.100000     227
-958.120000     352
-947.140000     433
-936.160000     387
-925.180000     150
-914.200000     41
-903.220000     2
-892.240000     0
8.120000        0
19.100000       0
30.080000       2731
41.060000       57726
52.040000       33758
63.020000       474