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probability of success in trading according to the risk-reward

  • #1
  • First Post: Jun 15, 2016 11:16pm Jun 15, 2016 11:16pm
Hello everyone

lately I've thought about the possibility of using semi - martingale in trading.

I have managed to make a table in excel, which takes into account the conditions of the broker (leverage, spread, comission, etc.) and the conditions of the trader (initial capital, risk, risk-reward ratio etc)

in some configurations like this (see picture)

Attached Image (click to enlarge)
Click to Enlarge Name: 8a92919da2e1185571d1aaa4936a09f3.png Size: 50 KB


I've tried using the demo broker accounts that the numbers are correct

but the question is ...

taking into account the risk-reward (1: 6), how likely lose 19 consecutive operations ??

hopefully someone can show me how I can calculate the probability

-Sorry for my bad English-
  • #2
  • Jun 15, 2016 11:35pm Jun 15, 2016 11:35pm
Quoting ekud87
Hello everyone lately I've thought about the possibility of using semi - martingale in trading. I have managed to make a table in excel, which takes into account the conditions of the broker (leverage, spread, comission, etc.) and the conditions of the trader (initial capital, risk, risk-reward ratio etc) in some configurations like this (see picture) {image} I've tried using the demo broker accounts that the numbers are correct but the question is ... taking into account the risk-reward (1: 6), how likely lose 19 consecutive operations ?? hopefully...
Hi,
ekud87
There are too many unknown variables to calculate the probability. Your trading style, how long you want to hold on to trades, each pair and its behavior, and whether you are trading manually or automated. If it is automated, do many backtests. If it is manual, look at some of the bigger swings on your chosen timeframe. Your entry points will be critical with such a tight stop loss.
You can lead a man to knowledge, but you can't make him think.
  • #3
  • Jun 16, 2016 9:39am Jun 16, 2016 9:39am
of course, many factors can influence the outcome.

say a random and automatic system pick a RR 1: 6.

if we have a random and automatic system pick a RR (1: 1), the probability of winning an operation should not be 50% ??

then what should be the probability of losing 20 consecutive operations of an automatic and random system looking for a RR of 1: 1

What should be the probability of losing 20 consecutive operations of an automatic and random system looking for a RR of 1: 6

???? I have no idea how to calculate this

  • #4
  • Jun 16, 2016 8:17pm Jun 16, 2016 8:17pm
in the following link


http://www.forexfactory.com/showthre...72#post7141572

the "Proximus" user shows the chances of losing 20 operations consecutively.

but with a RR 1: 6, as could apply the calculations ??

anyone have any idea?

  • #5
  • Jun 17, 2016 12:16am Jun 17, 2016 12:16am
If you assume no spread and a completetly random market, then a 1:1 trade has a 50% of winning, because there are 2 ways the market could move and winning is one of them, 1/2 = 0.5 = 50%. If you wanted to know the chances winning 5 in a row, it's 0.5 x 0.5 x0.5 x 0.5 x 0.5 which works out at 3.125% (you can use the excel function =POWER(0.5,5) to work this out). 3.125% means the chances of winning 5 coin tosses in a row are one in 32 (100 / 3.125).

WINNING 20 1:6 trades in a row is (in excel) =POWER(1/7,20) - "7" because you have 6 chances of losing and 1 chance of winning (assuming random behaviour)...and the answer in as close to zero probibility as matters

LOSING 20 1:6 trades in a row is (in excel) = Power (6/7,20) -"6/7" because there are 6 chances to lose out of 7 posibilities and is about 4.6% or about one in 22 (100 / 4.6)

However ..... you need to think about the fact you will have to factor in spread costs on each trade and that the market does not always exhibit purely random behaviour (trends exist, even if you may only know wabout them after they have happened). Hope that helped, if it did please take a look at one of my recent threads and consider taking part in the very short survey (please!).

Thanks
Chris
  • #6
  • Jun 17, 2016 2:03am Jun 17, 2016 2:03am
Quoting k6chris
WINNING 20 1:6 trades in a row is (in excel) =POWER(1/7,20) - "7" because you have 6 chances of losing and 1 chance of winning (assuming random behaviour)...and the answer in as close to zero probibility as matters LOSING 20 1:6 trades in a row is (in excel) = Power (6/7,20) -"6/7" because there are 6 chances to lose out of 7 posibilities and is about 4.6% or about one in 22 (100 / 4.6)
This does make one assumption namely that an R:R of 1:6 is six times less likely than an R:R of 1:1. That's quite a big leap to make without any supporting evidence.
The only way you can get a reasonable idea is from knowing the actual win/loss ratio. What you would need to do is to select a strategy, back/forward test it over a significant period of time covering trending, flat and choppy markets. Once you've got the actual ratio you can use Chris's formula to get an answer.

However there are two BIG caveats:

  1. Firstly past performance does not guarantee a similar result in the future.
  2. What ever the odds, sooner or later they WILL catch up with you.

David

Too tired to argue. Too old to care.
  • #7
  • Jun 17, 2016 7:32am Jun 17, 2016 7:32am
Hello everyone!

thanks for your answers, the truth is that they give me more clear about this idea.

regarding the response of k6chris, I have a question ( ¿ because 6/7 ? ):
Quote
LOSING 20 1:6 trades in a row is (in excel) = Power (6/7,20) -"6/7" because there are 6 chances to lose out of 7 posibilities and is about 4.6% or about one in 22 (100 / 4.6)



regarding David's response:
Quote
However there are two BIG caveats:

 

  1. Firstly past performance does not guarantee a similar result in the future.
  2. What ever the odds, sooner or later they WILL catch up with you.


of course, I am aware that past performance does not guarantee future results, I also know that using a modification of the martingale someday reach the final
the question is...

How much can be deferred that end?

I think if I can increase the number of operations that I can lose consecutively, I can postpone also (slightly) the end of the account
  • #8
  • Jun 17, 2016 8:19am Jun 17, 2016 8:19am
Quote
I have a question ( ¿ because 6/7 ? ):

Think of a dice that has 6 sides, chances of throwing a 'three' is 1/6 because there are 6 possible outcomes and only one of them is a 'three'. Chances of not throwing a 'three' are 5 out of 6, or 5/6 (0.83 or 83%). 6/7 says that (assuming a random market) a 6:1 RR trade will win 1 and lose size, so for every 7 trades, 6 wil be losers, hence 6/7. Sorry, not very well explained :-( Don't forget that in real life you pay the broker a spread for each of those 7 trades too!
  • #9
  • Jun 17, 2016 11:13am Jun 17, 2016 11:13am
ok I understand the point.

then (if I understood well) according to these calculations, there is a 4.6% chance of losing 20 operations consecutively?

therefore, correct would say that there is a 95.6% chance of winning one trade in a "round" of 20 trades?
  • #10
  • Jun 18, 2016 2:32am Jun 18, 2016 2:32am
Quoting ekud87
of course, I am aware that past performance does not guarantee future results, I also know that using a modification of the martingale someday reach the final the question is...
I probably should not have have added the caveats as it detracts from my main point namely that odds and RR are not necessarily equal.

As I understand it you are deriving a number by saying that an RR of 1:6 will be achieved once every seven times and presumably 1:5 every six times and so on.

IMHO that is a risky assumption. It is easy to imagine that the vast majority of trades are closed well before 1:6. If this is the case then the odds of the 1:6 would be much worse than you assume. To go back to Chris's dice analogy it is like having a dice loaded to come down preferentially on 1. It still has six sides but the chances of that 3 coming up are now much worse than 1 in 6.

To be fair I don't have any evidence either way for this premise, maybe you do. Ultimately of course it is up to you, its your money your risk and your decision and I wish you well and big profits with it.
Too tired to argue. Too old to care.
  • #11
  • Jun 18, 2016 4:36am Jun 18, 2016 4:36am
Quoting ekud87
Hello everyone lately I've thought about the possibility of using semi - martingale in trading. I have managed to make a table in excel, which takes into account the conditions of the broker (leverage, spread, comission, etc.) and the conditions of the trader (initial capital, risk, risk-reward ratio etc) in some configurations like this (see picture) {image} I've tried using the demo broker accounts that the numbers are correct but the question is ... taking into account the risk-reward (1: 6), how likely lose 19 consecutive operations ?? hopefully...
There is a stake calculator (search on google) which shows what are the odds of losing N traders in a row from M total trades. For example a probability of 9 losing trades in a row from 1000 total trades is....100%
That's why Martingale is all about random quits that save your money and start all over again.
  • #12
  • Jun 18, 2016 6:40am Jun 18, 2016 6:40am
Quoting leebsurag
{quote} There is a stake calculator (search on google) which shows what are the odds of losing N traders in a row from M total trades. For example a probability of 9 losing trades in a row from 1000 total trades is....100% That's why Martingale is all about random quits that save your money and start all over again.
I can not find the calculator ... you have the link?
  • #13
  • Edited 9:33am Jun 18, 2016 6:44am | Edited 9:33am
Quoting davidb
{quote} I probably should not have have added the caveats as it detracts from my main point namely that odds and RR are not necessarily equal. As I understand it you are deriving a number by saying that an RR of 1:6 will be achieved once every seven times and presumably 1:5 every six times and so on. IMHO that is a risky assumption. It is easy to imagine that the vast majority of trades are closed well before 1:6. If this is the case then the odds of the 1:6 would be much worse than you assume. To go back to Chris's dice analogy it is like having...
hello davidb

Well, I've done an backtest of just over a year in m1 (in the American session) during this course of time I have not come to find more than 17 consecutive bad trades of a simple system (divergences in favor of the trend)

It is for this reason that I am "encouraged" to apply the calculations of the table I did.

even so, I am aware that at any time can reach 20 ... 30 ... 50 consecutive bad trades, is only a matter of time
  • #14
  • Jun 18, 2016 7:34am Jun 18, 2016 7:34am
Quoting ekud87
{quote} Well, I've done an backtest of just over a year in m1 (
That's great, in that case you should be able to calculate the win/loss ratio from your backtest which then gives you the feed into Chris's formula and out pops the answer you want.

Job done!!

When you do get the answer please post it, I'd be interested to see what the result is.
Too tired to argue. Too old to care.
  • #15
  • Jun 18, 2016 9:36am Jun 18, 2016 9:36am
Quoting davidb
{quote} That's great, in that case you should be able to calculate the win/loss ratio from your backtest which then gives you the feed into Chris's formula and out pops the answer you want. Job done!! When you do get the answer please post it, I'd be interested to see what the result is.
of course; The idea behind is to analyze whether the result of my backtest was just "lucky" or if there was any real chance of expect those results.
rinse thoroughly as the result of the probability of excell table passeth to make a small EA to help us obtain more precise data on the number of consecutive bad trades.
  • #16
  • Jun 18, 2016 10:29am Jun 18, 2016 10:29am
the question is ... taking into account the risk-reward (1: 6), how likely lose 19 consecutive operations ?? hopefully someone can show me how I can calculate the probability -Sorry for my bad English-[/quote]


very easy in Excel, I have done this myself several times. You make an excel table with the random number generator in one column, in another column you say win or lose, with an 'if' function. Your 'if' function will see check each random number and say numbers less than 0.166667 is a win and greater than or equal to 0.166667 is a loss. You make both columns very long (I went for 2000 cells) and you will have a reasonably big sample.

You can then write another function to count how many losses you have in a row. Or you can do it manually. I did this just now and the first 23 'trades' were losses, and I had 6 sets of longer than 19 losses in a row before I had 400 trades, and the longest losses in a row was 27.

This is obviously all random and when you 'F9' (recalculate) then you will get new results which you can also check. The point of the exercise is for you to see how many losses in a row you are reasonably likely to see with your present risk-reward of 1 in 6 so you can set your position size accordingly.

However, the main problem you are likely to face is that you clearly don't have a strategy since you are looking at a risk-reward as your probability definer. That's not going to work because your strategy will set your win/loss probability. But you have an overall good idea, you just need to go about it slightly differently. You need to work out a strategy and try it many many times, and your strategy (which has its own risk reward ratio in it) should give you an overall over time win-loss ratio. And then you can use your win-loss ratio which you calculated based on real results as your win loss ratio for your excel spreadsheet, then you set up your table as I wrote above and decide on your risk percentage based on how many times you think you are reasonably likely to lose in a row.

It's a long process, but I don't think there is another way to get where you want to go.
  • #17
  • Edited 5:30pm Jun 18, 2016 3:26pm | Edited 5:30pm
Quote
very easy in Excel, I have done this myself several times. You make an excel table with the random number generator in one column, in another column you say win or lose, with an 'if' function. Your 'if' function will see check each random number and say numbers less than 0.166667 is a win and greater than or equal to 0.166667 is a loss. You make both columns very long (I went for 2000 cells) and you will have a reasonably big sample.

You can then write another function to count how many losses you have in a row. Or you can do it manually. I did this just now and the first 23 'trades' were losses, and I had 6 sets of longer than 19 losses in a row before I had 400 trades, and the longest losses in a row was 27.

This is obviously all random and when you 'F9' (recalculate) then you will get new results which you can also check. The point of the exercise is for you to see how many losses in a row you are reasonably likely to see with your present risk-reward of 1 in 6 so you can set your position size accordingly.

However, the main problem you are likely to face is that you clearly don't have a strategy since you are looking at a risk-reward as your probability definer. That's not going to work because your strategy will set your win/loss probability. But you have an overall good idea, you just need to go about it slightly differently. You need to work out a strategy and try it many many times, and your strategy (which has its own risk reward ratio in it) should give you an overall over time win-loss ratio. And then you can use your win-loss ratio which you calculated based on real results as your win loss ratio for your excel spreadsheet, then you set up your table as I wrote above and decide on your risk percentage based on how many times you think you are reasonably likely to lose in a row.

It's a long process, but I don't think there is another way to get where you want to go.

hello epidural

It is a good idea to use the random number generator, but could you explain to me why use smaller numbers to 0.166667?
  • #18
  • Edited 8:24pm Jun 24, 2016 2:14pm | Edited 8:24pm
Hello everyone

I made a small program in visual basic, the program takes "randomly" a value of stoploss.

acontinuacion also takes "randomly" takeprofit worth of.

then compares the value of stoploss and value of takeprofit to calculate whether the operation had been winning considering the R: R
the results I get for more than 40,000 samples, is that the probability of losing 20 trades in a row is 2%

Attached Image (click to enlarge)
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40,000 trades (2000 rounds of 20 trades), lost only 20 trades consecutively 28 times.
equivalent to 1.4%

Attached Image (click to enlarge)
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400.000 trades (20000 rounds of 20 trades), lost only 20 trades consecuutively 370 times.
equivalent to 1.85%
  • #19
  • Jun 24, 2016 8:42pm Jun 24, 2016 8:42pm
now ... having this pair of tools (excell sheet that allows me to set the boundaries of the semi-martingale, and the small software that gives me an idea of the probability of loss)

the question is, is it possible that the probability of loss of the merchant, is less than the calculated probability with random data?

the question arises that the trader does not make random entries, but rather operates with a clear system that allows you to set when to enter the market

  • #20
  • Jun 25, 2016 2:02pm Jun 25, 2016 2:02pm
if anyone is interested in the probability of loss calculator with martingale ... (in the folder bin/relase, this executable file)
Attached File(s)
File Type: zip Calculadora Probabilidad Martingala.zip   123 KB | 485 downloads
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