Chance of Consecutive Losses

I am trying to figure out what is a safe money management scheme (ie. how much to risk per trade) based on the statistical probability of a certain number of consecutive losses (assume a risk:reward of 1).

I know that the percent of the account remaining after a number of consecutive losses (CL) will be:

Percent remaining = 100*[(1+risk)^(-CL)]

So, if one risked 1% on every trade and had 20 losses in a row, you would have 100*[(1+0.01)^(-20)] = 81.9% of your account remaining.



Now one has to figure out what the chance of a certain number of consecutive losses is and this is where I need help/input. Does anyone know how to calculate this?


I'm pretty sure my following logic is incorrect:
It makes sense to me that if we have a system that has a 50% win rate, then the chance of 5 consecutive losses is 0.5^5= 0.03125 = 3%
If my system has a 97% win rate, then the chance of 5 consecutive losses is 0.03^5 = 0.00002% (which is 1 in 41,152,263). The chance of 2 consecutive losses is 0.09% (or 1 in 1111).

By solving for Risk in the first equation above, it would appear that if I was comfortable losing 20% of my equity due to 2 consecutive losses (1 in a 1000 trades in my 97% win rate system), I could risk up to 11.8% per trade (which seems insane).

Risk = 10^[log(0.8)/-2] - 1 = 0.118 = 11.8%

Any criticisms? I'm sure something isn't right with this train of thought/math...

http://www.forexfactory.com/showthread.php?t=236755

Here's the original work on the theory of runs. Seventy years old, but math results don't rust.

At the end, you'll find asymptotic results, which are useful when binomial coefficients get too big (n is too large).

http://projecteuclid.org/DPubS?servi...oms/1177731825

Thanks for the interesting article.

Lots of detailed maths to show that a large amount of random binomial (eg. forex) data will usually demonstrate a distribution that is asymptotically normal (not the same as an asymptotic function):

"The distribution of a statistic is said to be asymptotically normal if, as the sample size increases, the distribution of the statistic approaches a normal distribution"

Unfortunately, I haven't quite figured out how to use those equations to get a final numerical answer as to a probability of a certain outcome (eg. run of losses). It would be nice if these published theorems gave some examples!

You need some more info:

1) What level of confidence do you want your results to have? I would suggest 99.9% (1.0 confidence is not achievable), but you may be feeling "risky" and be comfortable with a lower confidence level.

2) What is the total number of trades in your trial? 1,000----5,000? Obviously, the more trades you make, the higher the chance of a bad run against you.

3) What is the win rate of your system?

Once you know these variables:

N= log(1-DC) / log(1-PW)

For example: 99.9% confidence with a win rate of 60%

N=log(1-.999) / log(1-.6)
N=7.5

For every 1000 trades you make, you would expect to hit at least 1 string of 7 or 8 losers in a row. No way to know when this string will happen. Also, you may hit more or less bad streaks in your limited experience. The results given here will conform to a large number of 1000 trade trials. In any 1 given 1000 trade trial, anything can happen....and often does. Hope this helps.

Regards,

ps. a 60% win rate for 1/1 rr is pretty high, if not impossible.

Do you have any idea where can I find the source or derivation of this formula? I think this could be a quick approximation method because it seems that the N consecutive losses is underestimated.

Go to your local university book store (or library) and browse through a graduate statistics textbook. Look for sections that relate to probability and random processes. The formula is correct, and the number N that it yeilds is correct. It "seems" like a low number because I used .60 as a win %. As I said, that number is not really realistic. I have an entire thread on that subject here: http://www.forexfactory.com/showthread.php?t=191230

In conclusion, I will tell you that I have been down the road you are on right now. I have spent many, many months trying to quantify the markets volitility and bracket it. It cannot be done. You are on the wrong road. It is a waste of time and energy and is an exercise in futility. There are roads to profitability in forex..... but this is not one of them. Good trades to you and all the best.

Regards,

N = log(1-DC) / log(1-PW)
=> N x Log(1-PW) = log(1-DC)
=> (1-PW)^N = 1-DC

This is no difference than how OP and most people calculate the chance of consecutive losses. It is just the way you present it in different way (same with acrary).

Btw market volatility has nothing related to this thread right? Anyway it is not that hard to quantify it, and this is not a waste of time.

When I get tired of trying to academically derive things like this, I just write a simulation program. Not elegant but gets the answers I want.

Thank you all for your thoughts. I agree it is exactly the same as the equation I used, just presented in a different manner (and useful in either format). For instance:
a) the likelihood of 2 consecutive losses on a 97% win rate is 1 in 1111 = 0.03^2.
b) the longest length of consecutive losses likely in 1000 samples is 1.96 = log(0.001)/log(0.03) = approx 2.

This thread is not a waste of time (and not about volatility) as I am not trying to come up with a money management scheme to beat the markets. I'm just trying to calculate what would be a comfortable amount to risk and how many pairs to trade.

So back to the original example in post 1 (97% win rate risking 11.8% per trade) and trading one pair, I know the chance of a run of 2 consecutive losses(and 20% drawdown) is 1 in 1111. Now, if I trade five pairs simultaneously, it would seem initially the chance of 2 consecutive losses is 5 times more likely. But after more thought, I think it is still 1 in 1111.

What do you think? Again, is there a flaw in my math?

(You would think if I risked 11.8% on every trade with one pair, I should lower my risk if I traded five pairs - say lower down to 11.8/5 = 2.4%. But then again, the next trade is independent of past trades and thus has a 97% chance to win regardless of how many pairs are traded)

If only I had the skills to write a simulation program...so I must depend on my think-tank to arrive at the theoretical statistical outcomes...which, at best, is still just a probability or confidence factor rather than any kind of certainty.

With 97% win rate, the probability to encounter at least 1 time of 2 consecutive losses is:

1. 0.8% for 10 trades.
2. 8.5% for 100 trades.
3. 36.2% for 500 trades.
4. 59.3% for 1000 trades.

With 97% win rate, the probability to encounter 2 losses in next 2 trades is 0.03^2 = 0.09%.

see the difference?

Yes, this is what I am looking for, FQ! I just wanted to better understand what to expect out of a certain system with specific win rate (so that risk could be better determined).

I understand that I was calculating the percent chance of 2 consecutive losses (CL) in the next two trades. Whereas the problem is combinatorial in nature.

That is, if you had 10 trades, one could have 2 consecutive losses on trade#1&2, #2&3, #3&4, 4&5, 5&6, 6&7, 7&8, 8&9, or #9&10. Nine different ways essentially.

So the chances would be a sum or (0.03^2)*9 = 0.09*9 = 0.81%
On 100 trades, you would have (0.03^2)*99 = 8.91%
On 1000 trades, (0.03^2)*999 = 90%
On 10000 trades, 900% (meaning it will likely happen 9 times in 10000 trades?)

I assume this is how you calculated, but please share what formula you are using if different? (as I see you got 59.3% not 90% for 1000 trades)

I bet that you never checked the link at post #2, isn't it?

Nevermind I figure out another simple explanation for those who do not know how to use binomial distribution.

In your example, there are 4 possible cases at every 2 trades. It is either:

C1. Win & Win = 0.97 * 0.97 =0.9409
C2. Win & Loss = 0.97 * 0.03 = 0.0291
C3. Loss & Win = 0.03 * 0.97 = 0.0291
C4. Loss & Loss = 0.03 * 0.03 = 0.0009

So the probability to encounter C1 or C2 or C3 = 0.9409 + 0.0291 + 0.0291 = 0.9991. Or you can simply use 1 - 0.0009 = 0.9991 because the sum of all possible cases is always equal to 1.

Since You may encounter 1 time, or 2 times, or 3 times .....or maximum 9 times of 2 consecutive losses within 10 trades. So probability to encounter at least 1 time of consecutive losses is actually equal to:

0.0009^1 x 0.9991^8 +
0.0009^2 x 0.9991^7 +
0.0009^3 x 0.9991^6 +
0.0009^4 x 0.9991^5 +
0.0009^5 x 0.9991^4 +
0.0009^6 x 0.9991^3 +
0.0009^7 x 0.9991^2 +
0.0009^8 x 0.9991^1 +
0.0009^9 x 0.9991^0 = 0.81%

Hey what about if I got 1000 trades? It's gonna to be a very long calculation! But no worry because there is another simple solution.

Since we know the sum of probability of all possible cases is always equal to 1, we now ask a question in reverse. What is the probability for 0 consecutive losses within 10 trades? It is 0.0009^0 x 0.9991^9 = 99.19%.

Therefore the probability NOT to have 0 consecutive losses within 10 trades = 1 - 99.19% = 0.81%

End of story

p/s: 1st rules of probability: The sum of success cases and failure cases is always equal to 1 (p + q = 1), which also means the max probability is 100%, in regardless of scenarios. So if your calculation shows more than 100% for different scenario (10000 trades in your previous example), that means your logic is incorrect.

p/p/s: Do not apply this into martingale type of money management system. It wont beat the market without real edge.

I did read post#2, but that was over two weeks ago and I had forgotten that information. I hate having things spelled out for me, but thank you for the detailed explanation. You raise yet another very interesting way to solve the problem: "what is the probability of getting zero consecutive losses"

This is gonna be ugly, but here goes a summary equation of your math:

Probability of n consecutive losses (nCL) over N trades = 1-[{1-(1-WR)^nCL}^{N-(nCL-1)}]

So, if I had a 97% win rate and wanted to know the probability of 2 consecutive losses over 1000 trades = 1-[{1-(1-0.97)^2}^{1000-(2-1)}]

= 1-[{1-(0.0009)}^{1000-1}] = 1-[{0.9991}^{999}] = 1-[0.4068] = 0.5932 = 59%


Another example...the chance of 3 consecutive losses over 10000 trades = 23.6%

FQ, how do you think trading more than one pair simultaneously affect the above calculations?

I think what several people are trying to say is.... that calculations like this are not really of any use. I know you think that they are, but they are not. Intresting, yes. Braincandy, yes. Help you make money.... no. Here is why. As a small retail investor, the probability of consecutive losses really does not matter. It only matters what can happen to you in you limited experience. Believe me, it can be a hell of a lot worse than those probabilities reflect. I have made thousands of trades, and I hit one string of 17 losers in a row. If you trade long enough, you to will have a similar streak... maybe even worse. I know a trader who had 26 losers in a row. It can happen. The point is, as a small retail investor, you do not have the time or resources to make enough trades to ensure that your results will approach mathmatical expectancy. More than likley, you will always be on one side of the bell curve or the other. Given that retail forex has built in negative expectancy, the odds are that you will be on the negative side of the curve for most of your journey. You seem genuinely inquisitve and intelligent, so I tell you this with sincere goodwill.... The day I stopped looking for answers in mathematics and started looking at what any child can see is the day things started turning around for me. As a mathmetican, that was hard for me to do and accept. But I am not working for a hedge fund or bank making 100's of flash trades per day. That is the environment that such probabilites can be useful. Sorry if this sounds like a sermon. I am only trying to help.

Kindest Regards,

Interesting info.

Now you know that at one point in your trading career you can get 7-8 consecutive losses assuming your hypothetically high win rate.

So when that does happen (as your maths proves will at some point in time), you will almost wipe out your account if you allowed each 2 consecutive losses to take off 20% of your account.

So in summary:
-No don't risk that much ever.
-Risk very little
-This probably means i'm not going to be rich tomorrow/next month/year.

Sad I know, but it does take a while to get there.

Yes I agree that this will not help in making money, but it does help not to dig further when you are in a hole. Traders always make a basic assumption that our future win% will always fluctuate around historical win%. But how do we know that the future win% is still fluctuate within the acceptable noise level? or it just begin to rot? There are many techniques can be used to determine it, the number consecutive losses is one of the technique that help us from further trading when thing goes bad.

Using this method, we can know what is the number of consecutive losses which is very unlikely to happen (<1% probability) within 1000 trades.

Loss% ---- number of consecutive losses that happen <1% probability
5% ---- 4
10% ---- 5
20% ---- 8
30% ---- 10
40% ---- 13
50% ---- 17
60% ---- 23
70% ---- 33
80% ---- 52
90% ---- 109
95% ---- 220

So if your historical win% is 50% and now you have encounter 17 or more consecutive losses, you should stop trading your system now because your assumption that the future win% will fluctuate around historical win% is no longer valid (<1% validity).

This method also shows that most traders underestimate their risk of consecutive losses. They think that their system with 60% winrate will have less than 0.1% chances to have 8 consecutive losses but the truth is 48% of chances to encounter 8 consecutive losses! They have underestimated it by 480 times! How could we just make a conclusion that math is useless in trading if you do it wrong?

Math is just a tool or technique for problem solving. Even if you know how to calculate does not mean that you can solve the problem if you dont know where is the problem. Sometime knowing how to ask a correct and creative question is much more important than ability to find out the answer. It is even worse if you use a wrong solving technique, which cause you make a wrong conclusion.

Regarding the negative expectancy because of spread, I just want to say that if you manage to ask a correct question, the spread cost can be overcome, with high school math, easily. And I believe that you have travelled 1000 miles and just one step ahead of your destination.

Anyway I still appreciate your input because it really made me think deeper.

p/s: Please be reminded that if the |Z-score| > 0.1 then I would not recommend this method to determine whether your system is begin to rot because the current trade result is no longer independent of previous trade.

Yes, because they are no longer independent unless these pairs have 0 correlation.

p/s: if you can prove that you have 97% winrate system (and R:R = 1) without using martingale or average down technique, I will happily to teach you how to get rich quick (very quick!) beyond your wildest dream, without risking too much.