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Calculating probability of losing streaks

  • #21
  • Edited 6:50pm Jan 27, 2016 6:31pm | Edited 6:50pm
Quoting bdfranca
{quote} Really nice conclusion. I really needed that formula! I`m already using it but I didn`t understand at all. ForexQuant, can you please elaborate it? Why the first argument is 0 and not 1? If I`m asking to much can you just indicate me a book though? Thanks already!
Just saw your question. I seldom login here nowaday and hopefully it is not too late.

The 1st argument is 0 because Zero means the event of consecutive y losses in x number of trades does not happen at all. Since we want to know the probability of such event to happen at least 1 time and sum of the probability is always equal to 100%, therefore it is easier to deduct it from 1.

Edit: Just noticed that I had answered a similar question in http://www.forexfactory.com/showthre...41#post4336241
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  • #22
  • Jan 27, 2016 6:36pm Jan 27, 2016 6:36pm
Quoting triphop
Any mathematicians able to help out here? Stats have never been my finest point so... Probability of a losing streak is, in Excel speak: =(1+(n-s)*p)*(POWER(l,s)) where n=number of trades, s=number of losing streaks, p=probability of winning, l=probability of losing. Ok, so say I have a strat 23% successful over 220 trades, I get figures like this (written number of losses followed by probability) 10 361.21% 15 95.49% 20 25.23% 25 6.66% 30 1.76% ... something is amiss - 361%?? Any ideas?
There is a massive flaw in approaching the markets using statistics like these.


Say you have a 50% winning system. If you lose 25 times in a row, the % chance of your next trade losing is still 50%. This is because your next trade does not give a damn about your previous trades and is not influenced by them in any way, shape or form. The only thing that has influence from trade to trade is your account balance.

The gambler's fallacy, also known as the Monte Carlo fallacy or the fallacy of the maturity of chances, is the mistaken belief that, if something happens more frequently than normal during some period, it will happen less frequently in the future, or vice versa.
  • #23
  • Jan 27, 2016 7:14pm Jan 27, 2016 7:14pm
Quoting ForexQuant
{quote} Honestly I do not know the logic behind this equation. Why (1-Win%)^(consecutive losses) = N?

I use 50000 trades (being a conservative Large numbers guy) in my calc of Max consecutive losses using (Ln(50000)/-Ln((1-Pwin))).

eg. Pwin = 49.3% = 16 consecutive losses. Pretty close to the mark from my testing so far with 13 consec losers in a row (ouch) from 1262 trades from my discretionary approach.
  • #24
  • Jan 28, 2016 3:56am Jan 28, 2016 3:56am
Quoting Sis.yphus
{quote} There is a massive flaw in approaching the markets using statistics like these. Say you have a 50% winning system. If you lose 25 times in a row, the % chance of your next trade losing is still 50%. This is because your next trade does not give a damn about your previous trades and is not influenced by them in any way, shape or form. The only thing that has influence from trade to trade is your account balance. The gambler's fallacy, also known as the Monte Carlo fallacy or the fallacy of the maturity of chances, is the mistaken belief that,...
Oh to be green again.
  • #25
  • Edited 12:41pm Jan 28, 2016 11:36am | Edited 12:41pm
Quoting triphop
{quote} Oh to be green again.
You're the one who is green.

If what I said wasn't true we could take ANY strategy and wait for enough consecutive losses to occur before taking the signal and it would be a 100% winner.

Also if what I said wasn't true we could simply just wait for say 5 red candles to form in a row before betting long. I mean that would make our probability of losing only 0.03125%(.5 x .5 x .5 x .5 x .5 = 0.03125%) ... Who wouldn't take that bet?

Too bad this is not how the market works.

I took about a years worth daily candles to see how they were distributed.

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If you subtract each row from the one above it, you will get how many wins you would have had trading the upper row.

If you bet after 1 red candle formed the next would be green, you would have won 156 out of the 281 trades you took(55.516014234% win rate -- Your math says it should be 50%).


If you bet after 2 red candle formed the next would be green, you would have won 63 out of the 125 trades you took(50.04% win rate -- Your math says it should be 75%).


If you bet after 3 red candle formed the next would be green, you would have won 30 out of the 62 trades you took(48.387% win rate -- Your math says it should be 87.5%).

If you bet after 4 red candle formed the next would be green, you would have won 12 out of the 30 trades you took(40% win rate -- Your math says it should be 93.75%).

If you bet after 5 red candle formed the next would be green, you would have won 12 out of the 18 trades you took(66.66% win rate -- Your math says it should be 96.875%).

etc.

Notice non of these win rates add up to they should according the kind of math you are applying to the market. This is because of the gamblers fallacy.

And the reason the win rates are different, but all hovering around 50% is because of variance and the sample size. If you had a million samples of these patterns you all of the win rates would be right around 50%
  • #26
  • Jan 28, 2016 12:53pm Jan 28, 2016 12:53pm
Quoting Sis.yphus
{quote} You're the one who is green.
Yeah yeah, no that's what you are that is, we could go on .

Risk of ruin is for position sizing, but feel free to create a straw man and argue against it.

And as for the next trade having no bearing on the last. True unless... *fill that in *

I won't be drawn on why, but that's a good 90% of the battle.
  • #27
  • Jan 28, 2016 1:00pm Jan 28, 2016 1:00pm
Quoting triphop
{quote} Yeah yeah, no that's what you are that is, we could go on .
What? I do not understand your grammar.


Quoting triphop
{quote} Risk of ruin is for position sizing, but feel free to create a straw man and argue against it.
Position sizing has NOTHING to do with the probability of taking an individual win or loss. It just affects how big the wins and losses are.

Quoting triphop
{quote}And as for the next trade having no bearing on the last. True unless... *fill that in *
It is always true as long as they are separate trades coming from separate signals.
  • #28
  • Jan 29, 2016 2:17am Jan 29, 2016 2:17am
Are you actually on the right thread or... hold on. Your 5 quid is up.

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  • #29
  • Mar 20, 2016 6:49am Mar 20, 2016 6:49am
I was trying to figure this out and found this thread. If you want to work it out by hand then watch this video from the Khan academy. It uses a basketball example with the probability of a winning free throw of 80% and five shots. It then goes through two examples. The first example is the probability of getting exactly 3 baskets. Then second example is the probability of at least 3 baskets.

All you do is substitute the probabilities and number of baskets with your win/loss rate (whatever you are trying to analyse) and the number of shots with the number of trades.

Try it out with a small number of trades like 5 or 10. You will need excel to do larger sample sizes (use the BINOM.DIST function that ForexQuant mentioned above).

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  • #30
  • Feb 6, 2025 1:36pm Feb 6, 2025 1:36pm
Quoting Sis.yphus
{quote} There is a massive flaw in approaching the markets using statistics like these. Say you have a 50% winning system. If you lose 25 times in a row, the % chance of your next trade losing is still 50%. This is because your next trade does not give a damn about your previous trades and is not influenced by them in any way, shape or form. The only thing that has influence from trade to trade is your account balance. The gambler's fallacy, also known as the Monte Carlo fallacy or the fallacy of the maturity of chances, is the mistaken belief that,...
You misunderstood the intent of the problem.
He did not say what the probability of a single trade happening next time was, but the probability of a cluster of consecutive trades happening. This takes into account the Monte Carlo fallacy.
Suppose the probability p of a single trade of mine is 75% (win rate), then the probability of losing is (1-p).
So the probability of 4 consecutive losing trades is:
(1 - p) ^ 4 = 0.39%
No one says a next trade where the probability of all 4 consecutive losing trades.
  • #31
  • Last Post: Feb 10, 2025 9:56am Feb 10, 2025 9:56am
Quoting sniper2023
{quote} You misunderstood the intent of the problem. He did not say what the probability of a single trade happening next time was, but the probability of a cluster of consecutive trades happening. This takes into account the Monte Carlo fallacy. Suppose the probability p of a single trade of mine is 75% (win rate), then the probability of losing is (1-p). So the probability of 4 consecutive losing trades is: (1 - p) ^ 4 = 0.39% No one says a next trade where the probability of all 4 consecutive losing trades.
9 years later! But bang on, yes
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