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A "hardcore" mathematics question

  • #21
  • Apr 2, 2010 7:06pm Apr 2, 2010 7:06pm
I think you have correctly seen the complexity of the problem. Cheers. The original question do not have a fix no. value. 30 trials is only a possible scenarios.

Quoting Sauron
I think he meant what is the chance of the drawdown to be at least 10 units after 30 trades.
The excel formula doesn't work because you consider 10 winning to be enough but you forgot to consider 10 losses at the beginning of the trial.

There's no magic formula. You must calculate every possible permutation of winning and losses which is some N. For every permutation you must calculate if it goes to 10 DD, the sum of them being some M. The exact result is of course M / N.

As it is very hard your only chance is to generate a few million permutations...
  • #22
  • Apr 2, 2010 7:13pm Apr 2, 2010 7:13pm
Yes. Cheers, you understand me.

Quoting Intu
That is great stuff, but...

I still believe his original question pertains to the possibility of a loss of 10 or more units of trading capital during a sequence of 30 trades with those parameters. So what is the probability that he will be down 10 units, not lose 10 trades during a 30 trade sequence. I may have misunderstood.


For example:

Assumes he has $1000 account. He trades with a 60% win rate and 40% losing rate RRR 1:1. He bets $100 every trade. Every 30 trades he resets his account to $1000 somehow. What are the...
  • #23
  • Apr 2, 2010 7:18pm Apr 2, 2010 7:18pm
Although you are not asking the exact question as I originally posted, the answer to your question is BINOMDIST(10,30,0.6,TRUE), and that is 0.285% (i.e.~=0.3%).

Quoting ForexQuant
LossTrades + WinTrades = 30

LossTrades - WinTrades >= 10

Solve the above equations then we know Loss >= 20 so that we can have at least 10 unit of losses.

so the cumulative probabability from 0 to 19 losses is BINOMDIST(9,30,0.4,TRUE), which is about 99.7%, so the cumulative probability from 20 to 30 losses is 100 - 99.7 = 0.3%

Anyone else can verify this?
  • #24
  • Edited 7:33pm Apr 2, 2010 7:23pm | Edited 7:33pm
Hi Craig,

Your info. seem to be the most promising one for solving my problem. Do you think that the link alone, if I dig deep enough can completely solve my problem?

Do you think I can simply construct the Pascal's Triangle and sum the columns I need, then I can complete my task?

cheers,

kk

P.S. Oh no. The prob. on each row on the Pascal's Triangle is not mutual exclusive, cannot solve the "having ever" problem.



Quoting Craig
It is a Binomial Distribution of some sort, depending on what the OP actually meant. If anybody actually cares to dig though the link I posted, the result which is derived is quite general, more so, the formula has something quite interesting to say on the nature of markets in general (hint: you can also measure the average deviation of market data from a starting point, how does this compare to the result in the link?).
  • #25
  • Apr 2, 2010 8:17pm Apr 2, 2010 8:17pm
Quoting artov
82,4%
This is wrong sorry, but good attempt (If you have walked on the same path as ForexQuant). Read the later post of ForexQuant on this thread, I have answered his question.
  • #26
  • Apr 2, 2010 8:26pm Apr 2, 2010 8:26pm
If on average you have a 60% win ratio , R:R = 1:1 in an average 30 trade set you can only go -10 on average with in the first 14 trades of each set, where you would experience -10 to -10/12 consecutive losers. Once you have 3 winners then on average you will not go -10 for the set.
  • #27
  • Apr 2, 2010 9:11pm Apr 2, 2010 9:11pm
The problem is harder than you think.

Quoting auxesis
If on average you have a 60% win ratio , R:R = 1:1 in an average 30 trade set you can only go -10 on average with in the first 14 trades of each set, where you would experience -10 to -10/12 consecutive losers. Once you have 3 winners then on average you will not go -10 for the set.
  • #28
  • Apr 2, 2010 9:22pm Apr 2, 2010 9:22pm
Quoting kk007
Hi Craig,

Your info. seem to be the most promising one for solving my problem. Do you think that the link alone, if I dig deep enough can completely solve my problem?

Do you think I can simply construct the Pascal's Triangle and sum the columns I need, then I can complete my task?

cheers,

kk

P.S. Oh no. The prob. on each row on the Pascal's Triangle is not mutual exclusive, cannot solve the "having ever" problem.
I can't answer for sure, the set up of the random walk is obviously equivalent to the problem you poised, however, whether you can derive the specific result you want (as opposed to the general one in the link) I'm not sure. I would Google around on the random walk subject and see if you can pick up any other related results or techniques which may help.

The only other way of doing it I can think of was mentioned earlier, somebody posted the link to the equity curve simulator, this effectively generates a random walk(s) with the probabilities you specified. If you are handy at programming you could easily write a program to generate walks and measure the rate of occurrence of your sought condition, over a number of runs this measurement should converge on the number you are after. This is a 'Monte-Carlo' approach to the problem.
The breaking of a wave cannot explain the whole sea.
  • #29
  • Apr 2, 2010 9:38pm Apr 2, 2010 9:38pm
This question is unanswerable in my opinion, it ask for a statistical distribution based on a random walk yet alters the w/l ratio to an arbitrary level. If the r/r was 1:1 and the market was truly random than you would have a 50/50 win/loss ratio, not 60/40; if you were to change the r/r to anything but 1:1 then your w/l ratio would be 1:3. Yet a 6/10 w/l ratio is defying the emh to begin with. If you want to keep that ratio then obviously your entries are based off of analytical skill, resulting in your chances of seeing 10 consecutive losses being completely arbitrary as your trade entry is arbitrary and based off of skill not statistics.
  • #30
  • Edited 11:02pm Apr 2, 2010 10:46pm | Edited 11:02pm
I took Craig's idea on Monte Carlo simulating it with a BASIC program.

Don't laugh at my ancient programming skills... it's been a long time, and I've forgotten most of the Fortran, Pascal, and C that I ever knew. BASIC sort of just stays with you... especially when you were around for the beginning of personal home computers.


10 rem x = number of 30 trade sequences
20 rem p = profit during a sequence of trades
30 rem a = number of trades
40 rem r = random number generator

100 cls
110 for x=1 to 10000
120 let p=1
150 for a=1 to 30
160
200 let r=int(rnd(1)*100)+1
210 if r>=61 then let p=p-1
220 if r<61 then let p=p+1


250 if p<-9 then print "Down 10 units or more at "; x; " sequence of 30 trades."
300 next a
310 next x


Here's a sample of what you get:

Down 10 units or more at 236 sequence of 30 trades.
Down 10 units or more at 599 sequence of 30 trades.
Down 10 units or more at 599 sequence of 30 trades.
Down 10 units or more at 599 sequence of 30 trades.
Down 10 units or more at 897 sequence of 30 trades.
Down 10 units or more at 897 sequence of 30 trades.
Down 10 units or more at 897 sequence of 30 trades.
Down 10 units or more at 897 sequence of 30 trades.
Down 10 units or more at 897 sequence of 30 trades.
Down 10 units or more at 897 sequence of 30 trades.
Down 10 units or more at 897 sequence of 30 trades.
Down 10 units or more at 897 sequence of 30 trades.
Down 10 units or more at 1726 sequence of 30 trades.
Down 10 units or more at 1726 sequence of 30 trades.

My little subroutine repeats itself as long as the net profit is -10 or more, it still rolls the random number and checks for all 30 trades. Those are the repeats. So unfortunately, you have to manually add the occurrences, just counting each repeated sequence once.

(I'm sure someone smarter than me could wrap another subroutine around mine to fix this.... it's late here by me... I just wanted some quick estimated answers.)

I ran it 5 times and manually added the occurrences counting them each once.

Results:

20/10,000
27/10,000
24/10,000
30/10,000
24/10,000

or an avg of 25/10,000 => so the answer is probably close to 0.25%

Which is strangely close to ForexQuant's and kk007's binomial distribution estimate!
  • #31
  • Apr 2, 2010 11:04pm Apr 2, 2010 11:04pm
That's pretty cool, it's been a long time since I've seen a BASIC program
The breaking of a wave cannot explain the whole sea.
  • #32
  • Edited 11:13pm Apr 2, 2010 11:04pm | Edited 11:13pm
Quoting kk007
Although you are not asking the exact question as I originally posted, the answer to your question is BINOMDIST(10,30,0.6,TRUE), and that is 0.285% (i.e.~=0.3%).
ops sorry i typed 9 instead of 19 here, it should be BINOMDIST(19,30,0.4,TRUE), not BINOMDIST(9,30,0.4,TRUE).

Anyway BINOMDIST(19,30,0.4,TRUE) + BINOMDIST(10,30,0.6,TRUE) = 1
  • #33
  • Apr 2, 2010 11:47pm Apr 2, 2010 11:47pm
I just made a simple montecarlo simulation and the binomial distribution chart in excel.

Have a look!
Attached File(s)
File Type: zip hardcoremath.zip   386 KB | 397 downloads
  • #34
  • Apr 3, 2010 3:26am Apr 3, 2010 3:26am
Thanks for the simulation. I am only interested in exact solution. Unfortunately, such an exact solution may not exist.
  • #35
  • Apr 3, 2010 3:41am Apr 3, 2010 3:41am
Quoting kk007
Thanks for the simulation. I am only interested in exact solution. Unfortunately, such an exact solution may not exist.
What make you think binomial distribution cant solve your question? In fact your question and all the above discussions enlighten me to solve a discrete probability distributions for asymmetry payout using simple calculation.
  • #36
  • Apr 3, 2010 3:49am Apr 3, 2010 3:49am
Quoting ForexQuant
What make you think binomial distribution cant solve your question? In fact your question and all the above discussions enlighten me to solve a discrete probability distributions for asymmetry payout using simple calculation.
I said that your question is not the same as mine. My question has the keyword "having ever".

Say a gambler has 10 golden coins, and in each trial he takes one for gambling. He may not have the chance to wait until the end of the 30 trials. He may lose all the 10 coins at 18th trial or 21st trial.

Put it another way, a result that "15 win and 15 loss after 30 trials" does not mean the gambler have not ever over-drawdown in the progress. If he firstly losses 15 times and then win back 15 times, his historical maximum drawdown is -15 not zero.

If you understand what I means here, you would know my actual question is much difficult to solve than the answer I offered to verify your 0.3% answer.
  • #37
  • Apr 3, 2010 4:02am Apr 3, 2010 4:02am
Quoting ForexQuant
What make you think binomial distribution cant solve your question? In fact your question and all the above discussions enlighten me to solve a discrete probability distributions for asymmetry payout using simple calculation.
Intu's BASIC simulation has addressed the issue correctly, while your excel simulator does not.
  • #38
  • Apr 3, 2010 4:14am Apr 3, 2010 4:14am
Quoting kk007
Intu's BASIC simulation has addressed the issue correctly, while your excel simulator does not.
Yes I misunderstood your idea the 1st time but i think i read it correctly 2nd time. IMO my excel sheet did the same thing as intu's BASIC program.

Maybe we should alter the winning probability by double confirm the value.
  • #39
  • Apr 3, 2010 4:38am Apr 3, 2010 4:38am
kk, no offence but i may just not understand what you looking for. I didnt mean to sound sacarstic.

Intu, can you pls help to double check with your simulation with mine? Thank you.

as usual I maybe wrong again.
Attached Image
  • #40
  • Apr 3, 2010 5:33am Apr 3, 2010 5:33am
I must admit that I'm wrong again.
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