I'm trying to isolate ST, i.e. ST = ...
PR = (1-((1-(1-(WR/100))^ST)^((TR*365*50)-(ST-1))))*100
Can anyone help me?
That monster looks to me like a financial formula, ST being the number of periods; yes?
As far as I can tell, cannot be solved directly for ST; numeric method required.
Anyhoooo: I suggest you simplify it as much as possible, a bit this way:
p=PR, w=WR/100, t=TR*365*50, s=ST
I reformatted this to make it easier to read. There is no easy way to isolate $ST$. You are likely to wind up with the Lambert W function in the solution. Let's give single letters to every double-letter variable. $ST = x, WR = y, TR = z, PR = w$. Then you have:
$$w = \left(1-\left[ \left(1-\left[1-\left( \dfrac{y}{100} \right) \right]^{x} \right)^{18250z-x+1} \right] \right)*100$$
Perhaps a computer algebra system would be able to chew through this and solve for $x$, but there is no easy way to do that. The problem is that you have $x$ in two separate exponents.
which is really:
PR/100 = 1-((1-(1-(WR/100))^ST)^((TR*365*50)-(ST-1)))
right?
Are these "reasonable" values:PR = probability of x consecutive losing trades,
WR = win-rate,
TR = number of trades per day
ST = streak of consecutive losing trades.
PR = 25 (25% probability)
WR = 65 (win 65% of time)
TR = dunno...perhaps >0 and <100?
ST = calculated using above
Can you supply a clear actual example, and ST's calculated value?
I think can be solved not directly but from a very simple looper program.....
Ideally, the probability (PR) should be around 0.1%, since it is the probability of losing a certain number of trades in a row which would lead to a trading-account drawdown of a certain percentage (in my case 20%).
Win-rate (WR) at 65% is good.
Trades per day (TR) is usually around 0.5 or so, since the setups that I trade don't occur every day -- and also since there are only 5 trading days per week.
The streak (ST) is calculated using the amount of risk per trade to find the number of losing trades that would lead to a maximum account drawdown of 20%. Depending on the strategy, the streak number is usually around 10-15.
TR*365*50 is intended to return the total possible number of trades in 50 years.
Thank you kindly for all your help!
So, using WR = 65, TR = .5 and ST = 12, then PR = ~3.033
PR = (1-((1-(1-(65/100))^12)^((.5*365*50)-(12-1))))*100 = ~3.033
...which means PRobability = approx. 3.033
Relating that to your .1% then means 3.033/100 = .03% (approx.) : yes?
What seems STRANGE to me is : ^((TR*365*50)-(ST-1))))
WHY have -(ST - 1) in there: seems meaningless; here's why:
PR = (1-((1-(1-(65/100))^12)^((.5*365*50)-(12-1))))*100 = 3.030328833
remove the (ST - 1):
PR = (1-((1-(1-(65/100))^12)^((.5*365*50)))*100 = 3.030364876
....a difference of .000036043
Sooooo: remove the darn thing!!
Get my drift?
I hope SlipEternal checks this out...in case I goofed!!
Hi DenisB,
My apologies for my delayed response; I've been traveling around in Europe and the Middle East. Currently on a break in the Iceland airport.
I tested out your hypothesis about simply removing -(ST - 1), and it does seem to make only a minimal difference, insignificant for my purposes.
I'm still trying to isolate ST. I've got this far but am having a mental block.
PR = (1-((1-(1-(WR/100))^ST)^(TR*365*50))*100
PR/100 = 1-((1-(1-(WR/100))^ST)^(TR*365*50)
-(PR/100 - 1) = (1-(1-(WR/100))^ST)^(TR*365*50)
(TR*365*50) = ln(-(PR/100 - 1))/ln((1-(1-(WR/100))^ST))
(TR*365*50)*ln((1-(1-(WR/100))^ST) = ln(-(PR/100 - 1))
Do you know how to isolate ST?
Any help is much appreciated!