# A Formula For A Winning Strategy...

#### charlesD

##### Member
53 3
Here is a formula to work out what it takes to grow your grubstake a thousand fold... this is a valuable formula that will help you work out how much you need to win and to set stops at a suitable level.

The inputs are:

E = how much you end with
W = number of winning trades
L = number of losing trades
U = average % of each winning trade
D = average % of each losing trade

For those not familiar with the mathematical syntax, * (asterisk) represents multiplication, and ^ (carat) represents to the power of, and / (forward slash) represents division

The formula is:

S * [(1 + (U/100)) ^ W ] * [(1 - (D/100) ^ L ] = E

So let's say you start with £1000, win on average 10% on each trade, and lose on average 3% on each trade. How much will you have after 100 trades with a win ratio of 30%? Then just enter the required inputs into the formula:

S = £1000
E = ?
W = 30
L = 70
U = 10
D = 3

£1000 * [(1 + (10/100)) ^ 30 ] * [(1 - (3/100) ^ 70 ] = E

~

£1000 * (1.1)^30 * (0.97)^70 = £2069.20

You can change the inputs for different scenarios:

Let's say you set stops which if triggered results in loss of 5% of your balance, and your win ratio is now only 20% (e.g 2 wins out of 10 trades), and your average win is still 10%, how much will you have now after 100 trades?

~

£1000 * [(1 + (10/100)) ^ 20 ] * [(1 - (5/100) ^ 80 ] = E

~

£1000 * (1.1)^20 * (0.95)^80 = £111.11!

As you can see, small changes can be the difference between winning and losing! In this case, a win ratio of only 2 out of ten and a slightly bigger stop loss of 5% results in large losses, whereas a win ratio of 3 out of ten and a 3% stop loss results in more than doubling your money after 100 trades. Both scenarios, however, assume you take 10% profit on each winning trade. You can change the values of W,L,U,D to see the changes in the final result.

Now you can use the formula to answer a question like: With a win ratio of 20%, How much do I need to win on each winning trade to break even after 100 trades using 5% stops?

We now want to find the value of U:

£1000 * [(1 + (U/100)) ^ 20 ] * [(1 - (5/100)) ^ 80 ] = £1000

Doing a bit of algebra:

Let (1 + (U/100)) = X

£1000 * (X)^20 * (0.95)^80 = £1000

(X)^20 * (0.95)^80 = £1000/£1000

(X)^20 * (0.95)^80 = 1

(X)^20 = 1 / (0.95)^80

LOG (X)^20 = LOG (1 / (0.95)^80)

20 * LOG (X) = LOG (1 / (0.95)^80)

LOG (X) = [LOG (1 / (0.95)^80) ] / 20

LOG (X) = 0.089105578

Inverse the LOG to get 1.227737663

£1000 * (1.227737663)^20 * (0.95)^80 = £999.99

X = 1 + (U/100) = 1.227737663

(U/100) = 1.227737663 - 1

U = (1.227737663 - 1) * 100 = 22.7737663

So we now know that we need to make 22.77% profit on each winning trade to break even after 100 trades with a win ratio of 20% and setting stops at 5%.

If you haven't got a scientific calculator with a LOG function you can't answer questions like this.

If you can spot any mistake please let me know. Knowing stuff like this is vital if you want to win!

#### NVP

##### Legendary member
37,767 2,101
sounds great .....shame a human Trader has to manage them all ....jees theres always a catch

N

#### Pat494

##### Legendary member
14,617 1,579
Why not toss a coin ? At least you get a win ratio of 50% !

#### counter_violent

##### Legendary member
11,284 3,005
Why not toss a coin ? At least you get a win ratio of 50% !

Not true !

#### WR1

##### Active member
206 9
K.I.S.S is another very good formula?
or is it too simple ?

#### Pat494

##### Legendary member
14,617 1,579
Not true !

You are splitting hairs CV are you not ?

#### counter_violent

##### Legendary member
11,284 3,005
You are splitting hairs CV are you not ?

Well, it's old ground to cover, but in a nutshell, a series of 1000 coin tosses should produce a roughly 50/50 split, heads tails. A series of 1000 trades may produce a similar result, except that there is a cost of 1000 commissions which must be overcome...ie we need an "edge". Therefore, no edge = no possible chance of winning.

Hopefully, Shakone will be along shortly to give us a proper explanation and some numbers in support.

Last edited:

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