Could someone plz tell me if this math is correct:
example 1
50% lose ratio
probability of 4 trades in a row being a loss
Probability : 0.5^4 = 0.0625
No. of trades needed for this to occur : 4 / 0.0625 = 64
example 2
80% lose ratio
probability of 6 trades in a row being a loss
Probability : 0.8 ^ 6 = 0.2621
No. of trades needed for this to occur : 6 / 0.2621 = 22.89
Is this correct ?
If not could someone please show how to calculate with working ?
thankyou
plz help me with my math: probability of N losses in a row
In both examples the first statement is correct but the second is wrong.
The number of trades needed for this to occur canÃ¢â‚¬â„¢t be calculated like that. I will give you an example with 2 losses in a row from a fair coin flip.
The sequence may look like this
Loss-loss we will need two coin flips
Win Ã¢â‚¬â€œ loss Ã¢â‚¬â€œ loss we will need three coin flips
Loss Ã¢â‚¬â€œ win Ã¢â‚¬â€œ loss Ã¢â‚¬â€œ loss we will need four coin flips and so on Ã¢â‚¬Â¦
Now I will describe the logic for finding this number x (the number of trades needed for this losing streak to occur).
There is 1/2 chance the first flip to be a win (remember we have a fair coin here) and in this case that number x should be increased by one and become x +1 (because we have spent one trial).
There is 1/2 chance the first flip will be a loss so either the second flip will also be a loss and the number x = 2 or the second flip will be a win so 2 trial are now wasted and we must replace x with x + 2.
In math notation now all the above become
x = 1/2 * (x+1) + 1/2 * [ 1/2 *2 + 1/2 * (x+2)]
Solving this we take the solution x = 6.
Similarly we can solve for 3 consecutive losses in the fair coin example and we find x = 14 by solving the equation
x = 1/2 (1+x) + 1/2*{ 1/2 *[ 1/2 *3 + 1/2 *(3+x)] + 1/2 *(2+x) }
And so on Ã¢â‚¬Â¦.
This job is kind of hard to do by hand and a computer algorithm may become handy.
The number of trades needed for this to occur canÃ¢â‚¬â„¢t be calculated like that. I will give you an example with 2 losses in a row from a fair coin flip.
The sequence may look like this
Loss-loss we will need two coin flips
Win Ã¢â‚¬â€œ loss Ã¢â‚¬â€œ loss we will need three coin flips
Loss Ã¢â‚¬â€œ win Ã¢â‚¬â€œ loss Ã¢â‚¬â€œ loss we will need four coin flips and so on Ã¢â‚¬Â¦
Now I will describe the logic for finding this number x (the number of trades needed for this losing streak to occur).
There is 1/2 chance the first flip to be a win (remember we have a fair coin here) and in this case that number x should be increased by one and become x +1 (because we have spent one trial).
There is 1/2 chance the first flip will be a loss so either the second flip will also be a loss and the number x = 2 or the second flip will be a win so 2 trial are now wasted and we must replace x with x + 2.
In math notation now all the above become
x = 1/2 * (x+1) + 1/2 * [ 1/2 *2 + 1/2 * (x+2)]
Solving this we take the solution x = 6.
Similarly we can solve for 3 consecutive losses in the fair coin example and we find x = 14 by solving the equation
x = 1/2 (1+x) + 1/2*{ 1/2 *[ 1/2 *3 + 1/2 *(3+x)] + 1/2 *(2+x) }
And so on Ã¢â‚¬Â¦.
This job is kind of hard to do by hand and a computer algorithm may become handy.
Our old reliable friend the Omega List contains an interesting answer that comes in the form of a table and a formula. It's 2 years old, but I don't think the mathematics of probability has changed very much since then.
http://www.purebytes.com/archives/omega ... 04142.html
http://www.purebytes.com/archives/omega ... 04142.html