<s>[INST] Probable a Binomial Probability Problem :) (not sure) I need to calculate the average profit in a month of a trading strategy.
My probability of a successful trade is 33%
My probability of a bad trade is 67%
The strategy has a Risk Return ratio of 1:3 meaning that I risk 1 to gain 3
So in a bad trade I loose one and in a good trade I gain 3.
I do 3 trades a day.
What is the expected return in a day and in a month, considering 22 days in a month?
My thought was:
$C\binom{3}{1} = \frac{3!}{1!(3-1)!}=\frac{6}{2}=3\\
P=C\binom{n}{k}\cdot p^{k}\cdot q^{n-k}\\
P=3\cdot \left( \frac{1}{3}\right )^1 \cdot \left( \frac{2}{3}\right )^{3-1}\\
P=3\cdot \left( \frac{1}{3}\right )^1 \cdot \left( \frac{2}{3}\right )^{2}\\
P=0.148$
However by doing this I do not take in consideration the risk return ratio so I thought that I should weight the equation with the risk return factor, and I am not sure if this is right.
When I introduce the Return R and Risk rsk I get
$C\binom{3}{1} = \frac{3!}{1!(3-1)!}=\frac{6}{2}=3\\
P=C\binom{n}{k}\cdot p^{k}\cdot q^{n-k}\\
P=3\cdot \left( \frac{1}{3}\right )^1 \cdot \left( \frac{2}{3}\right )^{3-1}\\
P=3\cdot R \cdot \left( \frac{1}{3}\right )^1 \cdot rsk \cdot \left( \frac{2}{3}\right )^{2}\\
P= 1.33$
Assuming that this is OK it means that after a day I will multiply the Value at Risk by 1.33.  A superb return ratio of 33% a day and only feasible in math concept not in real world.
After the consideration of Laars Helenius I changed to
Probability of no Success trade:
$C\binom{n}{k} = \frac{n!}{k!(n-k)!}\\
C\binom{3}{0} = \frac{3!}{0!(3-0)!}=\frac{6}{6}=1\\
P=C\binom{n}{k}\cdot p^{k}\cdot q^{n-k}\\
P=1\cdot \left( \frac{1}{3}\right )^0 \cdot \left( \frac{2}{3}\right )^{3}\\
P=0.29\\$
Probability of one success trade:
$C\binom{n}{k} = \frac{n!}{k!(n-k)!}\\
C\binom{3}{1} = \frac{3!}{1!(3-1)!}=\frac{6}{2}=3\\
P=C\binom{n}{k}\cdot p^{k}\cdot q^{n-k}\\
P=3 \cdot \left( \frac{1}{3}\right )^1 \cdot \left( \frac{2}{3}\right )^{2}\\
P=0.44\\$
Probability of 2 success Trades:
$C\binom{n}{k} = \frac{n!}{k!(n-k)!}\\
C\binom{3}{2} = \frac{3!}{2!(3-2)!}=\frac{6}{2}=3\\
P=C\binom{n}{k}\cdot p^{k}\cdot q^{n-k}\\
P=3 \cdot \left( \frac{1}{3}\right )^2 \cdot \left( \frac{2}{3}\right )^{1}\\
P=0.22\\$
Probability of 3 success trades:
$C\binom{n}{k} = \frac{n!}{k!(n-k)!}\\
C\binom{3}{3} = \frac{3!}{3!(3-3)!}=\frac{6}{6}=1\\
P=C\binom{n}{k}\cdot p^{k}\cdot q^{n-k}\\
P=1 \cdot \left( \frac{1}{3}\right )^3 \cdot \left( \frac{2}{3}\right )^{0}\\
P=0.037\\$
How to consider the outcome of each possible situation and finally find out the average profit in a day to use it after in composite interest rate?
Second EDIT
I trade 5% of my capital.  So in a \$2000 account I would trade \$100
Meaning that after each day my Value at Risk would be different.
 [/INST] Expected payouts per day:
A = 3 losses = -3 units
B = 1 win / 2 losses = +1 unit
C = 2 wins / 1 loss = +5 units
D = 3 wins = +9 units
Probabilities:
$\Pr(A) = \binom{3}{0}(1/3)^0(2/3)^3=8/27$
$\Pr(B) = \binom{3}{1}(1/3)^1(2/3)^2=12/27$
$\Pr(C) = \binom{3}{2}(1/3)^2(2/3)^1=6/27$
$\Pr(D) = \binom{3}{3}(1/3)^3(2/3)^0=1/27$
Expected Value:
$A\cdot\Pr(A)+ B\cdot\Pr(B)+ C\cdot\Pr(C)+ D\cdot\Pr(D)=1\text{ unit}$
Then the linearity of expectation tells you to expect to make 22 units over 22 trading days.
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