Probable a Binomial Probability Problem :) (not sure)

Question

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.

Answer 1

One way of analyzing it is to treat each trade as first you paying one unit. Then a loss is just not getting the money back, and a gain is getting 4 back (you get 3 times your stake, plus you get your stake back). Then your return is

4*(number of gains)-(number of trades)

Because expectation is a linear operator, you have

$E(4G-n) = 4E(G)-E(n)$

$n$, of course, is not a random variable, it's just the number of trades you performed, so you can replace $E(n)$ with just $n$.

So the expected return is $4E(G)-n$, and $G$ is a binomially distributed RV with $p = .33$ . The expected value of a binomial RV is $pn$, so you have $4*.33n-n = 1.32-1 = .32$.

This is assumes that you're risking the same amount each trade. If you're risking a proportion of your current balance each trade, then the analysis is easier when done in terms of logs. Let's say each trade you risk $r$ times your current balance. Then starting with $B$, a win will result in you having $B+3rB = (1+3r)B$, while a loss will result in you having $B - rB = (1-r)B$. Thus, a win will result in the log of your balance being $log[(1+3r)B] = log(1+3r)+log(B)$, while a loss will result in $log(1-r)+log(B)$. (BTW, I'm not specifying a base for the log; different bases don't change the essential nature of the analysis).

Assuming .33 is an approximation of 1/3, the expected value is

$$[log(1+3r)+log(B)](n/3)+[log(1-r)+log(B)](2n/3) =$$ $$ [log(1+3r)](n/3)+[log(1-r)](2n/3)+log(B)$$ $$[log(1+3r) + 2log(1-r)](n/3) +log(B)=$$ $$log \left[ B ((1+3r)(1-r)^2)^n\right]$$

Note that if $(1+3r)(1-r)^2<1$, then this will be decreasing.

$(1+3r)(1-r)^2<1$

$(1+3r)(1-2r+r^2)<1$

$r-5r^2+3r^3<0$

$r(1-5r+3r^2)<0$

$r(1-5r+3r^2)$ has three roots: $r=0$ and $r = \frac{5\pm \sqrt{13}}{6}$

$r=0$ corresponds to risking nothing, so of course your expected value is just B, and there will be no growth. $r = \frac{5+ \sqrt{13}}{6}$ gives 1.43, and you can't risk 143% of your balance. $r = \frac{5+ \sqrt{13}}{6}$ gives .2324. This means that once you risk more than 23.24% of your balance, the expected value of the log of your balance will be decreasing. Note that while the expected value of your balance is always increasing, the expected value of the log of your balance can decrease if you bet too much (as an extreme, if you bet everything, then no matter how many wins you have, one loss will wipe everything out). You can look up "Kelly Criterion" if you're interested in more information.

Answer 2

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.