a trading problem for mathematicians
- Thread starter trend2009
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I mean you are totally confused. How can you know when you are taking the trade whether the system will win? Oh my!
Can you realize what you did? The problem stated that there are two systems that generate signals and trades are taken when they both give a signal. How can you know whether that will be a gain or a loss?
If you have two systems with win rates w and a single position is taken when both generate a signal then the combined win rate is wxw, in this case 0.36.
If two separate positions are taken then the combined win rate is
1 - loss rate = 1 - (1-w)x(1-w) in our case 1 - 0.4 x0.4 = 0.84
This assumes same P/L for the two systems and that flat trades count as winners. As a explained in a different post, if the P/L is not the same the problem is more complicated.
The 0.69 number is wrong and based on the false assumption that the outcome of a random process is known in advanced.
Having re-read the question, I am going to agree with those who say that itâs 60%. Hereâs the logic.
Given a signal from the first system, call it signal A, and a confirming signal from the second system, call it signal B, we are looking for a joint probability distribution which is given by
P(A=True and B=True) = P(A=True | B=True) * P(B=True).
Now, the conditional probability P(A=True | B=True), the probability of A being True given that B is True, is equal to 1, since if the two systems give, say, a signal to Buy and it turns out true for signal B, it cannot turn out false for signal A, since two Buy signals are either both right or wrong. Thus P(A=True | B=True) = 1 and we have
P(A=True and B=True) = P(A=True | B=True) * P(B=True) = 1*P(B=True) = 1*0.6 = 0.6.
We only consider P(A=True and B=True), and not also P(A=True and B=False) and P(A=False and B=True) since the latter two are equal to 0, because the sets {A=False and B=True} and {A=True and B=False} are empty - we cannot have two Buy signals with one being right and another wrong.
Given a signal from the first system, call it signal A, and a confirming signal from the second system, call it signal B, we are looking for a joint probability distribution which is given by
P(A=True and B=True) = P(A=True | B=True) * P(B=True).
Now, the conditional probability P(A=True | B=True), the probability of A being True given that B is True, is equal to 1, since if the two systems give, say, a signal to Buy and it turns out true for signal B, it cannot turn out false for signal A, since two Buy signals are either both right or wrong. Thus P(A=True | B=True) = 1 and we have
P(A=True and B=True) = P(A=True | B=True) * P(B=True) = 1*P(B=True) = 1*0.6 = 0.6.
We only consider P(A=True and B=True), and not also P(A=True and B=False) and P(A=False and B=True) since the latter two are equal to 0, because the sets {A=False and B=True} and {A=True and B=False} are empty - we cannot have two Buy signals with one being right and another wrong.
Sorry asia it is you who clearly dont understand and the throwing out of insults confirms this.
If both systems produces equal profit and losses then the answer is 69.23% as I previously demonstrated. All the coin flipping analogies and such hold true.
However if the systems have different p/l's then the problem is unsolvable without that piece of information.
Brushing up on some basic statistics might help your trading.
asiaprop - Are you sure three whole posts is enough to properly convey your disgust with ET's stupidity...maybe another one just to be safe?
Way to take the high road.
Way to take the high road.
this thread should have started out with the three door monty hall problem.
only people who understood that you should make the switch should have been allowed to post.
this procedure would have been an excellent way to vet out the people who think they understand probability.
"Let's Make a Deal!
Imagine that the set of Monty Hall's game show Let's Make a Deal has three closed doors. Behind one of these doors is a car; behind the other two are goats. The contestant does not know where the car is, but Monty Hall does.
The contestant picks a door and Monty opens one of the remaining doors, one he knows doesn't hide the car. If the contestant has already chosen the correct door, Monty is equally likely to open either of the two remaining doors.
After Monty has shown a goat behind the door that he opens, the contestant is always given the option to switch doors. What is the probability of winning the car if she stays with her first choice? What if she decides to switch?"
I am serious as some of the responses are prima facie wrong particularly people who use the word elementary.
only people who understood that you should make the switch should have been allowed to post.
this procedure would have been an excellent way to vet out the people who think they understand probability.
"Let's Make a Deal!
Imagine that the set of Monty Hall's game show Let's Make a Deal has three closed doors. Behind one of these doors is a car; behind the other two are goats. The contestant does not know where the car is, but Monty Hall does.
The contestant picks a door and Monty opens one of the remaining doors, one he knows doesn't hide the car. If the contestant has already chosen the correct door, Monty is equally likely to open either of the two remaining doors.
After Monty has shown a goat behind the door that he opens, the contestant is always given the option to switch doors. What is the probability of winning the car if she stays with her first choice? What if she decides to switch?"
I am serious as some of the responses are prima facie wrong particularly people who use the word elementary.
Agreed. (Assuming R:R=1 and each system has an independent random distribution of wins and losses.)
Another way to look at it. Let's say each system gives a daily signal long or short with a target and stop of 10 points.
Trader A trades system A .He trades every day, wins 60%. He averages 2 points per day.
Trader B trades system B .He trades every day, wins 60%. He averages 2 points per day.
Trader C trades both systems. He trades 52% of days, wins 36% of days, loses 16% of days. He averages 2 points per day.
So,
If you said 36%, you're right that's the % winning days.
If you said 84%, you're right - a day off counts as a win in my book too.
If you said 69%, you're right that's the % winning trades.
If you said 60%, hmm, er, ok - that's the % winning trades you'd need per day to make the same profit.
Before we start, please let us know what you mean by probability. Your definition please.
I'm serious too. This isn't going to be an easy ride for you.
1 in 3 if sticking with the first choice, 2 in 3 if always switching.
Now that I've qualified to be here, I'd say that while the original question may seem interesting, it's not very relevant to the bottom line. The original poster seems to totally miss the fact that the "goodness" of a trading system can't be determined by the percentage of winning trades along.