a trading problem for mathematicians
- Thread starter trend2009
- Start date
Thats my whole point, system 1 makes 1k, system 2 loses 10 bucks, its an overall winner. System 1 makes 10 bucks system 2 loses 1k then overall its a loser. If system 1 makes 10 bucks and system 2 loses 10 bucks its a scratch. So if looking at a combined win % rate, you need to know the p/l ratio of the system. Looking at them individually you dont need the p/l ratio to calculate win %.
Thanks for clarifying. My answer is still 0.6923.
Here's the reasoning again (but this time corrected the for typos and omissions in the earlier post!).
New win rate = (new number of winning trades) / (new number of all trades)
New number of winning trades
= (A gives signal and wins) INTERSECTION (B gives signal and wins)
= N x P(A gives signal) x P(A wins) x P(B gives signal) x P(B wins)
= N x P(A gives signal) x P(B gives signal) x 0.6 x 0.6
= 0.36 x N x P(A gives signal) x P(B gives signal)
New number of all trades
= (A gives signal and wins) INTERSECTION (B gives signal and wins)
+(A gives signal and loses) INTERSECTION (B gives signal and loses)
= N x P(A gives signal) x P(A wins) x P(B gives signal) x P(B wins)
+ N x P(A gives signal) x P(A loses) x P(B gives signal) x P(B loses)
= N x P(A gives signal) x P(B gives signal) X ((0.6 x 0.6) + (0.4 x 0.4))
= 0.52 x N x P(A gives signal) x P(B gives signal)
SO, New win rate
= (new number of winning trades) / (new number of all trades)
= 0.36 / 0.52
= 0.6923
Over and out.
... except it's the case the OP was referring to (see OP's clarification)!
As the new system won't trade in these cases, they don't contribute to either the count of winning trades, or the count of losing trades. So, correct to omit them ...
Over and out (ooops! I wrote that above, as well ... well this time it's [probably] for real!)
Lets say there is a 500 loose quarters on the floor. It is known that quarters made in the year 1998 flip heads 60% of the time. It is also known that quarters made in the Philadelphia mint flip heads 60% of the time.
You find a coin made in Philadelphia in 1998. What is the probability that that coin flips heads? It would almost certainly be 60% in this case. If the number was greater than 60%, than there would be a correlation between the two.
You find a coin made in Philadelphia in 1998. What is the probability that that coin flips heads? It would almost certainly be 60% in this case. If the number was greater than 60%, than there would be a correlation between the two.
" New number of all trades
= (A gives signal and wins) INTERSECTION (B gives signal and wins)
+(A gives signal and loses) INTERSECTION (B gives signal and loses)
= N x P(A gives signal) x P(A wins) x P(B gives signal) x P(B wins)
+ N x P(A gives signal) x P(A loses) x P(B gives signal) x P(B loses)
= N x P(A gives signal) x P(B gives signal) X ((0.6 x 0.6) + (0.4 x 0.4))
= 0.52 x N x P(A gives signal) x P(B gives signal) "
This seems over-complicated and I do not see the logic in multiplying the P(win)'s.
The new number of trades = N x P(A gives signal) x P(B gives signal)
I don't see how this is disputable.
= (A gives signal and wins) INTERSECTION (B gives signal and wins)
+(A gives signal and loses) INTERSECTION (B gives signal and loses)
= N x P(A gives signal) x P(A wins) x P(B gives signal) x P(B wins)
+ N x P(A gives signal) x P(A loses) x P(B gives signal) x P(B loses)
= N x P(A gives signal) x P(B gives signal) X ((0.6 x 0.6) + (0.4 x 0.4))
= 0.52 x N x P(A gives signal) x P(B gives signal) "
This seems over-complicated and I do not see the logic in multiplying the P(win)'s.
The new number of trades = N x P(A gives signal) x P(B gives signal)
I don't see how this is disputable.
For two trading methods with no correlation to each other, the probability of winning when both signals are agreed upon is 60% that's because the probability of multiple non-correlated systems is the highest probability of those systems. For 60% + 60%, you get 60%. For 100% + 50%, you get 100%. You can't increase the probability of a system by using multiple signals. If it was this easy, everyone would be rich. 