a trading problem for mathematicians

[u21c3f6]

Quote from jhiro:

No, you bozo's each have half of the story.. back a horse in at 60%, if it wins what is the chance of your very next bet being a winner if you back another 60% chance?

This means you would only get the two signals lining up together 36% of the time

So why would you trade a 60% system in for one that still only has a 60% strike rate but generates the signals almost half as often?

In addition, your horse example is incorrect. If horse 1 has already won, the chance of the next horse winning is 60%. I think you are referring to: What is the chance of hitting two winners in a row? Then the correct answer would be: 36%.

If you don't understand the difference, you really need to study more.

Joe.

 

[SnakeEYE]

Quote from jhiro:



This means you would only get the two signals lining up together 36% of the time

Honey,

This means you would only get the two lips and tongue lining up together for my cock,100% of the time.

 

[u21c3f6]

Quote from jhiro:

Jesus Joe that was implied by the problem.. of course together.. aren't we talking about putting 2 systems together ffs.. back to the books for you Einstein


You are all breaking out your calculators fighting over 60% vs 36% when the actual answer is yes the strike rate stays at 60% but signals won't occur as often as a single system (60% vs 36%)

So to answer the poor bastards original question it won't improve your strike rate and you'll get less signals.. therefore the combined system is worse generating less 60% signals than a single system over time


Fucken morons

WOW! I am sure if you tried harder you could have included a few more insults. I have no idea where all your anger comes from but the original question was not whether or not the single system was better, the original question was: what is the winning rate?

Joe.

 

[SnakeEYE]

Quote from jhiro:

Was that your dick? I thought it was an overly large clit

don`t worry,keep going

 

[SnakeEYE]

There is no such thing as probability in trading and there is no such thing as probability diapason.The 'probability' is always 100%.For 2 different events you always have 100% 'probability' outcome at the same time.

Are you that dumb to get it

When you are flying,for e.g., there are 2 different outcomes which are both 100% ,either you crash, or you`ll reach your destination.

 

[SnakeEYE]

Confirmed

 

[trend2009]

I do not think this is a correct answer. think in this way: if the system B is random, ie., 50% winning rate, Applying B to the signals from A will not change the winning rate of A. thus A+random = A. now B is a 60% winning rate, A+random will less than A+B.

Quote from swingtrader123:

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.

 

[swingtrader123]

Here is my logic: A game of flipping coins. You have two coins to flip. If either coin turns up heads, you win. You flip the first coin, and its a head, you win. The second coin flip does not matter. Your first win has a 50% probability of winning. Second try, your first coin flip is a tail. You flip the second coin. Now what is the probability of a head on the 2nd coin? It's 50%. So, you still have a 50% chance of winning. Mathematically, for two coins, you have 2 chances to win, but with 4 possible outcomes. So, the probability is 2/4 or 50%.

The key here is that only one system must be correct. So, the system with the highest probability is the probability of success. The second system's probability does not contribute or remove any probability to the first.

Quote from trend2009:

I do not think this is a correct answer. think in this way: if the system B is random, ie., 50% winning rate, Applying B to the signals from A will not change the winning rate of A. thus A+random = A. now B is a 60% winning rate, A+random will less than A+B.

 

[Random.Capital]

Quote from trend2009:

Suppose I have two trading methods with no correlation to each other.

Do you mean they have had no correlation in the past, or that we know for a fact they will have no correlation in the future, under any circumstance?

 

[MathAndLogic]

Quote from euclid:

Let's say both systems (A & B) give long and short signals such that e.g a long signal means the market will hit +X before it hits -X.

It follows that on occasions where both systems gave a signal, we have the following probabilities:

Market moved X:
A won, B won = 0.6*0.6=0.36
A won, B lost = 0.6*0.4=0.24
A lost, B won = 0.4*0.6=0.24
A lost, B lost = 0.6*0.6=0.16

So, 36% win, 16% lose, 48% no trade because the systems disagreed.

This leaves us with a win rate for the combined signal of 0.36/(0.36+0.16) = 0.69. A small improvement in hit rate, but a big reduction in the number of trades.

We can check this with some other figures. e.g:

If both systems gave a hit rate of 50%, then we get 0.25/(0.25+0.25) = 0.50. No improvement because 50% is what you would get at random anyway. Neither system is providing an edge.

If one system gave 100%, then we get 0.6/(0.6+0.0) = 1.0. You can't improve on 100% win rate.

Here is a question: If you don't trade, how do you know (deterministically) that A wins and B loses (0.24 probability) or A loses and B wins (0.24 probability)? So, when you say "48% no trade," you know a priori that your next trade will result in no profit and hence you don't want to take the trade. May I have your crystal ball, please?