I'm getting a bit dumb on the probabilities. - page 10
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It's my turn to be quiet. I'm lost in three pines) There are 3 systems, working on the same instrument:
1) Predicts the colour of 4H candle with a probability 0.8
2) Predicts the colour of 1H candle with the probability 0.8
3) Predicts the colour of a 30M candle with the probability 0.8
By combining the signals of these three systems we could get a system which will give more than 0.8 probability of predicting the candlestick colour. And what is the probability of predicting the correct candlestick colour when working with those signals from 1 system, i.e. we open in case all 3 systems give the same colour for the next candlestick.
Is that what flash royal is called now?
So the question is - can this increased probability be justified by the fact that the GCG is giving out hundreds of hands per second online?
Never mind, the explanation is wrong. You also have to take into account how many games you've played, that's the point of statistics.
A flush royal is too rare an event, it can have a very large spread in terms of frequency. You have to play a lot of games, many times more than you have played, to be qualified to judge a meaningful frequency difference.
In theory, yes - for the last 30-minute candle of the 4-hour candle ))) we have to figure it out.
Simple task, but somehow the second day doesn't count) It's embarrassing that there are uncalculated and unaccountable interdependencies between the probabilities of the 3 systems, and they clearly exist...
Man, it's a simple task, but the second day doesn't really count)
It's my turn to be quiet. I am lost in three pines) There are 3 systems, working on the same instrument:
1) Predicts the colour of 4H candle with a probability 0.8
2) Predicts the colour of 1H candle with the probability 0.8
3) Predicts the colour of a 30M candle with the probability 0.8
By combining the signals of these three systems we could get a system which will give more than 0.8 probability of predicting the candlestick colour. And what is the probability of predicting the correct candle colour when working with those signals in 1 system, i.e. we open in case all 3 systems give the same colour of the next candle.
If the systems are independent, i.e. take information for TA from different independent sources, then according to Bayes theorem, we can. If they are 100% dependent, we cannot.
I.e. if we assume that each system takes incomplete information for TA from the same source, and this information is partly independent, then increasing the probability is possible.
In theory, yes - for the last 30-minute candle of a 4-hour candle )))).
Well, probably for any candle, but yes, it's probably easiest to calculate for the last one.
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Sergey, what about the probability in the tester?
Sergey, what about in the tester to estimate the probability?
Unsportsmanlike in the tester)
Tried it actually. The tester gives an increase in the probability of the resulting system, so I tried to back it up with calculations and got hung up.
It's my turn to be quiet. I'm lost in three pines) There are 3 systems, working on the same instrument:
1) Predicts the colour of 4H candle with a probability 0.8
2) Predicts the colour of 1H candle with the probability 0.8
3th one predicts the colour of 30M candlestick with the probability 0.8
By combining the signals of these three systems we could get a system which will give more than 0.8 probability of predicting the candlestick colour. And what is the probability of predicting the correct candle colour if we work with those signals in 1 system, i.e. we open in case all 3 systems give the same colour of the next candle.
if all predictions are independent, then the probability of correct if all 3 signals are in the same direction = 1-0.2*0.2*0.2=0.992
If there are dependencies, then through conditional probabilities.
But in reality these values are hardly constant. I.e. frequencies do not converge to probabilities
if all predictions are independent, then the probability of correct if all 3 signals are in the same direction = 1-0.2*0.2*0.2=0.992