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Renat's logic is that the price goes against the crowd.
I understood it differently: "the sliding is extended by those who peddle it".
I read you guys and there's a strong belief that this whole sliding thing is also 50/50
I understood it differently: 'the sliding is extended by those who peddle it'.
I read you guys and I get a strong conviction that this whole sliding deal is also 50/50
"They started to suspect something" ) I said from the start this pattern works under certain conditions, under which one can trade profitably without these tricks using simple MA's. i.e. the whole problem is determining when the conditions have started and when they are over, classic :)
I read you guys and there's a strong belief that this whole sliding thing is also 50/50
If we're talking about trading two pairs, then the probability of splitting would be 10% )))
If we are talking about trading two pairs, then the probability of a split would be 10% )))
If we are talking about trading two pairs, then the probability of splitting would be 10% )))
How can you tell? By eye?
How can you tell? By eye?
Sort of.) If I'm counting probabilities, it's like this:
The movement can be up, down and straight ) )) formally the probability is 0.33
If two pairs - probability of movement in the right directions = 0.33 x 0.33 = 0.108 )) Is that correct?
Another thing is that the probability of going straight in some cases tends to 0. But it is always there.
Then there may be preconditions for corrections for each direction, but they are different for each.
Assuming that the probability of two states of the same currency is the same, then the probability of two bifurcation states between such currencies is the same.
Regarding standing still:
It turns out that there will be a spread 2/3 of the time, but what is the probability of a currency standing in one place? There is also a spread the size of the spread. So there is a 50/50 chance of making a profit. No?
Assuming that the probability of two states of one currency is the same, then the probability of two bifurcation states between such currencies is also the same.
Regarding standing still:
It turns out that there will be a bifurcation 2/3 of the time, but what is the chance of a currency standing in one place? I think it is small. Also we will see spread sizes. The result is most likely 50/50. Right?