On the unequal probability of a price move up or down - page 168
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Don't play the guru, everyone knows what you are.
It started ...
Another one of you put on rose-colored sunglasses
Is it slammed?
What happened to the last trick?
Has it slammed?
And this is unknown because of the lack of different inputs. For example, the size of the initial deposit. As for the rest, I posted the charts above, but it's only the initial position, and there was also a fill-up (I think, I'm not sure). Too lazy to bother.
What happened to the last trick?
Slammed?
What happened to the last trick?
Slammed?
Seems like the right branch.
Here's an idea. Open a triangle of pairs (full, with equal lots). We do not set takeoffs and stops right away - each pair (currency) is targeted at 200 p, with a stop loss of 200. There may be 8 outcomes: TTT, ... CCC. There will be such points (times), at which Take of one pair will be reached (relative to an arbitrary point). Unclosed pairs relative to the same arbitrary point must close in zero (on the average). Is it possible to test it with mql4? If you observe and advise me, I will write an EA.The branch seems to fit.
Here is an idea. We open a triangle of pairs (full, with equal lots). We do not set Take and Stop right away - each pair (currency) is targeted at 200 p, and the StopLoss is 200. There may be 8 outcomes: TTT, ... CCC. There will be such points (times), at which Take of one pair will be reached (relative to an arbitrary point). Unclosed pairs relative to the same arbitrary point must close in zero (on the average). Is it possible to test it with mql4? If you observe and advise me, I will write an EA.How did you determine this? You cannot in the MT4 tester.
Unclosed pairs relative to the same arbitrary point must close at zero (on average).
let it be 1 take, 0 stoploss, then (1 triple digit of a closed in plus pair)
111
110
101
100
11+10+01+00 - average zero
below combinations discard
011
010
001
000
Unclosed pairs relative to the same arbitrary point must close at zero (on average).
Suppose 1 is a TP, 0 is a TP, then (1 place of 3 closed in profit pair)
111
110
101
100
11+10+01+00 - average zero
below combinations discard
011
010
001
000
Are you assuming that the probabilities of reaching the stop and the take will be the same? And why do you discard 4 combinations?
Are you assuming that the probabilities of reaching the stop and the take will be the same? And why do you discard the 4 combinations?
Rejection - let me try a picture, for example, the position (one of the 3 legs) is open to buy
point A is an arbitrary point. When takeprofit reaches this point the other pairs give minus in total (at this point we should set TP and SL relatively to point A). The arbitrary point is arbitrary because we choose it ourselves.