Q
The count as it stands now is:
sw1=5168-4830=338
sw2=4830-4930=100(29.5%)
sw3=4930 downwards --target 1.62*sw1=1.62*338=550=4380 in 5 waves
(i)=4930-4759=171
(ii)=4759-4868* having retraced 64% of (i) so far
Now the question is, what are the probabilities for this sw3(ii):
the chances of sw3(ii) going further up are there, but it can not go above 4930 as per this count.All shorts, therefore, must have a SL at 4931
UnQ
Clearly this count has become invalid as (ii) cannot retrace more than 100% of (i).
And we have now to look for an alternative count. And for that lets go one degree higher----
C (on a higher degree) in play from 5944 down as a 5 wave form,having played out:
w1=5944-5329=615
w2=5329-5740 as an expanded flat;
w3=5740-4720=1020
w4=4720-5168
w5=5168 downwards started on 22-9-2011 and most likely cover(=w1) 615 points.And this w5 which was to cover 615 points in 5 waves has already covered 409 points(5168-4759) and we are yet struggling to find valid fractals for it.
Lets take various probabilities one by one( not necessarily in order of preference):
Probability 1: w4 did not end at 5168
If so then A of w4=4720-5168(448) a 3 wave form
B=5168-4759=409 having fractals
sw1=5168-4830=338
sw2=4830-4930=100
sw3=4930-4759=171(50%) neither qualifies as a 5 wave(to consider it as B(a) nor as 3 wave
so this probability is ruled out.
Probability 2 : W5 is over at 4759
As 5168-4759=409 is more than 0.618*w1(=380) so this probability also comes in play---w5 ending above w3.
Least probable because w3(=1020) was only 166% of w1(=615) and w5 truncations occur mostly when w3 is hugely extended over w1.
Probability 3:
w5 in play and its sw1 instead of ending at 4830 has gone upto 4759 thus making
sw1=5168-4759=409 and currently sw2 to the up in progress.
For a wave at whose starting the tentative target was a length of 615, it is very difficult to digest that its sw1=409 as that gives a tentative length of 1000+ to the full wave.
Probability 4: w3 did not end at 4720 and the move from 4720-5168 and 5168 - 4759 be seen as continuation of w3.
I request my fellow EW enthusiasts to contribute to these probabilities---
Raghu
Dear Guruji,
ReplyDeleteOne more Alt count based on observ of ur Prob-1, pl see wid deep thought & advise:-
Q
If so then A of w4=4720-5168(448) a 3 wave form
B=5168-4759=409 having fractals
sw1=5168-4830=338
sw2=4830-4930=100
UnQ
Pls see C.3.3.4 as :-
A 4720 5169 449
B.a 5169 4911 -258
B.b 4911 5168 257 -99.6%
B.c 5168 4759 -409 158.5% (B=410)
C 4759 4980 221 ** in play
T1: C = 0.618xA = 277 + 4759 = 5037
T2: C = A = 449 + 4759 = 5208
Does this look OK ji ?
urs
SMO
Dear Shri,
ReplyDeleteThis has been mentioned as Probability 4 by me.
Working on it----
:-)
Guruji Sa-adar pranam
ReplyDeletecan W4 be in the form of A:B:C 3:3:3 form
UPS
Dear UPS,
ReplyDeleteC can not be a 3 wave