Name of a suitable count?

Aug 15, 2017
651
413
My goal is to hide the third card.
It would be great if no resets were required...like in an emsley count with 4 cards we always need to put the 'card-to-be-hidden' back in the 3rd position for performing the count consecutively? Preferably, a count without any such resets required.
Honestly speaking, I heard a count like that somewhere (started with an 'H'?)...
And if there is a count like that...
Any tips for performing counts naturally?

thanx!


PS:- Please check out my question in the cardistry forum, because I dunno why people don't really answer that forum quickly enough. Should I not have put a verbal notification like this here? Let me know if I should remove this postscript! :)
 
Aug 25, 2017
172
93
Pittsburgh, PA
Off the top of my head... I would look into the Jordan. It hides the 4th card though and at the end of it, it's already reset. I'll give it some thought and see if I can come up with an answer for the third card. Maybe someone will reply before that though.
 
Jan 26, 2017
2,173
1,338
23
Virginia
You can always do an emsley count and then brake the packet and flick a single card or somethi g and put it back on the bottom.
 
Aug 25, 2017
172
93
Pittsburgh, PA
Elsmley rhen Jordan. Then Elmsley. Then Jordan. Repeat. Again.
The only thing that kept me from suggesting this was the reset. Aside from the reset though, you're right, this combo works perfectly. The OP may need to sacrifice the reset or at least make it as natural as possible. I've been thinking about it all day and nothing has really stood out to me that would meet all the OP's requirements.
 
Aug 25, 2017
172
93
Pittsburgh, PA
I actually just did them consecutively just now to check it. It would depend on how you execute the Jordan. If you drop the remaining packet as 1, then it auto resets in Jordan position again. If taking the 3 as one in the second count, then you’re set for an Elmsley. Of course, dropping the last packet of 2 as 1, would practically constitute a 4 card Hamman lol.

Anyone who read that and got confused...just know that I knew what I was saying lmao
 

DavidL11229

Elite Member
Jul 25, 2015
589
314
Seattle
My goal is to hide the third card.
It would be great if no resets were required...like in an emsley count with 4 cards we always need to put the 'card-to-be-hidden' back in the 3rd position for performing the count consecutively? Preferably, a count without any such resets required.
Honestly speaking, I heard a count like that somewhere (started with an 'H'?)...
And if there is a count like that...
Any tips for performing counts naturally?

thanx!


PS:- Please check out my question in the cardistry forum, because I dunno why people don't really answer that forum quickly enough. Should I not have put a verbal notification like this here? Let me know if I should remove this postscript! :)
Do an Elmsley count, after the third card you have the fourth by itself. Just put it on the bottom of the pile instead of the top when you are done. You are reset for another Elmsley. Or just do Jordan as above.
 

DavidL11229

Elite Member
Jul 25, 2015
589
314
Seattle
I was thinking of it as an alternate version (or whole different count if you will) where the last card is placed on the bottom instead of the top. You gotta put it somewhere, so put it on the bottom. If you put it there during, instead of after the count it's just a different count, not a reset. At least that was how I was looking at it.
 
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