.... I want other half or almost half to be odds, and other evens. The card in odds/evens can be in any order.

RJH

There are 13 cards in each suit in a deck of playing cards. Each suite contains six even cards and and seven odd cards. That is if we are counting the Jack as 11, the Queen as 12 and the King as 13. In a 52 card deck as a whole there is only 24 even cards and 28 odd cards. No matter what you want you can not have half odd and half even in a full deck. Now you are saying “I want the deck to be other half odds, other evens. I don't care what order the other cards are.” “.... I want other half or almost half to be odds, and other evens. The card in odds/evens can be in any order.” Which is just what you have got when you open a new deck without shuffling them at all. What more do you want?

:)

Tommy,

typical case of not seeing the forest for the trees. Kudos for you to spot it.

Sincerily

I am guessing that RJH is not a native English speaker, and that he is trying to separate the odds and evens, having all of the odds together on the top or bottom, and all of the evens together on the bottom or top.

The shuffle sequence I worked out above does just that, placing all of the odds together on top, and all of the evens together on the bottom.

Andru

OK

Luvisi, thanks. Yes I suck at english.

RJH

Seems like there is a problem about accomplishing this so here how I would do it. In my opinion, a stripper deck would probably be the best way to go, although it does require some skill.

Simply decide which cards are to be odd and which even. Let’s say A, 3, 5, 7, 9, J, K are odd and 2, 4, 6, 8, 10, Q are even. Seven cards are odd and six are even because we cannot cut 13 cards in half evenly. Therefore the full deck will consists of 28 odd and 24 even. How do we correct this problem? There are two ways to do so.

The first is to take two of the odd cards and designate them as even. For example, the two red Kings would be designated even and the two black Kings remain odd. Or the two red sevens would be even and the two black sevens would remain odd. (This can be done with any two of the odd cards.) Now we have 26 odd and 26 even.

The second way if to remove four odds cards from the deck. For example, remove one 3, 5, 7 and 9 or remove say four Kings or four fives or whatever. Now the full deck consists of 48 cards of which 24 are odd and 24 even.

These are the only two possibilities. Once that decision is made, the deck must be “trimmed” to enable one to pull the odds from the evens.

Arrange the deck composition decided upon in new deck order, or whatever order you so desire, and seal it back into a new deck case (if opening a deck from a new box is required).

When the deck is opened, it can be washed (scrambling the cards face down on the table) and then shuffled on the square. Then strip out the evens from the odds, do a perfect butt shuffle (or perfect Faro), and the deck will alternate even and odd. It is just that simple but not necessarily that easy to do.

I would think wedge strippers or bellies would do the job nicely. With wedge strippers it would be unwise to scramble/wash the deck first unless you did so very carefully not to reverse any of the cards end for end. Whether you wash the cards or not, the legitimate shuffles and running cuts should suffice to demonstrate the deck is well mixed, then strip out the even from odd followed by a perfect butt shuffle or Faro.

In the alternative, according to my 227.463 hours of rigorous computer calculations, in order to have the deck alternate odd and even with shuffles from a new deck requires 8,432,653 perfect out-shuffles and 9,675,098 perfect in-shuffles performed in a specific sequence, without a miss, to produce the same results. The stripper deck method might be easier and faster.

Quite frankly this entire concept is quite “odd” to me. I don’t mean to be impolite, but did you pull this question out of your ear or do you have a legitimate or practical reason for asking?

Just curious.

I need this to do a trick. I will have odd and even packet, cards selected from each one, but if you taked the card from odd packet, it will be but to the even packet, secretly ofcourse. And then I will find the card.

RJH

All he wants is to separate the 28 odd cards from the 24 even cards.

And if you don't stop causing trouble Cag I am going to report you to the management again.

Seems like I am incorrigible and still unrepentant, but if you report me AGAIN, who will be left to “give good post?” LOL

So anyway now we know what trick RJH wants to do.

But we can't tell him how to do it, as that would be exposuer and we would get banned if we did.

I am not a magician but there are so many easy and simple ways to do a trick like this and I will make only one last comment. For starters, try separating the deck into red and black.

Thanks for asking RJH anyhow as I did not know there was more odds than evens in a deck until you asked and I thought about it.

Surely if your going to use a stripper deck you may as well use that to locate the card rather than an odd/even set up? Less effort I'd say, and you don't have to look at the cards when locating, so gives you much more flexibility.