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If you take the Jack, Queen, King and Ace of Diamonds, then the JQKA of Hearts, JQKA of Clubs and JQKA of Spades and place them in a pile in that order and then cut the pile twice in the middle, then deal them - one card at a time- to four people, they all end up with four cards of the same value: four Jacks, four Queens and so on. Does anybody know how this works?

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You start up with JQKAJQKAJQKAJQKA , cut once to something else, reverse the cut (ie pretend to cut but actually chose the exact place to end up with the same as you started with), deal:

 

1st 2nd 3rd 4th

player 1. j j j j

player 2. q q q q

player 3. k k k k

player 4. a a a a

 

Read down column one then two etc and you will see that the original order is still there.

 

In fact with this trick the cuts don't even have to be that clever. Just make sure that both time you always cut to an ace (last card you pick is an ace) and the order will be preserved

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Yes, but that would be sleight of hand. wouldn't it? It also works with 2 random cuts , roughly in the middle, without a reverse cut. Would the answer be the same without the reverse cut?

Edited by goldglow
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Yes, but that would be sleight of hand. wouldn't it? It also works with 2 random cuts , roughly in the middle, without a reverse cut. Would the answer be the same without the reverse cut?

The idea, is that they're ordered JQKA.

So the first card, will give the first person a jack.

THe second card will give a queen.

The third a king.

The final a Ace.

So when you get to the next Jack, you're back with the person who got the jack first. Giving each person 4 of the same type.

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Yes, but that would be sleight of hand. wouldn't it? It also works with 2 random cuts , roughly in the middle, without a reverse cut. Would the answer be the same without the reverse cut?

 

yes - as Raider said. The reverse cut means the first person gets Jacks etc.; but now I am not sure you actually specified that. if you do not require the first person dealt to gets jacks it works with random cuts:

 

If you just cut (at random) you do this

 

jqkajqkaj//qkajqka => qkajqkajqkajqkaj

 

qkajqk//ajqkajqkaj => ajqkajqkajqkajqk

 

Notice that whatever you do the first,fifth, ninth, thirteenth cards are the same!

 

You have 16 cards in an order; the sequence would be continued by looping back to the beginning - ie the next card after the 4th ace would need to be a jack. So whenever you cut you do two things:

 

1. You add the first half of the pack to follow on from the back half - and you already know that the sequence will continue

2. The cut automatically creates a new situation where the follow on card in the sequence to the last card would be the first card

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yes - as Raider said. The reverse cut means the first person gets Jacks etc.; but now I am not sure you actually specified that. if you do not require the first person dealt to gets jacks it works with random cuts:

 

If you just cut (at random) you do this

 

jqkajqkaj//qkajqka => qkajqkajqkajqkaj

 

qkajqk//ajqkajqkaj => ajqkajqkajqkajqk

 

Notice that whatever you do the first,fifth, ninth, thirteenth cards are the same!

 

 

Yes, that's it, and no, the first person doesn't actually have to get a Jack- the first card dealt could be any of the four in the pack. Great answer. Thank you.

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If you take the Jack, Queen, King and Ace of Diamonds, then the JQKA of Hearts, JQKA of Clubs and JQKA of Spades and place them in a pile in that order and then cut the pile twice in the middle, then deal them - one card at a time- to four people, they all end up with four cards of the same value: four Jacks, four Queens and so on. Does anybody know how this works?

But the main factor here is that you have to cut them exactly in the middle. And not one card off! So you need to do it yourself, because if you allow somebody else to cut, as is the custom in most card games and tricks, the almost certainly will not cut them evenly. And when that happens your magic will not work.

 

But if cut exactly and distributed from the top, all you are doing is equally redistributing the original set of cards.

 

With only four suits, and the same amount of people to deal to, the only possible distribution result is yo have each person get four of the same suit and face. Think of the cards being numbered from one to four and co!Or coded in four colors. Every fourth card begins a new color, so every person of four gets that same number, in all four colors.

 

When the cards are thought of in this manner....Colors and numbers...The trick seems far less magical.

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But the main factor here is that you have to cut them exactly in the middle. And not one card off! So you need to do it yourself, because if you allow somebody else to cut, as is the custom in most card games and tricks, the almost certainly will not cut them evenly. And when that happens your magic will not work.

 

But if cut exactly and distributed from the top, all you are doing is equally redistributing the original set of cards.

 

With only four suits, and the same amount of people to deal to, the only possible distribution result is yo have each person get four of the same suit and face. Think of the cards being numbered from one to four and co!Or coded in four colors. Every fourth card begins a new color, so every person of four gets that same number, in all four colors.

 

When the cards are thought of in this manner....Colors and numbers...The trick seems far less magical.

 

As I have explained above - if you wish the first person to receive Jacks, 2nd Queen 3rd Kings 4th Aces then you need to either reverse the cut (which cutting exactly in half is just one potential way), or even merely make sure that you always cut to an ace.

 

However the OP has confirmed that he merely wanted everyone to receive four of the same card (no matter who got what set) so in that case any two cuts will do. No trick/sleight of hand is necessary

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But the main factor here is that you have to cut them exactly in the middle. And not one card off!

 

 

 

 

 

Thanks V_B. What you say is correct: if the cut is exactly in the middle it works, but i have also had the same result when i took 9 cards off the top and put them underneath for the first split and then took 5 cards from the top and put them underneath for the second split. Again ,4 top cards for the first split and then 4 top cards for the second split worked as well, just as imatfaal has explained, and what was especially impressive, in this second example, is that i dealt each of the 4 people 4 cards together in turn, not 1 each at a time, and they all ended up with a Royal Flush ( minus the 10 ,of course). I believe that in Poker, the chance of getting a Royal Flush is 1 in 649,740! ( Thanks Google.) By the way, you don't have to use JQKA- it works with any sets of 4 consecutive cards.

Edited by goldglow
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