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A Simple Card Game


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Hi Everyone,

 

I just learnt this game and ever since have been thinking about it. But couldn't figure it out with that garbage bag over my shoulder. So, I decided to post it here, but was a bit confused with where to post it. So, here it is -

 

 

1. First take 52 cards.

 

2. Tell Someone to think of a card in his mind. (Don't seperate the card, just think of it)

 

3. Then make three sets of card - I mean two sets will have 17 cards and one will have 18. (Make the sets like when u distrubute cards when u r playing a game , don't just take 17 , then 17 and then 18. Distribute them)

 

 

1 1 1

2 2 3

3 3 3

..... ..... .....

..... ..... .....

..... ..... .....

15 15 15

16 16 16

17 17 17

1

 

4. Then take one set of cards. Face the cards towards ur friend and show him the cards. Ask him if the card was in that set. If not take another set and ask him again. If it doesn't have the card either, take the last one(It must have the card, so no need to show him).

 

5. Then take the set that has the card and place the set between the other two sets. ( Remember there is no rule regarding the sequence of the other two sets. All u need to care about is placing the card holding set between the other two sets).

 

6. Now, do the same thing again (Procedure 3,4 & 5) (Make three sets and bla bla bla...)

 

7. Now, do this one more time (Procedure 3,4 & 5).

 

(Now u have done the same thing three times)

 

8. Now, take the 52 cards and start showing him the cards one after one. Keep counting the cards. When u have reached the 26 th card, ask him if this one's the card ( if not the next card is the card). U can't miss it. It's either the 26th card or the 27th one.

 

 

As simple as that. Nothing can go wrong if u follow the procedure accordingly. If something goes wrong, u can kick me for failing to describe it clearly. But I hope I was clear enough. It's always either the 26th one or the 27th one.

 

Now, Here is my question. How did this happen? I mean u just do the same thing three times and the card is there! No, complex calculation or nothing! Anyone please say how it worked?

 

[ Remember, you don't have to care about the other two sets, just place the right set in between them. And do the same thing for three times. (By card hoding set or right set, I mean the set that has the card) ]

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because each time you place the cards and deal sequemtialy after asking your friend where his card is, you`re dividing the odds by 3, and by a proccess of ellimination by this placement and division, it cannot do other than show the card your friend picked (providing he told the truth all the time).

 

NEXT.....

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YT2095 said in post #2 :

because each time you place the cards and deal sequemtialy after asking your friend where his card is, you`re dividing the odds by 3, and by a proccess of ellimination by this placement and division, it cannot do other than show the card your friend picked (providing he told the truth all the time).

 

NEXT.....

 

 

Thanks. But didn't understand a single word of it. (I told ya that I have the brain of a seven years old)

 

Could you please elaborate your answer (only if u have time). Sorry. :embarass:

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I`m not at all sure I can explain it in any more simple terms tbh? :(

 

providing your friend always tells the truth about what pile his card is in, each time you place that pile and deal out sequentialy 1 2 3 1 2 3 1 2 3 etc..

every time he says it`s such and such a pile, the closer his card gets to being singled out.

divided by 3 each time the sequence is done, until eventualy 3 divided by 3 = 1 (his card)

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