lol so anyway, whats the maths principle behind this trick? i cant work out how it works
Well, you put the long column with the selection in to the middle of the three columns when you pick the cards up. This puts the selection somewhere in the center of the twenty-one cards--between 8-14.
So, when you set the card down again the selection will fall somewhere in either the 3rd through the fifth three-card row. When they point to the column again, you have now narrowed the selection to one of three cards (or two). When you repeat the steps, the selection will always be one of the three cards in the middle row.
Hmm..I thought the 27-card trick was similar to the 21-card one but apparently they're somewhat different.
For the 27-card trick and have the spectator select a card from one of the 27 and a number from 1-27 and have him/her tell you the number.
Deal the cards into 3 colums of 9, left to right, ask the spectator to indicate the column the chosen card is in, and collect back the cards. This step is repeated three times with the way the cards are collected back being as follows:
1st time: Divide the spectator's number by 3. If the remainder is 1, the column with the selected card goes on top of the pack (pack face down), if it is 2, it goes in the middle, and for 3, on the bottom.
2nd time: Add the digits of the chosen number. If the result is 1-3, column with the selected card goes on top of the pack, 4-6, in the middle and for 7-9, on the bottom. (There is one special case here if the chosen number is 19, it goes on top: 1+9=10, then 1+0=1)
3rd time: If the chosen number if from 1-9, column with the selected card goes on top of the pack, 10-18, the middle and 19-27, on the bottom.
When the cards are collected for the final time, the chosen card should be at the position of the chosen number and you can reveal it however you wish.
I'm not sure how this actually works but it does!