Count On

A mysterious deal

I came across this self-working piece of simple card magic in an old second-hand book on Patience last Christmas. Although no proof of its working was given, it's simple enough to show why it does. You'll need your 13 times table for this one.

Shuffle a pack of 52 cards. Instruct a member of the audience to deal the cards in the following way, but to make sure that you are out of the room while they do. Deal the top card face up. If it is, say, a 7 then 5 cards must be dealt face down on it to make 12; if it is an ace then 11 cards will be needed, any face card (which are to count as 10) require 2 cards, and so on. The cards dealt on top are to be placed so as not to totally obscure the face-up card. The process should be repeated until either all the cards are used up or (more likely) insufficient cards remain to deal on the current face-up card to make a complete 12. If this happens the facce-up card and the other remaining are to be placed in a separate spares pile. The diagram below shows the result of one such deal, with not enough cards to complete the pile on top of the 6.

Now the total of the scores on the face-up cards at the bottom of the complete piles is to be found and the piles squared off so that these cards are completely obscured. When you return to the room you will be given the spares, and all you will see is what you see is what you see in the diagram below. From this meagre information you will tell them their total.

Suppose that there are p piles and s spares.

Then we must have: (12 - c1) + 12 - c2) + … + (12 - cp) + p + s = 52

where c₁, c₂,…, c_p are the values of the face-up cards.

From this case we can see that:
c₁ + c₂ + … + c_p = 13p - 52 + s = 13(p - 4) + s.

So, on returning to the room you need to assess the number of piles and spares. Do this casually - examine the spare cards so that it's not too obvious that you're merely counting them. Then,

piles minus four times thirteen plus spares!