# Deck of cards - method for finding missing card - help please

[This thread was copied here from the old forum.]

jontsef 31 October, 2012 - 11:40

Disclaimer: Before you suggest this method instead: I am aware that the method of mutilating the images for each card that has been seen works best for these types of challenges, but I wanted to do one based on mental arithmetic.

I came up with a method for detecting the missing card in a deck (this only works when one card is removed not more) which is based on simple arithmetic, but I would appreciate some help with it.
One simple method would be to assign a different value to every card, from 1 to 52, and then subtract it from the total sum of 1 to 52 (1378) until you go through the entire deck and are left with a number between 1 to 52, which corresponds to the missing card. You could also add them if subtracting is trickier, and you’d be left with a number between 1326 and 1377, and the difference from 1378 would correspond to the card.
The problem with this method, assuming we already have the cards assigned to the numbers, is that keeping a running sum up to, or down from, 1378 isn’t easy to do quickly, and one slip will result in the wrong answer.

My improvement is to have 4 separate running totals, for each of the suits, each adding up to just 91 (sum of 1-13). So, let’s say you subtract each card you see from the corresponding total for that suit, then when you go through the deck you’ll have 3 suits with a total of 0 and one with a total of 1-13, giving you the missing card instantly (1 left in the clubs total = ace of clubs). Great, any second grader can accomplish that.

Here is the part where I need help though: what would be a good method to keep track of of the 4 totals at the same time? Is there a technique that would work for this? I want it to be based on arithmetic which would then give the missing card instantly, without having to create images and then go through the cards again.
Grandmaster George Koltanowski could play blindfolded against 56 chess players at the same time, so surely 4 running totals shouldn’t be a challenge…

Tarsier 31 October, 2012 - 13:22

In my opinion the classical method is easier, but your idea is interesting.
Remember 4 numbers speedly it’s not so easy, if anyone knows a technique, please post it, because I’me intersted on it also.
For your problem, if you change the values for a total of 2000 instead of 1378, you can add numbers until 100, and then restart from 0 each time. In this way you work with number composed by 2 cipher.
At the end you will have the complement of the last card.

greymatters 31 October, 2012 - 14:40

One of the major advantages of the classic mutilated image version is that it actually gets easier as more cards are removed. In addition, it actually gets easier as more cards are removed.

In professional card circles, the mathematical approach you’re working on is known as “clocking the deck” or “card clocking”. That might at least give you something to search for, and see what improvements others have made. Beware that many systems don’t give suits, however.

suncover 11 November, 2012 - 19:20

Does sounds like a lot of extra work, to me But an interesting approach, nonetheless!

NeilG 12 November, 2012 - 01:32

I like the idea of doing it mathematically, and the sums wouldn’t be hard. I would think it easier to keep one running total, but perhaps with individual cards valued 1-13 but suites 1000,2000 etc. The sums would be a little easier, and there’s less to learn.
E.g.
5 of diamonds 4005
7 of clubs 3007

Until you end up with a total of 130364 less the missing card values. (Or keep two totals, one for card values, one for suite totals - much easier than four totals)

[EDIT]

• as we mentally think “4 of spades”, not “spades-4”, reversing the suite and values I suggested above would probably be mentally fractionally easier/quicker, so 4 of spades 4001 etc, and ending up with a total of 364130 less the missing card value.

• the logic for putting one part in thousands is so that the two sums, suite values and card values never “interfere” with each other by requiring carrying, so that the arithmetic is a little easier.

[/EDIT]

jontsef 12 November, 2012 - 02:54

Thank you all for the helpful suggestions. I read a bit on the subject since and noticed that there have been quite a few proposals over the years (not surprising), but I’ve decided to place this challenge on hold for now and focus my efforts on other endeavors.
Neil, your idea is interesting; I think I’ll use something similar if I do decide to use the math method.

gojaejin 12 November, 2012 - 19:03

If the goal is a method for only a single card, then it seems a lot easier to add only the card ranks and track the suit by visualizing flipping two binary parameters. For example:

-The total is either GREEN or PURPLE and either written in SMALL or LARGE font.
-A red card flips COLOR; a black card does nothing.
-A heart or spade flips SIZE (because they kinda look like up or down arrows); a diamond or club does nothing.