Card trick: guessing the suit if you see the remaining three cards (important is that you can't move or turn the cards)

Question

Suske, Sidonia, Lambik, and Jerom are in a room. Wiske has them each draw a card of their choice from an open standard deck of cards and lay it face up on the table. The remaining cards are set aside. There are four card suits: hearts, diamonds, clubs, and spades. Wiske then carefully looks at the cards on the table, takes one away, and calls Professor Barabas in. Everyone is surprised when, after looking at the remaining cards on the table and which person they belong to, he knows the suit of the removed card (without Wiske showing him the card or the remaining cards). How is this possible? You may assume that Wiske and Barabas have agreed on a strategy beforehand.

Answer 1

Step 1:

Assign each of the players a unique number from the set {0, 1, 2, 3}.

Step 2:

Assign each suit a unique number from the same set. For example, D=0, C=1, H=2, S=3.

Step 3:

Wiske sums the card values and takes the result modulo 4 (call this result x). Then, Wiske removes the card of player x.

Step 4:

To decode the result, Barabas sums the remaining cards, and knows that one more card must be added and that the total sum modulo 4 must be x. The suit of the missing card can then be uniquely determined.

Example:

Suske, Sidonia, Lambik, and Jerom are respectively assigned 0, 1, 2, 3, and they respectively draw heart, heart, club, spade. Using the example suit assignment in Step 2, the sum is 8, which taken modulo 4 is 0. Therefore, Wiske removes Suske's card.

When Barabas walks in, he sees heart, club, spade and knows Suske's card is missing. The sum of the shown cards is 6, and the total sum modulo 4 is 0. As such, the last card's value must be 2, as the previous/next available values would imply card values of -2 and 6 (impossible). Therefore, the last card must be a heart.

Answer 2

Just wanted to throw out what I hope is an easier explanation for ApexPolenta's answer...

The important information being communicated to Professor Barabas by Whiske is a single number from 0-3. The puzzle could be changed so that Whiske chose a person to give up their card and then left the room and let the four people swap cards. As long as the same person gets rid of a card after swapping them around so that Professor Barabas can see who the person selected by Whiske is it would work out the same.

The key is to use modulo arithmetic. Assign each suit a value from 0 to 3 and each person a value from 0 to 3. Wiske will add the values for the suits of all 4 cards and take it mod 4 to get a number between 0 and 3, then remove the card of the person associated with that number, not caring what the card is. When Professor Barabas comes in, he totals the three remaining cards takes the total mod 4 to get a number between 0 and 3. The missing card must right to move that total forward (mod 4) to get to the number assigned to the person that had the card removed.

Examples:

For examples I'll use ABCD for the people (where A=0, B=1, C=2, D=3) and 0-3 for the suits. The actual numbers assigned to the people and suits must be agreed upon ahead of time by Whiske and Professor Barabas, but the actual numbers don't matter as long as each suit and person has aunique number from 0 to 3.

| 1st| 2nd| 3rd| 4th| 5th| 6th| 7th|
|----|----|----|----|----|----|----|
A (0) | 0 | 0 | 3 | 2 | 3 | 3 | 3 |
B (1) | 0 | 1 | 3 | 3 | 3 | 2 | 3 |
C (2) | 0 | 2 | 3 | 3 | 3 | 3 | 2 |
D (3) | 0 | 3 | 3 | 3 | 2 | 3 | 3 |
Total | 0 | 6 | 12 | 11 | 11 | 11 | 11 |
Mod 4 | 0 | 2 | 0 | 3 | 3 | 3 | 3 |
Chose | A | C | A | D | D | D | D |
-----------------------------------------------
Remain | 0 | 4 | 9 | 8 | 9 | 8 | 8 |
Remain Mod | 0 | 0 | 1 | 0 | 1 | 0 | 0 |
Required | 0 | 2 | 3 | 3 | 2 | 3 | 3 |

So...

In the first example Whiske sees the total mod 4 is 0, so they choose 'A' which represents the total mod 4 of all the cards. Professor Barabas sees a total of 0 and that person A (where A = 0) had the missing card so the expected total is 0. Therefore the missing card had a value of 0. If it didn't, then Whiske would have chosen a different card.

4th through 7th are basically the same with 3 people having cards with the suit worth 3 and one person having a card with the suit worth 2, just a different person each time. The total is 11 which if you take off 4 twice is 3, so Whiske chooses D (because D=3) each time.

When D has one of the '3' cards, Professor Barabas sees two '3' cards and one '2' card for a total of 8, which mod 4 is 0. Since Whiske picked the person worth 3, they must have had a 3 to go from 0 (what Barabas sees) to 3 (which Whiske picked). When D has the '2' card, Professor Barabas sees three '3' cards for a total of 9, which mod 4 is 1. Since Whiske picked the person worth 3, they must have had a '2' card to go from 1 (what Barabas sees) to 3 (which Whiske picked).