BULLET BRAIN BUSTER
Father Noel Furlon's Picnic
At a work picnic, Father Noel Furlong announces a challenge to his coworkers. The Brain and Arja Stark are selected to play first. Father Noel Furlong places £100 on a table and explains the game. The Brain and Arja will each draw a random card from a standard 52card deck. Each will hold that card to his/her forehead for the other person to see, but neither can see his/her own card. The players may not communicate in any way.
The Brain and Arja will each write down a guess for the colour of his/her own card, i.e. red or black. If either one of them guesses correctly, they both win £50. If they are both incorrect, they lose. He gives The Brain and Arja five minutes to devise a strategy beforehand by which they can guarantee that they each walk away with the £50.
The Brain and Arja complete their game and Father Noel Furlong announces the second level of the game. He places £200 on the table. He tells four of his coworkers  Ava, The Brain, Furious Pete and P!nk—that they will play the same game, except this time guessing the suit of their own card, i.e. clubs, hearts, diamonds or spades. Again, Father Noel Furlong has the four players draw cards and place them on their foreheads so that each player can see the other three players' cards, but not his/her own. Each player writes down a guess for the suit of his/her own card. If at least one of them guesses correctly, they each win £50. There is no communication while the game is in progress, but they have five minutes to devise a strategy beforehand by which they can be guaranteed to walk away with £50 each.
For each level of play – 2 players or 4 players– how can the players ensure that someone in the group always guesses correctly?
Answer
The 2person case is simpler. Let r denote a red card and b denote a black card. Let’s list The Brain’s card’s colour first. Then the 4 distinct possibilities are rr, bb, rb, and br. The following strategy is foolproof:
The Brain guesses that his card is the same colour as Arja’s (covering the cases rr and bb). Arja takes the opposite approach and assumes that her card’s colour is different from The Brain’s (covering the cases rb and br). One of them will necessarily be correct. If The Brain and Arja properly employ these strategies, they cannot lose the game.
Before diving into the 4person case, let’s reexamine the 2person case. We could interpret a red card as a 0 and a black card as a 1. Then The Brain’s strategy, as described above, handles the cases 00 and 11. His strategy could be interpreted as “the sum of our cards is even.” (0+0=0, 1+1=2) In this language, Arja’s strategy would be “the sum of our cards is odd.” Again, one of them must be right, no matter the cards. Note that The Brain and Arja have divided the sample space of all possible outcomes into 2 cases, based on the sum of the cards.
The 4person case is more general and depends upon the notion of modular arithmetic, and in particular, remainders when dividing by 4. Rather than purely guessing “the suits are the same,” or “the suits are different,” the players must be more clever. Let’s map each suit to a number. Clubs are 0, diamonds are 1, hearts are 2, and spades are 3. So, if Arja looks out and sees hearts (The Brain), hearts (Furious Pete), and spades (P!nk), she interprets this as 2+2+3=7. Suppose that Ava’s card’s suit is clubs, which she obviously doesn’t know. Consider the following strategies:

Arja guesses her suit so that the total sum is a multiple of 4. She sees hearts, hearts, spades and guesses diamonds (2+2+3+1=8).

The Brain guesses his suit so that the total sum is 1 more than a multiple of 4. He sees clubs, hearts, spades and guesses clubs (0+2+3+0=5).

Furious Pete guesses his suit so that the total sum is 2 more than a multiple of 4. He sees clubs, hearts, spades and guesses diamonds (0+2+3+1=6).

P!nk guesses her suit so that the total sum is 3 more than a multiple of 4. She sees clubs, hearts, hearts and guesses spades (0+2+2+3=7).
In this example, P!nk guesses correctly because the total sum is 7, which is 3 more than a multiple of 4. For any combination of suits, one of them must be right. After all, the total sum must be a whole number. And all whole numbers have remainder 0, 1, 2, or 3 when divided by 4. As long as each of them sticks to his/her strategy, the group can ensure they won’t lose the game. Interestingly, it doesn’t matter which numbers they choose to use, as long as they all have different remainders when divided by 4. So, choosing any four consecutive whole numbers is a sufficient mapping.