$\textbf{Guess the Card}$
At a work picnic, Todd invites his coworkers, Ava and Bruce, to play a game. Ava and Bruce will each draw a random card from a standard $52$-card deck and place it on their own forehead. So they can see the other's card, but not his or her own. Meanwhile, they cannot communicate in any way. Then they will each write down a guess of his or her own card's color, i.e. red or black. If at least one of them guesses correctly, Todd will pay them $\$50$ each. If both guesses are incorrect, they shall each pay Todd $\$50$. If Ava and Bruce are given a chance to discuss a strategy before the game starts, can they guarantee to win?
After this game, Todd invites two more colleagues, Charlie and Doug, to join a new game. These four players will each draw a card and place it on their own foreheads so only others can see. What is different this time is that instead of color, they should guess the suite, i.e. spade, heart, club, and diamond. If at least one of them makes a correct guess, Todd will pay each of them $\$50$. Otherwise, they should each pay Todd $\$50$. Can these four co-workers guarantee to win if they are given a chance to discuss a strategy before the game starts?
$\textbf{Solution}$
For the two-player game, Ava can write down her guess by assuming the two cards have the same color, and Bruce can write down his guess by assuming the two cards have different colors. This strategy will guarantee their win.
For the four-player game, they can assume that spade is $0$, heart $1$, club $2$, and diamond $3$. Then Alice write down her guess so that the sum of the four (the three she knows, plus her guess) is a multiple of $4$. Bruce can make a guess so that the sum is a multiple of $4$ plus $1$. Charlie's guess makes the sum a multiple of $4$ plus $2$, and finally, Doug makes sure his sum is a multiple of $4$ plus $3$. This will be a guaranteed winning strategy.
$\textbf{Analysis}$
We will begin by analyzing the first game in an intuitive way. Then we will explain the second game using the remainder method. Finally, it will become clear that the two solutions are intrinsically the same with the different divisors used ($2$ vs $4$).
In the first game, because there are only two colors, colors of these two cards must be either the same or different. There are a total of four possible combinations: $B+B$ and $R+R$ for the same color combinations, $B+R$ and $R+B$ for the different color combinations where $B$ and $R$ indicate a black and a red card, respectively, and first color associates with Ava's card and the second one corresponds to Bruce's. However, two of these four combinations will be eliminated because each player can see the other's card. For example, assuming Ava sees Bruce's card is $R$, then the combination can only be $R+R$ or $B+R$, depending on whether her card is red or black. In this case, according to the strategy described above, Ava will write down $R$ which covers the case of $R+R$. If her card is $R$, then she will guess correctly. Otherwise, if her card is $B$, Bruce will write down $B+R$ because he sees Ava has a black card. In this case, Bruce's guess is correct.
The second game can be replaced as each one is randomly assigned with a number from $0$ to $3$, inclusive, according to our strategy described above. The remainder of their four numbers' sum divides $4$ can only be one of $0$, $1$, $2$, and $3$. Our strategy covers all these four cases, therefore one of them must be correct. To see this, let's elaborate with one example. Assuming Ava sees the other three numbers are $1$, $2$, and $3$. Then, if Ava's number is
- $0$, then the remainder is $2$ and Charlie will be correct. This is because he sees $0$, $1$, and $3$, in order to make the remainder be $2$, he will guess $2$.
- $1$, then the remainder is $1$. In this case, we can verify that Bruce will make the correct guess.
- $2$, then the remainder is $0$. In this case, Ava's guess will be correct.
- $3$, then the remainder is $3$ and Doug will be able to make a correct guess.
Finally, it can be seen that the first game's strategy is consistent with the second one. We can assign red as $0$ and black as $1$. Then the same color combination is equivalent to having the remainder as $0$, and different color combination is equivalent to having a remainder as $1$.