# Maximize likelihood of at least one correct guess with incomplete information

A team of four players, $\{ A, B, A', B'\}$ is to plan strategies and then play a game with the following rules:

The players will be separated into two rooms, with $A$ and $B$ in one room and $A'$ and $B'$ in the other. No communication between the rooms is possible.

Each player will then receive a random integer uniformly distributed on $\{1,2,3\}$. The players will see their own number as well as that of their roommate.

Then each of the four players will guess the number assigned to his/her counterpart. That is, $A$ will guess the number assigned to $A'$ and vice-versa, while $B$ will guess the number assigned to $B'$ and vice-versa.

The team wins if at least one guess is correct. (Thus the order of guessing really makes no difference.) The objective is to maximize $P$, the team's probability of winning.

What is the maximal probability of winning, and describe a strategy which achieves that probability?

It is trivial to find a strategy that achieves $P=\frac{65}{81}$ since uncorrelated uniform random guesses will win that often. I can also produce a strategy that achieves $P=\frac{72}{81}$, and this will serve as an example of the sort of strategy that might be used:

$A$ always guesses that $A'$ has the same number as $A$. $A'$ always guesses that $A \equiv A'+1 \pmod{3}$. $B$ and $B'$ do likewise. The team would win unless $A \equiv A'+2 \pmod{3}$ and $B \equiv B'+2 \pmod{3}$, which hppens $\frac19$ of the time.

Can somebody improve on this, or prove it is optimal?

• Not sure that the partners signify....just seems like $(A,A')$ and $(B,B')$ are playing disconnected games, no? $A'\equiv A+i\pmod 3$ so guessing is equivalent to guessing $i$...the two can either say the same $i$ or they can say two different ones. Saying the same is foolish, saying two different ones is the same as your strategy. – lulu Jan 5 '17 at 20:58
• If we use mod 2 or a coin toss then they are sure to win whereas we could think A has probability 1/2 to loose and B too so both together would loose with probability 1/4 ? – Julien Pitteloud Jan 5 '17 at 22:36
• If we use mod 2 or a coin toss then the pair $(A,A')$ alone has a sure-win strategy. – Mark Fischler Jan 6 '17 at 16:05