In the famous Three-Door-Game Monte cannot help to win all the time by signaling location of the prize, using only the freedom he allowed to use. No matter which strategies played, there is always at least one door where the prize will not be found. However, already in the game with four doors cooperative Monte can reveal two useless doors in sequence (leaving two doors unrevealed), to inform Conie about location of the prize, so enabling her to beat the only-switching strategies and win all the time.
Monte and Conie play the famous Three-Door Game. The quiz-team hides the prize behind one of the doors. Conie, who does not know where the prize was hidden, is asked to choose one of the doors as a first guess. Monte, who saw where the prize was hidden, will reveal then one of the doors which does not conceal the prize, but never the door chosen by Conie. Finally, Conie will be offered to either hold her choice or to switch to another yet unrevealed door. Conie wins if her final choice is the door which hides the prize.
See the book by Rosenhouse for history and variations of the problem [4]. The Three-Door-Game in proper sense of the game theory, as interaction of two actors, appeared before in [1], [2], [3]. In this note we shall focus on a combinatorial aspect of the game and a possibility of cooperative play for a certain design with four doors.
Let us label the doors 1, 2, 3 and think of four moves in the game. The first move is simple: the quiz-team hides the prize behind door p. On the second move Conie chooses door x. On the third move Monte offers a switch to door y = x by revealing a door which is not p. On the fourth move Conie chooses z from x and y: if she decides to hold her initial choice then z = x and if she decides to switch then z = y. She wins if z = p.
The strategy of the quiz-team is just the action of hiding the prize. We shall think of this move as a move of nature.
What can Conie do? On move two she chooses x, and on move four makes her finial choice z which depends on both x and y. For instance, she may first guess x = 1 and then decide to hold if y = 2 and to switch if y = 3. Her strategy can be labeled like e.g. 2hs which means the algorithm “first choose door 2, then hold if a switch to door with smaller number is offered and switch if a switch to a door with larger number is offered”. There are twelve such strategies 1ss, 1hh etc. It is a good exercise to write down them all! What can Monty do? He knows p, and on the second move observes x. If p = x then his decision is forced to offer a switch to another door, and if p = x he can choose out of two options: to offer a switch to a door with smaller or larger number. We can label his six strategies by sequences like 212 which indicate the switch offer y as reaction on match of the first Conies guess with door p = 1, 2, 3, respectively. Monte has six strategies in total.
It is important to understand that when Conie and Monte fix the way they will play the game, that is choose their strategies, the course of the game is completely determined by p. We can consider then the course of the game as the work of computer program which has input parameter p. In particular, for given profile of two strategies, one of Monte and one of Conie, the value of p determines if Conie wins or not.
One, two, three That’s how elementary it’s gonna be… Conie has strategy 1ss which wins for p = 2, 3 and loses for p = 1, so wins in two cases out of three, no matter how Monte plays. More generally, each always-switching strategy xss loses in case p = x and wins in two other cases.
It is intuitively obvious that no Conie’s strategy can win in all three cases p = 1, 2, 3 for any given Monte’s play. One explanation for that is the following The Unlucky Door Theorem. For every strategy S of Conie there exists at least one door u = u(S), which depends neither on p nor on Monte’s strategy, such that Conie loses when p = u for every strategy of Monte.
The proof is simple. If S = 1ss, then the prize is not won for p = 1 and we can take u = 1. If S = 1hh then Conie holds x = 1 whichever happens and so we can take u = 2 or 3. If S = 1sh then Conie will not switch to y = 3, but for p = 3 = 1 = x Monte will offer precisely y = 3, so we can take u = 3. Similarly, u = 2 for S = 1sh. The general principle to find the unlucky door is: there is always a door u which is never Conie’s choice at the forth move, whichever strategy of Monte.
We note in passing, that the number of winning cases may depend on Monte’s play. For instance for strategy 1hs Conie wins for p = 1, 3 versus any strategy 2 • • • , and wins only for p = 3 versus Monte’s 3 • • • . The complete matrix of winning cases is found in [2], [3].
Using unlucky doors we can readily show that every strategy of Conie is weakly dominated by some always-switching strategy. For strategy S one just takes uss. Since uss loses only for p = u, and S loses then too, each time S wins the strategy uss wins as well.
We see that whichever Conie does, Monte cannot play a strategy to make Conie sure winner, for all locations of the prize. In particular, if Conie plays some always-switching strategy, e.g. 1ss, she wins in 2 cases which is the maximum possible.
It is a well-known and obvious (and at the same time counter-intuitive) fact that, if the quiz-team places the prize at random by rolling a three-sided symmetric die then every always-switching strategy wins with probability 2/3, for every strategy of Monte. The
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