aa r X i v : . [ m a t h . HO ] J un The doors
Sasha Gnedin
October 29, 2018
Abstract
We emphasize the dominance in the Monty Hall problem, both in theclassical scenario and its multi-door generalization. This is used toshow optimality of the class of always-switching strategies for nonuni-form allocation of the prize and arbitrary door-revealing mechanismin the event of match.
To switch or not to switch . . .
You have two tries to find a prizehidden behind one of three doors. You are first asked to choose a door butnot open it yet. Then one of the unchosen doors with no prize behind it willbe revealed, and you will be offered a second try. You win if the final choicefalls on the door concealing the prize.
Does it matter if you switch?
This is the famous Monty Hall problem. It is commonly assumed, oftenimplicitly, that the prize is hidden “uniformly at random”, meaning that it isequally likely to be behind each of the doors. There are two basic argumentsshowing that switching is better. The arguments are among the best knownpieces of elementary probability theory (see e.g. the article [5] ), but we needto sketch them here to show our point.The cases argument goes as follows. Suppose the prize is behind door θ .If you choose door 1 and then never switch, you win if θ = 1. If you choosedoor 1 and always switch, you win when θ ∈ { , } . The cases are mutuallyexclusive, therefore the probability to win with switching is 2 / onditional odds rigorously requires a further assumption on the randommechanism selecting a door to reveal in the event of a match , when the initialchoice falls on the door hiding the prize. If the randomization is performedby tossing the same fair coin, the conditional odds in favor of switching are2:1 with certainty. This is the second, odds argument .If the randomization is done by tossing a biased coin, with bias allowedto depend on door θ ∈ { , , } at which a match has occured, the odds arestill never unfavorable for switching. See Wikipedia pages on the Monty Hallproblem as a general source of references, [3] for discussion of assumptions,[4] for a summary of other arguments and their interconnections and [10] formany variations of the problem.In the nonuniform case, denoting p θ the probability to find the prizebehind door θ ( θ = 1 , , q the conditional probability to leave door2 unrevealed in the event of match at door 1, the odds in favor of switchingfrom door 1 to 2 are p : p q , hence switching is advantageous when p q < p .The inequality holds always for p < p but depends on q if p ≥ p . Forinstance, if p = 4 / , p = 3 / , p = 2 / q < /
4. Despite disadvantage in some situations, switchingcannot be devaluated: in the example the initial choice of door 1 was notoptimal, and you would get a higher chance to win the prize by first choosingdoor 3 then switching all the time as the second try is offered. When theprobabilities p θ are not equal it is useful to evaluate the switching action incombination with the initial choice. . . . how can you know to which? Without assigning probabilitiesto the doors the question
Does it matter if you switch? is sometimes regardedas not well-posed mathematically (e.g. [2], p. 56). Nevertheless, the questionmakes sense if we consider the decision-making as a two-step process andcompare the class of strategies that always switch with all other strategies.It turns that the problem has dominance that speaks in favor of always-switching strategies.To see the dominance we need to “switch the door” in the two basicarguments. Let us start with the cases argument, and compare two constant-action strategies A =“choose door 1, never switch” with B =“choose door 2,always switch”. They might seem incomparable, because when A is playeddoor 1 cannot be revealed, and when B is played door 2 cannot be revealed.However, this is irrelevant and, obviously enough, A wins for θ = 1 while2 wins for θ ∈ { , } . Thus B is never worse than A and if door θ = 3 ispossible (in some sense), then B is even better (in this very sense).The dominance reasoning parallel to the odds argument is slighly morecomplicated. To increase generality suppose D is a finite set of doors withat least three elements. You wish to guess door θ hiding the prize. Youfirst choose door x and then you are offered a switch to door y ∈ D \ { x } .Both x and y can be θ , all other doors have been revealed as useless. Nowyou need to decide whether θ and x match or mismatch, that is whether x = θ or x = θ . A strategy is therefore a combination of x and a decisionfunction a ( x, y ) with values in the two-point set of actions { match , switch } ,with match meaning staying with x and switch meaning switching from x .We denote a ∗ the always-switch decision function, with the only action a ∗ ( x, y ) = switch for all y = x .Your strategy is rewarded by means of the win-or-nothing payoff function W defined as W ( θ, x, match ) = 1( x = θ ) , W ( θ, x, switch ) = 1( x = θ ) , where 1( · · · ) equals 1 if · · · is true, and equals 0 otherwise. Two actionsmight seem permutable, but the symmetry is fallacious, because for every θ there is only one x = θ and at least two x = θ .The way we write the payoff function does not involve explicitly the de-pendence on the door to which the switch is offered. It is this feature whichwill enable us to compare the always-switching strategies with others. Whenswitching is offered to some door y = x , the inequality x = θ means, ofcourse, θ = y as we know that other doors have no prize. Nevertheless, theintepretation of the win with switch as a mismatch of the prize door withyour initial choice is more insightful and general. We may think, for example,of the game in which less doors are revealed as empty and switching is offeredto a subset of D \ { x } , with the convention that the action is successful if θ is there.The following key lemma is obvious from the definitions, and the domi-nance is a consequence. Lemma
For all θ and distinct x, x ′ ∈ D W ( θ, x, match ) = 1 ⇒ W ( θ, x ′ , switch ) = 1 . ominance Theorem Suppose for some fixed x = y ′ a decision function a ( x, · ) satisfies a ( x, y ′ ) = match . Then W ( θ, x, a ( x, y )) ≤ W ( θ, y ′ , switch ) for all θ and y = x , meaning that the strategy ( x, a ( x, · )) is weakly dominatedby ( y ′ , a ∗ ) .Proof Strategy ( y ′ , a ∗ ) always wins unless θ = y ′ , when W ( y ′ , y ′ , switch ) = 0,but then W ( y ′ , x, a ( x, y ′ )) = W ( y ′ , x, match ) = 0 as well. (cid:3) One insightful way to explain the phenomenon of dominance is to observethat for every strategy ( x, a x ( · )) there exists a ‘unlucky door’ u such that thestrategy misses the prize when it is behind u , no matter in which admissibleway the switching option is offered. Then the strategy ( u, a ∗ ) is weaklydominating ( x, a ( x, · )).We display next the payoff structure for all strategies in the 3-door case, D = { , , } . The notation must be self-explaining, but keep in mind thatthe value of variable y in the upper row is given in the case of match x = θ (otherwise y = θ ). For instance, let us check the entry 2 ms /2,1: the notation2 ms encodes the strategy x = 2, a (2 ,
3) = switch , a (2 ,
1) = match , so if θ = 2 , y = 1 this is a win, W (2 , , match ) = 1. For entry 2 ms /1,3 we havethe same decision function, θ = 1, in the notation of column 1,3 digit ‘3’ isirrelevant since we have a mismatch, so y = θ = 1 and W (1 , , a (2 , W (1 , , match ) = 0. 4 , y = 1,2 1,3 2,1 2,2 3,1 3,21 ss ms sm mm ss ms sm mm ss ms sm mm / weak , in the sense that dominated strategycannot be strictly improved for all θ , it is a serious ground to discard domi-nated strategies in many settings of decision making. Think for example ofguessing the right door when the prize is hidden by some algorithm. Youwould not use a strategy if there is another one performing at least as goodand in some situations even better. To maximize your score . . .
In the Bayesian setting of the guess-ing problem probabilities are assigned to all values of the variables out ofyour control. One random variable is the door with the prize Θ, with somespecified probabilities p θ for each value θ ∈ D . Another random variable,the door offered for switching Y is defined conditionally on the value Θ = θ and your choice x . In the event of mismatch Θ = x we have Y = Θ, andgiven Θ = x the variable Y assumes each admissible value y ∈ D \ { x } withsome probability q x,y . Think of biased roulette wheels, distinct for each door x and each having D − Y .5he probability to win with strategy a ( x, · ) is equal to the expected payoff E W (Θ , x, a ( x, Y )) = X θ ∈D\{ x } p θ W ( θ, x, a ( x, θ )) + X y ∈D\{ x } p x q x,y W ( x, x, a ( x, y )) = X θ ∈D\{ x } p θ a ( x, θ ) = switch ) + X y ∈D\{ x } p x q x,y a ( x, y ) = match ) = X y ∈D\{ x } (cid:8) p y a ( x, y ) = switch ) + p x q x,y a ( x, y ) = match ) (cid:9) = X y ∈D\{ x } (cid:8) p y a ( x, y ) = switch ) + p x q x,y (1 − a ( x, y ) = switch ) (cid:9) = p x + X y ∈D\{ x } (cid:8) ( p y − p x q x,y )1( a ( x, y ) = switch ) (cid:9) . In particular, the winning probability is equal to 1 − p x for the always-switching a ∗ , and is equal to p x for always-matching a ( x, · ) ≡ match , as itwas clear without computation. . . . try first to miss the door! Finding the strategy with thehighest winning chance (aka
Bayesian strategy) seems cumbersome from theabove explicit formula for E W (Θ , x, a ( x, Y )). The odds in favor of switchingfrom x to y are p y : p x q x,y , and these could be arbitrary, so the sign of p y − p x q x,y depends on q x,y if p x > p y .The dominance helps. Discarding by dominance all strategies except( x, a ∗ ) we are left with maximizing 1 − p x over x ∈ D . Let θ ∗ be the least likelydoor to conceal the prize (such door need not be unique), p θ ∗ = min θ ∈D p θ .The always-switching policy ( θ ∗ , a ∗ ) yields the highest winning probability1 − p θ ∗ , that ismax x ∈D , a ( x, · ) E W (Θ , x, a ( x, Y )) = P (Θ = θ ∗ ) = 1 − p θ ∗ , whichever the random rule to reveal the doors in the event of match.Switching from the least likely door is always beneficial since p y ≥ p θ ∗ ≥ p θ ∗ q θ ∗ ,y . This could be concluded without evaluation of odds, directly from the op-timality of the strategy ( θ ∗ , a ∗ ), which instructs that to get the prize it isoptimal to first miss it with the highest possible probability.6 nstances of the problem with nonuniform distribution (3-door case) appear in[10],[11], [7].Dominance was used in some multi-door game variations of the problem [1].Apparently, the dominance (in which sense?) in the standard MHP was conjec-tured by columnist John KayIf my conjecture has been right switching would be a (weakly) dominantstrategy, i.e. you can never lose by switching but you may gain ( FinancialTimes , 31 August 2005).
Acknowledgments
The author is indebted to Richard Gill and Jim Pitmanwhose polar views on the author’s engagement in the MHP generated thisnote.
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