The Monty Hall conundrum is a brain-teaser based on the TV game show “Let’s Make a Deal” in which a contestant is given a choice of three doors, with a prize behind one of them. The host then opens one of the remaining two doors* to reveal no prize, then gives the contestant the opportunity to switch to the remaining door. It has been shown that human nature is to stick with the original choice, but probability calculations prove that it is better to switch.

One of the most interesting aspects of this problem is that it is quite difficult to convince people that switching is better. The traditional explanation is this: At the beginning, the contestant has a 1/3 probability of winning and, therefore, a 2/3 probability of losing. What many people call “common sense” would say that, after the host opens one of the other doors, the probability is 50/50, but this is incorrect. Because the host will never open a door to reveal the prize (that is, the probability of the host “winning” is zero), the odds of the contestant winning remain 1/3 with the original door and, therefore, 2/3 by switching. This seems to be a difficult concept to grasp. I therefore propose an intermediate step in the explanation.

Assume that, instead of opening a door to reveal nothing, the host offers the contestant the opportunity to take **both** of the remaining doors instead of the original door. Clearly, having two doors instead of one increases the odds of winning. A little thought should convince you that there is no difference between switching to both remaining doors and switching to the one remaining door after the host opens the other. After the contestant switches, both doors will be opened. It does not matter in which order they are opened, nor whether the host or the contestant does the opening. And, believe it or not, it does not matter whether the decision to switch is made before or after the “losing” door of the two remaining is revealed. The odds remain 1/3 by staying with the original choice and 2/3 by switching.

*In 2/3 of the cases, the contestant has not chosen the door with the prize, and the host has no choice but to open the other door that has no prize. In 1/3 of the cases, the contestant has chosen the door with the prize, and the host must make a choice of which door he reveals. Per the discussion in the comments, the host’s choice of which door to open in this case must be random (that is, unpredictable) to avoid giving the contestant more information and thereby changing the odds. If this choice is predictable, there is a possibility that the contestant may know with certainty that he must switch to win. With this scenario eliminated, the odds for the remaining scenarios become 50/50 between staying or switching. A random choice means that no additional information is given (we already knew that one of the doors unchosen by the contestant had no prize), and the odds do not change.

## 27 comments

Comments feed for this article

March 4, 2014 at 9:02 pm

PalmerEldritchI don’t think your explanation is particularly convincing since the statement

“…there is no difference between switching to both remaining doors and switching to the one remaining door after the host opens the other.” isn’t necessarily true. Depending on the algorithm Monty uses for opening a goat door switching (after the host opens a door) can have the same chance as sticking, whereas switching before the host opens a door always doubles your chance .

Say for example the host always opens the middle door if he can (i.e. you didn’t pick it and it contains a goat), then if he does open the middle door the 2 remaining doors are equally likely – 50/50. In order for it to be 33.3/66.7 if the host has a choice of 2 (goat) doors to open, he must pick one at random.

Most definitions of the problem (including yours) omit this necessary condition

March 6, 2014 at 7:52 am

Bill GascoyneThe key point in Monty’s “algorithm” is that he

neverreveals the prize. The odds of Monty winning arezerosince he intentionally “loses”. Nothing else matters. When you choose the first door, your odds are 1/3. One of the other two doors must conceal a goat. Thisdoes not changejust because that goat is revealed. The entire point is that Montyisn’tacting randomly, and when Monty opens a door, the odds for your doordo not change. Two-thirds of the time, your original choice is a goat, Monty knows it, and his choice of which door to open is made for him since he cannot reveal the prize. One third of the time, your original choice is correct, Monty knows it, and which of the two goat doors he opens does not matter. This is why switching after Monty opens the door is always identical to switching to both other doors before any door is opened. I don’t know how much clearer I can make it.March 6, 2014 at 9:56 am

Bill GascoyneHere’s another way to think about it: what’s important is not which door Monty opens, it’s which door Monty

doesn’topen. Suppose instead of three doors, there are 100. You choose one, and Monty opens all but one of the remaining doors to reveal 98 goats. Do you still think the odds between your door and the one Monty didn’t open magically transform to 50/50?March 6, 2014 at 3:16 pm

PalmerEldritchThe odds of Monty winning are surely the opposite of your odds of winning?

There’s nothing magical about how the odds can transform to 50/50 – it depends on whether Monty has a bias for opening one door over another (when he has a choice of goat doors)

Say you pick Door1 and Monty opens Door2, then the probability that the the prize is behind Door3 is 1/(1+p) and the probability it’s behind Door1 = p/(1+p) where p = Probability Monty opens Door2 if the prize is behind Door1.

If p =1 (Monty is guaranteed to open Door2 if it contains a goat) then there is no advantage in switching to Door3, both Door1 and Door3 are equally likely. So your initial odds can change (although over multiple trials you win on average 2/3rds of the time by always switching), and that’s why switching after Monty has opened a door isn’t necessarily identical as switching before Monty opens a door where you always have a 2/3 chance of winning.

March 6, 2014 at 4:16 pm

Bill GascoynePerhaps you are too young to ever have actually seen the game? The host, Monty, wants to give away the prize. Monty’s only bias is to not win. His show is more popular if the contestants actually win, but he can’t just give away the prize, the contestants have to at least appear to win the game by luck or skill. Monty knows where the prize is. It is Monty’s job to

notwin the prize. So, no, the odds of Monty winning are absolutelynotthe opposite of the contestant’s odds. Once again, the odds of Monty winning arezero. And, sorry but no, the odds do not change unless the contestant decides to switch. Then the contestant’s oddsdobecome the opposite, that is, 2/3 instead of 1/3. Please read my other two comments on the blog, in case you’re reading this in email.March 6, 2014 at 6:54 pm

PalmerEldritchI never watched Let’s Make a Deal, but it’s a game show and contestants don’t always win the big prize – when they don’t the show keeps the prize.

Anyway that’s beside the point, the MHP is a mathematical puzzle, and mathematically (using Bayes Theory) the probability of the contestant’s original door CAN change from 1/3 to 1/2 after a goat door is (deliberately) opened. It all depends on what value you assign to ‘p’ in the equation Prob(CarDoor3) = 1/(1+p) (see my previous comment)

March 7, 2014 at 7:22 am

Bill GascoyneI’ll concede that you are a better mathematician than I, but all the math in the world doesn’t help if you’re adding up the wrong things. Don’t take my word for it, the pros have actually done the experiment.

March 7, 2014 at 4:00 pm

PalmerEldritchBill,

If you’d read my previous comments you’d have seen I said “although over multiple trials you win on average 2/3rds of the time by always switching”, which is exactly what Mythbusters (and others) proved.

That doesn’t automatically mean that over every ,or any, individual trial you have a 2/3 chance of winning by switching as I’ve already explained, which is why I also said “switching after Monty has opened a door isn’t necessarily identical as switching before Monty opens a door when you always a 2/3 chance”.

That’s why (apart from not being an accurate description of the problem) I don’t think the “pick one door or two doors” approach is a better explanation.

March 7, 2014 at 4:55 pm

Bill GascoyneOK, you wrote, “… it depends on whether Monty has a bias for opening one door over another (when he has a choice of goat doors)”. You also wrote, “… the MHP is a mathematical puzzle, and mathematically (using Bayes Theory)…”.

Bayes Theorem applies to calculating probabilities in the absence of information. Since Monty is not lacking

anyinformation, Bayes Theorem does not apply to anything Monty does.It is the contestant who is operating with a lack of information, not Monty. The contestant’s first choice is random, meaning it has a 1/3 probability of being correct. If the contestant chooses incorrectly, Monty know it and he is trying to give away the prize by inducing the contestant to switch. If the contestant chooses correctly, Monty know it and he is trying to induce the contestant to lose in order to keep the game honest. In either case, Monty’s actions are fixed, he is not acting randomly, and the method of inducement (opening a door or offering both other doors) is irrelevant. From the contestant’s point of view, the odds of sticking vs. switching remain 1/3 vs. 2/3 just as they were from the beginning.

March 7, 2014 at 5:49 pm

PalmerEldritchThe MHP is a classic example of conditional probability and as such is ideal for solving by Bayes Theory (Google : “Monty Hall Problem Bayesian Probability” for 93,400 hits) .

Monty is neither trying to “induce the contestant to switch” or “induce the contestant to lose” (contradictory statements surely). In 2/3rds of all cases Monty’s actions are fixed, in 1/3 of all cases he has choice: which goat door to open?

Take 300 games of the MHP, say the contestant always chooses Door1 and Monty will ALWAYS open Door2 if it contains a goat.

A) In 100 games the car is behind Door1 and Monty opens Door2

B) In 100 games the car is behind Door3 and Monty opens Door2

C) n 100 games the car is behind Door2 and Monty opens Door3

In the 200 games where Monty opens Door2, you win 100 by staying and 100 by switching. So in this setup if Monty opens Door2 it makes no difference if you stay or switch.

In the ‘standard’ setup where given a choice Monty picks a door at random, then when the car is behind Door1 Monty opens Door2 50 times and Door3 50 times, so in the 150 games in total that Monty opens Door2 you win 50 by staying and 100 by switching – that’s the difference.

March 7, 2014 at 9:15 pm

Bill GascoyneI did the Google search you recommend, and looked at most of the top results. They all agree with me that the odds say switch. (Rather than providing a count, you might want to point out a specific link that agrees with your position.) The sentence I found that seems most relevant is, “So the [new] information is given by the fact that he [Monty] cannot open the winning door.” The consensus is that switching after Monty opens a door increases your odds from 1/3 to 2/3. Obviously, switching to both doors without having Monty open one also increases your odds from 1/3 to 2/3.

Now, you seem to be trying (desperately) to concoct a scenario where the contestant’s odds change from 1/3 to 1/2 when Monty opens a door. In your comment, you start by saying, “Take 300 games of the MHP, say the contestant always chooses Door1 and Monty will ALWAYS open Door2

if it contains a goat.”Then you artificially eliminate from consideration Scenario C, in which you win by switching. You’re not taking 300 games, you’re selectively taking 200 games. Totaling up all three makes winning 1/3 by staying and 2/3 by switching.And, BTW, I never said Bayes Theorem wasn’t applicable to the problem as a whole, I said it was not applicable to any of Monty’s actions. It is applicable only to the actions of the contestant.

March 13, 2014 at 11:49 pm

PalmerEldritchBill,

I don’t know what you’re disagreeing with, everything I’ve posted is correct (perhaps you can point out the relevant comments where you think I’m being inaccurate). You need to re-read what I actually said, not what you think I said. And I’ll say this again (for the 3rd time now) ” over multiple trials you win on average 2/3rds of the time by always switching”

My point (which you choose to ignore) is that in any single game the probability that the prize is behind the door not opened by Monty is only 2/3 IF Monty picks a door at random when he has a choice of 2 goat doors. In other words your statement “…there is no difference between switching to both remaining doors and switching to the one remaining door after the host opens the other.” isn’t necessarily true, it all depends on Monty’s door opening algorithm or method (which is what I said to begin with)

The odds on your door CAN change,because it DOES matter which door Monty opens when your door contains the car, as the example I gave shows.

And btw, Bayes Theory is applicable to Monty’s actions (that of opening the goat door).

You can look at “http://educ.jmu.edu/~lucassk/Papers/MHOverview2.pdf” pages 5 and 6, and I quote “…any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete”

Your “better explanation” is incomplete because of that reason.

March 14, 2014 at 10:29 am

Bill GascoyneI looked at the paper you cite. They explain it better than you. I think we’ve been talking past each other.

First of all, please be aware that my goal is to offer a layman’s explanation of why switching is better. I have no hope nor wish of being mathematically rigorous. My characterization of my explanation as being “better” is not meant to imply mathematical rigor, but only that I believe it is more understandable to most people. I said “better” (in a subjective sense), I never said “complete.”

Second, yes, I am making the assumption that Monty’s choice is random. It pretty much has to be in the real world. If it were not, and Monty’s algorithm becomes known to the contestant, then there is a scenario (your scenario “C” above) where, by opening the non-preferred door, Monty in effect tells the contestant that his original choice was wrong and that switching will guarantee winning. In the “real” game (Monty chooses randomly), the odds do not change because

realMonty gives no new information to the contestant by opening a door to reveal a goat. It is already known that one of the other doors held a goat, andin this casethere is no difference between offering the contestant the remaining door and offering both doors without opening one. In the variation, the odds change (to either 50/50 in 2/3 of the cases or 0/100 in the other 1/3) precisely because hypothetical Montyisgiving more information by choosing a preferred or a non-preferred door. It’s not that hypothetical Monty “uses a non-random algorithm,” it’s that he’s deliberately changing the odds by giving out new information. This change means that the game is no longer “fair” (and again, I concede that “fair” in this case is a subjective valuation which I have no intention of defending mathematically).Consideration of such variations (wherein one changes the game by assuming non-random choice by Monty) may add to the completeness of a mathematical analysis, but in my view they add nothing to an understanding of the actual game. My goal is simply to offer an intuitive sense of why switching is an overall better strategy.

August 7, 2014 at 3:30 pm

Freddie OrrellIn the post above containing the line: “Take 300 games of the MHP, say the contestant always chooses Door1 and Monty will ALWAYS open Door2 if it contains a goat.” three scenarios are given (A, B and C), each one occuring 100 times during the 300 games. The results are:

A) switching loses,

B) switching wins,

C) switching wins.

Switching wins 2 in 3 of these 300 games. The only way to arrive at the claimed 50/50 (1 in 2) chance is by pretending that the games when scenario C) happened, and Monty could not open Door2, never took place.

August 7, 2014 at 5:12 pm

PalmerEldritchWhat’s your point Freddie?

In any single game the probability the door Monty didn’t open contains the car = 1/(1+p) where p is the probability Monty chose the door he did given he has a choice of 2 (goat) doors to open.

The average or unconditional probability is 2/3 as I’ve already said (several times)

August 8, 2014 at 11:52 am

Freddie Orrell“What’s your point Freddie?”

I disagree over the role taken by Monty’s selection method when revealing a goat. I think their probability of a car remains 2/3 no matter how Monty chooses which of two goat doors to open, and never becomes 1/2 as a subset of an overall average of 2/3.

Reframing the hypothetical setup: “Take 300 games of the MHP, let the door the contestant chooses be subsequently labelled Door1 and let the goat-reveal door Monty chooses (by whatever means) be subsequently labelled Door2”.

Adding “On the occasions Monty has a choice between two goat doors, let the one he chooses be subsequently labelled Door3 instead of Door2” has no effect because those doors both have the same factual chance of a car (0/3) and the same theoretical chance of a car (2/3).

I believe that the chance in any single game is the same as that over a series of games, and does not vary according to Monty’s selection method when he chooses between two goat doors.

August 8, 2014 at 10:05 pm

PalmerEldritch“I think their probability of a car remains 2/3 no matter how Monty chooses which of two goat doors to open, and never becomes 1/2 as a subset of an overall average of 2/3”

This is simply untrue.

Take the example I gave ‘the contestant always chooses Door1 and Monty will ALWAYS open Door2 if it contains a goat’

If Monty opens Door3 it is 100% guaranteed that the car is behind Door2 (otherwise he would have opened Door2). The probability Monty opens Door3 is 1/3, and since the overall probability for switching is 66.7% then : (1/3 * 100%) + (2/3 * ?%) = 66.7%. So ?% = 50% , i.e. no benefit in switching when Monty opens Door2.

The probability the car is behind Door3 (if Monty opens Door2) is 1/(1+p), and the probability the car is behind Door1 is p/(1+p) where p is the prob. Monty opens Door2 when Door1 contains the car. Only if p = 1/2 do you get an answer of 2/3 and 1/3. If p=1 then you get 1/2 and 1/2

August 9, 2014 at 1:09 am

Freddie Orrell“Take the example I gave ‘the contestant always chooses Door1 and Monty will ALWAYS open Door2 if it contains a goat’”

Your scenario will occur one in six (I think) times the genuine Monty Hall game is played, but the probability of winning will remain stick = 1/3, switch 2/3 not 50/50. The probability will not change just on those occasions the scenario occurs.

You have described a new game where the contestant has a strategy which Monty knows. This does not have any bearing on the probability in the genuine MHP (where Monty does not know how how the contestant made their first choice and cannot base his strategy upon it).

Raises an interesting slant though – did Monty use his charisma to influence their final choice in a way that would increase dramatic tension and so improve ratings?

August 9, 2014 at 2:01 am

PalmerEldritchIt’s not a new game. It’s the MHP where Monty has a preference for opening a particular door when he has a choice of 2 goat doors. It has nothing to do with the contestant’s strategy but with Monty’s strategy, and in that strategy the probability that the door Monty didn’t open contains the car is either 1/2 or 1 (and conversely the probability that the door you picked contains the car is either 1/2 or 0)

On average it’s 1/3 and 2/3, but in an individual game it isn’t. I’m not wrong.

Your statement “where Monty does not know how the contestant made their first choice and cannot base his strategy upon it” is wholly irrelevant, the contestant’s 1st choice is academic, and the ONLY strategy at Monty’s disposal is which goat door to open when he has a choice of 2

August 9, 2014 at 2:43 am

Freddie Orrell“It’s not a new game. It’s the MHP where Monty has a preference for opening a particular door when he has a choice of 2 goat doors. ”

But Monty doesn’t do that in the MHP.

August 9, 2014 at 3:31 am

PalmerEldritch“But Monty doesn’t do that in the MHP” .How do you know?

In many (if not most) statements of the MHP Monty’s strategy is not defined.

The most reasonable assumption to make, and the one practically everybody makes without even realising it, is that he picks a door at random – which gives 1/3 and 2/3 for each game.(Oddly though most people don’t apply this same reasoning to the “I have 2 children at least one is a boy ….” problem).

It’s not however the only interpretation of the problem that can be made (as the blog writer has noted in his opening comment)

August 9, 2014 at 4:44 am

Freddie Orrell“How do you know?”

Exactly. As observers, we do not know what strategies the contestant or host may be using (although we know they are not colluding, as any suggestion of that is omitted from the rules and therefore not material to them).

This means that as observers, we do not know which scenario has led to Monty opening Door2. Being unable to separate out the individual occasions when it might be 50/50, we are left looking at a series in which it is 1/3 (stick) plus 2/3 (switch) to win the car.

August 9, 2014 at 4:46 am

Freddie Orrell“… we are left looking at a series in which it is 1/3 (stick) plus 2/3 (switch) to win the car.”

… plus 0/3 (chose Monty’s open goat door).

August 9, 2014 at 7:54 am

Bill GascoyneMany thanks to Freddie for taking on Palmer. I’ve found it entertaining. I shall attempt to clarify: Palmer is citing a very specific (and, I agree with Freddie, rather obscure and all but irrelevant) variation in which Monty’s criteria for opening a “goat” door are both non-random and known to the contestant. In Palmer’s scenario, Monty has a preferred door, and his only reason for not opening his preferred door is that it will reveal the prize. If Monty fails to open his preferred door, the contestant gains information about where the prize is (and it’s not behind his own door), making his odds of success 100% by switching and leaving the cases where Monty opens his preferred door to 50/50. In either case, Monty’s opening of the door gives the contestant more information because the contestant knows Monty’s non-random strategy. The original calculation of “it’s always better to switch” is based on the premise that Monty’s action of opening a door does not change the 1/3 odds on the contestant’s originally chosen door because the contestant gains no information about his original choice.

August 9, 2014 at 2:34 pm

Freddie OrrellI’m also taking on Ablestmage over at “50/50 Is King”, where the entertainment potential is also nicely buoyant. I hope Palmer puts in another appearance there – I’m not sure I am up to it without some help.

August 9, 2014 at 4:22 pm

PalmerEldritchIt is entertaining over at “50/50 Is King” isn’t it. It takes a special kind of stupid to continue to argue his position. I’m ‘Marley52’ over there btw.

August 9, 2014 at 4:59 pm

Freddie OrrellGlad to hear it, as I’m simultaneously tying myself in knots arguing it’s a 2-door problem at betterexplained.com and insisting it’s a 3-door problem at 50/50 while trying to say the same thing in both places.