# Monty Hall all over again?

Often, when people ask the Monty Hall problem, they omit some important details, such that the problem becomes ambiguous, or they add more details, such that the problem changes drastically. For example, what is the answer for the following problem?

You are participating in a show. There are three doors in front of you. There’s a car behind one of them; there’s a goat behind another; and there’s a sheep behind the third. To the best of your knowledge, every assignment of the three prizes to the three doors is equally probable. You do want the car, badly, and you don’t care about either of the animals at all. You can choose one door, and whatever is behind that door will be awarded to you.

So, without loss of generality, you choose the leftmost door.

After that, to your complete surprise, the host opens the middle door, and you observe a sheep behind it.

The host then offers you a chance to change your choice. Based solely on the information you have, without trying to guess the motives of the host, is it beneficial for you to change your choice?

EDIT: to make the question a little more specific, what is the ideal strategy for you, that maximizes your chances of getting the car in the worst case, no matter what the hosts motives are.

• How is this different than the Monty Hall problem? – JLee Apr 27 '15 at 0:15
• Well, this is precisely the point of the question -- to figure out if this problem is different from Monty Hall or not. – Ishamael Apr 27 '15 at 0:17
• We are not told that the host opens a non-car door. – Caleb Apr 27 '15 at 0:19
• By the way, this question is perhaps more elegantly presented here: puzzling.stackexchange.com/questions/3625/… – Ben Aaronson Apr 27 '15 at 17:54
• The question is not well formed. Reasoning "without trying to guess the motives of the host" or allowing an a priori model for its probability is like asking someone the expectation of a Bernoulli distribution without knowing the parameter p. It might be reasonable to assume E(p)=0.5, but there's no empirical reason that 0.6 is any worse a choice. – Joe Lee-Moyet Apr 27 '15 at 19:21

Here's the formal expression of what we seek. Let $C$ be the event "chose the door with the car" and $R$ be the event "revealed the door with the car". Now, assuming that the door opened is random (but not the one the player chose), we have

$$Pr(C) = \frac13\\ Pr(R|C) = 0\\ Pr(R|\bar C) = \frac12\\ Pr(C|\bar R) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+Pr(\bar R|\bar C)Pr(\bar C)} = \frac{1/3}{2/3}=1/2$$ And therefore the probability of winning by switching or by staying are equal.

This differs from the Monty Hall problem through the value of $Pr(R|\bar C)$, which is zero for the Monty Hall problem. In that case, we have $$Pr(C|\bar R) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+Pr(\bar R|\bar C)Pr(\bar C)} = \frac{1/3}{1}=1/3$$ and thus, switching doubles the chance of winning. This is not true if the host can open the door with the car, where the probability of winning is equal in either direction.

It's tempting to think that the situation is the same (the door that is opened is one without the car) between this situation and the Monty Hall one, but the balance of probabilities leading to the situation is different.

It's a mistake involving the confusion between counting the number of arrangements and determining the actual chance of each event. In the Monty Hall problem, if the player chooses a wrong door, which has a probability of 2/3, the host will necessarily open the other wrong door, and thus the chance of winning by switching is 1. In this problem, if the player chooses a wrong door, there's a 50% chance that the car will be opened. This halves the number of times that the player will choose the wrong door and have an opportunity to switch.

As a result, while the probability rules around the actual situation are the same, the underlying probabilities are shifted - there's only a 1/3 chance that the player will choose a wrong door and the host will open the other wrong door, and a 1/3 chance that the player will choose the right door. In the Monty Hall problem, the probability of the former is 2/3. So even though the situation itself looks the same, the chance of being in that situation while having chosen the wrong door is lower, with the chance of each of them being 1/2 (given that the situation arose), rather than 1/3 and 2/3.

Note that the probability that the player chose the car given that the car wasn't behind the door the host revealed depends on only two parameters. If we express our equation for $Pr(C|\bar R)$ slightly differently, we have $$Pr(C|\bar R) = \frac{Pr(C)}{Pr(C)+\frac{1-Pr(R|C)}{1-Pr(R|\bar C)}(1-Pr(C))} = \frac{A}{A+B(1-A)}$$ where $A=Pr(C)$ and $B=\frac{1-Pr(R|C)}{1-Pr(R|\bar C)}$ - as such, these are the only two values that are relevant. In the Monty Hall case, $A=1/3$ and $B=2$. In our case, $A=1/3$ and $B=1$. If the host is incredibly malicious, almost certainly set to reveal the car if the player doesn't choose it, then $A=1/3$ and $B\approx 0$, meaning that the player is practically guaranteed to win by staying.

If we were to allow the host to open other doors, then we have to handle those distinctly. As such, it is better to consider $\hat R$, which is the event "the host opened one of the other two doors, and it contained an animal" (or specifically "the sheep" - it's equivalent). For the three-door-only case, $\hat R \equiv \bar R$. So the general expression would be $$Pr((C|\hat R) = \frac{Pr(C)}{Pr(C)+\frac{Pr(\hat R|C)}{Pr(\hat R|\bar C)}(1-Pr(C))} = \frac{A}{A+B(1-A)}$$ where $A=Pr(C)$ again and $B=\frac{Pr(\hat R|C)}{Pr(\hat R|\bar C)}$. In this expression, it is easily seen that adding extra doors that aren't among the three we care about doesn't change the result. With an extra door, we have $Pr(\hat R|C) = 2/3$ and $Pr(\hat R|\bar C) = 1/3$, and we again have $B=2$.

Note - the assumption was described above as "assuming that the door opened is random (but not the one the player chose)". This is for the purposes of the calculations. The true assumption is actually that the host could have any possible motive with equal probability, and looks at the expected probability. It just happens that the two assumptions are equivalent.

Update: As I noted, taking the host's motive as "random" is the same as randomly choosing the host's motive. It's worth explaining properly how this works, and explain an easy pitfall in determining it.

Suppose that $X$ is a uniformly-distributed random variable (between 0 and 1) representing the probability that the host reveals the car given that the player didn't choose the car's door. That is, $$Pr(R|\bar C\land X=x)=x$$ Because $C$ and the value of $X$ are independent of each other, we can deal with these separately - that is, we can treat it as $(R|\bar C)|X=x$ or $(R|X=x)|\bar C$, and nothing will be wrong.

And here's where the pitfall comes in. It can be tempting to calculate $$Pr(C|\bar R\land X=x) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+Pr(\bar R|\bar C\land X=x)Pr(\bar C)} = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+(1-x)Pr(\bar C)}$$ and then integrate from there. However, there's a problem - $\bar R$ and $X$ are not independent, and thus we cannot treat $C|\bar R\land X=x$ as $(C|\bar R)|X=x$, which would be necessary to determine $C|\bar R$ by integration.

Instead, we need to calculate it through the full process. That is, we must get $$Pr(C|\bar R) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R)}$$ and to calculate the denominator, we must integrate. That is, $$Pr(\bar R) = \int_0^1 Pr(\bar R|X=x) dx$$ where $$Pr(\bar R|X=x) = Pr(\bar R|C)Pr(C) + Pr(\bar R|\bar C\land X=x)Pr(\bar C)\\ = Pr(\bar R|C)Pr(C) + (1-x)(1-Pr(C))$$ and this quickly gives us $$Pr(\bar R) = Pr(\bar R|C)Pr(C) + \frac12(1-Pr(C))$$ at which point the equivalence with assuming the host chooses at random becomes clear.

• Comments are not for extended discussion; this conversation has been moved to chat. – user20 Apr 29 '15 at 20:04

Without assuming anything more? No, we have absolutely no idea whether it is beneficial to switch or not.

If the host knows what is behind each door and is malicious, then he will try to trick us. But if we know he will try to trick us then we can use that against him. Unless he knows that we know, and so on and so on. In the end that means there is no difference from switching when we don't know more about the host.

If a door with an animal is ALWAYS revealed, then we should switch, as per original monty hall.

If a door is revealed at random it doesn't matter if we switch, the chance is 50/50. That is we wouldn't get a chance to switch if the revealed door had happened to be the car. The chance is 50/50 both when the random reveal can be all 3 doors or when it can only be the 2 we did not pick, doesn't matter.

If the host only opens a door when we have already picked the car, then of course we should stay.

Summation: We need to know the rules for the host opening a door to make an informed decision.

• To make it really clear, if we could answer this question without reasoning about the hosts' algorithm for opening the door and choosing to open the door, then any answer we generate would be correct regardless of the hosts' algorithm. As there are algorithms the host follows that makes "stay on your door" have 100% chance of getting the car, and the same for "switch to another door", this is clearly not the case. – Yakk Apr 27 '15 at 18:14
• This is technically the best answer because the question is so vague and assumptions must be made that drastically change the possible answers. – JLee Apr 27 '15 at 20:45
• If the door picked is random it is not the same as the original Monty hall. If the door is picked at random and no car is revealed the odds are 50/50. – Taemyr Apr 28 '15 at 10:22
• Hmm, yeah you are right, let me edit that. – RHawkeyed Apr 28 '15 at 12:39

I will stay with the door I have. I am banking on a malicious host who takes advantage of my Monty Hall familiarity and only offers me a switch when I am pointing to the car.

• This is precisely why I clarified at the end of the question that you should not try to guess the motives of the host :) – Ishamael Apr 27 '15 at 0:24
• Ah, ok. I took that to mean that we shouldn't assume the host has the original Monty Hall motives. I see the true meaning now. – Caleb Apr 27 '15 at 0:25

The accepted answer conflates "the host's knowledge and motivations are unknown" with "the host's actions are uniform and random", which is at best a debatable assumption. I actually think the correct answer to this question is that there is not enough information to answer it as phrased.

For example, if I flip a coin outside of your view and ask what are the odds it landed tails, the correct answer is "I don't know", not 50%. The coin might not be fair or it might be two-headed, etc. Probability itself is defined based on distributions. If you know nothing at all about the distribution, any discussion of probability is meaningless. (As @Yakk points out in a comment, bounds on the distribution can lead to bounds on the probability. But there is nothing like that in this question.)

But I want to ignore that debate and just consider the case where the host opens a closed door at random, since a lot of people get this well-defined question wrong. Should you switch doors in this version of the game?

Just to be very clear: You will pick a door. The host will open at random one of the two doors you did not pick. Then he will give you the option of switching to the other unopened door.

Imagine playing this game 198 times with the strategy "always switch". Just for descriptive purposes, let's color the sheep red and blue.

1) In 66 instances of the game, your initial guess is correct, the host opens one of the other doors at random, you switch, and you lose.

2) In another 66 instances, your initial guess has the red sheep. In 33 of these instances, the host opens the prize door at random, you switch to the blue sheep door, and you lose. In the other 33 of these instances, the host opens the blue sheep door, you switch to the prize door, and you win.

3) In the final 66 instances, your initial guess has the blue sheep. Symmetrically with (2), in 33 of these instances the host opens the prize door and you lose. In the remaining 33 instances the host opens the red sheep door, you switch, and you win.

Adding it all up, you win in just 33 + 33 = 66 instances, or 1/3 of the time you play. This is exactly the same number of wins as if your strategy were "never switch".

Since "always switch" and "never switch" glean the exact same number of wins, there is no advantage to switching doors.

In my opinion, this is the sort of conditional probability question that most people intuitively think is being asked in the classic Monty Hall problem, which is why so many people get it wrong (where the correct answer is "always switch"). Changing the host's actions to random completely changes the problem and the solution.

[Update]

Note that it does not actually matter what you are allowed to do when the host opens a door with the prize, because that did not happen in the question as stated.

A full, correct analysis is to throw out the 66 instances above where the host opens the door with the prize, because by assumption we are not in a world where that happened. In the remaining 132 instances, you win half the time and lose half the time whether you switch or not.

[Update 2]

Here is an analogy that some have found convincing...

Suppose we play a game with an ordinary deck of 52 cards. We shuffle the cards randomly, then you pick one without looking at it. I look at the remaining 51 cards and reveal 50 (i.e. all but one) of them, none of which are the ace of spades. What is the probability that you are holding the ace of spades?

If we play this game many times, 1 time out of 52 your original card will be the ace of spades. The other 51 out of 52 times, I will need to deliberately dodge the ace in order to avoid revealing it. That second scenario is 51 times more likely, so the answer to this question is that the odds are 51-to-1 against your holding the ace.

This is exactly like the classic Monty Hall problem, but with 52 doors instead of 3. And just like that problem, we have to make a whole bunch of assumptions that are often omitted in the statement of the question; e.g. the shuffling is uniform, I always reveal 50 non-ace-of-spades cards regardless of which card you choose, etc. But subject to those assumptions, the odds are 51-to-1 against your holding the ace of spades, just as in the classic Monty Hall problem, the odds are 2-to-1 against your original door containing the prize.

Now let's change the question slightly. Suppose we shuffle and you pick a card randomly as before, but now I reveal 50 other cards without even looking at them. And suppose, just by chance, that none of those revealed cards is the ace of spades.

Granted, this is unlikely; in fact, it will only happen 1 time out of 26 that we play this game. But supposing it does happen, what then are the odds that your card is the ace? (This is what "conditional probability" is all about.)

Obviously, the card I avoided flipping over blindly is completely random, just like the one in your hand. So the answer to this modified problem is that the odds are 1-to-1.

This modified problem is exactly analogous to the modified Monty Hall question, where the host opens a door at random. And the answer is the same: If the door is opened at random, the fact that you do not see the prize tells you nothing about where the prize is, except that it is not behind that particular door.

• If you ignore the problem with the accepted answer (that the accepted answer is assuming the door will be opened uniformly at random, a result that is not demonstrated by the puzzle, and in fact is very hard to prove in the real world), yes, the accepted answer is correct. The core of the problem with the accepted answer is that assumption. – Yakk Apr 27 '15 at 18:12
• @Yakk: Well, yes, that is a good point... "I have no idea" and "the chances are equal" are not the same thing. It is a shame this question is conflating the two, because lots of people really do not understand the "uniformly at random" case. – Nemo Apr 27 '15 at 18:33
• @Yakk: I have edited my answer a bit. Let me know what you think. – Nemo Apr 27 '15 at 18:47
• The "philosophical" position can actually be analyzed mathematically: we do get certain probabilities (mainly bounded away from 0 and 1). In theory, you could generate a situation where there is an "unknown probability" but despite that you can still get enough information to bound your success chance, regardless of what the "unknown probability" is (and not assuming it is uniform and random). I don't think this is the case here: but the existence of such cases implies that "unknown probability" can be reasoned about without being philosophical. – Yakk Apr 27 '15 at 18:53
• @Yakk: Yes... But as you say that is not applicable to this question. I have updated my answer again. – Nemo Apr 27 '15 at 19:00

The question is vague, and assumptions must be made that drastically change the possible answers.

EDIT: Analysis added below. Am I missing something?

• Comments are not for extended discussion; this conversation has been moved to chat. – user20 Apr 29 '15 at 20:07

Let's assume that the Host randomly decides whether to reveal a goat and offer a switch: if your first door has a car, he reveals with probability $p$, and if it was first a goat, then with probability $q$. This is the best we can do: we don't know whether the Host planned to open a door or not. Note that $p=q=1$ gives the original Monty Hall scenario.

Let $C$ be the event you originally picked the car, and $R$ be the event that he reveals. Then $$P(C|R)=\frac{P(R|C)P(C)}{P(R|C)P(C)+P(R|\neg C)P(\neg C)}=\frac{p\cdot \frac13}{p\cdot\frac13+q\cdot\frac23}=\frac{p}{p+2q}$$ If you knew the probabilities $p,q$, then the best strategy would be to stay the above probability is at least $\frac12$, namely, when $p\ge 2q$. But we don't know the Host's personality, so we can't deduce what the best strategy is.

Essentially, there is too little information given to determine what strategy maximizes your probability of getting the car. The only thing a rational game show guest can do is figure out how to maximize this probability in the worst case.

• If you always stick with the door you start with, you win $\frac13$ of the time, no matter what.

• If you always switch, a malicious host could conspire to make you never win (by offering the switch when you start with a car, and not offering it otherwise).

Thus, the strategy that guarantees you the best odds, no matter what personality the Host has, is to stick with what you initially pick.

• Good reasoning. But you make an assumption that host intentionally opens a door with a goat. What if he opened a random door and it just happened to have a goat. – Ishamael Apr 27 '15 at 1:59
• It turns out to not matter. If you want, you could redo the analysis, where $p$ and $q$ are the probabilities of revealing a goat, when you initially pick a car or goat respectively, and let $s$ be the prob. of revealing a car when you start with a goat. We only care about P(you have car|goat revealed), and you will find this doesn't depend on $s$. – Mike Earnest Apr 27 '15 at 2:05
• @Ishamael Even without making assumptions about the motives of the host, staying with your original choice is regret-minimizing. – Corvus Apr 27 '15 at 6:18
• I think to be precise you have to say, "If you always stick with the door you start with, you win at least 1/3 of the time." – cfh Apr 27 '15 at 14:10
• You can even do the math for the "we only know what we have been told". The bayes ends up not working, because we cannot bound it away from 0/0, if I did the napkin math right. (we don't know if the host's decision to open a door is conditional on us picking a car is the key part: if the host has a 0% chance of opening a door if we did no pick a car, and a 100% chance of opening a door if we did pick a car, that is consistent with our observations and is the "do not switch". If we flip those chances, switching gives us a 100% chance of getting a car. And we cannot distinguish between.) – Yakk Apr 27 '15 at 18:17

Well in this case the game becomes more or a Poker game because you don't know if the host followed a rule or not. There are two (already well-known) cases:

The host chose the other door randomly (he didn't know what was behind that one): in this case it doesn't change anything if you change or not: each door has a 1/2 chance.

The host chose on purpose a door without the car: in this case changing leads a a higher chance of getting the car: 2/3

Now, if we assume that the host followed one of these rules, it is still better to change. In the worse case it doesn't change anything and in the best case it improves your chance.

I am though doing one hypothesis on the host: he ALWAYS opens another door. If he weren't doing that, it could be a meta-trick to make you change doors (opening a door and letting you change only if you picked right in the first place).

• You make one further assumption, which OP specifically exluded; the host never opens the door that has the car. – Taemyr Apr 28 '15 at 10:19
• Well the formulation is subtle here:"to make the question a little more specific, what is the ideal strategy for you, that maximizes your chances of getting the car in the worst case, no matter what the hosts motives are." The fact is you don't know if the host got the door by chance or not. I just said the results considering the two options. Since you can't know it is better to assume he does since you're not losing anything by changing in the worse case. – meneldal Apr 28 '15 at 10:51
• You are losing something in the worst case. If the hosts strategy is "always choose the car if he is able" the strategy to always switch has a 0% chance to win. – Taemyr Apr 28 '15 at 10:56
• Well I guess the host could do that indeed. In this case it's just a poker game to know what the host is thinking so you can't have a superior strategy. I just considered when you leave the two commonly admitted options to the host (and not the malevolent one) – meneldal Apr 28 '15 at 12:18

If he always opens one of the doors you did not pick at random, then the door he opens should contain a car 1/3 of the time, a sheep 1/3 of the time, and a goat 1/3 of the time.

If the animals and car were placed at random, we had the following possible cases to consider at the start, each of which had equal probability:

S G C
S C G
G S C
G C S
C S G
C G C


Since the above has all the cases, we can make it easier to think about by saying that the first column represents the door you picked, the second represents the door he opened, and the third is the option you are given. Thus, after he reveals the sheep in the door he opened, we are left with the following cases:

G S C
C S G


Since the probabilities of each case are equal, your odds are 50/50 with either door.

• P.S. The first possibility reminds me of one of my all time favorite shows :) – Briguy37 Apr 27 '15 at 17:13
• "Since the probabilities of each case are equal" - This statement is nontrivial. Compare with the original monty hall. – Taemyr Apr 28 '15 at 10:23
• @Taemyr - Yes, in the original Monty Hall problem (where the host always reveals a door without a car) the probabilities are different. However, for the OP's question ("every assignment of the three prizes to the three doors is equally probable", "without trying to guess the motives of the host"), the probabilities are equal. – Briguy37 Apr 28 '15 at 14:46

EDIT: to make the question a little more specific, what is the ideal strategy for you, that maximizes your chances of getting the car in the worst case, no matter what the hosts motives are.

You can't sweep the host's motives under the rug like some small, unimportant detail. The player's strategy is completely dependent on the host's strategy.

HOST STRATEGY 1: The host always opens a door containing an animal. In this case the problem is the usual Monty Hall problem, and you should switch doors.

HOST STRATEGY 2: The host always opens one of the two remaining doors with equal probability. In this case, each door will have a 50% likelihood of having a car. It doesn't matter if you switch or not.

HOST STRATEGY 3: If you pick an animal, the host's strategy is to open the car door 99% of the time, and open the other animal door 1% of the time. If you pick the car, the host opens a random animal door with 50% probability. If he opens a door with an animal, then (from Bayes theorem) the odds are overwhelmingly likely that your door has a car. In this case you should NOT switch doors.

The point is with a sample size of one, and with zero information about the host's strategy, it doesn't matter if you switch or not, since some host strategies favor switching while others favor not switching. If you saw him open one hundred doors, assuming he had the same strategy each time, you could conceivably form a model for his strategy and fit certain parameters to guide your actions in future trials. But on the very first trial, there simply isn't enough information.

• This response to the "edit" about the host's motives nailed it. – Beska Apr 28 '15 at 12:25
• How did you interpret "in the worst case" in the question? – JiK Apr 28 '15 at 12:35

EDIT: to make the question a little more specific, what is the ideal strategy for you, that maximizes your chances of getting the car in the worst case, no matter what the hosts motives are.

If I understood your question correctly, "in the worst case" is the key here. So the correct answer is a strategy where you maximize your winning chance over all possible host's strategies. Considering worst case is identical to considering the situation where the host knows your strategy and wants you not to win.

• If your strategy is "never change", your probability of winning is $1/3$ no matter what the host's strategy is. Because you have a strategy with result $1/3$, your result with the optimal strategy is at least $1/3$.
• If the host opens a door with a goat and offers you a possibility to change only if you chose the car in the first try, you can win only if you choose the car in the first try, so your probability of winning is at most $1/3$. So whatever your strategy is, the host can limit your winning probability to $1/3$. So your result with the optimal strategy is at most $1/3$.

Therefore, your chances of winning are exactly $1/3$ with optimal strategy, and this can be achieved by never changing.

• This is the only answer that takes game theory into account appropriately. The other answers complain that there is no way to know because we do not have enough information, but this isn't a result of poor puzzle asking. Someone could actually be put into this situation in real life and have to decide what their best strategy is. If you decide to follow a minimax strategy (maximize your minimum gain), you must stay. – David Stone Apr 28 '15 at 15:00
• What if the host's strategy is to try to help you win by showing you the door you picked if you picked "car" [instant win] and otherwise by letting you change? Entirely plausible depending upon the show's sponsorship deal with the automaker. In that scenario, switching would offer a 100% win probability, while not switching would only offer a 33.3% probability – supercat Apr 28 '15 at 15:46
• @supercat What part of my answer do you disagree with? Do you interpret "in the worst case" in a different way? – JiK Apr 28 '15 at 16:20
• @JiK: The combination of "In the worst case" and "regardless of the host's motives" is inconsistent. If the questioner meant "assuming the host acts with hostile motives", the question should have said that. If the producer told the host "The automaker called and says their tax paperwork has already written off this car and it must be given away as a prize, but our other sponsor insists their car be shown alone as next week's prize. Do all you can within the rules of the game to give away this car or you're fired", that would establish a worst-case winning probability for switching of ~100%. – supercat Apr 28 '15 at 16:47

You should swap.

When you have three to choose from, you have a $\frac{1}{3}$ chance of picking the car. Any event that happens after that point can't change the chance of you being correct initially.

When one of the other doors is revealed to be a sheep, you're still just as likely to have won. So switching to the remaining door has a $\frac23$ chance of winning.

And even if the chance is 50/50, there's no harm in swapping.

• If the host intentionally decides to open the door and allow you to change only if you chose the car, this strategy will ensure you'll never win the car. – JiK Apr 28 '15 at 10:52
• @JiK, this is true. I was going on the basis that one of the doors was always opened just which one was unknown. – Holloway Apr 28 '15 at 11:11
• Even with the assumption that one door was always opened, the key question is whether it's better to stay or swap given that one door is opened to reveal the sheep. And if there's a 99% chance that the host reveals the car if the player doesn't pick the car, then the player should stay, swapping is a bad idea (it's highly unlikely that the car is behind the third door, because the host probably would have revealed it rather than the door he opened). – Glen O Apr 28 '15 at 17:00