Monty Hall: does it matter if the host knows?

Factual Questions
1

I know and understand that in the original Monty Hall problem, you have about a 67% chance to win the car behind one of the three doors (the other two have goats) when you pick a door and always switch IF Monty always picks a goat. That is all well and good. My question is what if Monty also picks a random door? That means that the player will win outright about 33% of the time without even having the chance to switch. 67% of the time the player has the opportunity to switch. Should the player make the switch?

Since the player will choose a goat 67% of the time, when the player switches the chance to win should be the same as in the original scenario. So the idea that Monty has to know where the prize is seems to be irrelevant. Thoughts?

2

The argument in the original Monty Hall problem rests on the premise that the probability of the candidate picking the winning door to begin with is 33%. This means the probability of it being in either of the other two doors is 67%. When Monty offers the candidate to switch to another door, he opens one door, which is - in the original scenario - always empty. This means the probability of the door that was just opened drops to 0%, so the probability of the remaining door being the right one rises to 67%. Hence, the candidate should switch.

But if Monty doesn’t know which door is the correct one, it’s possible that he chooses the right one by accident when he makes the offer to the candidate. In that case, the probability of either of the other doors being the right one drops to 0%, and the game is over - the candidate neither wins nor loses from switching, since the probability of winning will be 0% in either case. If Monty picks a wrong door when making his offer, however, then the same thinking as in the original scenario applies. In this case, the candidate should switch, thereby increasing his probability from 33% to 67%. That does not mean, however, that the overall probability of the candidate winning the prize is the same as in the original scenario, since you have to take into account the possibility that Monty picks, accidentally, the right door when making the offer. This possibility is absent in the original scenario.

3

I understand that the overall probability does change. But if we only concentrate on the act of switching, whenever I see discussions of the Monty Hall problem it’s usually stated that it’s somehow important that he knows where the prize is and always chooses a goat. If whether he knows or not doesn’t change the probability of winning when switching then why does is have to be mentioned so often? It seems like a red herring to me.

4

Actually I’m not sure how the probability in my scenario is different. We still only have a 33% chance of losing since the only way we can lose if we the chose the prize door in the first place (since we always switch).

5

I’m not sure how you think the player wins outright. In you scenario, is Monty randomly choosing one of all three doors? That still won’t have the player win 33% of the time, but he will see what his fate is 33% of the time.

If you mean he randomly picks one of the two remaining doors, he will lose outright if Monty picks the door with the prize, and that will happen 33% of the time. If he doesn’t lose immediately, it will be 50/50 that he’ll win, whether he switches or not.

6

If Monty doesn’t know, then in senario A where he reveals the car you should obviously switch. In scenario B where the door he chooses is empty it remains a 50-50 shot, as neither door is lmore likely than the other to hold a car.

Assuming you pick your first door at random and so does Monty there are 6 equally likely scenarios

AB
AC
BA
BC
CA
CB

first being your pick, second being revealed door.

If we assume the car is behind door A we can only be sure we haven’t encountered BA and CA
so 4 likely alternatives are
AB
AC
BC
CB

In half the cases the car is behind the door you origanally picked.

7

It is relevant.

If Monty’s pick is random, switching does NOT matter.

If Monty’s pick is NOT random, switching DOES matter.

For simplicity, let’s say that if Monty opens the door with the prize, you automatically lose. Then you automatically lose 1/3 of the time. Of the remaining 2/3, you have a 50/50 shot whether you switch or not.

Here’s the win/loss table for that.

Your Choice Monty Reveals Switch Doors Keep Door
#1 #2 Lose Win
#1 #3 Lose Win
#2 #1 Lose Lose
#2 #3 Win Lose
#3 #1 Lose Lose
#3 #2 Win Lose

Of these 6 possible scenarios, you lose 2 times automatically when Monty opens the door with the prize.

Of the remaining 4 scenarios, you win 2 times by switching doors. You also lose 2 times by switching doors. It’s a 50/50 shot if he doesn’t open the door with the prize.

Overall, you have a 1/3 chance of winning.

The analysis is the same if you win the prize if Monty opens the door. Then, you have an overall 2/3 chance of winning, and a 50/50 shot if he opens a door with a goat. Switching door doesn’t change this at all either way.

Basically, your choice doesn’t matter either way if Monty chooses totally randomly.

But if Monty does NOT choose randomly and only opens a door with a goat, you turn your 1/3 chance of winning into a 2/3 chance of winning by switching doors. That’s NOT true if Monty chooses randomly.

8

Yes I’m assuming Monty will randomly pick one of two remaining doors. I also assume if Monty opens the door with the prize the player wins.

9

To simplify:

If Monty chooses randomly, your chances of winning don’t change whether or not you change doors. They’re the same either way. It’s overall 2/3 (1/3 when he opens the door with the prize + 1/2 of 1/3 when you have to choose from the remaining 2 doors).

You might as well give the player the prize 1/3 of the time and just flip a coin the rest of the time.

Also notice that if you lose when he opens the door with the prize, your overall chances of winning are 1/3. So, yes, the rules do matter.

But if Monty knows which door has the prize and only opens a door with a goat, you can actually improve your chances of winning (from 1/3 to 2/3).

In that case, Monty has extra information of which some is passed to you by which door he chooses to open. If he’s picking randomly, there is no extra information, and you can’t improve your chances of winning by switching or not.

Just because two sets of rules end up with the same overall chances of winning (2/3, in this case) doesn’t mean they are equivalent situations or that the circumstances don’t matter. In your version of the game, switching doors is irrelevant. In the version typically stated, switching is totally relevant.

10

In the 2nd line of my post above, that should be “+ 1/2 of 2/3”.

11

I don’t think this is in the spirit of the puzzle. If he opens the door with the prize, you still have the opportunity to switch and you always should, obviously. For the remaining 2/3 of the cases you have a 50/50 chance of having chosen correctly at the start, so switching doesn’t change your prospects in those cases.

The end result is that you still win 2/3 of the time by switching, but half of those are certain. I agree with all your math, just slightly disagree on way to look at the puzzle.

12

Meh. I did note that the analysis is not fundamentally different if you made that assumption.

I did think of a better way of illuminating the big difference:

Classic version of Monty Hall Problem:
1/3 chance of winning by never switching
2/3 chance of winning by always switching

yarblek version of Monty Hall Problem:
2/3 chance of winning by never switching (except when Monty opens the prize door)
2/3 chance of winning by always switching

I’d say that’s a big, fundamental difference between the two versions, even if the outcome is the same if you always switch. It does show that Monty’s knowledge and whether he picks truly randomly or not is a key consideration.

13

But the puzzle didn’t include the scenario where Monty opens a door with the prize, and on the show it certainly wasn’t the case that you could switch to a door where Monty has already showed you the grand prize. If you picked Door 1 with the goat (but not seen it yet), and Monty shows you Door 2 with the car, then the game would be over and you’d own a new goat. Then he’d show you Door 3, a year’s supply of Brillo Pads.

And Great Antibob’s analysis is right - if Monty doesn’t know, and it just happens that on your turn he revealed a non-winning door, then you don’t improve your odds by switching. Either door, in this situation, has a 50/50 chance.

14

I’ve never seen the show, and only heard about “Let’s Make a Deal” in the context of this problem. Did Monty really reveal the car on the actual show? Discussions about the “Monty Hall Problem” always assume it doesn’t happen.

15

The other key point, often overlooked in this problem, is whether Monty is obligated to open a door. On the actual show, he wasn’t, which meant that he’d sometime just leave you stuck with the first door if you chose a loser, but he’d always try to get you to switch if your first door chosen were a winner. It essentially becomes a game of Rock-Paper-Scissors, with the theoretical optimum play for both players involving random choices. Except that, due to experience, Monty was much better at the game than the players, and won more often than chance would dictate.

16

I never saw that, or heard of it. It would make for lousy television. :slight_smile:

17

On the real show, Monty may have revealed the car after your choice, but just to show what you could have got. He was a fast-talking wheeler-dealer, and the whole point of the show was to put people in suspenseful situations involving choices for cash and prizes. It was pretty entertaining.

Something typical for him to do would be to reveal one of the doors you didn’t pick, which had a nice prize, then offer you the choice of keeping what you picked, or the $500 wad of cash in his hand. If you said “no” he may up it to $600 or $700 (but maybe not). It was all very unpredictable.

Here’s an example that was pretty typical of the show.

18

Let’s suppose the player picks a door, then Monty has to open another door and offer a “stay or switch?” choice. But let’s hold off for now on pinning down exactly what the process is by which Monty chooses which door to open.

On any given occasion, one of three things happens: [ul]
[li]The player initially picks the prize door, then Monty opens one of the goat doors. Let’s call the proportion of times when this happens A.[/li][li]The player initially picks a goat door, then Monty opens the other goat door. Let’s call the proportion of times when this happens B.[/li][li]The player initially picks a goat door, then Monty opens the prize door. Let’s call the proportion of times when this happens C.[/li][/ul]

The proportion of time when Monty opens a goat door is A + B. Out of those times, the relative frequencies of the stay door having the prize vs. the switch door having the prize are A to B. [If the only information you have is this probabilistic model and that Monty opened a goat door, and your goal is to make the choice with the highest probability of getting the prize, then you should stay or switch according as to whether A or B is bigger].

But we haven’t actually answered the question: Out of A and B, which is bigger?

In fact, we haven’t said much at all yet about what A, B, and C are.

Of course, one constraint is A + B + C = 1. Typically, we assume the player’s initial choice has no particular correlation to the location of the prize and goats, so that A = 1/3 while B + C = 2/3.

But we haven’t pinned down exactly what B and C are beyond that. If B accounts for more than half of B + C, then B will be larger than A. If B accounts for less than half of B + C, then B will be smaller than A. If B accounts for precisely half of B + C, then B will equal A.

So there you go.

If the player initially picks a door at uniform random and Monty then opens a non-player goat door whenever possible, then A = 1/3, B = 2/3, C = 0, and, over all the cases where Monty opens a goat door, the stay door has the prize half as often as the switch door.

If the player picks a door at uniform random and Monty then opens a non-player door at uniform random, then A = 1/3, B = 1/3, C = 1/3, and, over all the cases where Monty opens a goat door, the stay door has the prize equally as often as the switch door.

19

[quote=“yarblek, post:3, topic:704375”]

I understand that the overall probability does change. But if we only concentrate on the act of switching, whenever I see discussions of the Monty Hall problem it’s usually stated that it’s somehow important that he knows where the prize is and always chooses a goat. /QUOTE]

Yes, because otherwise he might open the door with the car (or correct prize.)

Well that is how the information appears to help improve the odds.
The removal of the goat gives the information that the switch will improves the odds from 1/3 to 2/3… by informing that best of two doors is this door here !