Monty Hall

I’m fascinated by this one.

The analogy is used, to describe the probability, that it’s like Monty’s offering you Door #2 AND Door #3, or just Door #1.

But if he’s opening Door #3 anyway, isn’t your choice, effectively, Door #1 + #3, or Door #2 + #3?

No, it’s (assuming you chose Door 1) Door 1 against Doors 2 and 3. If you choose to switch, you win if the prize is behind either Door 2 or Door 3, and you lose if it’s behind Door 1. If your strategy is to switch, the three outcomes are:

The prize is behind door 1. Monty opens door 2 or 3, you switch to the one he doesn’t open, and lose.
The prize is behind door 2. He opens door 3, you switch to door 2 and win.
The prize is behind door 3. He opens door 2, you switch to door 3 and win.


You select a door–let’s say #1. Now, if you stick with this door to the end, you have a 1/3 chance of getting the gold, non? Now, one of the other doors–#2 or #3–or possibly both, doesn’t have the gold behind it. Monty always opens the door with the goat; this is part of the problem statement. You now have information about the total set of doors (#2 and #3) that you didn’t open; in effect, if you switch you get to see the contents behind both Door #2 and #3. Since you have only a 1/3 chance of getting the right prize with Door #1, switching must offer you a 2/3 chance of success. The key here is that your knowledge of the system changes after you’ve made the initial selection, and which door you select initially will affect that change. In other words, if the gold is behind Door #2, if you select Door #1 Monty will open #3; if you select #3 he’ll open Door #1. If you select #2, he’ll open either door and you’ll be wrong if you switch, but that only occurs 1/3 of the time. So ultimately you make your decision to either stick with your initial selection (1 door), or your nonselection (2 doors).

Regarding Cecil’s weasel: the problem states that Monty always opens another door. However, if Cecil’s strategy is adopted (where Monty only opens another door half of the time when he’s wrong and always when he’s right) the odds are 1/2 when switching. This goes from being a straight probability problem to a game theory problem, albeit one with a trivial (but nonsymmetric) zero-sum solution; the odds end up being
((1/31 + 2/31/2) + (1/3*0 +2/3 * 1/2))/2 = 1/2


The key point is that Monty systematically revealed the wrong door. Because you knew one of the doors you didn’t pick was wrong, systematically revealing that one didn’t change the odds of your original choice being right. (In contrast, if Monty randomly revealed one of the doors you didn’t choose, then the fact that it turned out to be wrong makes it more likely your door was right – basically, you’re taking a random sample of the doors you didn’t choose to gather evidence of whether or not they collectively contain the prize.)

How does this relate to your point about choosing between 1&3 or 2&3? Well, the reason the problem is equivalent to being offered either 1 or 2&3 is because the reveal doesn’t change the odds of the prize being in either of those sets. This is because it gives you no new information about either of those sets. (You already knew at least one from 2&3 was empty, so that’s not new information.)

It does change the odds of it being in 2, however. Because Monty systematically chose a door that didn’t have the prize, the fact that he didn’t choose door 2 is evidence that it does have the prize. (It’s not evidence for door #1, because Monty didn’t have the option of picking that one.) So the odds for door 2 go up, while the odds for door 1 stay the same.

Now, is the problem equivalent to being offered 1&3 or 2&3? For that to be the case, his choice must reveal nothing about either of those sets. As stated above, it reveals nothing about 2&3. You knew one of those was wrong, and you still do. You might say the same about 1&3 – but there’s a difference. While it’s true you knew one of them was wrong, you didn’t know that the one Monty could choose to reveal – door 3 – was wrong. If door 3 had been right, Monty couldn’t have revealed a door from that set. So while the fact that Monty was able to pick a wrong door from 2&3 tells you nothing new, the fact that he was able to pick a wrong door from 1&3 does tell you something new. Thus the odds of 1&3 being right are cut in half, from 2/3 to 1/3, whereas the odds of 2&3 being right are still 2/3.

We studied this one in school, and had to write a program to demonstrate it. SOAT is absolutely right. If you never switch, your odds of picking right are 1/3. If you always switch, your odds of getting it right are 1/2. It still boggles my mind a bit, but it’s true.

I’m wondering how that affects the lottery question referenced in Cecil’s article. Your initial odds were 1/1000, now your odds are 1/2. I wouldn’t pay out lots of money on a 50/50 shot, but to trade the lottery tickets outright? I’m not sure how I see how that’s different from the original.

Anyone ever watched the (incredibly drawn out and over-produced) show Deal or No Deal? There’s a bit of that game in this question…

Did you mean 2/3 here?

I hope so, or there’s a 1/6 chance of the entire universe vanishing in a logical contr

To elaborate, you can group the doors however you want, it doesn’t change the answer. The key point is that the odds of the prize being behind a certain group of doors only change if Monty reveals information about that group.

The other thing to remember is how Monty reveals information. If Monty systematically avoids revealing the prize, then whether or not he reveals a door from a particular group is evidence of whether or not the prize is in that group – but only if there was initially some chance that he would reveal a prize from that group and some chance that he wouldn’t (given that he has to reveal one and can’t reveal the one you pick).

Since the numbering doesn’t matter, let’s stick with the assumption that the door you picked is #1, and the door Monty revealed is #3.

Odds for the prize being behind each group of doors (before reveal):
1: 1/3
2: 1/3
3: 1/3
1&2: 2/3
1&3: 2/3
2&3: 2/3
1&2&3: 1

Odds for the prize being behind each group of doors (after reveal):
1: 1/3
2: 2/3
3: 0
1&2: 1
1&3: 1/3
2&3: 2/3
1&2&3: 1

-The odds don’t change for 1, since Monty didn’t have the option to reveal this door.
-The odds change for 2, since Monty could potentially have revealed this door, but didn’t. This is evidence for it being the right door (assuming Monty systematically avoids revealing the right door.) Its odds go from 1/3 to 2/3
-The odds change for 3. Monty revealed that it’s wrong, so its odds drop from 1/3 to 0.
-The odds change for 1&2. Monty would have had to reveal one of these doors if they didn’t contain the prize. (Because he couldn’t reveal 3 if it has the prize.) So the fact that he didn’t reveal one proves this set of doors contains the prize – its odds go from 2/3 to 1.
-The odds change for 1&3. Monty could potentially have left these doors unopened, but he opened one – evidence that the prize isn’t behind this group of doors. Its odds drop from 2/3 to 1/3.
-The odds don’t change for 2&3. Monty had no possible choice but to reveal one of these (assuming he has to reveal a door), so the fact that he did so tells you nothing.
-The odds don’t change for 1&2&3. Monty had to reveal one of these, so the fact that he did tells you nothing.

Note that the odds are always consistent – e.g., the odds of the prize being in 1&2 are always the sum of the odds for 1 and the odds for 2. This is true both before and after the reveal.

Oh God, not this thing again! Threads from the past six months alone on this problem:

Deal or No Deal Mathematical Probabilities

Monty Hall (edited title)

Monty Hall question

Deal or No Deal/Monty Hall Problem

Optimal Strategy for playing Deal or No Deal

No matter how often it’s explained, the incorrect answers never die. I wonder if Copernicus spent much of his life after De Revolutionibus explaining why the common-sense idea that the sun revolves around the earth is wrong…it’s no wonder he died 13 years later!

With the lottery, trading is as good as not trading. One simple way to see this is the fact that if you can construct an argument that trading increases your odds of winning, then the guy with the other ticket can make the exact same argument. Trading can’t possibly increase both your odds (since the sum of your chances of winning must add to 100%), so the argument must be wrong.

Likewise, any argument that not trading helps your chances must be wrong, since the other guy could say the same thing.

The origin of this symmetry is the fact that you looked at all but the last two tickets (yours and the other guys), and it happened that they were all losers. So yours and the other guys were treated the same. That’s not the same as systematically looking at all the tickets except yours, and removing all the losers (except one, possibly). In that case, the other guys ticket was selected in a different way than yours, so the symmetry is broken.

In summary:
If you simply pick two tickets (yours and the other guys) and reveal all others, then yours and the other guys have to have the same odds.

If you have a method of picking the other guys ticket, then its odds are determined by that method. Suppose the method is pick the right ticket if your ticket is wrong, or pick a random wrong ticket if your ticket is right. With that method, the odds of the other guy’s ticket being right are exactly the same as the original odds of yours being wrong – in other words, he’s very likely to be right (since you were initially very likely to be wrong.) In this case, the odds for your ticket stay as bad as they ever were.

I think I’m going to emphasize that last point, since it captures the essence of the problem:

**If two doors were picked at random, then their odds must be the same.

If one door was picked at random but the other was chosen systematically, then the odds aren’t necessarily the same.

In the original Monty Hall problem (with Marilyn’s assumptions), Monty chose the offered door systematically. His system is to offer the correct door in every case where the contestant initially chose wrong.**

There’s a game on that show? You could have fooled me.

I remember this column from the original Parade magazine article. The “trick” that did it for me was the reply FROM A MATHEMATICS PROFESSOR AT MIT (um, that’s the Massachusetts Institute of Technology for you knuckleheads out there …). MIT is the MOST prestigious math school in the country, if not the world.

He said that Marilyn was right … that probabilities indicate that you should switch doors.

Mathmatics is NEVER wrong … ask Albert Einstein (well, if he were alive).


Yeah, I think that’s the part that had me hung up - If I used the same logic in my simulation program to wind up with two tickets in which ONE was the winner, and mine was the one I originally picked, I would have artificially eliminated the random choice up front of the other guy’s random pick. To correctly simulate this, you’d have to run lots and lots of runs, and only pick the ones where random uncovering of the other 998 tickets did NOT reveal the winning ticket, ensuring articificially picking the winning ticket for the other guy didn’t enter into the equation.

The problem with the inumerable threads on this topic is that the underlying assumptions are typically not stated, not understood or people are fooled by complex but inherently flawed arguments.

If you start from a simple assumption, which in fact may be a requirement of game show fairness, Monty Hall has no more idea as to what is behind each panel than the contestant and the position of the good prize is random, the the following is true.

The odds are always 1/3.

The initial choice is 1 of 3 doors.

After choosing, the 2nd choice is often stated as “stay with your choice or choose one of the other two doors.” Of course, this is still (remember my assumption), choose one of 3, just stated differently.

You could be offered to change infinite times and the odds would not change so long as the randomness assumption remains and no door is opened between subsequent choices. Why??? Because NOTHING HAS CHANGED so the odds cannot change. Until something has changed, this remains a question of logic not math.

Now, if you put all sorts of other assumptions into play, the odds will change. But, I’ll leave others to argue the specific mathematics.

Here’s a simple way to intuitively understand it.

There are a thousand doors. Behind one is a prize. The rest have goats.

You pick one at random.

Monty, who knows where the prize is, opens 998 of the remaining 999 doors to reveal goats.

Do you trade the door you picked for the one remaining door that Monty didn’t open?

Of course you do! There’s only a one-in-a-thousand chance you guessed right the first time. The prize was almost certainly behind one of the doors that you didn’t pick. And Monty has just eliminated all of those doors but one.

It works the same way with three doors. The odds just aren’t as extreme.

I think you’ve missed the whole point here. You’re correct, but vacuously. The problem you’ve stated has nothing to do with the Monty Hall problem: there aren’t “all sorts of other assumptions”; there is one assumption, that is a part of the problem as stated, and for that matter is kind of required for the problem to make any sense. Namely, it is that Monty Hall always opens the door (of the two that the contestant did not choose) with the goat behind it. In this case, Monty’s revelation adds information to the problem, and makes the remaining door a more likely shot at the winner than the one chosen originally.

Here’s the explanation that I finally understood:

If you never switch, your odds are just the 1/3.
If you always switch, then your odds to win are 2/3 because:
If you picked the correct door the first time (odds: 1/3), and you switch, you lose.
If you picked the wrong door the first time (odds: 2/3), and you switch, you win.

As with many multi-step probability questions sometimes it’s best to start from the losing perspective.

In this case (with the following assumptions)

  1. Dealer always picks a goat

IF you picked a goat right off (2/3 of the time)
the dealer will then eliminate the remaining loser and you will pick the winner by switching.

IF you picked the prize right off (1/3 of the time)

the dealer elimination does not help you and you switch to a goat.

ergo this strategy wins exactly 2/3rd of the time and loses 1/3rd of the time.
… also … note that if you don’t switch you will lose 2/3 of the time and win 1/3 of the time as the dealer display is meaningless.

That is to say by switching every time, your goal is to pick a goat first off to win.

I love that one post where the guy says it loses 1/2 of the time and wins 1/3 of the time. 1 out of 6 times apparently the game crashes to the desktop.
The reason that this works is subtle indeed though, as there is more information given when a dealer eliminates a choice from a smaller subset of an overall set than there is if he eliminates from the overall set, even if the elements are removed randomly. Because it alters the probability differently for other members of the same subset and nonmembers.

here’s a macro example

let’s say we have a jug with volume 3A (2 goats and a prize) with 66% water (goat) 33% juice (prize) in it.
split it into to jugs of volume 1A and 2A

Precipitate out exactly 1A of water (remove one goat after separating) of the larger jug

you are left with

1 jug volume 1A with 33/66 and 1 jug volume 1A with 66% juice 33% water.

if you need to pick a molecule of juice at random you certainly need to pick the second jug (in fact it would be second nature to anyone participating in this inane experiment)


Here’s the way I finally understood it.

If you are asked, “What is the probability that one of the two remaining doors contains the money” then the odds are 1/2. But that isn’t what is asked. What is asked, and what you are betting on, really, is:

“What are the odds that the first door you chose is correct?”

Those odds are always going to be 1/3, because you had 3 doors to select from. Changing your answer will always give you odds of 2/3.