I was an original 50/50 thinker until a few minutes of discussion lit the bulb in my brain. The reveal of the door with a goat is really meant to confuse and get people to think it’s a 50/50 chance and therefore stick to their original decision. But think about this, after you make the initial pick you know for a fact that one of those doors has a goat, right? You don’t need him to show you, you already know that one of them must contain a goat. So you are basically trading a pick of 1 door for 2 regardless if he shows you or not. If he gave you the option of trading your single door pick for both doors would you take it, of course you would 1/3 vs 2/3, well it’s the exact same decision when he shows you the door with a goat, you already know it’s there.
@Mike
What you say also applies to an ignorant Monty (who doesn’t know where the car is), but in this case after Monty randomly reveals a goat then it becomes 50/50 between the 2 remaining doors .
The key is the difference between Monty intentionally and randomly revealing a goat.
- Let’s see: catch bus, get off, walk down road, key in door, kettle on, boot up computer … Richard, are you home from the course yet? It’s week three and you’ve had nothing to say. Did the MHP come up yet? Do tell.
Great explanation as always. Thank you very much!
- "Don’t believe me?"
No.
It is a probability problem, the solution that switching doors is a 2/3 chance is proven by Bayes theory (look it up), and the assumption in the proof ((if not already stated in the problem definition) is that given a choice between 2 goat doors Monty picks one at random according to the Principle of Indifference.
Even if Monty doesn’t pick at random when he has a choice, the ‘overall’ or ‘average’ probability over multiple trials is still 2/3.
Your comment doesn’t explain HOW Monty makes his decision
"The reveal of the door with a goat is really meant to confuse …"
Taking this to refer to the puzzle rather than the host’s behavior, certain assumptions are necessary for the solution. The greatest assumption is perhaps that we are presented with a mathematical rather than a psychological puzzle (ie Monty’s intention is neutral). The MHP states that ‘the host knows’ but not whether he always reveals a door, always reveals a goat or always tries to help (or hinder) the contestant. For example, could Monty have opened the contestant’s own door to reveal a car and offered a switch, leaving them 100% chance of winning (by sticking). If most of Monty’s conceivable behaviors result in the same or increased chance by switching, the solution still holds. The assumptions implied by the given solution are that he always reveals one goat making the advantage 2/3.
@tommyt
Say you exhibit a certain medical sympton, you look it up and find that 100% of people who have Disease A exhibit this sympton , but only 50% of people who have Disease B exhibit the same sympton. You’re more likely to think you’ve got Disease A than Disease B right?
I wonder if it would be helpful to imagine that you had to choose your stick/switch policy before you chose your door? You have to tell Monty if you’ll be sticking or switching first, and only then does he let you choose a door. If you’re a sticker then you know you’ll have to get it right the first time, and you only have a 1/3 chance. If you’re a switcher then you know you’ll have to avoid the car on your first pick in order to leave it available for the switch. You’ve got a 2/3 chance of winning this way. Maximise your odds, always switch.
@Mike 840. "If he opens the car then the games ends and the choice isn’t given."
If Monty randomly opens one of the 2 doors not picked AND it is not the car, the only new information is that you are not in the 1/3 of games where Monty reveals the car. You have no further information about the remaining doors’ chances. Hence 50/50. If you know the rules are being followed and Monty reveals only a goat and only a door you did not choose first, you have 2/3 chance of being in a game where Monty has to avoid revealing the car. Hence 2/3 by switching.
@tommyt 844."But you lost me on how to get from there to the conclusion that scenario a) is twice as likely to occur as scenario b)."
Expanding 843:
(Assuming you pick D1) When Monty opens D3 you are left with D1 and D2, where either
a) (i) D1(First Goat) and D2(Car)
a) (ii) D1(Second Goat) and D2(Car)
b) D1(Car) and D2(Either Goat)
There are two ways a) could play out and one way b) could play out.
@Gary Oh. My. Gosh. I read the whole freakin’ article and STILL didn’t understand the concept. (Not to any fault of the author). BUT I read your beautiful little comment and it came to me! Thank you! I might be being a little dramatic but I thought I’d have to live the rest of my life without understanding it.
@Palmer, not necessarily. Even if Monty randomly opens one of the 2 doors not picked the decision to switch if he opens a goat is still 2/3 vs 1/3. If he opens the car then the games ends and the choice isn’t given. Most people given the choice mistakenly rationalize that it is a 50/50 choice and stick with their initial gut reaction and stay which has the lower 1/3 success rate.
@Mike
It is demonstrably true that if Monty doesn’t know where the prize is and picks a door at random, then it’s 50/50 if he reveals a goat. Analyse the outcome of 300 games, you win 100 by switching, 100 by staying and lose 100 when Monty reveals the car.
I understand that this is not a 50/50 proposition after Monty opens a door. (It was the analogy to picking a specific card from a deck that made it clear to me).
What I’m struggling with is the math.
Probability is defined mathematically as the number of favorable outcomes divided by the total number of all possible outcomes
Before a door is opened there are three possible outcomes, one of which is favorable, resulting in a 1/3 probability
D1-D2-D3
C—G---G
G—C---G
G—G---C
However, once Monty opens a door revealing a goat (let’s say it’s D3 in this case) one possible outcome is eliminated.
D1-D2-D3
C—G---G
G—C---G
How can the probability of a favorable outcome still be 1/3 when the total number of all outcomes has been reduced from 3 to 2, one of which is favorable?
I’ve looked at Bayes’ Theorem and it’s all Greek to me. I desperately need someone to show me some actual math that explains this! Thank you.
@tommyt
"Probability is defined mathematically as the number of favorable outcomes divided by the total number of all possible outcomes" is only true if each ‘favourable outcome’ is equally likely - in the MHP they’re not.
(Assuming you pick D1) When Monty opens D3 you are left with D1 and D2, where either
a) D1(Goat) and D2(car) OR
b) D1(Car) and D2(Goat)
These 2 scenarios ar not equally likely because Monty is twice as likely to open D3 when the car is in D2 than he is when the car is in D1 (since he’s just as likely to open D2 instead). So scenario a) is twice as likely to occur as sceanrio b) - which gives relative probabilities of 2/3 and 1/3 .
@Palmer
Thank you for the comments. I understand your point that Monty opening D3 is twice as likely if the car is in D2 than it is if the car is in D1. But you lost me on how to get from there to the conclusion that scenario a) is twice as likely to occur as scenario b). Sorry to be so slow on the uptake on this one…I feel that your explanation has taken me right to the edge and all I need is that final piece to finally understand this darn thing.
@tommyt
OK, (You’ve picked D1) If a) the car is behind D2 then it’s 100% certain Monty opens D3. If b) the car is behind D1 then it’s only 50% certain Monty opens D3. Therefore the event of Monty opening D3 is twice as likely to happen when the car is behind D2 than when it’s behind D1. You should conclude therefore that it’is twice as likely the car actually IS behind D2 than D1.
I don’t know if that explains it any better,
at the start you have a 2/3 chance of getting it wrong and a 1/3 chance of getting it right. if you get it wrong (which happens 2/3 of the time) then there will be a right and a wrong door left, the wrong door will be revealed and if you switch you will be at the right door. if you choose right at the start however ( which happens 1/3 of the time) then there will be 2 wrong doors left, one will be revealed and if you switch you will be at the right door. that is why there is a 2/3 chance of winning if you switch 
- Adam
I think the people that don’t get it need to do a role-reversal and play Monte Hall.
Here’s a way:
I’m going to guess your birthday. June 1.
Your job, as host, is to give me one other date, so that either June 1 is your birthday or the other date you give me is your birthday.
Now, I have the option of going with June 1 as your birthday or the other date you gave me. I think I’m going to go with the other date your gave me.
We can replay this with your father’s birthday, your mother’s birthday, your best friend’s birthday. Here are my guesses: Jan 15 for your dad, Nov 20 for your mother, Sept 15 for your best friend.
Now give me 3 other dates so that either my guesses are correct or the dates you gave me are correct.