It doesn't matter. Because the problem explains what happens: he opens a door and reveals a goat.
That is crucial information. There's now only one goat and one car left. But the choices have not been shuffled: you're definitely still pointing at your original choice.
What were the odds that your original choice is a goat? Two in three.
If you picked a goat, what is guaranteed to be behind the other door? A car.
So what are the odds that behind the other door is a car? Two in three.
It's actually counter-intuitive. I was going to argue on your side, and then I wrote up a quick program that proved me wrong[0]. Let's go through the scenarios, with Goat A, Goat B, and Car C. In the scenario where Monty picks a door purposefully, always selecting a goat, the scenarios are:
You picked A, Monty showed you B, and you switch to get C.
You picked B, Monty showed you A, and you switch to get C.
You picked C, Monty showed you either A or B, and you switch to get the other goat.
So a 2 / 3 chance of getting the car if you always switch.
If Monty is choosing randomly, we have the following scenarios:
Initial Choice | Monty's choice | Remaining Door
A | B | C
A | C | B
B | A | C
B | C | A
C | A | B
C | B | A
But we know in the problem statement that Monty hall showed us a goat, so we can eliminate possibilities 2 and 4 to get:
Initial Choice | Monty's choice | Remaining Door
A | B | C
B | A | C
C | A | B
C | B | A
Whether you switch or not, you have a 50/50 chance.
I'm not great with probabilities, but the major difference I can see is that in the first scenario, if you pick the car, Monty will either show you the goat A or B with equal probability as a part of the same 1/3 scenario. So you have really:
1/3: You picked A, Monty showed you B, and you switch to get C.
1/3: You picked B, Monty showed you A, and you switch to get C.
1/6: You picked C, Monty showed you A, and you switch to get B.
1/6: You picked C, Monty showed you B, and you switch to get A.
But in the second scenario, each of those options is actually 1/4, because he was choosing randomly. Most importantly, each option was a 1/6, but two options where you selected a goat were eliminated because those were ones where Monty selected the car.
You confused yourself. There are four possible outcomes at the end, but they are not equally likely. So not 50/50.
If Monty always shows a goat, it is undeniably a 2/3 chance to win on switch.
The nuance here being discussed is whether you can assume Monty would have shown you the goat had you chosen a different initial door just because he showed you one this time. If you don’t know that, then you don’t learn anything.
Edit: actually I misread. You just chose to ignore the cases where Monty revealed a car. Which is correct although most people chalk that up as a win or loss.
I'm covering the problem statement. I'm ignoring the cases where Monty revealed a car because it didn't happen. You can't chalk it up as a win or a loss, because it didn't happen.
It's always "You choose a door, the host reveals a goat behind another door. Do you switch?" The scenario does not include him revealing the car.
If he intentionally chooses a goat, switching gets the car 2 times out of 3.
If he chose randomly and just happened to choose a goat, switching doesn't matter. It's 50/50.
Of course, these aren't the only scenarios. As others have mentioned, if Monty can decide whether to reveal what's behind a door, he can act maliciously and only reveal a goat if you've already selected a car, and switching will always make you lose. Without knowing these constraints, a single answer isn't knowable, as you mentioned before.
Yes you’re correct. You’re just the first person I’ve ever seen cover the true random Monty variety who doesn’t force the situation where he shows a car in as a win or loss.
Incorrect. You have no way to know if he would only show you a goat if you picked the car.
If that were the case then switching is a guaranteed loss. Simply knowing that he opened a door and showed a goat does not mean he would have done this regardless of your choice.
You cannot assume that your context would apply from all starting conditions!
That’s why this problem is kind of bad. It does not describe the behavior of the host. It describes the perspective of a contestant halfway through the game.
Dumb example. Host flips a coin 9 times in a row and lands heads each time. If you believe this to have happened by chance then it’s probably rigged because that’s insane and it’ll likely be heads again.
If it’s ALWAYS heads then you haven’t learned anything.
Yes but in fact Monty always shows you a goat. There's always a goat for him to choose regardless of your choice, he knows where it is, and he's not going show you the car or the goat you already picked.
Showing you the goat is the event that, as you say, guarantees that the other door has the opposite of your original choice behind it. Because nobody closed the curtain to shuffle the choices.
One of the things I think people struggle with -- and I struggle with -- is that probability isn't about hypothesising about a single event that happened and how it might have happened. It's about encapsulating all the possible ways a single specified scenario can play out in a single expression.
Monty shows you a goat this time, but this means Monty always shows you a goat. There's no scenario where he is unable to show you a goat. Just like you always only pick one door. And there's always only two goats and one car.
No. I am not a mathematician but I got it once I understood that this is not drawing coloured balls randomly from a bag (which is how all my school probability problems seemed to go).
That is:
- the setup of the system matters.
- The state of the system at the point of the decision to switch matters.
- The choices don’t get re-randomised.
So the probabilities assigned to the original choice (and the remaining alternative) still count.
If the host had closed a curtain over the stage and randomised the remaining doors, then it would be 50:50.
But he didn’t. So you’re still in the probabilities of the original choice.
One of the goats has been removed. The car and one goat remain: you know this for sure.
You are being offered a door knowing that behind it must, necessarily, be the opposite of your original choice, and the probabilities have not been reset.
If you originally picked the goat, that door absolutely has a car behind it. And there's a 2/3 chance you picked the goat originally. So by inference there's a 2/3 chance the door has a car behind it. You should switch.
I didn’t get it until I had written a simulation to see it for myself though!
> Without additional assumptions this original wording is not clear
The only problem you are being asked to solve is the one described only in the words of the problem.
That's actually what this is all about. This is a learning exercise, not just about probability, but about comprehension of what the problem as described exactly says.
> But she wasn't really criticized for being a woman
Nobody is saying she was criticised for being a woman, though. What they are observing is that it was the basis of and the tenor of attacks on her character, intelligence, certainty etc.
It was nasty and it really happened, is all I'm saying. Are we better? I hope so. But I don't know that I think so.
Exactly. Would the tone of the letters been different if the column was said to be have written by a man? Likely they would be different.
This is what makes discussions about discrimination so hard- it can be very subtle. And it can include situations like this one, where we don’t have a test environment where we can compare responses to the article written by a woman to the one written by a man.
> Of the cited hate mail Savant received, I have to wonder how many were setup with poor writing on the topic.
Have you spent at least equal time considering, given the tenor of much of it, how many were instead willingly exposing their simple misogyny?
You might be a bit younger than me, but I remember this well, and I remember this being one of my first introductions into the idea that men who are wrong will lash out with misogyny if it is a woman who proves their ignorance. Even academics.
Interestingly I’ve correctly grow up more recently such that I instead see internet simps here rather than academic misogny as notable, seeing as the tables have largely socially turned since your (I assume by the username) birth year of 1875.
> I, like many, stated “50-50” because the odds of getting one of two doors right is in fact 50-50.
Only if (in fact) it's an independent choice. It's not.
If the prizes had been juggled behind the curtain after one door had been opened, then you'd be right. But the "weird setup", which is the point of the puzzle, makes it clear.
Let's be real here: the point here for most of us is to think "oh it's 50:50", be proved wrong, enjoy being humbled, learn something new about how to interrogate a problem, and enjoy setting this puzzle to younger thinkers in the future.
Not to attack the (very closely-worded) description of the problem because you're upset to be wrong.
It's a probability puzzle and a character test for thinkers.
Most replications of the puzzle are incredibly bad such that one is led to the independent branch with little indication that there was an intent to examine the totality of circumstance.
I find the problem interesting for its historically poor examples floating on the internet and their application socially beyond a math puzzle.
