Always switch
I’ve never understood the logic behind switching.
It might be helpful to extrapolate this out into a higher number of doors. Say there are 10 doors instead of 3, with still just the one prize. So instead of a 33% chance of getting the right answer off the bat, you have a 10% chance.
Now, after making your choice, Monty, being the good guy that he is, opens 8 of the doors that contain no prize, leaving only the door you picked, and one other closed door. Originally, your chance of getting the right door was 10%. However, that's just because you didn't know what was behind the other doors. Now, you know what was behind the other doors. Now, the number of doors that could contain the prize has shrunk from 10 to 2.
The prize is definitely behind one of the 10 doors. It could be behind the one you picked at random at the start. But that only occurs 10% of the time. 90% of the time, it's behind one of the other doors, and Monty has just shown you which of those doors it is by eliminating the other 8 possibilities. So 10% of the time, if you stick with your original choice, you're going to get a prize. But 90% of the time, you won't. So it's way better to switch when given the opportunity.
Does that make it any clearer?
You switch because new information has been added... If the host didn't reveal what was behind one of the doors, you wouldn't switch (or do, it wouldn't matter the chances are the same).
Host shows you something that removes things from the table... And that impacts the probabilities.
ponder on the fact that the host is not just opening one of the doors you didn't pick at random, they are specifically opening one you didn't open and that had the inferior prize. this reveals new information. if you did choose one of the two bad prizes initially (a two in three chance, right?), then the one that you can switch to must have the good prize.
i thought it would be helpful to make a comment on why the probability after switching is 66%, since it's very counterintuitive. the main reason has to do with why a certain door was chosen to be opened.
in more detail: it helps to instead think about a situation where there are 100 doors, 99 of which have a goat behind them. (original problem was about goats and a car). you choose one of 100 doors. there's a 1% chance you picked the door with the car. in other words, there is a 99% chance the car is behind a door you did not pick. put differently, 99 times out of 100, you are in a situation where the car is behind a door you did not choose.
afterward you pick your door, the host picks 98 doors with goats behind them and opens them. (this is another crucial detail, the host can only open doors that don't have a car behind them.) it is still true that 99% of the time, the car is behind a door you did not choose. this is because the doors were opened after you made your choice. but now, there is only one door you did not choose, so that door has a 99% chance of having a car behind it.
Your hypothetical strapped 495 people to train tracks, you absolute monster! Statically, about forty of those 495 have serious chronic back pain, too. You obviously don't care about the disabled. Be more careful next time when changing the number of doors in a Monty Hall problem!
On the other hand, statistically, the chance of anyone dying is much much lower
I can kind of understand the logic behind it, if you assume your door can't be affected by the probability of it, but the thing that still stumps me about this is how the probability for your door is "locked in."
You picked a door out of a set, and by opening any number of doors, the host has altered the set. The other door remaining went from being a 99/100 chance of having a goat behind it to being in a set of 98 knowns, and 2 unknowns. While the host can't choose it if it has a car, he also can't choose yours. You wind up with 2 identical doors and X number of open doors, with each door having a 50/50 chance given the re-evaluation.
I know this is supposed to be the wrong answer, but I can't see why it's wrong. If you have an explanation, I'd love to finally be able to understand this problem.
It's because the host has knowledge of the situation. Each door has 1% chance. You lock one door closed that the host can't touch, goat or car. The other rule for the host is they have to open all of the doors except one and they also can't open the car.
The door they leave is all possibilities from all doors at the begining minus your door (since the host couldn't mess with it), which is 1%. Your original choice determined that door would stay closed by the host so the host can't effect it's odds with their knowledge. There is only a 1% chance you locked the car behind that door picking randomly. Either you got the 1% and locked the car door or you didn't. The host will remove 98 goats without random chance and which will leave a car or a goat left. There's a 99% chance you made them leave the car with your first choice being 1% and a 1% chance you made them leave a goat. It's still the original odds, but it flips because the host has knowledge and either was forced to removed all possible bad choices and leave the car or you managed to hit the 1/100 chance at the start. There's a 99% chance you didn't, so switching has a 99% chance the host left it there.
that's a really good question. the probability getting locked in is very counterintuitive. i think it helps to think about what happens in each case.
let's say you pick one of the 99 doors with goats behind them. this happens 99% of the time. the host is tasked with opening 98 doors. of the 99 doors you didn't choose, one has a car behind them. the host does not have a choice about which doors to open, his hand is forced. in this sense, the 99 other doors are tied together: since you originally chose a door with a goat behind it, the host is forced to leave only the door with a car unopened. so, you switch and you get a car.
next, let's say you get lucky and pick the door with a car behind it. this happens 1% of the time. now the host gets a choice about which doors to open: he gets to pick one door with a goat to leave unopened. in this scenario switching gets you no car.
so, 99% of the time, switching gets you a car. i hope this is helpful!
Your door has 1% chance at the start and will always have 1%. The other door "inherits" the probability of all other doors the host opened. So that door has its original 1% as well as the 98% of the other doors, resulting in 99%.
Like sure, your door could have been correct from the start, but the chance is 1%. The other door has 99%. Betting on a 1% chance is just not the smart thing.
Way easier to explain it this way:
Switching turns a correct guess into an incorrect one, and an incorrect one into a correct one. Your initial guess was more likely incorrect.
I never understood why in the 100-door case, the host opens 98 doors, and not just one door. That feels like changing the rules.
I fully understand the original problem with 3 doors; I know the win probability is 2/3 if you change. But whenever I hear the explanation for 100 doors case, it just makes everything confusing. By opening 98 doors, it feels like the host wants you to switch to the other door. In 3 doors case it's more natural.
Because the problem is explicitly about the choice between two doors. You have to eliminate all but two choices.
But even then, you'd still have a better chance by switching.
Your intuition about the change is the whole point - it exposes why the result is what it is.
In both cases the host opens every door but one.
Oh hey, it's that Statistics problem from undergrad I haven't thought about in over a decade.
looks like i have to teach you all kindergarten level statistics
What kind of kindergarten is teaching the Monty Hall Problem? No wait, what kind of kindergarten is teaching stats?!
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this never made sense to me till I took a probability class then it clicked lol
I feel like there is some overthinking of the monty problem here. The answer no makes the most sense. Consider this, what is behind the doors is predetermined, right? So either the smallest number of people you can hit is 0 or 5. The most is 10. Chosing either a or c guarentees at least 5 people hit. Choosing B limits the number who can be hit to 5 (if they're behind the door). So you should always choose B and always keep the choice.
It says "one of the doors you didn't choose". So if you choose B, it's always 0 people.
I believe you're the onenoverthinking it. The first choice is 1/3. After this someone with knowledge of the situation eliminates a bad choice that you didn't choose. Your choice is now 1/2 if you take it. If you don't take it then it's still 1/3 because the odds of the first choice don't change.
Overthinking it is exactly what leads to the solution "no" imo. If you say no every time, then you can only possibly be correct 1/3 of the time since it's a 1/3 chance of picking the correct door. Hence 2/3 of the time you picked incorrectly and hence should switch. By design if you pick incorrectly then the switched door will be the correct solution
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Choose B, door on trolley opens, reveals 5 people, wave at them, leave
