You participate in a game show where you must choose between 3 doors. Behind 2 doors is a goat, but behind 1 door is a car. You pick a door, say door no. 1, but before you open the door, the host opens another door, say door no. 3, which has the car behind it. "Oops," says the host.
You lose.
Well, yes, a host would never choose randomly, for that reason (and that would be hilarious), but...even if he did choose randomly, but "randomly" chose a non-prize...it does still make sense to switch. Wild, but true.
It's not the knowledge that changes the odds.
I guess what I mean is that a key feature of the simulation is the host always reveals a non-prize.
These days, since I already have a car without government mandated shut-off switches, I'd rather the goat.
True.
No, if he chose randomly and didn't reveal the car by pure chance then both doors remain equally likely to have the car behind them and switching doesn't matter. Most teachers are bad at explaining/don't understand the monty hall problem themselves and they take shortcuts when explaining it so that's an understandable confusion though.
The point is that the host choosing to reveal a door has more knowledge than you and he divulges new information by elimination when he opens a door that lets your make a more favourably informed choice in the second round.
But lot of people try to explain the system where the probability of your first choice is just frozen in time to before the outcome is revealed, and you get to chose a second time at 1/2 odds simply because there happen to be only two doors left. The leads to the nonsense outcome where if host knows nothing and accidentally reveals the car at random, then you definitely shouldn't chose to switch doors because there's still a 1/3 chance you win a car if you stick, but a 0/2 chance if you switch. When in reality the choice is meaningless.
No, actually both explanations you’ve given are wrong as I understand it. Here’s the solution as I know it:
When you picked, you had a 1/3 chance to pick “right.” Since probabilities sum to 1, that means there is a 2/3 chance that the prize is behind a door you did not pick. One of those doors is then eliminated and you are shown that door did not contain the prize. It doesn’t matter whether that door was eliminated at random or purposefully—what matters is that you know the prize was not behind that door, and that out of the selection categories remaining, [door you originally picked, 1/3rd chance of having a prize] and [all doors you did not pick, 2/3rd chance of having a prize] you get to repick your category. If it helps, you may think of repicking as opening every door in the latter category, since that’s effectively what will happen.
If you switch, you have a 2/3rd chance of hitting the right door, because that’s the necessary inverse of your original 1/3rd chance, now that the other possibility of a dud door has been removed.
The door the car is behind is fixed, no amount of door opening will change its actual position, but your knowledge is probabilistic and the "smart" choice leverages that knowledge to maximise your chances with your knowledge at that time.
You have to consider the whole probability tree at any given time, so if you're saying the host has no knowledge of the car's location the branch points for switching are that 1/3 times you picked the car first time and switching is automatically losing, or that 2/3 times you choose a goat, then 1/2 of those times (i.e. 1/3 of the total possibilities) the host accidentally reveals the car and you lose immediately, or 1/2 he reveals a goat and switching will win. Bringing you back down to 1/3 total chance of winning if he only accidentally reveals a goat and you switch. Just because it didn't happen doesn't mean it wasn't part of the probability space and without the host having additional knowledge, to your knowledge you have no way knowing if you're on the now equally likely "chose right the first time and switching loses" or "chose wrong the first time and the host got lucky" branch, so switching is meaningless
But if the host knows which door has the car and can't reveal it, then 2/3 times you chose wrong first and 1/1 times he uses that knowledge to essentially show you by omission which door the car is definitely behind and gives you the chance to open it, or 1/3 times you chose right and he reveals none of that knowledge by randomly picking one of the two wrong doors and gives you a chance to open a door that the car is definitely not behind. That's where the inversion from 1/3 to 2/3 if you switch happens.
You're right. I understand my error now. Thanks for explaining.
I think, for me, the easiest way to visualize this is this.
1 )Instead of 3 doors, you have, lets say 10.
You first pick a door. You have a 1/10 chance of been right the first time. This chance doesn't change at all.
Host eliminates 8 of the remaining doors with no car, leaving just 1. We are back the "50/50" situation, except its not truly 50 /50. Your initial chance of 1/10 has not changed.
Ask yourself this - Do you stick with your initial chance of 1/10? He already removed 8 bad choices for you. Remember the host knows where the car is always. Given the new information, you should definitely positively switch.
No, I was actually wrong, sadly. It does matter if it's random or intentional. It get's pretty hard to grasp, but there are ways to prove it.
In that example, if the choice is random, and he doesn't reveal the car...he had a 50% chance of the 66% chance to reveal the car, and didn't...meaning the remaining door is also 33% chance, just like the door you already chose. Or, since they're equal chance, now each is 50%, and it doesn't matter if you switch.
If he intentionally removes a known non-car, it is your example; 33% vs. 66%, and you should switch.
It can be a bit much to wrap ours heads around, since the whole concept of predetermined but random outcomes isn't really standard fare, but it does make sense.
Where it gets crazy is, to make your decision on whether you should switch doors, you need to know how the host made his decision in eliminating doors. It gets pretty mindfucky.
Yeah, I think I was wrong. I was thinking about it before I left for work, but didn't have time to edit. It's weird to think about, and still hard to completely grasp, but I do believe intent matters. It's just like how something crazy improbable can happen, and it doesn't affect the next result. You can flip 99 coins, and have them all turn up heads...you're already in the crazy universe, but the next flip is still a 50/50 chance.
So, you could have ten doors, and the host could randomly remove eight losers...the remaining two doors are still equally likely, you just got a really unlikely outcome...which I suppose means logically if not mathematically, you actually have more likelihood of the car being behind the already chosen door.
So, if the host chooses randomly, and you know that, I suppose you're actually better sticking with your choice. The more doors he reveals (again, if randomly, and you know it's random) without revealing the car, the more likely the car isn't in the doors that were unchosen.
Ugh. This gets more and more confusing, because you do have to know he's not choosing randomly, to be able to reach the "answer." If you don't know how he's choosing, it gets into more mind games than logic or math.
So I suppose I was a little hasty. I was still right that all those people on Twitter were retards and didn't even get the basics, but there were more assumptions you have to make than I took into account, initially. You have to know how the host is removing doors to be able to calculate whether or not you switch.
It's made a lot simpler by the fact that switching doors is meaningless rather than detrimental if the host has no extra knowledge. It basically means you're still better off switching in any instance where there's a non-zero chance the host knows the answer, as there's no harm to doing it if he doesn't know, so you don't have you have to balance anything against the chance he does know.
Yup. I hadn't done the full rationale, but it does seem to always break even. If chosen randomly, remaining doors minus no-prize doors, you're still left with an even chance versus your initial pick. In the three-door scenario, it's 1/3rd vs 1/3rd, and the last third is missing, so equal chances, i.e. 50/50.
And, that's a good point, even if you don't know how he's choosing, you always switch, because if he is choosing randomly it's a coin flip, and if he isn't you're increasing (sometimes very drastically depending on number of doors) your chances, so it's an overall gain.
The odds only change if what's behind the doors are fixed ahead of time.
If the host chooses a door, the game show rolls 1 out of ndoors chance of car behind it, and then puts the car or goat behind it before opening it then the odds don't change at all even if the host didn't know.
Switching your choice is the classical hidden-information answer, staying with your first choice is the quantum changes when-observed answer.
Yeah, but the assumption is that there is a right door. You're changing the fundamental rules of the problem and saying "guys, when you say the whole thing works differently, it works differently!" That's true, but it's not exactly insightful or useful.
You're saying your assumption is a fundamental rule... really? The insightful thing here is don't confuse assumptions with rules.
It's an entertainment show that's giving away free cars for advertisement. If they give contestants 50/50 they lower their prize payout, or if they give 2/3 odds maybe they get more audience investment. Maybe it's more work for them to keep giving out goats than cars because they have to train them to be calm on set. I have no idea how game shows prioritize these things.
Okay, man.
Anyway, here's the actual question:
I bolded the parts where it explicitly says it works the way I told you it works.
Source
Yup.
If everything is fake and gay, it doesn't matter if you switch or not, the funniest outcome will happen, if someone has their hand in it. i.e., you switch, and the car was "always" behind the door you initially chose. Oops, you lost!
It's also pointless, though, because if everything is in flux, your choice doesn't matter. So the fixed version is the only one that really makes sense, if the "problem" is expected to be solved.
It could be worse. You could open a door and there be 3 doors behind it.
It doesn't have to be 'fake and gay'.
Typically in regulated gambling, like with slots, they tell you your odds of winning. Slot machine doesn't know what the roll will be before you roll it.
Same principle. The game show may give all contestants a 50/50 chance of winning regardless of what doors they pick.
In which case it doesn't matter. The only way the problem makes sense is if the prizes are already fixed.