Doesn't disallow it either. That's the problem: if you describe the problem without all necessary details, it doesn't work like the Monty Hall Problem is supposed to. If you fill in the details yourself (host never showing the prize in this case) then of course the problem becomes complete and then works like Monty Hall Problem again.
"But he could have" but he didn't.
"But does he always" but he did.
What he always does or could do in an alternative scenario are irrelevant, you are given one scenario: He reveals the goat. Don't overthink it, it's not hard, switch and double your odds or stick with it and keep the 1/3rd chance.
Your chances improve from 1/3 to 1/2 when the host opens the empty door, but whether you stay on the first door or not doesn't matter.
One of three doors is selected.
One of the other two is opened, revealing a goat.
Should you switch to the unopened door? Yes, it doesn't matter how that goat was revealed, it could've kicked the door open itself. The odds are better to switch.
If the host knows where the prize is, the odds change when the host picks a door, because we know the hosts strategy. If it's just random, then the odds don't change and changing the door doesn't matter, because each unopened door has the same chance of revealing the prize (1/3 before a door is revealed, 1/2 after).
Look up "other host behaviors" on the Wikipedia article, especially the table with "Ignorant Monty", if you don't believe me.
No, I said the player's odds of winning improve if they switch (to the unopened door).
If the host randomly revealed a goat or deliberately revealed the goat is immaterial. A goat was revealed, that's all that matters.
The player has 3 doors to choose from. 1 is selected, 2 are not. That means the other two have a combined 2/3rds chance of holding the prize. The selected door has a 1/3rd chance of holding the prize. A goat is revealed behind one of the two unselected doors. The remaining closed, but unselected, door has a 2/3rds chance of being the one with the prize.
The selected door's odds of being the winning door remains 1/3rd after the reveal. The player's odds of winning improve if they switch.
The original problem:
>The host acts as noted in the specific version of the problem.
>Switching wins the car two-thirds of the time.
The random door pick:
>"Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car.
>Switching wins the car half of the time.
Let’s try to simplify the problem. Let’s assume there are three numbered doors: 1, 2, and 3. A car is randomly placed behind one door, and goats behind the other two.
To make it easy, I will always guess door 1. In the ‘ignorant host’ scenario, we can just have the host always reveal door 2 (the math doesn’t change if he randomly selects which door to open or always opens the same door, so keeping it always door 2 makes it easier to understand).
Ok, so we guess door 1, and the host opens door 2.
1/3 of the time, the host reveals a car. Obviously we would switch in this case, since we know exactly where the car is.
2/3 of the time, the host will reveal a goat. When this happens, 1/2 the time the car will be in door #1 (our door) and 1/2 the time the car will be behind door #3. If the car is not revealed behind door #2, we have a 50% chance whether we switch or not.
With this random reveal style, we end up winning 1/3 (the chance of revealing the car behind #2) + (2/3 * 1/2) = 2/3rds chance of winning… which makes sense, because we get our original 1/3 chance plus the free 1/3 chance it is behind the revealed door. Crucially, though, it doesn’t matter if we switch or not. You always win if it is behind door #2 (because you are shown it) and win half the time if it is behind #1 or #3.
Contrast this with the case of the host always revealing a goat. In that case, you again have a 1/3 chance of it being behind your door #1 choice. However, in this case, you can essentially guess that your choice was wrong (2/3rds chance) by switching your guess, which will make your guess correct if it is either in door #2 or #3 (since the host will reveal the one which it isn’t behind if it is one of 2 or 3.)
In both the ignorant host and non-ignorant host situations, you end up with a 2/3rds chance of getting the car. However, in the ignorant host situation, you don’t need to switch unless the host shows the car in his reveal to get the 2/3rds odds.
Of course, it never hurts to switch in either case, so always switching is not a bad strategy, just not necessary if the host knows nothing.
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Assumptions about the strategy or decision rule of the host are not irrelevant, but crucial. Suppose the hosts' decision rule (unknown to the candidate) is
IF (candidate's initial choice is correct): open one of the remaining goat doors ELSE: don't open anything
then it's of course not advantagous to switch. The problem with the original Parade formulation is that it only describes what the candidate sees and what the host actually did in one particular scenario, not what rules the hosts must obay. So you don't know if the hosts is playing adversarial games, but you can't rule it out either.
That's why (as per linked article) in later formulations the "standard assumptions" (that the hosts will always open a door with a goat) are usually given explicitly (in which case the solution is also a lot more obvious).
We can come up with a million what-ifs, but none have a basis in the problem statement so there's no point in them.
But in the setting of an actual repeated game show, I think the more realistic case to keep the show interesting is that the host will mix up his strategy, i.e. sometimes he opens another door, sometimes he doesn't. If it's a long running daily or weekly show, if he always opens a door with a goat, the public will work out pretty quickly that it's advantageous to always switch.
The game described in Parade magazine is different from the TV show game, and the assumptions (about the host's behavior) are incomplete/ambiguous enough that switching is not necesarrily the optimal strategy. This, like almost everything else you can think about the Monty Hall Problem, is also described in TFA[2].
[1] https://en.wikipedia.org/wiki/Let%27s_Make_a_Deal#Format [2] https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...
I agree, but this subthread is discussing a misinterpretation of the problem statement, and asking the question about whether switching still matters if the host is also picking randomly. This is showing that it does in fact matter whether the host is picking deliberately or randomly.
> I was told about this problem and the person telling it didn't say anything about that, just that you pick a door and the host opens another one which doesn't have a prize behind it. So I thought the probabilities don't change at all because the host just picked a random door. [emphasis mine]
It doesn't matter if it was at random or deliberate. It matters that a non-prize (goat in most versions) is revealed. The question is about one very specific scenario, even in what antris was originally given: The goat is revealed. In that scenario, you switch.
The OP misunderstood the problem because they forgot (as many others have in this discussion) that it's about one scenario, not about all times. If Monty selects a door at random and reveals the prize, then it's not this same scenario anymore so the odds of that situation are necessarily different over all scenarios. And if we include all possible scenarios (really, two) then the odds of winning drop, but only because you have a 100% chance of losing when he reveals the prize (or 100% chance of winning and the odds increase).
This is the fundamental misunderstanding in many of the comments. If I tell you, "We select, each day, either a prize or a tiger to put behind this door at random. Today it's the tiger. Do you open it?" You'd be a fool to open it. And over all scenarios you'd have a (if chosen uniformly) 50/50 chance of getting the prize or the tiger. But if today I tell you, "It's the prize" should you open it? There's a 100% chance it's the prize, I've told you it's the prize.
Now we can get into adversarial situations, whether I'm honest or not. But that's diving into something entirely different and makes it impossible to consider the problem statement because we'll be chasing nonsense after a few minutes. I've just told you, 100% chance it's the prize today. You should open it.
The Monty Hall problem is the same. It is one scenario, a goat is revealed. The reason or mechanism of that reveal is irrelevant. He could know it, he could not know it. He could have revealed the goat by dumb luck. Or maybe because it started kicking at the door and was too obvious not to reveal. But today, for this game show he revealed a goat. So your odds are better if you switch. And even antris' original statement presented that scenario, they just answered a different question.
You can just go back to the simulation numbers and check. If you eliminate the games where Monty "wins" (i.e., reveals the prize), you are in the specific scenario you described. You have chosen a door and Monty has revealed a goat, and you win 3309 times by staying and 3307 times by switching.
If you rewrite the simulation to follow the correct problem statement (that Monty knows and always picks a goat), you will see that switching is in fact the correct strategy.