I don't think the host's knowledge is important. Consider the alternative where:
1. You pick a door.
2. The host reveals a different door at random. The host has no knowledge.
3. The door the host revealed is a goat.
Should you switch or stay?
1. You pick a door.
2. The host reveals a different door at random. The host has no knowledge.
3. The door the host revealed is a goat.
Should you switch or stay?
In the example you gave it's just random chance. In the Monty Hall puzzle it works out that statistically you should swap.
"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.
Swapping doesn't affect the outcome, it's a 50/50 whether you swap or not.
The obvious answer is the right one - it's not possible and Monty's knowledge has no effect on the game or the strategy the player should use. Monty knowing where the prize is and opening a bad door is the same as Monty not knowing where the prize is and having randomly opened a bad door. In either case, you should switch, Monty's knowledge doesn't change things.
Another way of thinking about it - how would the participant know if Monty opened a door at random or knowingly opened a goat door? What if they thought Monty knew, but Monty had actually forgotten and just coincidentally opened a goat door? Does any of this matter? No, because Monty's knowledge doesn't effect the game or strategy.
I also wrote a quick simulation in my Javascript console which confirms what I'm claiming here.
The "pick and stay" strategy wins about a third of the time while "pick and switch" wins about two thirds - same as the original problem. Writing the code emphasizes that it is basically the same thing - Monty coincidentally reveals a goat versus knowingly reveals.
Different ways of host operation changing the outcomes are listed here: https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...
>"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.
You’ll see that that changes it by discarding scenarios where switching is good (the prize is shown to you) but not ones where staying is bad.
But given they do open a door with a goat, the probability is the same as the regular Monty Hall problem. So the parent's point stands: it's not the host's knowledge that makes the game "work". He's leaking information by not opening your door, thus "concentrating" the 2/3 chance it wasn't behind your door into the one remaining door.
correct = numpy.random.choice(3, 10000)
(correct[correct != 1] == 0).mean()
This gives 0.49 or similar when I run it.
- 1/3 it is behind your door
- 1 host picks a goat
- switching always loses -> 1/3 chance of losing
- 1/2 host picks the goat
- switching wins always -> 2/3 * 1/2 = 1/3 chance of winning
- 1/2 host picks the prize
- result undefined -> 1/3 chance of ending up here
On reflection, this makes sense. The host is indeed leaking information by not be able to pick the around the prize. If the prize is not behind your door, his choice is constrained. So his revealing it supplies information about that constraint. If he is not constrained, then you are both picking doors at random. I'm sold.
You are playing the classic Monty Hall problem. You pick door 1. Monty reveals a goat behind door 2. Monty asks if you want to switch. You know the correct strategy is to switch, so you are about to say "Yes - switch" when suddenly, the lights dim and announcer's voice booms over the speakers "You thought Monty was knowingly revealing something to you, but actually Monty just revealed a door at random. He had no foreknowledge." Are you now ambivalent about switching or staying?
If you still want to switch, and you should, that's because obviously it does not matter what Monty knew going into the problem. If you don't want to switch, please explain how the contents of Monty's brain affect the probability of which door conceals a car and which a goat.
Let’s turn this around: explain why I should switch doors, but starting from scratch with the new problem, instead of by reference to the original. I think you’ll end up thinking the original solution is wrong, based on your no-telepathy rule, or you’ll see how they are different.
Or try this out. Let’s have another variant: before you pick the door, the host picks one that turns out to be a goat. Now you pick a door, and _then_ you have the option to switch. Do you still switch? Does that make sense? The situation is exactly the same (a goat behind one door and two closed doors). If both actions so far are random, it doesn’t matter what order they go in.
(def doors {0 :goat 1 :goat 2 :prize})
(def choose (rand-nth 3))
(defn game []
(let [choice (choose)
remaining (dissoc doors choice)
monty (rand-nth (vec (keys remaining)))]
{:choice (when (= (get doors monty) :goat) (get doors choice))
:monty (get doors monty)
:switch (when (= (get doors monty) :goat) (first (vals (dissoc remaining monty))))}))
(def ten-thousand-games (repeatedly 10000 game))
(frequencies (map :choice ten-thousand-games))
;; {:goat 3309, :prize 3307, nil 3384}
(frequencies (map :switch ten-thousand-games))
;; {:goat 3307, :prize 3309, nil 3384}Monty's knowledge has no effect on the problem or the player's choices. What has an effect is the information the player learns from seeing the revealed door. If the player sees a goat behind the revealed door - that is the relevant information. Whether Monty was surprised or not by the revealed goat is irrelevant.
And yes, your simulation should include that Monty always shows a goat. That is given in the problem statement.
I feel like I've said this thirty times now, I'm specifically discussing a misinterpretation of the problem where Monty does not know what he's opening, so he sometimes opens a prize door. A game where Monty always opens a goat door is the original, correct Monty Hall Problem, and switching is the correct strategy.
If you are talking about some other scenario - then why? I've just explained what this thread is about. Why would you invent an alternate scenario, not tell anyone about your alternate scenario, and argue that correct interpretations of the scenario I laid out in [2] are wrong because they don't fit with your interpretation of your private variation?
Bayes’ theorem is the relevant piece of information here. It tells us the relationship between a conditional probability and the information you have. Specifically, the probability of P(door 3 has the prize | host opened door 2 and found a goat) depends in a very important way on P(host opened door 2 and found a goat). You are acting as though it does not, that only the specific current state matters. This is the cause of your confusion: everyone else is working out the conditional probabilities and you are declaring them irrelevant.
The key is that if the host doesn’t know where the prize is, he can’t avoid picking the door its behind. This changes the probability that he would open that door and reveal a goat in the first place, and how that relates to the various places the prize could be. It would be really strange if the host’s behavior were in some way controlled by the location of the prize and this didn’t give you additional information on that location.
To spell it out:
You choose door 1, the prize is behind door 3, host opens door 2
Prop A: the prize is behind door 3
Prop B: the host opens door 2 and found a goat
P(A|B) = P(B|A)P(A) / P(B)
P(A) = 1/3 in either scenario
In regular Monty Hall, the host will open door 2 always if the prize is behind 3, and half the time it is behind 1. So:
P(B) = (1/2)(1/3) + 1/3 = 1/2
P(B|A) = 1
Thus P(A|B) = (1/3)/(1/2) = 2/3. You’ll recognize this as the answer to Monty Hall. Note that conditional probability is foundational to how we got that answer.
Let’s try with the new rules. The host can pick either door, whether it has a goat or not. He can do this because he doesn’t even know which door is which. The host opens 2 half the time. 2/3 of the time he picks that, the prize is not behind it, so he finds a goat.
P(B) = (1/2)(2/3) = 1/3
P(B|A) = 1/2
P(A|B) = (1/3)(1/2)/(1/2) = 1/2
If you scroll down to "Monty Fall," the description of the problem is "The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car" and the solution is "Switching wins the car half of the time." (This is the result demonstrated in my simulation)
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.
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_...
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.
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.
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.
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.
If you picked the prize P(A = 1/3), then the random open is going to be Goat P(1). if you didnt pick the prize P(B = 2/3) then the random open is going to be Goat P(1/2) or Prize P(1/2)
P(Goat reveal) happens (1/3)(1) + (2/3)(1/2) = 2/3 times
P(Prize reveal) happens (1/3)(0) + (2/3)(1/2) = 1/3 times
P(you picked prize, given goat reveal) = (1/3)(1) / (2/3) = (1/2)
Everything else is overthinking.
Yes, it does. this is the entire point behind the popularity of the Monty Hall problem. It is counter-intuitive and feels like overthinking, but it is truth.
I described the situation pretty clearly (imo). I did not need to think about it much at all, though it did take some thought to describe it clearly.
I can tell you, without much thought, having been trained thoroughly in probability and statistics, it is 50-50 if a door is opened randomly to reveal a Goat. You can switch if it makes you feel better - because it is 50-50 so you wont be hurting yourself.
https://en.wikipedia.org/wiki/Conditional_probability
there are other examples that make it more clear why condition matters.
Whether you end up switching or not, the other opened door is 2/3 likely to contain the prize and the door you picked originally has a 1/3. Whether you switch or not, doesn't change the odds.