Hi Reddit, settle an argument.
My friends and I are playing a guessing game. There are three options remaining, and we have two guesses. I assert that this is essentially the Monty Hall problem, and ask my friend to pick a door. I pick a door he does not pick. I guess a door, and am incorrect. Should my friend switch his door.
I'm sorry if I violate any etiquette with this post. I am relatively new to posting.
Edit:
As a clarification, I understand that it is not the Monty Hall problem if I guess correctly. I'm asserting that it is should my guess be wrong. If that helps.
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Switching does not matter here.
I always find it easiest to explain Monty Hall (and related problems) by making an explicit list of all cases.
Let's say the prize is behind door 1.
Then these are the options:
Friend: 1, You: 2. Not switching wins.
Friend: 1, You: 3. Not switching wins.
Friend: 2, You: 1. No opportunity to switch.
Friend: 2, You: 3. Switching wins.
Friend: 3, You: 1. No opportunity to switch.
Friend: 3, You: 2. Switching wins.
And these are all the possible options and they all have the same probability to occur.
We could also list the cases for the prize behind door 2 and door 3, but the following will always be true:
You win immediately in 2 cases.
Your friend wins by not switching in 2 cases.
Your friend wins by switching in 2 cases.
So switching wins in 2/4 cases where your friend has the opportunity to switch at all. That makes it a 50/50. So it does not matter whether they switch or not.
To compare to actual Monty Hall:
Let's assume the prize is behind door 1 again.
You: 1. Monty shows 2 or 3 (doesn't matter). Not switching wins.
You: 2. Monty shows 3. Switching wins.
You: 3. Monty shows 2. Switching wins.
As you can see, switching wins in 2/3 cases here.
These two situations are not the same. Because Monty knows which door to open. So his action is what gives more information on the probabilities because it is an informed action. Your random guess at another door is not.
Edit:
I've written some python code to simulate this whole situation with a knowledgeable Monty (classic Monty Hall) and a randomly guessing Monty (OP's problem).
Here it is for you to run in your browser: https://www.online-python.com/FhbMJxe5Bi
The results are pretty clearly in support of my argument.
I don’t see any justification why in the first example, you count friend picking 1, and you picking 2 or 3 as two events, and in the Monty Hall case you count friend picking 1, and Monty revealing 2 or 3 as only one event. Both cases are identical from friend’s point of view. Friend has three options. If he picks 1 (correct door), it doesn’t matter when you pick, he loses by switching. If he picks 2 or 3, and your happen to randomly pick a “wrong” door, he wins by switching. So he wins in 2 out of 3 cases. Not 2 out of 4. Like someone already said, if it were 52 doors, and you were unlucky enough to pick 50 wrong ones, it seems quite obvious that it would be better to switch. Reminds me a bit about the early quantum physics discussions where there were fighting over whether an interference pattern would disappear only if a human was looking at it (being an observer), and then even if it had to be a physicist or if a cleaner (who doesn’t understand what he sees) would be enough.
Because each of these events have the same probability as all the others in their respective situation.
Let's break it down again, but with probabilities attached:
OP's situation. Prize behind door 1:
Friend has 1/3 chance to pick 1. OP then has 1/2 chance to pick 2. Total probability: 1/6. Win by staying. Friend has 1/3 chance to pick 1. OP has 1/2 chance to pick 3. Total probability to occur: 1/6. Win by staying.
Friend has 1/3 chance to pick 2. OP then has 1/2 chance to pick 1. Total probability: 1/6. Always lose because OP wins. We know that this could have happened but did not.
Friend has 1/3 chance to pick 2. OP then has 1/2 chance to pick 3. Total probability: 1/6. Win by switching.
Friend has 1/3 chance to pick 3. OP then has 1/2 chance to pick 1. Total probability: 1/6. Always lose because OP wins. We know that this could have happened but did not.
Friend has 1/3 chance to pick 3. OP then has 1/2 chance to pick 2. Total probability: 1/6. Win by switching.
In total, the friend wins by switching in 1/6 + 1/6 = 1/3 cases.
The friend wins by staying in 1/6 + 1/6 = 1/3 cases.
Now, since we know that certain outcomes could have happened but did not, they can be disregarded.
But we can clearly see that the remaining possible cases were equally likely. So whether the friend wins by switching or staying must also be equally likely.
Monty Hall. Prize behind door 1:
Player has 1/3 chance to pick 1. Monty then has 1/2 chance to reveal 2. Total probability: 1/6. Win by staying.
Player has 1/3 chance to pick 1. Monty then has 1/2 chance to reveal 3. Total probability: 1/6. Win by staying.
Player has 1/3 chance to pick 2. Monty then always reveals 3. Total probability: 1/3. Win by switching.
Player has 1/3 chance to pick 3. Monty then always reveals 2. Total probability: 1/3. Win by switching.
In total the player wins by switching in 1/3 + 1/3 = 2/3 cases.
The player wins by staying in only 1/6 + 1/6 = 1/3 cases.
The difference to OP's situation is that certain outcomes are impossible due to the a priori knowledge of Monty. There can not be a case where Monty reveals the door with the prize.
In OP's situation, that was a possibility but we now (after the fact) know it didn't come to pass.
These two are not the same.
Hmmm.... I think you're right but I'm not sure if I agree with the vehicle. My understanding of the Monty Hall problem is that initially you have a 1/3 (hereafter referred to as 33% because it's easier for me), or a 2/3rds chance of getting it wrong (66%, same reasons). The reason you switch is because his revealing the door does not change your percentage of 'getting it wrong,' meaning the other door still has a 66% chance of being correct.
I was thinking on it last night and I realized that because the two of us were picking the doors with interests aligned, we actually start in the 66% chance group. I'm wondering if the right answer is actually he should not switch, and I've messed up the problem by telling people to assume I got it wrong. (In the actual scenario, I just happened to guess right but I didn't think the problem still existed if I got it right, and my friends and I were arguing over this for a literal hour afterwards.)
Now it could be that we are saying the same thing and I'm just not connected enough with the math to understand that. I was always better at English. Does this probability split to 3rds because my answer is forced incorrect, or is my entire premise a false misunderstanding?
People in this thread seem mostly confused by / disagree with the interpretation of your statement
Because there is, to my understanding, a subtle difference between this and the Monty Hall problem.
Monty knows which door holds the prize. Therefore, he will never reveal the door with a prize. That is part of the rules.
You do not know which door holds the prize. Therefore, you could reveal the door with the prize. We're just not looking at the cases where you did (because that robs your friend of his choice).
Is that correct?
The only reason that is so in Monty Hall is because Monty's actions depend on his knowledge of the situation. He will ensure that you can't learn that you got it wrong (which he would if he opened the prize door). That is why your probability to get it wrong remains the same.
I'm afraid we're not saying the same thing.
There is a difference in probabilities resulting in whether you intentionally answered incorrectly or just happened to answer incorrectly by chance.
The former is the Monty Hall Problem, the latter is not.
In Monty Hall, there is just one decision point that matters for the probability: Which door the player picks. If the player picked wrong, Monty must reveal the other wrong door, so that's no decision. If the player picked right then it doesn't matter which door Monty reveals because they're identical. So that's no meaningful decision either.
In your case, you do have a second decision to make. And it matters. Because if your friend picked wrong, you could still find the prize.
Sure, you happened to pick wrong. But you did have that option. Monty never does.
That's why Monty's player has a 66.66...% chance to win when switching. Your friend only has a 50% chance either way.
It’s a well reasoned answer, and it does give me pause. It’s good to remember that back in the day there was a big discussion about this topic - with mathematicians on both sides.
My argument against your case is that you remove the cases that we know didn’t happen too late, only at the end when counting the sum of portabilities. I think they schuld be eliminated immediately.
So, friend picks 2 (p = 1/3), I always pick 3 (because we know I didn’t pick 1), so total probability for this branch stays 1/3, not 1/6.
On an intuition level I would find it very weird that from the friend’s perspective it makes a difference whether the wrong door was revealed consciously or not. But that’s just a feeling, that’s not my argument. Probabilities can be counter intuitive, I get that.
But that's the thing, you don't always pick 3. We just ignore the times that you did.
The rules of Monty Hall state that he can not show the prize. You can. You just randomly happened to not do that.
In Monty Hall, this was never an option to begin with. In our case it was, it just didn't happen.
That's why it's disregarded from the start in Monty Hall but is only discarded at the end in this case.
It’s much closer to “if you pick 2 or 3, Monty shows you 3 or 2 both times”.
Because you pick randomly, you pick door 1 the same number of times as door 2 and 3. Using the method of dividing the probability space into potential outcomes and then counting the outcomes, you have to split evenly at each decision point: 1/3 of the time picking each door, and then if you pick door 1 Monty determining which door to show counting as another decision and giving two outcomes to picking door 1 initially means that you have to give two outcomes each to picking door 2 and door 3, despite those outcomes having no other decision points.
Because it doesn’t matter which door Monty “picks”. The Monty Hall problem only cares about which door YOU pick, because once you pick a door, Monty is a machine. His action is already determined. 1/3 of the time, he’ll pick a random door and you lose by switching, and 2/3 of the time, he’ll pick the other wrong door and you’ll win by switching. Again, change it to 100 doors and it’s easier to see.
There are 6 possibilities of which door you pick vs which one Monty picks.
You pick 1, he picks either 2 or 3. Both of these possibilities have a 1/3 x 1/2 = 1/6 chance.
You pick 2 or 3 (1/3 chance). Monty is 100% going to pick the other wrong door, so these have a 1/3 chance of happening.
You pick 2 or 3 (1/3 chance) and Monty picking 1 will NEVER happen, so that’s your 100% sum there.
This is the answer to your first question- why you picking 1 can be collapsed into a single event.
This is not correct. The whole crux of the Monte Hall problem is that when you first pick, you’re picking randomly one of three potential correct options. So 33% chance of being correct.
If one incorrect solution is taken away and you do not switch, it’s still the original 33% chance of being correct (and importantly, 66% chance of being incorrect). So switching guesses to the other option changes your pick to the 66% chance of picking the right one.
He says he guesses a door and it's incorrect, if that's always the case, it is the Monty hall problem isn't it? Cause he's essentially playing the role of Monty
No, because Monty's knowledge of which doors are correct doesn't just eliminate some options. It collapses them onto ones where switching is correct. Thus adding their probabilities together.
That's what makes switching more probable to be the correct choice.
Just removing those options does not.
See the first set of options I've listed. Out of the 6, we know that we are not in #3 or #5. But that knowledge does not make #4 or #6 any more likely. We just end up with 4 equally likely options, two of which reward switching and two of which do not.
I assume the guess is revealed to the friend and then he has the opportunity to switch at least by the way which op's question is phrased, that would remove one of the options and thus group the probabilities together.
Yes, it removes an option. No, it does not group the probabilities together. Why would it?
I've written some python code to simulate this whole situation with a knowledgeable Monty (classic Monty Hall) and a randomly guessing Monty (OP's problem).
Here it is for you to run in your browser: https://www.online-python.com/FhbMJxe5Bi
The results are pretty clearly in support of my argument.
You can extend and generalise the Monty Hall problem by for example adding more doors and more guessed.
The important part is that someone plays the role of Montt and does something that adds information in the end.
Just randomly revealing what is behind one or more doors won't work though. The Monty has to have knowledge and use that knowledge in his action and thus reveal it to you by for example revealing what is behind one or more doors he knows doesn't contain the prize.
The whole issue becomes more obvious with larger numbers.
An extreme example:
There are 100 doors, 1 contains a prize and the rest nothing. You pick one at random. Monty then reveals what is behind 98 of the doors that he knows not to contain the prize.
You then get the choice of either staying with your original guess or picking the one other Monty hasn't revealed yet.
This makes it a rather obvious choice: staying with your original guess which had 1% of being right or going with the remaining door which will contain the prize in the 99% of cases where you didn't pick the right one originally.
All of this is true but it doesn't answer OP's question. Monty knows which door has the prize so he will always intentionally open the one(s) without prizes. That is not true in OP's question.
But the problem with 100 doors is: if friend picked wrong, it’s very unlikely that you picked wrong 98 times with the right door still an option. While if friend picked right, of course you did. So you randomly picking wrong 98 times is a strong clue that friend picked right in the first place. While Monty opening 98 doors is not new information about the initial pick
No, it is not like the Monty Hall problem. For your game, there is a 1/3 chance that it's behind the door your friend chose, a 1/3 chance that it's behind the door you chose, and a 1/3 chance it's behind the last door. Switching does not help.
Compare that to the Monty Hall problem. There is a 1/3 chance it's behind the door you chose, a 0/3 chance it's behind the door Monty chose, and a 2/3 chance it's behind the remaining door. The difference is Monty knows the answer while making his choice, but you don't
Yep, Monty would be proud of you. Getting your pick wrong adds information making the unchosen door more desirable.
Exactly the same as Monty showing the goat before asking if you want to change your door.
No, it doesn't. The only information gained is that it's not the door OP picked. It does not add additional information on the unpicked doors because the person who picked the door (OP) did not have additional information. Monty would have that information, though.
Edit:
I appreciate the downvotes, but would like an actual counterargument. Or, even better, math showing why I'm wrong. See my top-level comment for my full breakdown.
Edit2:
I've written some python code to simulate this whole situation with a knowledgeable Monty (classic Monty Hall) and a randomly guessing Monty (OP's problem).
Here it is for you to run in your browser: https://www.online-python.com/FhbMJxe5Bi
The results are pretty clearly in support of my argument.
This is factually false and you can practice it with it someone for 5 minutes to make it obvious why.
Try framing the original Monty problem this way: what door do you have to pick in order to follow the prescribed switching protocol and fail to win? With the original question, you can only lose if you start with the winning door. With the question in the OP, this is not the case.
Guess again.
Comment deleted by user
It's actually completely different depending on which of these situations you're in. If the host is just guessing, the Monty Hall result doesn't apply and switching is not advantageous.
Comment deleted by user
Yes, it actually is important whether the wrong door was revealed intentionally.
The first door chosen has a 1/3 chance of being correct. If the host then reveals a door knowing that it will be wrong, whichever door he chooses has a 0 chance of being correct, leaving 2/3 unaccounted for, which must lie with the final door. But if the host is just guessing, there's a 1/3 chance that the door he reveals is correct, leaving only 1/3 left for the final door.
There are a number of other ways to show this (you can enumerate the possibilities, or write a simulation, for example) but, if done correctly, they all result in the same answer. Switching is of no benefit if the "Monty" figure doesn't intentionally pick an incorrect door to show you.
Doesn’t matter.
After you pick a card, flipping 50/51 wrong cards by accident has a 1/51 chance of happening. Flipping 50/51, if you know where the right card is, has a 100% chance. In other words, your friend’s pick is independent of where the right card is. Monty’s “pick” (because most of the time, it isn’t actually a pick) is not independent of where the right card is. That’s the difference.