Underrated KZread channel.

Would love more similar content, loved the explanation.

It's conditional probability. Every door has 1/3 chance of the prize. One door/goat is eliminated by the host, who is not going to reveal the car until the final selection. If you unknowingly picked the car ( which occurs 1/3 of the time ) and switch, you lose ( as you switch to a goat ). However if you unknowingly picked a goat (which occurs 2/3 of the time) and one door/goat is eliminated, then when you switch you win ( as you switch to the car ). Thus 2/3 of the time switching wins and 1/13 of the time switching loses.

@JoeyBlogs007

6 ай бұрын

If you pick randomly all along it's 50/50. If you don't switch it's 33.33% and if you switch it's 66.66%

This is way better an explaination than a "debunk" video i just saw

What a great explanation :)

Finally got it. Thanks!

The Monty Hall problem is presented as a single instance of someone on a game show. We are told what has already happened and we are asked a question where we calculate the probabilities to find the answer. No rules of this game are given and we don't need any for the single game we are looking at because the host has already finished his presentation. According to the question: After the contestant picks a door, the host knowingly (he knows what's behind the doors) opens _another door_ to reveal a goat. The rules you stated were satisfied (though not specifically mentioned). In that scenario, the probabilities can only be 1/3 to stay and 2/3 to switch. A statement of these rules only becomes necessary when we want to explain the probabilities or to look at additional games... these games must follow the same rules that the original scenario did in order to have the same probabilities. Most of the people that disagree with the solution present a different scenario than what the question describes (the host eliminates a door before the contestant picks or opens the contestant's door, etc.) They sometimes claim the rules must be known by the contestant before the start of the game but that is not true either... they only need to be able to tell that the host revealing the goat from one of his doors was deliberate. Anything else is a different game. Nice video :)

@ProgThoughts

Жыл бұрын

Yes, if specifically stated that he would pick "another door to reveal a goat" then those conditions are implicitly satisfied. However, I find it rather common explanations overlook these conditions. This happens to the point where the explanation given can be wrong as it assumes 2/3 chance must be true as long as the host picks a different incorrect door without taking into account the rules.

@freddieorrell

Жыл бұрын

@Hank254 Quite so, we don't need the rules given here to solve the MHP. These rules are only needed to code the simulation. The program has to be told what scenario to analyze, and be prevented from analyzing non-compliant scenarios. Perhaps this has led to the correct answer being demonstrated by the wrong method. And yes, the one thing not emphasized yet crucial, is that the host knows.

@ProgThoughts

Жыл бұрын

@@freddieorrell The code is equivalent to the expression "pick another door to reveal a goat". The expression says the host will not pick the door you chose, and will only pick an incorrect door. This is simply laid out logically in code to demonstrate the reality. I'd recommend learning proper logic and perhaps even some programming.

@freddieorrell

Жыл бұрын

@@ProgThoughts "The code is equivalent to ... etc" Yes, that's what I said. The code needs a mechanical representation of the MHP as it can't read. But the contestant solver can already see that the host has not revealed their door or the prize. They don't need to be told this is not the scenario.

@ProgThoughts

Жыл бұрын

​@@freddieorrell "But the contestant solver can already see that the host has not revealed their door or the prize. They don't need to be told this is not the scenario." This is incorrect. As stated in the video, it's not enough that the host does not reveal their door or the prize door. If the host "happens" to reveal a door that was not the prize or the door they chose, the odds are not 2/3 for the remaining door. This is only true if the host is explicitly following those rules. Put simply, "the host revealed a different door containing a goat" as an action is not the same as "the host must reveal a different door containing a goat". One says that's what the host did, the other is what the host must do. If the host, by pure chance or on a whim, chooses a door. It doesn't matter if that door met the criteria, because the host was not following the rules. Hence, his decision had nothing to do with your decision, so there's no information to obtain from his choice.

Nice.. :) Thanks man ..!

Finally somebody explained to me why it isn't 1/2 & 1/2

@harryrussell154

Жыл бұрын

It is 1/2 &1/2. They say that when you pick door #1, because there are two other doors, you picked the one door that could have a prize of the three doors that could have the prize. So your #1 pick has a 1/3 chance of having the prize, and the other two doors have a 2/3 chance so they have a greater chance TOGETHER of having the prize. They are saying this means you have a greater chance of picking the door with less chance so you door was without the prize, for some reason. When the second door was eliminated they say this means that because you were wrong the first time, door #3 must have the prize, for some reason. They are wrong. By eliminating the second door you also eliminate the 2/3 chance of having picked the wrong door to begin with. Right? So the two remaining doors have an equal chance of having the prize. Right? Right. You are right, they are wrong.

@BritCap1

Жыл бұрын

​@@harryrussell154 Yeah. That's exactly what I thought when I first learned about the Monty Hall problem. The difference is that in this video, he added another rule: they can't open the door you chose. So, they reveal the door that is wrong, and that is not your door. In the beginning, you have a 2/3 chance to pick the wrong door. If you picked a wrong door, then the left one is correct, because another wrong door eliminated. Therefore, there's a 2/3 chance, that the prize is behind the other door. It isn't about the eliminated door, as we both first though. It's about that they can't reveal your door. To summarize, if you chose the correct door in the beginning (1/3 chance), then after they revealed one of the left incorrect doors, the door that's left is also incorrect. If you chose the wrong door (2/3 chance) AND they can't reveal your door, then the other one must be the one with the prize, since he can't reveal your incorrect door and the door with the prize.

@harryrussell154

Жыл бұрын

@@BritCap1 This explanation is only true if it is ok to automatically assume your pick is the wrong door. Once it boils down to only two doors how can one still assume you have still picked the wrong door? An example of how academian overfactoring can lead out of reality. Like the teaching that it is the sun's gravitational pull that is responsible for the earth's orbit around it. Every body in space has a gravitational pull that PULLS TO THEIR CENTERS. If the sun had the earth in its grasp it would have pulled the earth into its center long ago. Another theory from the scientific overthinkers.

@BritCap1

Жыл бұрын

​@@harryrussell154 we don't assume that you picked the wrong door, but the chances that you did are higher. And if you chose the door with the prize you definitely shouldn't switch. But since you don't know what's behind the door you chose, we need to come up with a solution that would work better than random. Also, it seems like you don't understand how orbits work. It's okay. Sun does pull Earth towards it. The Earth doesn't fall on it because it's moving too fast to the direction almost perpendicular to the gravitational force. The way Sun affects Earth's movement is that it change its direction and velocity. Other objects do affect Earth's orbit too, but it's nothing, compared to Sun. By the way, just so you know: I don't look down on you or anything. I like curious people and I'm curious myself also. It looks like you were studying, and have gotten some questions that nobody could answer. You maybe you just were confused about something. It's okay to. I ask a lot of question. (that's why I watched this video: because "Why isn't it 50/50??") I wish you to stay curious

@harryrussell154

Жыл бұрын

@@BritCap1 "Sun does pull Earth towards it. The Earth doesn't fall on it because it's moving too fast to the direction almost perpendicular to the gravitational force." Perpendicular to the gravitational force after being turned toward the sun at the end of its outgoing orbital path? If you need these type of fallacious explanations to support your position you have no viable position. Why does the earth not increase in speed on its incoming orbital path, and decrease in speed on its outgoing orbital path if it is being pulled by the sun? "It looks like you were studying, and have gotten some questions that nobody could answer" Yes, like maybe the logical reason why with only two places left for the prize to be, neither door could have a 2/3 chance? Because there no longer is a 3 in the equation? Denying reality does not make it go away.

Spoiler: The prize is behind Door 13.

@jakejones5736

4 ай бұрын

Now THAT sounds like a lucky number....

Two thoughts: a) extremely few people will understand this b) I see you making only one mistake. The Monty Hall problem as usually presented doesn't declare a rule at all and doesn't exclude the host from opening your door. So the real MH problem is 50/50, as in your second scenario after denying him to open the prize door. Other than that, kudos for understanding why people disagree whether it's 2/3 or 1/2.

@ProgThoughts

Жыл бұрын

Thanks! I actually wanted to make this video after arguing with a mathematician on the subject so I definitely made this for people who are not new to the Monty Hall Problem. Since the Monty Hall Problem says the outcome should be 2/3, the rules given must apply - though usually never stated. Numberphile and Vsauce are the only channels that mention this rule, but they gloss over it and don't show the importance of it.

@signeCS

Жыл бұрын

The monty hall problem is not 50 50, the rules are implied

@insignificantfool8592

Жыл бұрын

@@signeCS yes, some people really do see the situation described in the MH problem as a rule. Mistakenly, I would say, because it's not made clear that it's a rule at all. Some people, when describing the problem, even add things like "suddenly" or "the host decides", or use two different losing prizes. All these things show that they didn't interpret the MH problem as a rule.

@klaus7443

Жыл бұрын

"So the real MH problem is 50/50, as in your second scenario after denying him to open the prize door." No it isn't.

@signeCS

Жыл бұрын

@@klaus7443 he's confused, don't mind him

At 1:23, there is no need in the MHP for a rule preventing the host from revealing the prize. This is superfluous, since in the MHP he has not. Also, at 1:42 there is no need for a rule preventing the host from revealing the chosen door. Again, in the MHP he has not. However, these rules ARE needed in coding your simulation at 5:11. Here, they are necessary to tell the program which scenario to analyze and prevent it analyzing the wrong scenarios. There is no need for these rules in the contestant's calculation of the probability. A prerequisite for this calculation is that the host knows what is behind the doors, which you imply throughout but do not stipulate.

@ProgThoughts

Жыл бұрын

This makes no sense, as the host can know which doors contain what and still decide to open your door or the prize door. If there's nothing preventing the host from doing so, your odds are not 2/3. In fact, if the host opens the door you chose, you don't even get to decide whether you want to switch doors or not.

@freddieorrell

Жыл бұрын

@@ProgThoughts If the host does open your door or the prize door, your odds of having the prize are indeed not 2/3. They are 3/3 or 0/3. However, even if the host is not prevented from opening your door or the prize door, yet nevertheless does not do so and reveals a non-chosen goat, your odds are 2/3. Those rules are not necessary. If the host does not know, or the contestant does not know the host knows, it's 1/2. It must be explicit that the host knows.

@ProgThoughts

Жыл бұрын

​@@freddieorrell I replied to this point in your other comment and I'll repost it here with greater depth: This is incorrect. As stated in the video, it's not enough that the host does not reveal their door or the prize door. If the host "happens" to reveal a door that was not the prize or the door they chose, the odds are not 2/3 for the remaining door - it's 50/50. In fact, this is the first scenario I show in the video. The remaining door is only 2/3 odds if the host is explicitly following those rules. Put simply, "the host revealed a different door containing a goat" as an action is not the same as "the host must reveal a different door containing a goat". One says that's what the host did, the other is what the host must do. If the host, by pure chance or on a whim, chooses a door. It doesn't matter if that door met the criteria, because the host was not following the rules. Hence, his decision had nothing to do with your decision, so there's no information to obtain from his choice. Your very statement is one I proved wrong to a mathematician I was debating with and I showed in the last scenario with the code. The last scenario for the code is where I remove the rules the host has to obey, then I only check scenarios in which the host, by pure chance, did not reveal the prize door or your chosen door. This led to a 50/50 chance between your door and the remaining door. Again, this is because the host's decision does not give any information unless they are specifically following the rules. This is also related in-depth to what probability REALLY is, but beyond the scope of this comment.

@freddieorrell

Жыл бұрын

​@@ProgThoughts If the contestant knows the host is acting by chance, you're right it's 50/50. However the contestant does not know that. They do know the host knows, so must be acting deliberately. The probability that the host is acting by chance deliberately (ie choosing to act randomly by eg throwing a dice) is 1/2, according to the (lack of) information available to the contestant. The other 1/2 is that the host is deliberately not acting by chance (principle of indifference). The solution is still that it is to your advantage to switch. No rules needed.

@ProgThoughts

Жыл бұрын

​@@freddieorrell The video is about why switching is a 2/3 chance, not whether or not you should spam switching doors. If we took your pragmatic approach, we could also say that you shouldn't switch because you may have picked the prize and they want to cajole you into switching since they don't want to hand over an expensive prize. However, this approach ignores that this is simply a placeholder situation for a statistical problem. You would not get any marks on a math problem if you were to give your reasoning.

This explanation is not accurate, stick or switch, because the context changed when you paired doors #2 and #3 together. It went from a 'door' probability to a 'sides' probability. One side now has only one door-Side A, and the other side, Side B, now has two doors. Since there is only one prize, both doors in Side B cannot each have the prize, so since there is only one prize only one door on Side B could have the prize. This means that, although there are two doors seen, one of them MUST have a goat and not be in the equation, so there is in reality only one door on Side B that can have the prize. Now since each side only has one door in the game, each SIDE has a 'One in Two Chance' of having the prize. The only way Side B could have a 'Two in Three Chance' of having the prize is if there were two prizes, and both doors has them. There aren't two prizes, only one. Two doors, or Two Million doors, only one could have the prize, and the rest goats. Also, when you omit the second door, there remains only two doors left, each having the possibility of having the prize. So again it remains each SEPARATE entity door has an equal chance for the prize, 'One of Two Chances.'

@Araqius

Жыл бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. Basic math/logic kids understand, idiot among idiots doesn't.

@ProgThoughts

Жыл бұрын

Let me bend the laws of physics real quick so you can be correct :)

@harryrussell154

Жыл бұрын

@@ProgThoughts The mistake that is being made is in grouping doors 2 and 3 together saying they represent a two out of three chance of having the prize. With two doors and only one prize there has to be a goat behind one of them, so how do they have a two in three chance if we know one has to have a goat? And when the one with the goat is removed from the equation, it goes back to having only two doors left, each with a one in two chance of having the prize. With only two doors there is no longer a mention of any number 3.

@ProgThoughts

Жыл бұрын

@@harryrussell154 "Grouping" the doors in your head is something you can do to help understand it, but it's not mathematically relevant. I'll try to explain it even simpler: When the host reveals a door, his decision was affected by your decision. The fact that the host's decision to reveal a door was affected by your decision, along with the fact that the door the host reveals is different than your decision, mathematically links the door you chose and the opened door. If you were to simply ignore this new information and decide it's a 50/50 chance, you'll be doing yourself a disservice. The program I wrote shows that my explanation is true mathematically.

@klaus7443

Жыл бұрын

@@harryrussell154 You're simply trolling because you don't have the math skills to solve the problem. If the host was opening a door that has a 1/3 chance of revealing the car but it revealed a goat instead then the 1/3 probability is equally shared among the two remaining doors. You are taking a conditional probability problem and ending up with the result as if it were an unconditional one.

The situation is that "decision without results" and "choice with actual benefits" are confused and regarded as the same meaning. It has been clearly decided. If there are really two choices, shouldn't we get two things? In fact, it is The rules change from one of three choices to one of two choices, and mislead you into thinking that you have already chosen once, but in fact you can't get any of the so-called choices. When you finally make the real choice, in order to prove how smart and knowledgeable you are, of course there are only two. The choice should also be regarded as 2/3. What a verification of the Goebbels effect. If scholars and education circles have not corrected the extension of the Monty Hall problem and still think that 2/3 is correct, what is the point of educational scholarship? ?It was destroyed.

@ProgThoughts

5 ай бұрын

Your decision affected the outcome of the host's decision. Giving you more information to use in your second choice. You can choose to disregard this new information, but that only hurts you. Moreover, Goebbel's effect wouldn't explain why the program I coded (which simulates this situation) finds that switching doors works 2/3 of the time.

@dawyer

5 ай бұрын

@@ProgThoughts I'm not good at using english so translated by google is in used. First, It's unfair for you to shift the discussion issue to me. This is not my decision in this problem to discuss. Hurts?? if you say so, The same goes for misunderstanding information. No matter what the player decides, the player does not get anything from this decision. If there is no result, the probability is not calculated as a choice. On the other hand, no matter what the player decides, the host will still open the door with sheep, which means that the player's decision It does not affect the moderator's change of the new rule from one of three to one of two.

@ProgThoughts

5 ай бұрын

@@dawyer You said, "No matter what the player decides, the player does not get anything from this decision" This assumption is wrong, hence your argument built off this assumption is wrong. The player's decision affects the host's decision 2/3 of the time.

@dawyer

5 ай бұрын

@@ProgThoughts To change the host's terminology, there are three pairs of doors, two sheep and one car. If you choose the car, you will win. I will give you a prediction before choosing. After you tell me, no matter what, I will take away the pair of doors with the sheep, but the remaining I will give you a choice between two pairs of one sheep and one car. Can you tell me whether the previous prediction decision has any effect? ​​Does it make it easier for you to choose the door with the car?

@ProgThoughts

5 ай бұрын

@@dawyer The host MUST remove a door that does NOT contain the prize behind it AND NOT the door you chose. 2/3 times, you will choose a door that does NOT contain the prize behind it. Therefore, 2/3 times, the door the host opens will be a FORCED decision as it will be the ONLY door he could have opened. Therefore, your first choice has affected the host's decision, which, therefore, changes your odds.

The real crux of the problem is that you make a choice of three options when in fact there are only two because no matter what you pick...one of the choices is going to be removed leaving only two. To solve this problem it's best to not think of the doors as 1,2,3 ...but of 'Middle' and 'Ends'...which makes it a binary choice and if you always choose the 'Middle'...the host will always be forced to remove one of the 'Ends' for you leaving only the two that you chose from. Meaning you made a choice from two which is a 50% chance rather than the 33% the original way the game is structured gives you and is why you are at a disadvantage. Increasing your initial odds more than offsets the supposed advantage of always switching and running simulations choosing the middle door only will show a 50/50 split as expected. Make a coin toss and good luck!!

@ProgThoughts

Ай бұрын

That makes absolutely no sense

@stephenolan5539

Ай бұрын

Nope having two options is NOT 50/50. Look at it this way. If you do it right you can choose two doors. What you do is pick two doors. You tell the host that you choose the other door. The host then eliminates one of yhe doors that you want. Then you tell him that you switch. As for thinking that two choices is 50/50. Get two identical boxes an object, and two dice. Have someone roll the dice. If it is snake eyes ( two ones) have put the object in box one. Do you really think that you have a 50% chance of guessing right?

You're full of crap. Since the original door is 1/3 it is ALWAYS 1/3. Thus switching always means a 2/3 chance of winning! It can NEVER be 50-50 as you suggest.

@ProgThoughts

4 ай бұрын

I did not say that it's a 50/50 chance

FAILURE. The ONLY thing that these 2 rules determine is that the odds for round 1 are meaningless because the round will NEVER be resolved and it will be forced into round 2, where the odds remain 50/50. Nothing in round 1 changes this. Please learn a _little_ bit about probability before dumping garbage like this on KZread. I can prove, beyond any doubt (to anyone who is intellectually honest) how this video is complete and utter bullshit. What is presented here is called, in mathematical and scientific nomenclature, a "model". There are 2 ways to prove a model: direct testing and observation and applying it to similar yet slightly different scenarios and having it be consistent. Since testing isn't possible in this venue, I will prove this model is wrong by the second method. Core Scenario: A contestant is presented with 3 doors. They get to choose 1. For simplicity, we'll say they pick door 1. Each round, the contestant get to switch to a different door or stay with the door they have. There is only 1 door that has a car, but 1 door _always_ has a car. So, at this point, each door has a 1:3 chance of winning. If the round is resolved, the contestant has 1:3 chance of winning because they only got to choose 1 out of the 3 doors. If anyone disagrees with this, please comment below. The host opens one of the doors the contestant didn't choose (door 3). If the car is behind the door the host opened, the contestant loses. If the host opens a door with a goat, then we move to round 2. The contestant is allowed to switch to door 2 or stay with door 1. I think we are all in agreement up to this point. According to the model presented here, the odds of the eliminated door gets shifted over to door 2. This is based solely on the fact that the contestant picked a door in round 1 (an action that played absolutely no role in establishing the odds in the first place). According to the standard model, the odds of the eliminated door get evenly distributed between the existing doors. This is based on the number of doors remaining. In order for a given model to be correct, the outcome must equal 100%. There is a 100% chance that the contestant will win or lose. All odds needs must equal 100%. So, in round 2, according to the presented model, there is a 1:3 chance you win if you stay with door 1 and a 2:3 chance you win if you move to door 2. This equals a 3:3 chance, which is equivalent to 100%. We're good so far. But what about the standard model? In round 2, with an eliminated door, since there are 2 doors, the odds of either door having the car is 1:2 (also described as 50%). Since each door has a 1:2 chance, that adds up to 2:2 or 100%. So, both models are seemingly viable at this point. So, let's apply these models to 2 slightly different scenarios: Alternate Scenario 1: In round 2, the round isn't resolved. The host opens door #2, which has a goat. What are the odds of the door the contestant has picked having the car? Let's go back to our models: In the presented model, remember that the ONLY criteria of the odds for being correct are the doors that the contestant DIDN'T pick. So, according to this model, if the contestant stayed with door #1, there is a 1:3 chance of winning. But the 2 other doors, having been opened and showing goats, is clearly 0:3. What does that add up to? 1:3 + 0:3 = 1:3. So, if the car isn't behind door 2 or 3 and is only behind door 1 1/3 of the time, where is the car the other 2/3 of the time? The total for winning and losing here is 33%. So this model fails. What about the standard model? Once door 2 is eliminated, the odds of each remaining door get recalculated and distributed evenly. So, if there is only 1 door, that means that the chance of that door being correct is 100%. So the standard model is superior and proven correct so far. But let's move to alternate scenario 2 and see what happens: Let's play the exact same game except with 2 contestants: In the 1st round, contestant A chooses door 1. Contestant B chooses door 2. The host opens door 3 with a goat). Where do those odds get shifted? In the presented model, the odds can't be shifted because there is no door that hasn't been picked, so the odds are: Door 1 -- 1:3 Door 2 -- 1:3 Door 3 -- 0:3 (shown to be a goat) This adds up to 2:3 or 66.6%. So this model fails because it doesn't add up to 100%. 1/3 of the time contestant 1 wins, 1/3 of the time contestant 2 wins, and 1/3 of the time they both lose. If they both lose, where is the car? *_OR_* The contestants both switch to the other door. According to the model, "the other door" gets the additional odds. Okay, so the odds end up like this: Door 1 -- 2:3 (because contestant 2 didn't pick it, so it gets the odds from door 3) Door 2 -- 2:3 (because contestant 1 didn't pick it, so it gets the odds from door 3) Door 3 -- 0:3 (exposed and eliminated as a possibility) So, the odds (2:3 + 2:3 + 0:3) equals 4:3, or 133%. Once again, a failure. 1/3 of the time contestant 1 wins, 1/3 of the time contestant 2 wins, and 1/3 of the time they both win. We've established that there is only 1 car behind 1 door, so it is impossible for both contestants to win. So, what does the standard model say? The odds get evenly distributed to the remaining doors. How do those odds look? Door 1 -- 1:2 (one out of 2 remaining doors) Door 2 -- 1:2 (one out of 2 remaining doors) Door 3 -- (not in the calculation because it was eliminated by being exposed as a losing door) 1:2 + 1:2 = 2:2 or 100%. Once again, the standard model is victorious. So, in both of these scenarios, the presented model fails miserably and the standard model stands. Can anyone come up with an alternate scenario where the presented model works, but the standard model fails? I doubt it, but please present it if you can think of one.

@ProgThoughts

Жыл бұрын

You said testing wasn't possible. However, the code I wrote is the equivalent of a test. But, let's put that aside. You said, "The host opens one of the doors the contestant didn't choose (door 3). If the car is behind the door the host opened, the contestant loses. If the host opens a door with a goat, then we move to round 2." This is not possible as it goes against one of the rules which I explicitly stated in the video, "the host cannot open the door containing the prize, and cannot open the door that you've chosen". So I'm going to assume that the host always picks a door that is not the prize and is not the originally chosen door - so the game ALWAYS goes into "round 2". "According to the model presented here, the odds of the eliminated door gets shifted over to door 2. This is based solely on the fact that the contestant picked a door in round 1 (an action that played absolutely no role in establishing the odds in the first place)." No, not really. The odds getting "shifted over" is a way people like to look at it, but not mathematically accurate and I did not state it anywhere in this video. What really happens is that the odds that the remaining door is correct is now the same as the odds of your initial decision being wrong (totals to 100%). Moreover, you yourself stated the rule that the host can't open the door you selected, therefore the initial selection DOES affect what the host can do. To say that this would in no way affect the odds shows you haven't thought about it at all. Once the host eliminates a door, your door is still only a 1/3 chance. I'm disappointed, as it seems to me that YOU are being intellectually dishonest. Let's look at your scenario 2 where you ask, "Where do those odds get shifted?" First of all, if we're following the same rules as before, it's clear why you said, "If the car is behind the door the host opened, the contestant loses". Understand that if the host is free to open the door with the prize, then the probability is completely different than what I stated in the video. I explicitly stated the host CANNOT do that. Second, if we assume we're following the proper rules, we need to look at all possible outcomes to determine the probability. If the host cannot open the prize door and cannot open the two doors the contestants chose, that means that 2/3 times, the host is unable to reveal a door at all. This means we cannot "always" go to round 2. This completely invalidates the probability analysis. The situation is now so different that if you were to write a truth table for two contestants would look nothing like the truth table for one. All you did was twist the situation then proceeded to say that since it doesn't work for two contestants, it can't work for one. This is like saying that a winning strategy in chess is actually not a winning strategy, because what if I suddenly turn it into a 3-player chess game and completely change the board? BUT - LET'S DO IT! Let's change your scenario a tiny bit to show that this still works with your 2 contestants. Instead of 3 doors, we need more! In fact, this scenario WILL work with 4 or more doors. Each contestant picks a door, the host reveals the door (or doors) they didn't pick and does not have the prize, and, then, the door with the highest probability of winning will be the door NEITHER contestant selected - the final remaining door. The probability is over at or above 50% for the remaining door, depending on whether the contestants can choose the same door or not. But whether or not they're allowed to chose the same door, the probability will continually increase on the remaining door as the total number of doors goes up. At 100 doors, the probability by switching is roughly 98%, whether or not they're allowed to make the same choice (just because it's so unlikely that they will). I would encourage you to watch the entire video. At the end, I show a program I coded using the logic I explained, with the outcome clearly showing that switching gives you a high probability of winning.

@ProgThoughts

Жыл бұрын

@@dienekes4364 Please, put all your thoughts into one comment or this discussion will get messy. You don't need to trust my testing, you can look at the code and evaluate it yourself if you have any coding knowledge. You'll see it's a perfectly valid representation. "If the model fails, the model fails" If I take the model which tells us how planets form and try to use it to explain why cancer is more prevalent in old age, it will fail. Now I can go to the scientific community and tell them their model failed? I'm not interested in having a "model", I'm interested in the probability of one specific situation - which I've proven mathematically. "Go back and re-read it with the intent of UNDERSTANDING what I'm saying" I've read it over several times. Your argument is that if you have two contestants, the probabilities of each person's door would be doubled, hence the probabilities would not add up to 100%. I showed you it's a stupid argument as that's not how probability would work for those scenarios simply because there are not enough doors to always allow a "round 2". And you cannot address this by saying there will always be a round 2 by eliminating the correct door if they both choose wrong - that would go against the rules. If I misunderstood, restate your position. While I'm open minded, the fact is that my logic and coded program are logically and mathematically sound, you *are* wrong. "Is it possible that it's because you have an emotional interest in ignoring that reality?" Ah yes, I'm very emotionally invested in whether or not opening a door will give me a cash prize. I'm expected on a T.V. show later today.

@ProgThoughts

Жыл бұрын

​@@dienekes4364 Is this what insanity looks like? If you can gather your main points and put them into one comment, I'll reply again. Shouldn't be too difficult for a programmer of your caliber.

@dienekes4364

Жыл бұрын

@@ProgThoughts Fine. _"You said testing wasn't possible."_ -- It's actually 2 things: I don't trust the testing, but it is irrelevant. If the model fails, the model fails. Period. And this model failed. TWICE. _"This is not possible as it goes against one of the rules"_ -- That is correct, which is EXACTLY what I pointed out. So thanks for confirming exactly what I said. _"Once the host eliminates a door, your door is still only a 1/3 chance."_ -- Once again, exactly what I said this model predicts. So, once again, thanks for agreeing with me. _"I'm disappointed"_ -- You should be, since you clearly didn't read what I wrote. Go back and re-read it with the intent of UNDERSTANDING what I'm saying, not just looking for something to strawman my position with. _"If the host cannot open the prize door and cannot open the two doors the contestants chose"_ -- I SPECIFICALLY addressed this. Go back and re-read my post. _"I would encourage you to watch the entire video."_ -- I've watched several of these videos and they all show the exact same flaw as yours: your assumption that the odds are based on the contestant picking a door instead of what they are REALLY based on: how many doors are available in any given round. I'm curious about something. Why is it that you completely ignored the 2 scenarios that I showed that proved this model wrong? Is it possible that it's because you have an emotional interest in ignoring that reality? _"No, not really. The odds getting "shifted over" is a way people like to look at it, but not mathematically accurate and I did not state it anywhere in this video."_ -- Really? So, when door 3 gets opened, and door 2 inherits the 1/3 odds from door 3, why isn't it door 1? Because... wait for it... THE CONTESTANT PICKED DOOR 1? But you didn't say that? Really? Can you explain why door #1 DIDN'T get that 1/3 odds, if not because the contestant picked it? _"What really happens is that the odds that the remaining door is correct is now the same as the odds of your initial decision being wrong (totals to 100%)."_ -- But door #3, along with its odds, is eliminated. Before the door is opened, what chance does door #2 have of being correct? Are you saying that BOTH doors 2 and 3 EACH have 2/3 of a chance of being right? So you are saying that, between all doors, there is a 5:3 chance of any given door being correct? Really? _"Moreover, you yourself stated the rule that the host can't open the door you selected, therefore the initial selection DOES affect what the host can do. To say that this would in no way affect the odds shows you haven't thought about it at all."_ -- Again, go back and read my post. I said nothing of the kind. I don't care what the host does, it is completely immaterial, other than the premise that the host always forces the game into round 2. If the host doesn't know which door is correct and still opens an incorrect door, the outcome is exactly the same with the exception that the contestant might lose in round 1. But the premise is that the host always opens an incorrect door, which has the effect of forcing the game into round 2 where there are only 2 doors remaining. _"Once the host eliminates a door, your door is still only a 1/3 chance."_ -- Yes, I understand the premise. Too bad I proved it wrong. _"it's clear why you said, "If the car is behind the door the host opened, the contestant loses". I explicitly stated the host CANNOT do that."_ -- As I clearly pointed out. Why are you ignoring THAT part of my post? I was simply laying out all possible scenarios. At no time did I say or even imply that you said this could happen. But, once again, it is a moot point. AS I SAID IN MY POST. _"Second, if we assume we're following the proper rules, we need to look at all possible outcomes to determine the probability."_ -- Correct. The rules say that the host will eliminate a door and force the game into round 2. That's all that happens with that. As I clearly pointed out. _"This means we cannot "always" go to round 2. This completely invalidates the probability analysis."_ -- Wait, didn't you just say this is against the rules? Why are you bringing it up? _"The situation is now so different that if you were to write a truth table for two contestants would look nothing like the truth table for one."_ -- Let's examine that, shall we? If the host is going to open a random door: Car is behind door 1: Host opens door 2: forced to round 2 Host opens door 3: forced to round 2 Car is behind door 2: Host opens door 2: player loses Host opens door 3: forced to round 2 Car is behind door 3: Host opens door 2: force to round 2 Host opens door 3: Player loses Now let's look at the proposed scenario: Car is behind door 1 Host opens door 2: forced to round 2 Host opens door 3: forced to round 2 Car is behind door 2 Host opens door 3: forced to round 2 Car is behind door 3 Host opens door 2: forced to round 2 So, they look EXACTLY the same except that the player can't lose in round 1 because the host specifically doesn't choose a door that wins, so the round is never resolved. But you say they look "nothing like" each other? Who, exactly, is being dishonest? _"BUT - LET'S DO IT! Let's change your scenario a tiny bit to show that this still works with your 2 contestants. Instead of 3 doors, we need more! In fact, this scenario WILL work with 4 or more doors."_ -- How, EXACTLY does this disprove the standard model? Sure, the numbers work with your model (even though they are wrong), but I wasn't asking for another scenario where YOUR model worked, I was asking for a scenario where the STANDARD MODEL DOESN'T WORK. _"If I take the model which tells us how planets form and try to use it to explain why cancer is more prevalent in old age, it will fail."_ -- Okay, hold on. You HONESTLY think that the proposed scenario is SO different from the 2 scenarios that I laid out that it's like comparing planetary movements to cancer? And you expect people to take you seriously? I have to say, it's pretty clear that you know you're wrong. I don't know why you are trying to defend your position. You are straw-manning the crap out of my position and completely ignoring the 2 scenarios that I proposed as absolute proof that your model is complete garbage. But you think you can salvage this conversation by diversion? You know people are going to read BOTH of our posts, right? Your credibility is getting trashed with every post that you prove you are intellectually dishonest.

@dienekes4364

Жыл бұрын

@@ProgThoughts Here's the thing: We can go back and forth forever with you trying to defend your model by ignoring everything I say or using obvious strawman arguments. But here's how to shut me up and prove me wrong: Explain how your model works with the 2 scenarios I posted and how I'm wrong. If you can't do that, you lose this debate. You can try to show a scenario where your model works *and the standard model breaks down* to prove your model is superior, but we both know that's not going to happen. So try to show a little intellectual integrity and just simply show how your model works with the 2 scenarios. If you can't, just admit you're wrong. It's not the end of the world for you.

At no point in time do you have even a 1% probability to get Monty’s ⅓ or 33.3% no-car probability. It's not about the door Monty opens, it's about the door you will never get to pick. As one of the leaders of the ⅓-⅔ Cult once proudly proclaimed that probabilities are space, you don't actually see them. He's absolutely right. Though you see two doors that you didn't pick, they both combined to contain only ⅓ or 33.3% no-car probability to you. That means each door only has a ⅙ or 16.6% no-car probability. Bottom line - the Monty Hall Problem is a 50/50 guess to stay or switch from your ⅓ or 33.3% probability to get the car and your ⅓ or 33.3% probability to not get the car. The ⅓-⅔ Cult made the mistake that just by seeing two doors you must have twice the advantage to not pick the car. Well that’s why the doors are there, to mess with you, on a game show. But what did you expect from someone who was NOT a mathematician or a university math professor? Because she had a high I.Q.? That’s human calculator both candy bar and compact car each costing $100 real situations distorted Rain Man smarts. Back then we didn’t have social platforms for mass reach and instant debates. Biased media thirsting for a feel-good story just ran with it while blatantly ignoring THOUSANDS of actual mathematicians and university professors disagreeing with this false claim. Well, it’s never too late to get it right. First is to call out all these drones of the Emperor’s New Clothes to stop with the nonsense. You now know where you’ve made the mistake, which I'm 66.6% certain that most of you knew it all along. 🤙

@ProgThoughts

21 сағат бұрын

Please don't strain your last 2 brain cells

@TristanSimondsen

12 сағат бұрын

@@ProgThoughts And does you brain cell tell you that you actually CAN pick Monty's ⅓ or 33.3% no-car probability? 🤔

@ProgThoughts

8 сағат бұрын

@@TristanSimondsen Just because you have two options doesn't mean the odds are 50/50. Some intuitive examples: You're either dead or alive right now. Since you were alive 4 hours ago, you're more likely to still be alive than dead right now - not 50/50. A bag of marbles with red and blue colors. Only two options - so 50/50? No, there could be 99 blue marbles and 1 red marble. The list could go on forever. The host knowing which door you picked forces the host to not be able to reveal that door. This changes the odds. Simple.

@TristanSimondsen

4 сағат бұрын

@@ProgThoughts Yeah those are completely incomparable analogies. We are in the confines of three rows of three doors and one car is in front of each row as your entire overall probabilities but you think it can still be a 99-1 split when there are two of something. 👌 And you completely avoided answering the simple question, which I don't blame you. The entire ⅓-⅔ Cult's explanation rests on the foundation that the extra ⅓ no-car probability from Monty is not only accessible, it is the reason why you should switch. I appreciate your time.

@ProgThoughts

4 сағат бұрын

@@TristanSimondsen "The entire ⅓-⅔ Cult's explanation rests on the foundation that the extra ⅓ no-car probability from Monty is not only accessible, it is the reason why you should switch" No, this is a common way of *thinking* about it. Probabilities are how we look at the universe without access to all the information, so we need to generate likelihoods using the information we have. The the 1/3 probability being "accessible" doesn't matter. "Yeah those are completely incomparable analogies" Not really. Here's my proposition to you. I'm prepared to take an hour of my time to simulate a Monty Hall game with you. You can be the host and you'll select a random door out of 3 to have the "prize". I will simply select a door. Then, following the rules that you can't eliminate the correct door or my chosen door, you will eliminate a door. Then, I'll switch doors when you offer. We can do this 10-20 times. By the end, you will see that I switch to the correct door around 2 out of 3 times. This means switching doors gives you a 2/3 chance of winning, while staying at your original door is a 1/3 chance of winning. In fact, this is the entire reason people say you can "access" the 1/3 probability of the door that was eliminated. Because your door had a 1/3 chance of being correct. This *DOES NOT* change when the host eliminates a door. Your door still only has the 1/3 chance of being correct as in the beginning. This means the other door *MUST* have a 2/3 chance of being correct. The entire reason the probability of your door does not change is this rule: The host CANNOT eliminate the door you chose. This rule preserves the original probability of your door at 1/3. If this rule was not in place, the odds would then be 50/50, whether the host *happens* to reveal your door or not (as shown and fully explained in my video). And in fact, we do not have to do take the time to play the game together, because this exact scenario is already simulated in the C++ Program I wrote and showed at the end of the video.