So you have a 1/3 chance to pick the car door, so when he opens a door with a goat, there was a 2/3 chance the original door you picked was a goat, so switching would give you a higher chance (2/3) of getting a car
@@nikolasmakarios904you ran a statistically insignificant sample size. At 100 it’s likely you will get a less accurate result. Two doors does not imply that the odds of those doors are even, that is simply incorrect.
I understand the logic behind the explanation but I’ve always had an issue with it: 1. You choose 1 out of three (33.3% correct) 2. Host opens one door revealing a goat 3. And this is where I’m puzzled - at this point for you (and actually anyone watching from the beginning), the door you’ve picked at the beginning still has 33.3% vs. 66.6% for the other remaining door, but if another person enters the room at this point, having no knowledge of what just happened and this person is asked to pick a door, he would evaluate both remaining doors to be at an equal 50% chance to win the car. So basically, two people at the same time and place should theoretically have different chance of winning the car solely based on their knowledge of the past. It gets even more extreme if say originally you pick one out of 100 doors and the host then opens 98 revealing goats. At this point you should have a 99% chance of winning the car if you switch doors, but the “new guy” would still be at 50%? Not sure I totally agree, or am I getting this all wrong?
You are getting it 'sort of wrong'. The puzzle needs to be thought of in terms of the probabilities of the doors, not the probabilities of the contestant(s). After the host presents the game, the two remaining doors have different chances of having the car. The person who saw the host's performance can determine those chances. The new person would have to pick a door randomly but if they pick the same door the contestant picked, they would have a 1/3 chance of getting the car. If they choose the door the host kept, they will have a 2/3 chance of getting the car. If you play many such games, the statistics of the 'new guys' would average out to 50%. Does that make sense?
Let's say there are 4 doors. Your first pick is door #1 and the last door is door #4. The chance for door #1 to be a car is 25%. The chance for door #4 to be a car is 75%. The chance the second player pick door #1 (50%) and get the car (25%) is 12.5%. The chance the second player pick door #4 (50%) and get the car (75%) is 37.5%. Thus, the chance the seccond player make a random pick and win the car is 12.5% + 37.5% = 50%.
@@Hank254 Yup. No different than buying a lottery ticket. If the first purchaser knows which store of two has the winning ticket, they have a better chance of getting the winning one. The second purchaser has no idea which store has the ticket which makes their odds only HALF that of the first person. Gaming establishments love players that don't know the odds of a bet because they actually REDUCE their own odds of winning by shear ignorance.
The host must know where the car is and always reveal a goat left over from one of the two doors you didn't pick. If the car is behind Door 1 and you picked either Door 2 or 3 he will simply show you the other goat and switching wins. So picking either of the two wrong doors will win by switching and staying can only win if you picked the right one originally
There are doors A, B, and C. Let's say you pick door A. You will pick right 1/3 of the time, so you'll pick a door with a goat 2/3 of the time. If you pick a door with a goat, the other two doors combined will contain a goat and a car. These doors together have a 100%*2/3 chance of containing a car. Now Monty shows which of those two contains the goat. The only object left behind that set of doors is the car, while the chance that that set of doors contains a car hasn't changed, and thus still is 2/3. We already know that the first door will contain a car 1/3 of the time, so we can now conclude that: there is a 1/3 chance that the car is in your door, and there is a 2/3 chance that it is in the other door. You can use this logic to conclude that this works for N doors and a chance of (N-1)/N.
EDIT: I concede to being wrong. I designed a C++ program to simulate the whole selection process, and finally had to concede that the algorithm could be shortened to the following lines: iPrize=rand()%iNumberOfDoors; iMyPick=rand()%iNumberOfDoors; if (iPrize==iMyPick) iNoToChange++; else iYesToChange++; As iNumberOfDoors becomes larger, the difference increases. The initial pick has a (1/ iNumberOfDoors) probability and you are switching to (iNumberOfDoors-1)/(NumberOfDoors) probability by switching to the one remaining door in the group of doors. The probability of the door initially picked remains the same, while doors in the group increase probability as doors are eliminated. OLD TEXT: ---------------------------------------------------------------- João Honrado da Silva, and you shouldn't understand this, because it is humbug!!! I don't agree with this logic. The past is irrelevant. It is still 50/50 chance for picking a car out of the two remaining doors. Let's do a mind experiment and pretend in a parallel reality in your mind you first picked the other door that is left. Well, then you would according to your logic have to switch to the door you first picked in this reality. But what is the difference between these 2 parallel realities? None!!! The only difference is you picked in your mind 2 different doors, which does not affect the reality which is the same in both parallel worlds. But if you are assuming it is not random and the game is being rigged, it matters what the riggers of the game think, but that has nothing to do with random probability. Perhaps the establishment, by teaching us this flawed logic, is making us think in predictable patterns so they can beat us in game systems like the stock market. Or let's say there is an uprising and the rebels follow this flawed but predictable logic, then they can easily be beaten. So, the correct strategy is to be unpredictable. Sometimes you switch, other times you don't.
EDIT: I concede to being wrong. I designed a C++ program to simulate the whole selection process, and finally had to concede that the algorithm could be shortened to the following lines: iPrize=rand()%iNumberOfDoors; iMyPick=rand()%iNumberOfDoors; if (iPrize==iMyPick) iNoToChange++; else iYesToChange++; As iNumberOfDoors becomes larger, the difference increases. The initial pick has a (1/ iNumberOfDoors) probability and you are switching to (iNumberOfDoors-1)/(NumberOfDoors) probability by switching to the one remaining door in the group of doors. The probability of the door initially picked remains the same, while doors in the group increase probability as doors are eliminated. OLD TEXT ----------------------------------------------- You are saying mine has been saved for last because I am holding on to it, whereas the probability increases for the last one in the other group as the all the goats are revealed. But probability also increases for my pick as the goats are revealed. The motivation for not opening my door until the end because it is my pick, is irrelevant We are still left with one goat and one car.
Klaus 74, I concede to being wrong. I designed a C++ program to simulate the whole selection process, and finally had to concede that the algorithm could be shortened to the following lines: iPrize=rand()%iNumberOfDoors; iMyPick=rand()%iNumberOfDoors; if (iPrize==iMyPick) iNoToChange++; else iYesToChange++; As iNumberOfDoors becomes larger, the difference increases. The initial pick has a (1/ iNumberOfDoors) probability and you are switching to (iNumberOfDoors-1)/(NumberOfDoors) probability by switching to the one remaining door in the group of doors. The probability of the door initially picked remains the same, while doors in the group increase probability as doors are eliminated.
I think you are comparing the chances, but i still don't understand your perspective. If the host revealed one of the door that has a goat behind, your chances of choosing the right door will be 50-50, sure its better than choosing between 1/3 doors but at the end we won't know the right answer
It is not 50% for each door; it is 1/3 for the one you selected first and 2/3 for the other. The reason is because the other was not left randomly, but instead it was left by the host, who knew the locations of the contents and was not allowed to reveal the car. Since you only manage to pick the correct option 1 out of 3 times on average, then he is who has to leave the car hidden in the other door that he avoids to reveal the remaining 2 out of 3 times. So, always two doors remaining, but the one left by the host tends to be correct with more frequency than yours.
Your math is faulty since total adds up to only 2/3. The initial pick is 1/3 which is the same as 2/6. You then claim that the remainder is 1/2 which adds up to 3/6. Add them together, and you have a grand total of 4/6, which is 2/3! So, where did the other 1/3 go? Perhaps this will help you to understand the correct odds: Since the contestant's door has a 1/3 chance of winning, the remaining two doors MUST have a 2/3 chance of winning. 1/3 + 2/3 = 3/3, which is 1. Remember, all equations must add up to 100%. Since the host will ALWAYS eliminate a goat, the contestant is really trading their one door for TWO! Hope this helps.
Now, say if you were presented with 3 doors with one of it already opened in front of you that has goat. Now if u r asked to pick one door of the 2 closed ones, is your chances of getting a car still 2/3???
@@Hank254 Exactly, because there is no third door in the original scenario as it has been eliminated right out the gate. It is 50/50 because the game actually begins with one car, one goat, and two doors. Odds are based on number of goats, cars, and doors BEFORE any choice is made.
Easy. In the second selection there are two closed doors: the one you chose at first and the other that the host decided not to reveal. Now, which do you think is more likely to have the car? Take into account that you chose a random door from the three doors, so you were only 1/3 likely to hit the car. On the other hand, the host knew the positions and was forced by the rules to always reveal a goat from the other two doors that you didn't select. The fact that he must reveal a goat means that he has the responsability of keeping the car hidden. Since you select a goat door in 2 out of 3 games (most of the time) and the car must remain hidden anyway, the host is who leaves the car hidden in the other closed door in that same most of the time. So, you can think about it in this way: who left the car hidden, the contestant or the host? You know the contestant does it 1/3 of the time and the host 2/3, that is, with more frequency.
EDIT: I concede to being wrong. I designed a C++ program to simulate the whole selection process, and finally had to concede that the algorithm could be shortened to the following lines: iPrize=rand()%iNumberOfDoors; iMyPick=rand()%iNumberOfDoors; if (iPrize==iMyPick) iNoToChange++; else iYesToChange++; As iNumberOfDoors becomes larger, the difference increases. The initial pick has a (1/ iNumberOfDoors) probability and you are switching to (iNumberOfDoors-1)/(NumberOfDoors) probability by switching to the one remaining door in the group of doors. The probability of the door initially picked remains the same, while doors in the group increase probability as doors are eliminated. OLD TEXT ---------------------------------------- Anshul Agrawal, and you shouldn't understand this, because it is humbug!!! I don't agree with this logic. The past is irrelevant. It is still 50/50 chance for picking a car out of the two remaining doors. Let's do a mind experiment and pretend in a parallel reality in your mind you first picked the other door that is left. Well, then you would according to your logic have to switch to the door you first picked in this reality. But what is the difference between these 2 parallel realities? None!!! The only difference is you picked in your mind 2 different doors, which does not affect the reality which is the same in both parallel worlds. But if you are assuming it is not random and the game is being rigged, it matters what the riggers of the game think, but that has nothing to do with random probability. Perhaps the establishment, by teaching us this flawed logic, is making us think in predictable patterns so they can beat us in game systems like the stock market. Or let's say there is an uprising and the rebels follow this flawed but predictable logic, then they can easily be beaten. So, the correct strategy is to be unpredictable. Sometimes you switch, other times you don't.
Try to understand it from the perspective of the host. The player provides his prediction to the host. The host opens a door with sheep, leaving only two pairs of doors, one sheep and one car for the player to choose. If the player predicts the door of the car is correct, what will the host do? of couse Want the player to change their choice. The host asks the player to calculate the odds with misleading the player already have once choice before. player swapping his choice, and saying 2/3 chance will win. On the contrary, the player predicted the door with the sheep is correct. What will the host do to make the player still choose the door with the sheep? you can tell. Then..... if you are the player, do you think have any advantage from changing your decision? ........ Dawyer's door problem, calculate the chance of the host winning.
*I can prove the true nature of this 'problem'!* .. they trick you into thinking you have 'chosen' the first door. Think of it this way: the first round of the game is the fake 'choice' where you don't actually choose, you 'reserve', to protect one door and let one goat be eliminated from the game, 1st round is now done! Now they start the second game where the REAL choice is allowed, there are no longer any 'reserved' doors and the game parameters have now changed to allow 50/50 odds because you get to play this game from scratch with two doors and for the FIRST time get to actually 'choose' instead of 'reserve/protect', which is very different than the first game's round of choices..it's TWO different games NOT one, so if you do not account for this by creating two separate math problems then you are NOT accurately representing the true nature of the game, as you have to change the odds for all 'unknowns' every time you eliminate possibilities. This logic applies to different examples of this 'problem' as well. The key is understanding that the only real game is when the final choice is made, and that everything before that is just changing the parameters, you have to make the math adjustment for the new parameters as they change, it all comes down to the *state of the parameters* when you *actually choose* and NOT when you are simply 'negotiating' the parameter changes, ..thanks and you're welcome;)
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. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots. Let's say the game start with 1 door (1 car, 0 goat). After you make a pick, the host add a goat door and ask whether you want to switch door or not. What is your winning chance if you stay with your first pick? The answer is obviously 100%. Super easy, right? J: They trick you into thinking you have 'chosen' the first door. Think of it this way: the first round of the game is the fake 'choice' where you don't actually choose. J: 1st round is now done! Now they start the second game where the REAL choice is allowed and the game parameters have now changed to allow 50/50 odds because you get to play this game from scratch with two doors and for the FIRST time get to actually 'choose'. J: It's TWO different games NOT one J: I am a genius. Hoooraaay!!!
@@Araqius exactly, the illusion comes in when you just look at what is actually in front of you at the time and disregard the past process. In either case you are looking at, what appears to be a binary equation with the statement 'it HAS to be inside one or the other' being 100% true, so some people assume: "how could two identical scenarios have different properties? it has to be 50/50 no matter what!" I like to say:"think of splitting a shuffled Deck of cards into two piles, but then you remove half the Spades from one pile and balance the amount out using just Diamonds from the other half. One pile has a lot more red cards in it, so it is more likely one pile will give you a red card and the other will give you a black card, even though there are only two piles" eliminating the door in the game is what 'takes the spades away', so lol, yeah, everyone is trying their best to describe being right so i took the other-side and did my best at describing the logic 'in being wrong'..did i win? (P.S. people who actually don't get this should never gamble)
@@jaybefaulky4902 "(P.S. people who actually don't get this should never gamble)" Well, you still can't understand 2 doors are holding a car twice as likely as 1 door does, yet you are trying to claim two doors should hold a car equally likely as one door does, so you are the one who should never gamble, or write about probability which you can't even grasp.
@@max5250you didn't read what i said .. i was playing at showing the logic of 'being wrong', afterwards i wrote: , the *illusion* comes in when you just look at what is actually in front of you at the time and *disregard* the past process. In either case you are looking at, what *appears* to be a binary equation with the statement ''it HAS to be inside one *OR* the other'' being 100% true, however some people (wrongly) assume: "how could two identical scenarios have different properties? it has to be 50/50 no matter what!"(except the scenarios *are* different) I like to say:"think of splitting a shuffled Deck of cards into two piles, but then you remove half the Spades from one pile and balance the amount out using just Diamonds from the other half. One pile has a lot more red cards in it, so it is more likely one pile will give you a red card and the other will give you a black card, even though there are only two piles" eliminating the door in the game is what 'takes the spades away', so lol, yeah, everyone is trying their best to describe being right so i took the other-side and did my best at describing the logic 'in being wrong'..did i win? (P.S. people who actually don't get this should never gamble) .. did this help you understand where i was coming from?
@@jaybefaulky4902 " they trick you into thinking you have 'chosen' the first door. " This is plain wrong. There is no trick, and you are actually choosing the "first door", and that's the most important thing since it determines outcome of the entire games. "and disregard the past process" We cannot disregard "the past process" since we live in a reality that is dependant on the "cause-consequence" relation. "1st round is now done! Now they start the second game where the REAL choice is allowed" When there are 2 doors only, you cannot choose even if you want, since these two doors are holding the opposite content, and you already own one of these two doors, although you don't know what is inside, but you still know how likely it holds a car, and that's enough to make a decision (to stay or to switch), and not to make a "choice (randomly pick between two doors).
🤐 This will go down as the biggest hoax in mathematics history. 🤫 What you just explained are 3 different games where the car didn’t move, and player picks Door A in game 1, Door B in game 2 and Door C in game 3. When in reality, each game player only has the capability to pick one door. ⅔ switch is actually ⅔ of the time you won’t pick the car so switch to the door with the car. The two doors you didn’t pick are (1-X). Can’t use (1-X) as X to then say you are twice as likely to pick these doors so switch. Player picks ONE door per game. He only has one door that he’s able to stay at or switch from. He doesn’t have two doors to stay at or switch from. In order for player to be able to switch from a no-car door, that door needs something tangible behind it. Otherwise a no-car door will always be recognized and calculated as the complement of 1. Replace the two no-car doors with a dog and a horse so you’re not letting two goats confuse what is the probability behind each single door. Sims, i.e. Python, require very specific language and prompts to get precisely what you’re looking for. “Picking” a door is different from being able to then “get the item behind that door”. “Eliminating” a door is different from the door being “removed entirely”. P.O.V. is also sensitive. Asking for player’s probability to the item is different from the probability of an item behind a specific door. Paul Erdős, the “Albert Einstein of Mathematics”, along with thousands of mathematicians and university math professors did not believe it was ⅔ always switch. Paul only relented because the simulations kept showing him ⅔. In 2024, there are ChaptGPT and other AI for everyday people like myself to play around with the language and isolate the flaws in those sims. The only way for you to clearly see what is happening, you need to solve for 9 Xs - finding player’s probability to pick a specific door, being able to open that door to get the car, the dog or the horse, all at the same time. Let’s call this player “Max”: 1. What is Max's probability to pick Door A, open Door A to get the car? 2. What is Max's probability to pick Door B, open Door B to get the car? 3. What is Max's probability to pick Door C, open Door C to get the car? 4. What is Max's probability to pick Door A, open Door A to get the dog? 5. What is Max's probability to pick Door B, open Door B to get the dog? 6. What is Max's probability to pick Door C, open Door C to get the dog? 7. What is Max's probability to pick Door A, open Door A to get the horse? 8. What is Max's probability to pick Door B, open Door B to get the horse? 9. What is Max's probability to pick Door C, open Door C to get the horse? In ONE game, this will be the result if you code it correctly: Max picking Door A, opening Door A and getting the car behind Door A ½ Max picking Door B, opening Door B and getting the car behind Door B 0 Max picking Door C, opening Door C and getting the car behind Door C ½ Max picking Door A, opening Door A and getting the dog behind Door A ⅓ Max picking Door B, opening Door B and getting the dog behind Door B 0 Max picking Door C, opening Door C and getting the dog behind Door C ⅓ Max picking Door A, opening Door A and getting the horse behind Door A ⅓ Max picking Door B, opening Door B and getting the horse behind Door B 0 Max picking Door C, opening Door C and getting the horse behind Door C ⅓ As you can clearly see, in this one sim, Door B was removed entirely, which means Max randomly picked either Door A or Door C. Probability to get the car in either door is the same whether he stays or switches. If Max picked Door A and stays at Door A, he has a ½ probability to get the car. And if he switches to Door C, which is the other door that wasn't removed entirely, he has a ½ probability to get the car. If Max picked Door C and stays at Door C, he has a ½ probability to get the car. If he switched to Door A, he has a ½ probability to get the car. If you are getting a different answer just ask yourself, and AI, does Max have ANY probability to pick, open and get something behind a door that is no longer even there? Gotta love 2024 for you to find these simulation mistakes. Even AI will tell you, it’s made a mistake. I’m sure the programmers for these current AIs are all ⅔-always-switch cult members. In THREE games or more, this will be the result if you code it correctly: ⅓ (Max's probability of picking Door A, opening Door A to get the car) ⅓ (Max's probability of picking Door B, opening Door B to get the car) ⅓ (Max's probability of picking Door C, opening Door C to get the car) ⅓ (Max's probability of picking Door A, opening Door A to get the dog) ⅓ (Max's probability of picking Door B, opening Door B to get the dog) ⅓ (Max's probability of picking Door C, opening Door C to get the dog) ⅓ (Max's probability of picking Door A, opening Door A to get the horse) ⅓ (Max's probability of picking Door B, opening Door B to get the horse) ⅓ (Max's probability of picking Door C, opening Door C to get the horse) In these games, the probability is ⅓ because Max's initial pick is random, and Monty’s elimination process does not alter Max’s chances of picking, opening a door, and getting a specific item behind it. Vos Savant said of the thousands of letters that came in to dispute this claim, 65% were from mathematicians and university math professors. Right; they're ALL wrong. Now I know why the sims are always 2/3. Hey Paul - this is for you. I’d take your word regarding math over a Sunday columnist for a celebrity newspaper any day.
Each sheep is also an independent and distinct individual. This game cannot mislead players by treating every sheep as the same thing. There are three options, and two of the sheep also exist independently. If the player can distinguish them, there will be no chance of winning like 2/3. Is the reality of quantum mechanics more logical than what the math? The player never knows which goat is the one revealed.。Don't let the jews fool you.. If life is an endless choice, then if we face it with this way of dealing with things, then what aspect of human nature is eternal? Some people always want to control the world and act as the so-called "God"? In fact, this is just an ideology produced under the Goebbels effect.
FAILURE. This is very disappointing for this channel. The ONLY thing that the idea that the game is played out this way 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. I can prove, beyond any doubt (to anyone who is intellectually honest) how this video is complete and utter garbage. 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.
"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." That part is wrong, the host opens a door to show a goat... he doesn't open the door with the car. The rest of your post is analyzing the wrong problem.
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.
@@Araqius _"Assume you stay with your first pick."_ -- Puking the original premise to me doesn't do anything but show how mindless you are. If you can't answer how the model applies to the 2 situations I present, you've lost the argument and no amount of babbling bullshit will salvage that.
@@dienekes4364 Dinkass, when you pick a door with a goat the host reveals a goat and leaves the car for you to switch with. So the onus is upon you to show how the probability of picking a goat is 2/3 but the probability of the host revealing a goat and leaving the car isn't 2/3.
@@klaus7443 It is mathematical fact that it's a 1/3 chance when you choose a door. The other two doors equate to a 2/3 chance which adds up to 3/3. Regardless of which door the host opens, that does not change the odds that those 2 doors hold a 2/3 chance. In other words, the 2/3 holds true for BOTH of the remaining doors. This is why you should ALWAYS switch, because the host just eliminated one of the two 2/3 chance doors meaning the other door now holds the entire 2/3 chance! Another way to look at it is if you divide the doors into two groups (1door and 2 doors) the 2-door group will always have a HIGHER chance of containing the car. In the beginning, you are always forced into a less that 50/50 chance. So why not change it to the better odds.
Nicely done! This is my favorite way to explain the problem to others!
So you have a 1/3 chance to pick the car door, so when he opens a door with a goat, there was a 2/3 chance the original door you picked was a goat, so switching would give you a higher chance (2/3) of getting a car
Its 50/50
@@nikolasmakarios904 neat how do you run a sim
@@nikolasmakarios904you ran a statistically insignificant sample size. At 100 it’s likely you will get a less accurate result.
Two doors does not imply that the odds of those doors are even, that is simply incorrect.
@@SailingQuicksilverincorrect.
I understand the logic behind the explanation but I’ve always had an issue with it:
1. You choose 1 out of three (33.3% correct)
2. Host opens one door revealing a goat
3. And this is where I’m puzzled - at this point for you (and actually anyone watching from the beginning), the door you’ve picked at the beginning still has 33.3% vs. 66.6% for the other remaining door, but if another person enters the room at this point, having no knowledge of what just happened and this person is asked to pick a door, he would evaluate both remaining doors to be at an equal 50% chance to win the car.
So basically, two people at the same time and place should theoretically have different chance of winning the car solely based on their knowledge of the past. It gets even more extreme if say originally you pick one out of 100 doors and the host then opens 98 revealing goats. At this point you should have a 99% chance of winning the car if you switch doors, but the “new guy” would still be at 50%? Not sure I totally agree, or am I getting this all wrong?
You are getting it 'sort of wrong'. The puzzle needs to be thought of in terms of the probabilities of the doors, not the probabilities of the contestant(s). After the host presents the game, the two remaining doors have different chances of having the car. The person who saw the host's performance can determine those chances. The new person would have to pick a door randomly but if they pick the same door the contestant picked, they would have a 1/3 chance of getting the car. If they choose the door the host kept, they will have a 2/3 chance of getting the car. If you play many such games, the statistics of the 'new guys' would average out to 50%. Does that make sense?
Let's say there are 4 doors.
Your first pick is door #1 and the last door is door #4.
The chance for door #1 to be a car is 25%.
The chance for door #4 to be a car is 75%.
The chance the second player pick door #1 (50%) and get the car (25%) is 12.5%.
The chance the second player pick door #4 (50%) and get the car (75%) is 37.5%.
Thus, the chance the seccond player make a random pick and win the car is 12.5% + 37.5% = 50%.
@@Hank254 Yup. No different than buying a lottery ticket. If the first purchaser knows which store of two has the winning ticket, they have a better chance of getting the winning one. The second purchaser has no idea which store has the ticket which makes their odds only HALF that of the first person.
Gaming establishments love players that don't know the odds of a bet because they actually REDUCE their own odds of winning by shear ignorance.
How do we come up with that result at 2:11, the two thirds? I can't figure it out
The host must know where the car is and always reveal a goat left over from one of the two doors you didn't pick. If the car is behind Door 1 and you picked either Door 2 or 3 he will simply show you the other goat and switching wins. So picking either of the two wrong doors will win by switching and staying can only win if you picked the right one originally
There are doors A, B, and C.
Let's say you pick door A.
You will pick right 1/3 of the time,
so you'll pick a door with a goat 2/3 of the time.
If you pick a door with a goat, the other two doors combined will contain a goat and a car. These doors together have a 100%*2/3 chance of containing a car.
Now Monty shows which of those two contains the goat. The only object left behind that set of doors is the car, while the chance that that set of doors contains a car hasn't changed, and thus still is 2/3. We already know that the first door will contain a car 1/3 of the time, so we can now conclude that:
there is a 1/3 chance that the car is in your door, and
there is a 2/3 chance that it is in the other door.
You can use this logic to conclude that this works for N doors and a chance of (N-1)/N.
EDIT: I concede to being wrong. I designed a C++ program to simulate the whole selection process, and finally had to concede that the algorithm could be shortened to the following lines:
iPrize=rand()%iNumberOfDoors;
iMyPick=rand()%iNumberOfDoors;
if (iPrize==iMyPick)
iNoToChange++;
else
iYesToChange++;
As iNumberOfDoors becomes larger, the difference increases. The initial pick has a (1/ iNumberOfDoors) probability and you are switching to (iNumberOfDoors-1)/(NumberOfDoors) probability by switching to the one remaining door in the group of doors. The probability of the door initially picked remains the same, while doors in the group increase probability as doors are eliminated.
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João Honrado da Silva, and you shouldn't understand this, because it is humbug!!!
I don't agree with this logic. The past is irrelevant. It is still 50/50 chance for picking a car out of the two remaining doors. Let's do a mind experiment and pretend in a parallel reality in your mind you first picked the other door that is left. Well, then you would according to your logic have to switch to the door you first picked in this reality. But what is the difference between these 2 parallel realities? None!!! The only difference is you picked in your mind 2 different doors, which does not affect the reality which is the same in both parallel worlds. But if you are assuming it is not random and the game is being rigged, it matters what the riggers of the game think, but that has nothing to do with random probability.
Perhaps the establishment, by teaching us this flawed logic, is making us think in predictable patterns so they can beat us in game systems like the stock market. Or let's say there is an uprising and the rebels follow this flawed but predictable logic, then they can easily be beaten. So, the correct strategy is to be unpredictable. Sometimes you switch, other times you don't.
EDIT: I concede to being wrong. I designed a C++ program to simulate the whole selection process, and finally had to concede that the algorithm could be shortened to the following lines:
iPrize=rand()%iNumberOfDoors;
iMyPick=rand()%iNumberOfDoors;
if (iPrize==iMyPick)
iNoToChange++;
else
iYesToChange++;
As iNumberOfDoors becomes larger, the difference increases. The initial pick has a (1/ iNumberOfDoors) probability and you are switching to (iNumberOfDoors-1)/(NumberOfDoors) probability by switching to the one remaining door in the group of doors. The probability of the door initially picked remains the same, while doors in the group increase probability as doors are eliminated.
OLD TEXT
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You are saying mine has been saved for last because I am holding on to it, whereas the probability increases for the last one in the other group as the all the goats are revealed. But probability also increases for my pick as the goats are revealed. The motivation for not opening my door until the end because it is my pick, is irrelevant We are still left with one goat and one car.
Klaus 74, I concede to being wrong. I designed a C++ program to simulate the whole selection process, and finally had to concede that the algorithm could be shortened to the following lines:
iPrize=rand()%iNumberOfDoors;
iMyPick=rand()%iNumberOfDoors;
if (iPrize==iMyPick)
iNoToChange++;
else
iYesToChange++;
As iNumberOfDoors becomes larger, the difference increases. The initial pick has a (1/ iNumberOfDoors) probability and you are switching to (iNumberOfDoors-1)/(NumberOfDoors) probability by switching to the one remaining door in the group of doors. The probability of the door initially picked remains the same, while doors in the group increase probability as doors are eliminated.
I think you are comparing the chances, but i still don't understand your perspective. If the host revealed one of the door that has a goat behind, your chances of choosing the right door will be 50-50, sure its better than choosing between 1/3 doors but at the end we won't know the right answer
It is not 50% for each door; it is 1/3 for the one you selected first and 2/3 for the other. The reason is because the other was not left randomly, but instead it was left by the host, who knew the locations of the contents and was not allowed to reveal the car. Since you only manage to pick the correct option 1 out of 3 times on average, then he is who has to leave the car hidden in the other door that he avoids to reveal the remaining 2 out of 3 times. So, always two doors remaining, but the one left by the host tends to be correct with more frequency than yours.
@@RonaldABG Exactly. The key is that the host knows what's behind each door and will never reveal the prize. Crucial detail often left out!
Your math is faulty since total adds up to only 2/3.
The initial pick is 1/3 which is the same as 2/6. You then claim that the remainder is 1/2 which adds up to 3/6. Add them together, and you have a grand total of 4/6, which is 2/3! So, where did the other 1/3 go?
Perhaps this will help you to understand the correct odds: Since the contestant's door has a 1/3 chance of winning, the remaining two doors MUST have a 2/3 chance of winning. 1/3 + 2/3 = 3/3, which is 1. Remember, all equations must add up to 100%. Since the host will ALWAYS eliminate a goat, the contestant is really trading their one door for TWO!
Hope this helps.
Now, say if you were presented with 3 doors with one of it already opened in front of you that has goat. Now if u r asked to pick one door of the 2 closed ones, is your chances of getting a car still 2/3???
No.
@@Hank254 Exactly, because there is no third door in the original scenario as it has been eliminated right out the gate. It is 50/50 because the game actually begins with one car, one goat, and two doors. Odds are based on number of goats, cars, and doors BEFORE any choice is made.
i did not understand this
Easy. In the second selection there are two closed doors: the one you chose at first and the other that the host decided not to reveal. Now, which do you think is more likely to have the car?
Take into account that you chose a random door from the three doors, so you were only 1/3 likely to hit the car. On the other hand, the host knew the positions and was forced by the rules to always reveal a goat from the other two doors that you didn't select. The fact that he must reveal a goat means that he has the responsability of keeping the car hidden. Since you select a goat door in 2 out of 3 games (most of the time) and the car must remain hidden anyway, the host is who leaves the car hidden in the other closed door in that same most of the time.
So, you can think about it in this way: who left the car hidden, the contestant or the host? You know the contestant does it 1/3 of the time and the host 2/3, that is, with more frequency.
EDIT: I concede to being wrong. I designed a C++ program to simulate the whole selection process, and finally had to concede that the algorithm could be shortened to the following lines:
iPrize=rand()%iNumberOfDoors;
iMyPick=rand()%iNumberOfDoors;
if (iPrize==iMyPick)
iNoToChange++;
else
iYesToChange++;
As iNumberOfDoors becomes larger, the difference increases. The initial pick has a (1/ iNumberOfDoors) probability and you are switching to (iNumberOfDoors-1)/(NumberOfDoors) probability by switching to the one remaining door in the group of doors. The probability of the door initially picked remains the same, while doors in the group increase probability as doors are eliminated.
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Anshul Agrawal, and you shouldn't understand this, because it is humbug!!!
I don't agree with this logic. The past is irrelevant. It is still 50/50 chance for picking a car out of the two remaining doors. Let's do a mind experiment and pretend in a parallel reality in your mind you first picked the other door that is left. Well, then you would according to your logic have to switch to the door you first picked in this reality. But what is the difference between these 2 parallel realities? None!!! The only difference is you picked in your mind 2 different doors, which does not affect the reality which is the same in both parallel worlds. But if you are assuming it is not random and the game is being rigged, it matters what the riggers of the game think, but that has nothing to do with random probability.
Perhaps the establishment, by teaching us this flawed logic, is making us think in predictable patterns so they can beat us in game systems like the stock market. Or let's say there is an uprising and the rebels follow this flawed but predictable logic, then they can easily be beaten. So, the correct strategy is to be unpredictable. Sometimes you switch, other times you don't.
@@RonaldABG
Thank so much for explaining again I understand probably
@@CCLGamer
Yes you are right bro but it's your choice switch or not it's have same chance 2/3 and other 2/3
What if I really like goats?
Try to understand it from the perspective of the host. The player provides his prediction to the host. The host opens a door with sheep, leaving only two pairs of doors, one sheep and one car for the player to choose. If the player predicts the door of the car is correct, what will the host do? of couse Want the player to change their choice. The host asks the player to calculate the odds with misleading the player already have once choice before. player swapping his choice, and saying 2/3 chance will win.
On the contrary, the player predicted the door with the sheep is correct. What will the host do to make the player still choose the door with the sheep? you can tell.
Then..... if you are the player, do you think have any advantage from changing your decision?
........ Dawyer's door problem, calculate the chance of the host winning.
😳😵😱🤷
This is actually an EASY problem, I solved it in my mind in 5 minutes with no difficulty 🤷♂
no one cares, bro, but congratulations
*I can prove the true nature of this 'problem'!* .. they trick you into thinking you have 'chosen' the first door. Think of it this way: the first round of the game is the fake 'choice' where you don't actually choose, you 'reserve', to protect one door and let one goat be eliminated from the game, 1st round is now done! Now they start the second game where the REAL choice is allowed, there are no longer any 'reserved' doors and the game parameters have now changed to allow 50/50 odds because you get to play this game from scratch with two doors and for the FIRST time get to actually 'choose' instead of 'reserve/protect', which is very different than the first game's round of choices..it's TWO different games NOT one, so if you do not account for this by creating two separate math problems then you are NOT accurately representing the true nature of the game, as you have to change the odds for all 'unknowns' every time you eliminate possibilities. This logic applies to different examples of this 'problem' as well. The key is understanding that the only real game is when the final choice is made, and that everything before that is just changing the parameters, you have to make the math adjustment for the new parameters as they change, it all comes down to the *state of the parameters* when you *actually choose* and NOT when you are simply 'negotiating' the parameter changes, ..thanks and you're welcome;)
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.
It's just basic math/logic kids understand.
Sadly, it's far too hard for idiots.
Let's say the game start with 1 door (1 car, 0 goat).
After you make a pick, the host add a goat door and ask whether you want to switch door or not.
What is your winning chance if you stay with your first pick?
The answer is obviously 100%.
Super easy, right?
J: They trick you into thinking you have 'chosen' the first door. Think of it this way: the first round of the game is the fake 'choice' where you don't actually choose.
J: 1st round is now done! Now they start the second game where the REAL choice is allowed and the game parameters have now changed to allow 50/50 odds because you get to play this game from scratch with two doors and for the FIRST time get to actually 'choose'.
J: It's TWO different games NOT one
J: I am a genius. Hoooraaay!!!
@@Araqius exactly, the illusion comes in when you just look at what is actually in front of you at the time and disregard the past process. In either case you are looking at, what appears to be a binary equation with the statement 'it HAS to be inside one or the other' being 100% true, so some people assume: "how could two identical scenarios have different properties? it has to be 50/50 no matter what!" I like to say:"think of splitting a shuffled Deck of cards into two piles, but then you remove half the Spades from one pile and balance the amount out using just Diamonds from the other half. One pile has a lot more red cards in it, so it is more likely one pile will give you a red card and the other will give you a black card, even though there are only two piles" eliminating the door in the game is what 'takes the spades away', so lol, yeah, everyone is trying their best to describe being right so i took the other-side and did my best at describing the logic 'in being wrong'..did i win? (P.S. people who actually don't get this should never gamble)
@@jaybefaulky4902
"(P.S. people who actually don't get this should never gamble)"
Well, you still can't understand 2 doors are holding a car twice as likely as 1 door does, yet you are trying to claim two doors should hold a car equally likely as one door does, so you are the one who should never gamble, or write about probability which you can't even grasp.
@@max5250you didn't read what i said .. i was playing at showing the logic of 'being wrong', afterwards i wrote: , the *illusion* comes in when you just look at what is actually in front of you at the time and *disregard* the past process. In either case you are looking at, what *appears* to be a binary equation with the statement ''it HAS to be inside one *OR* the other'' being 100% true, however some people (wrongly) assume: "how could two identical scenarios have different properties? it has to be 50/50 no matter what!"(except the scenarios *are* different) I like to say:"think of splitting a shuffled Deck of cards into two piles, but then you remove half the Spades from one pile and balance the amount out using just Diamonds from the other half. One pile has a lot more red cards in it, so it is more likely one pile will give you a red card and the other will give you a black card, even though there are only two piles" eliminating the door in the game is what 'takes the spades away', so lol, yeah, everyone is trying their best to describe being right so i took the other-side and did my best at describing the logic 'in being wrong'..did i win? (P.S. people who actually don't get this should never gamble) .. did this help you understand where i was coming from?
@@jaybefaulky4902
" they trick you into thinking you have 'chosen' the first door. "
This is plain wrong.
There is no trick, and you are actually choosing the "first door", and that's the most important thing since it determines outcome of the entire games.
"and disregard the past process"
We cannot disregard "the past process" since we live in a reality that is dependant on the "cause-consequence" relation.
"1st round is now done! Now they start the second game where the REAL choice is allowed"
When there are 2 doors only, you cannot choose even if you want, since these two doors are holding the opposite content, and you already own one of these two doors, although you don't know what is inside, but you still know how likely it holds a car, and that's enough to make a decision (to stay or to switch), and not to make a "choice (randomly pick between two doors).
🤐 This will go down as the biggest hoax in mathematics history. 🤫
What you just explained are 3 different games where the car didn’t move, and player picks Door A in game 1, Door B in game 2 and Door C in game 3.
When in reality, each game player only has the capability to pick one door.
⅔ switch is actually ⅔ of the time you won’t pick the car so switch to the door with the car. The two doors you didn’t pick are (1-X). Can’t use (1-X) as X to then say you are twice as likely to pick these doors so switch.
Player picks ONE door per game. He only has one door that he’s able to stay at or switch from. He doesn’t have two doors to stay at or switch from.
In order for player to be able to switch from a no-car door, that door needs something tangible behind it. Otherwise a no-car door will always be recognized and calculated as the complement of 1. Replace the two no-car doors with a dog and a horse so you’re not letting two goats confuse what is the probability behind each single door.
Sims, i.e. Python, require very specific language and prompts to get precisely what you’re looking for. “Picking” a door is different from being able to then “get the item behind that door”. “Eliminating” a door is different from the door being “removed entirely”. P.O.V. is also sensitive. Asking for player’s probability to the item is different from the probability of an item behind a specific door.
Paul Erdős, the “Albert Einstein of Mathematics”, along with thousands of mathematicians and university math professors did not believe it was ⅔ always switch. Paul only relented because the simulations kept showing him ⅔.
In 2024, there are ChaptGPT and other AI for everyday people like myself to play around with the language and isolate the flaws in those sims. The only way for you to clearly see what is happening, you need to solve for 9 Xs - finding player’s probability to pick a specific door, being able to open that door to get the car, the dog or the horse, all at the same time.
Let’s call this player “Max”:
1. What is Max's probability to pick Door A, open Door A to get the car?
2. What is Max's probability to pick Door B, open Door B to get the car?
3. What is Max's probability to pick Door C, open Door C to get the car?
4. What is Max's probability to pick Door A, open Door A to get the dog?
5. What is Max's probability to pick Door B, open Door B to get the dog?
6. What is Max's probability to pick Door C, open Door C to get the dog?
7. What is Max's probability to pick Door A, open Door A to get the horse?
8. What is Max's probability to pick Door B, open Door B to get the horse?
9. What is Max's probability to pick Door C, open Door C to get the horse?
In ONE game, this will be the result if you code it correctly:
Max picking Door A, opening Door A and getting the car behind Door A ½
Max picking Door B, opening Door B and getting the car behind Door B 0
Max picking Door C, opening Door C and getting the car behind Door C ½
Max picking Door A, opening Door A and getting the dog behind Door A ⅓
Max picking Door B, opening Door B and getting the dog behind Door B 0
Max picking Door C, opening Door C and getting the dog behind Door C ⅓
Max picking Door A, opening Door A and getting the horse behind Door A ⅓
Max picking Door B, opening Door B and getting the horse behind Door B 0
Max picking Door C, opening Door C and getting the horse behind Door C ⅓
As you can clearly see, in this one sim, Door B was removed entirely, which means Max randomly picked either Door A or Door C. Probability to get the car in either door is the same whether he stays or switches. If Max picked Door A and stays at Door A, he has a ½ probability to get the car. And if he switches to Door C, which is the other door that wasn't removed entirely, he has a ½ probability to get the car. If Max picked Door C and stays at Door C, he has a ½ probability to get the car. If he switched to Door A, he has a ½ probability to get the car.
If you are getting a different answer just ask yourself, and AI, does Max have ANY probability to pick, open and get something behind a door that is no longer even there?
Gotta love 2024 for you to find these simulation mistakes. Even AI will tell you, it’s made a mistake. I’m sure the programmers for these current AIs are all ⅔-always-switch cult members.
In THREE games or more, this will be the result if you code it correctly:
⅓ (Max's probability of picking Door A, opening Door A to get the car)
⅓ (Max's probability of picking Door B, opening Door B to get the car)
⅓ (Max's probability of picking Door C, opening Door C to get the car)
⅓ (Max's probability of picking Door A, opening Door A to get the dog)
⅓ (Max's probability of picking Door B, opening Door B to get the dog)
⅓ (Max's probability of picking Door C, opening Door C to get the dog)
⅓ (Max's probability of picking Door A, opening Door A to get the horse)
⅓ (Max's probability of picking Door B, opening Door B to get the horse)
⅓ (Max's probability of picking Door C, opening Door C to get the horse)
In these games, the probability is ⅓ because Max's initial pick is random, and Monty’s elimination process does not alter Max’s chances of picking, opening a door, and getting a specific item behind it.
Vos Savant said of the thousands of letters that came in to dispute this claim, 65% were from mathematicians and university math professors.
Right; they're ALL wrong.
Now I know why the sims are always 2/3.
Hey Paul - this is for you. I’d take your word regarding math over a Sunday columnist for a celebrity newspaper any day.
Each sheep is also an independent and distinct individual. This game cannot mislead players by treating every sheep as the same thing.
There are three options, and two of the sheep also exist independently. If the player can distinguish them, there will be no chance of winning like 2/3.
Is the reality of quantum mechanics more logical than what the math? The player never knows which goat is the one revealed.。Don't let the jews fool you.. If life is an endless choice, then if we face it with this way of dealing with things, then what aspect of human nature is eternal? Some people always want to control the world and act as the so-called "God"? In fact, this is just an ideology produced under the Goebbels effect.
FAILURE. This is very disappointing for this channel. The ONLY thing that the idea that the game is played out this way 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.
I can prove, beyond any doubt (to anyone who is intellectually honest) how this video is complete and utter garbage. 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.
"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."
That part is wrong, the host opens a door to show a goat... he doesn't open the door with the car. The rest of your post is analyzing the wrong problem.
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.
@@Araqius _"Assume you stay with your first pick."_ -- Puking the original premise to me doesn't do anything but show how mindless you are. If you can't answer how the model applies to the 2 situations I present, you've lost the argument and no amount of babbling bullshit will salvage that.
@@dienekes4364 Dinkass, when you pick a door with a goat the host reveals a goat and leaves the car for you to switch with. So the onus is upon you to show how the probability of picking a goat is 2/3 but the probability of the host revealing a goat and leaving the car isn't 2/3.
@@klaus7443 It is mathematical fact that it's a 1/3 chance when you choose a door. The other two doors equate to a 2/3 chance which adds up to 3/3. Regardless of which door the host opens, that does not change the odds that those 2 doors hold a 2/3 chance. In other words, the 2/3 holds true for BOTH of the remaining doors. This is why you should ALWAYS switch, because the host just eliminated one of the two 2/3 chance doors meaning the other door now holds the entire 2/3 chance!
Another way to look at it is if you divide the doors into two groups (1door and 2 doors) the 2-door group will always have a HIGHER chance of containing the car. In the beginning, you are always forced into a less that 50/50 chance. So why not change it to the better odds.