Pseudo mathematics

smart

this is the only acceptable explanation in the entire youtube

@@basiert7701no he is right

It's a 2 door proposition 1 car and 1 goat 1:1 odds

Yeah same here ,that makes it way easier to understand , but at the same time feels wrong , cause we are adding a made up condition that the dealer wants the car. Idk it seems like this problem is just based on perspective or utterly specific conditions

Brooooo i ve been looking everything and it just didnt make sense only helpful comment is yours thanks

Listen, Monty gives nothing away. That's why his name is Monty. 50-50 at the start; 50-50 at the end. www.onofframp.blogspot.com/2014/09/monty-hall-door-goat-and-car.html

John D Give it a rest, you're wrong. Mathematically, logically and empirically it's NOT 50/50 between staying and switching. You're just making yourself look foolish,

+Peter Thai oh shit, that makes sense now... so you are more likely to choose a goat in the first round, then if you switch in the second round, you are just as likely to get a car (if your first choice was in fact, a goat).

EXACTLY! I have described it that way many times.

Amir Aribokill No. In that case, it's 50-50. The host of the quiz knows where the car is, so possibility of the host's opening the door where there's a car is zero. On the other hand, the friend knows nothing about what's behind the doors, which means possibility to open the door without booby traps is 1/3.

But if a door is revealed to be a booby trap by a friend, doesnt it become the same as Monty showing you the goat?

@@Zaekr2111 once a door is revealed to have the booby trap, then you should switch because your door always only has 33 percent chance of a good outcome. Switching doubles your chances to get a door without traps.

That really helps me to understand this.

Only way I was able to understand it, thank you

Yes.

Does your 'algorithm' always bet on door "one"🚪(was it a 'constant' choice in the experiment) or every time a random (different) door was chosen to bet on, before you check what's behind the 2 remaining🚪?

I feel bad for him... I hope he's not making a full out of himself. I would have been probably the same way, if I hadn't learned about it on my own... lol

Your stupid. it is only 2 choices left 50% chance 3rd grade math

When Bob has done the cards, Bob is allowed to make comments. This is the purpose of the cards.

Kirk Bocek haha I've already done the cards 20 times 11 times switching worked 9 times it didnt . doesn't seem like switching is 2/3 chance to me. You are going by group mentaliy, everybody says switching increases chancesso you lose common sense and just agree haha there are only 2.

Kirk Bocek LOL I am an Actuary went to Northwestern if you don't know what that is look it up. I think I wud know probability haha.

Well you don't know the probability here. Go read the comments on the KZfaq posting. Seriously, don't think you can just *figure* this out. Do the cards. You'll be surprised. I promise.

Because they only do it 100 times and a freak occurence happens and they declare they don't trust it and quit.

I dont understand how intention can affect the probability. Could you explain it a bit more?

Nipun, your question makes a valid point. Many, many persons on these Monty Hall videos make the claim that if Monty forgets where the goats are, takes a guess, and luckily picks a goat door, that that makes the option to switch a 50/50 proposition. Those who make that claim are absolutely wrong. It is still to the contestants clear advantage to switch. The problem above about booby-trapped doors is identical to the Monty Hall problem. The person trying to find the exit should switch when his friend shows him a bad door.

IE Only. Didn't work in Chrome for me. Didn't try Firefox. Very nice BTW. An excellent way for people to easily do this programmatically. Is your simulator an argument for 50-50? I received precisely the results expected. This is a simple problem with very well defined parameters. There are indeed *two* choices here. This first is which of three doors to choose. You can choose any of the three. That choice can result in a goat or a car. The second choice is to switch or not. Don't try to somehow remove the first choice. It's really there. If you don't switch, your odds are 1 out of 3. So if you *do* switch, how can that magically change you from a 1 out of 3 probability universe into a 1 out of 2 probability universe?

Kirk Bocek It says it doesn't work, but just click and then it does. The odds remain 1/3 for the first card, and increase to 2/3 for the unpicked card, the Monty Hall Problem is true! I spent quite a while arguing against it, but I was wrong. The "key" is that it is NOT random, the removed card is known, thus the second round isn't "random chance" at all!

5Cats After, what, 3 years, you are the first, the very first to allow his mind to be changed and to admit it. You are a god among men, an angel among demons. You are the very force of order in the universe. But I gush after having been hammered here for so long and watched others receive the same treatment. You are correct. The door being opened is not a random event. Monty knows which doors have goats and shows you one of them. There is always one available even if you have selected one with your initial choice. But again, it is not "removed." It is simply opened. It is information given to you. There are still three doors. That's why the odds remain 1/3 to 2/3.

Yes! The "it's not random" part threw me for quite a while. If it were random, the odds would be 50-50, BUT you'd get knocked out 1/3 of the time! Since you cannot get knocked out, the odds do indeed carry over from round 1 to round 2, giving you 1/3 and 2/3. The "Gambler's Fallacy" only applies to random games of chance, not the MHP, this also confused me. I am indeed a cat among the mice, Lolz!

5Cats At the outset you don't get knocked out. Guaranteed. So there is no probability of getting the first round wrong. Or right. It's neutral. It's a non-event with a guarantee that Monty will do what he always does-turn over a goat first. It exerts no control on the second choice, and does not filter down conditionally. Judgment is not passed from an event that's 100% not to knock you out. These are not linked events. After the first flip, there are only two doors to consider. You're guaranteed to get there because of the way Monty plays. No need for a pad and paper. Or cards.

It has been three years and I do not know if you keep thinking the same thing, but just in case. Think of the odds as the approximate fraction of times an event will be fulfilled after numerous repetitions. For example, a coin has probability 1/2 to fall heads or tails, so after 1000 attempts it should have left about 500 times heads, and about 500 times tails. The margin of error must be small. When we say that the probability of winning in the Monty Hall problem is 1/3, we refer when your choice NEVER changes, so after many attempts you would have won about a third of the time maintaining that same strategy. You will always keep the door of the initial election, you will not make again a random selection among the remaining two doors, and I think that is the greatest confusion. The probability of being successful at first was 1/3. If anyway you will not change the choice, in the end you have to win the prize many times as if the presenter had never opened the other door. That action does not affect the result. It's like if you have a box with 99 white balls and only 1 black; grab one without seeing it and keep covered in hand. The probability of having caught the black is 1/100. If someone removes all other balls in the box leaving only one, but you will not have the opportunity to change which is already in your hand, then the probability of having caught the black will not increase to 1/2. In the Monty Hall problem is the same: if you will never change, the proportion of times you win must be the same as if they never give you the opportunity to change. If this is not clear, another way to look at it is thinking that the probability of an event is equal to the number of favorable cases divided by the number of possible cases, only if the cases are equaly likely. For example, if a dice was not balanced, the probability to be out wouldn't be 1/6 for each number, despite they are still six possible outcomes. In this case, the fact that gives greater probability weight to a gate than the other is that yours was completely randomly selected from three possible doors, while the other was selected with the obligation that the car must remain covered. Think about this: when opening the other, does it give you additional information about your door? The answer is no, you could have chosen one with a goat ore one with a car, and anyway it was going to be an additional goat, so seeing that in another door there is a goat does not make yours more likely to have the car. Instead, you effectively have new information about the other door closed. Suppose you had chosen the door 1 and the presenter opened the 3. You know he is forced to leave the car covered. If he did not open 2, it is probably because the car was behind door 2 and not behind 3, and so he was forced to open 3. Or maybe the election between 2 and 3 was irrelevant because both had goats, but this is more difficult, because that implies you chose the correct door at first, which has probability 1/3. Confusion usually comes because some people see the two stages as independent games, when in fact they are not. If at first you had chosen the car (1 case), changing will make you lose. If at first you had chosen a goat (2 cases), changing will make you win. As seen, if changing makes you win or not depends on your first election. If in the second stage the choice was randomly, then the probability of winning would be 1/2. To understand why this doesn't contradict previous conclusion is that it is good to think the odds as a fraction of times you succeed after numerous attempts. Making randomly, about half of the time you pick the same door you chose at first, and about half of the time you pick the door the presenter left closed. This is because randomly there is nothing that forces to pick one door more times than the other. In the half you pick the same door you chose at first, 1/3 you should win, but that represents 1/2 * 1/3 = 1/6 of the total, and you should lose 2/3 of the time (2/6 of the total). In the half you chose the presenter's door, you should win 2/3 (2/6 of total) and lose 1/3 (1/6 of the total). C: Changing N: No changing N C Win---> 1/6 + 2/6 = 3/6 = 1/2 Lose--> 2/6 + 1/6 = 3/6 = 1/2

As I've said here, again and again, don't try to do this puzzle in your head. *Actually* do the cards and see what results you get. *Then* come back and tell us what you get.

Kirk Bocek done with paper, actually :-|. my point isn't that the monty hall problem is wrong or that kinda stuff. what i want to say is the trick is actually in the 1st assumption (en.wikipedia.org/wiki/Monty_Hall_problem#Standard_assumptions); and only through that assumption, the whole thing gets a paradoxical look. and when u get what this assumption does to this theory, sadly enough, all charms will just be wiped away.

Md. Abdullah Al Muyid Yes, but if you leave out those assumptions, it is no longer the "Monty Hall problem", and so it is irrelevant what the chances without those assumptions would be. You should not really think of those points as "assumptions", they are the rules which HAVE to be followed for any discussion to make sense. Same as that the host has to actually KNOW what is behind each door, so that can can ALWAYS show a goat and does not accidentally open the door with the car behind it.

Oh! I missed that part! That's kind of funny, eh? Just came to look at that recent comment. It still "bends my brain" a little but it's still 1/3 and 2/3 after the 'extra' card is removed. It was 1/3 for your card in BOTH rounds? Perhaps it's easier to understand it like that? If it's 1/3 with 2 other cards, then each card is 1/3. If it's still 1/3 with ONE other card? That card other must be at 2/3 odds. That makes sense? I hope it can help someone grasp this slippery problem.

5Cats I think you have it. The mistake that I and most others make is to "remove" the first door when it is opened and somehow think that now only two doors are involved when we make our choice. Three doors are always involved.

Yes, only when you choose prize in your initial door, host will have opportunity to choose between two doors with the goat, while in other situation, when the goat is behind your door, host can only offer you to switch to the door with a prize. And that (host left with no option but to offer you door with the prize) happens twice as much (66%) than situation where host can choose between two doors with a goat behind (33%).

Good point.

lol

You are a complete idiot.