Monty Hall Problem

When I was initially exposed to Monty Hall, what confused me the most is that I didn't know that the host would always open a door with a goat, rather than just picking one of the remaining doors at random. In the latter case (i.e. the case where the host can pick a car), even provided that you are in the universe where he picked a goat, switching and staying both yield a 50% chance of success.

I have fun trying to explain this problem. Let me see if I can give an explanation that will help you.So lets say you have just picked a door in the beginning. You know you have a 1/3 chance of being right.If I then tell you, "I will give you two options... you can either bet you are right, or bet that you are wrong"You would obviously choose to bet you are wrong, correct? Because you know you only have a 1/3 chance of being right with your guess, which means you hav

I'll try one more time. Let's imagine there are two worlds, A and B.In A, Monty Hall behaves like you think: always opens a goat door, always gives the option to switch.In B, he behaves like I described: opens a goat door and gives the option to switch, but only when the player has chosen the car door. Otherwise he does not let you switch.And then we find ourselves in this situation:> Suppose you’re on a game show, and you’re given the choice of three doors: Behind one d

Nope. Here are two ways to see how this works.First, let's make it clear exactly what variation of the game we are playing.1. Prize is assigned to a random door with each door being equally likely. Neither you nor Monty know which door.2. You pick a door. Because the prize was assigned randomly and you don't know where it is, it is irrelevant how you pick your door. Without loss of generality (WLOG) we can assume you always pick door #1.3. The host picks a door and opens it

Understanding this problem means you can differentiate between the two scenarios below:(1) If the host reveals an empty door at random, your odds are 1/3 and you should switch.(2) If the host reveals a door at random that happens to be empty, your odds are 1/2 and it doesn't matter.The reason that so many people get confused is that they read the words and their minds automatically jump to scenario 2.

I don't know what to tell you at this point. From the wikipedia article:> Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice. This would be true if the host opens a door randomly, but that is not the case; the door opened depends on the player's initial choice, so the assumption of independence does not hold.Random means there's a possibility the host opens the door containi

The assumption is that Monte will only reveal the one of the two unopened doors that has the goat behind it, as opposed to picking a door at random (which may be the car or may be the door the participant chose, which itself may or may not be the "car door").The distinction is at which point Monte, assuming he has perfect knowledge, decides which door to reveal.In the former, the chance to win is 2/3, in the other 1/2. However in any case, always (always meaning: in eac

The way he presented it there's a 50% chance that opening another door reveals a goat. I've of the other doors will always contain a goat The key piece of information that is merely implied by the Monty Hall problem but not explicitly defined is that the host knows which door is the winning door and deliberately opened a losing door. If the host does this purposefully then you should switch your answer because two thirds of the time you will have chosen a goat and the

Switching doesn't change where the prize is more likely to be. That's not how the Monty Hall Problem works. It's the host giving you information about what's happening behind the scenes, by choosing a door to open and you knowing how the host operates.Whether you end up switching or not, the other opened door is 2/3 likely to contain the prize and the door you picked originally has a 1/3. Whether you switch or not, doesn't change the odds.

I am not able to follow the authors proposition.If the host picks randomly, then on some occasions he reveals the car and the game cannot proceed to conclusion. In other words the door picked by the host reveals a goat, by definition. So what the host knows is irrelevant except to permit the game to proceed. This leads to the contestant facing consistent odds.My understanding pertains to the Monty Hall problem as described by Wikipedia, in which game it always makes sense to take the host&