261–270 of 331 posts
I'm no way an expert, so even if Ms. Savant beautifully explained it; my puny brain thinks... "how is it different from starting the game with 2 doors?". If you choose again, how is the probability not 1/2
There’s a bit of historical revisionism at play in the article. Here’s an accurate representation by Aaron Brown of what transpired: https://www.quora.com/Why-do-some-PhDs-argue-against-Marilyn... Quoting for those who don’t want to go off-site: “You have to have lived through this ancient travesty to care about it. It’s probably best forgotten. Marilyn Vos Savant published an incorrect answer to the Monte Hall probl…
Earlier quoted context omitted.
Can you help me understand - in my view the choice to switch doors or keep the same door is irrelevant, because even if you keep the same door you're making a choice that is now 2/3 of the right answer. If you switch or keep, you're still choosing from two doors that contain a car and a goat. The other door is no longer relevant and doesn't affect the new state at all. It's your perspective (narrowing the choice down…
Here's another thought experiment. Forget the opening of the door. Start with picking a door at random, you have a 1/3 chance of having picked the car. On that I think we all agree. Now let's say that the host offers to let you switch from the door you picked, to the other _two_ doors combined. He hasn't opened any doors, they are all closed, you're allowed to stick with your initial guess of one door, or switch to a…
Let's say we buy four scratch-off tickets. The clerk selling the tickets assures us that one of the four is a winner. You choose two tickets, and I take the remaining two.
I scratch one of my tickets, revealing that it is a dud. At this point, I suggest we should swap your two tickets for my two tickets. You start to object, but I point out that this is just like the Monty Hall problem.
When we made our initial choice, there was a 2/4 chance the winning ticket was in the set you picked, and a 2/4 chance that it was in the set that I picked. The fact that I revealed a ticket didn't change that. So you should be willing to trade, as both sets of tickets are equally likely to contain the winner.
You refuse, obviously. The difference is that whether there's two goats behind his doors or one, Monty will always choose a goat. He provides no information about the door you've already chosen, as the probability of him choosing a goat is unchanged based on whether you picked correct or not. I, on the other hand, could have revealed the winning ticket, but the odds of me doing so would be lower if you already had it in your possession. So, my dud updates the probability of all remaining tickets.
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It's great that you're so sure of yourself, just a shame that you're wrong. If the host picks randomly, your chances are in fact 50-50. Think of it this way: imagine every time the hosts picks the car the universe resets. Now imagine you see the host picking a goat. There's a 2/3rds chance in your universe you picked the car, and a 1/3rds chance you picked a goat, and so switching would seem like the bad option. This…
It's funny to me that there always seem to be an over representation of rude people among those who get this problem wrong. There is absolutely no need to attack my person. As for the problem, you are just plain wrong. We are talking about are regular old TV-quiz, there is not "universe resetting button" to save your logic. In the regular situation, the host always remove a goat. because otherwise the quiz show is ki…
When the host does not know which door to pick, there's a 1/3 chance that you picked right, a 1/3 chance that the host opens the door with the car, and a 1/3 chance that you should switch. The host opening the door with the car ends that branch, so you're left with 2/3 of the original probability, and your odds for either branch remaining are 50:50.
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That is literally not explicit (where by "literally" I mean literally, not figuratively, and by "explicit", I mean explicit, not implicit). It is sort of hinted at, but it is not explicitly said that the host will mechanically reveal a door that has a goat. It is quite conceivable that the host picks a door randomly, or in fact that he picks the door with the car with a certain probability (saving the show quite some…
> It is literally explicit in the 1990 rendition. > You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat Remove the parenthetical example "say #3", and the parenthetical "who knows what's behind the doors" and that sentence reads: "...and the host opens another door which has a goat". The "has a goat" is not a hypothetical example. It's a (both literal…
It's not explicitly saying the former even when stating the result, because everything is written as a certainty in retrospect. You can say, "the roulette wheel stopped on 00," but it was still a random event when it happened.
I think this is what trips many readers up, given that we tend to associate impartiality with game show hosts. A better statement of the problem might be something like "The producers have instructed Monty in advance that after the initial selection, he must open one of the unselected doors to reveal where a goat is (under such instruction he will never open the door with the car.)"
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> In this: "the host will open the door at random and in this example it happen to have a goat" vs "the host will never open a door with a car behind it".) If the host opens a door with a goat, then it doesn't matter whether or not it was intentional.
Yes it does. Or more precisely, it matters whether the host could be counted on to do so reliably; the mechanism for that doesn't matter. There's a difference here, that our language obscures, between procedure and hypothetical.
If the host has revealed a goat door, and the contestant then has to decide what to do, the intentions of the host for having chosen the door are irrelevant.