Because the probability doesn't get shifted onto the door.
Probability exists in the mind, not in reality. In reality, two of the doors have 0% chance of having the car while one of the doors has 100% chance of having the car. A coin doesn't have 50/50 chance of coming up either side, it has a 100% chance of coming up the way it will come up and a 0% chance of coming up the other way.
Only your state of knowledge,* as influenced by the evidence presented to you, can have probabilities different from 0 and 100%, because you don't have all the evidence.
So, to start with, in your picture: Door 1 has .33 chance of having a car, and so does door 2, and so does door 3.
Picking any door does not change this, although it is helpful at this point to conflate doors 2+3 to collectively .66 chance of having a car, and 1.0 chance of having at least one goat.
Now, a goat is revealed in group 2+3. What is this evidence for?
Basically nothing. You already knew there would be at least one goat in group 2+3.
Since you have no new evidence, you have nothing to update probabilites on, so your probabilities are still .33 for your door and .66 for the other two doors.
*And quantum outcomes, if the collapse interpretation is correct.
The general rule is the probability of winning when switching is (n-1)/n for n > 2.
Not sure why you see that the rule applies for n = 100, but not for n = 3. Of course if n = 2 the host either reveals you've chosen the right door (you win) or that you've chosen the wrong door (so you switch), giving a 100% win rate.
>>6658278 its actually a very poor explanation, because it just reframes the exact situation with a different number, hence the same question is asked with no new information.
in other words, it is just as easy to assume that the two remaining doors have an equal chance of being right. considering they both started out with 1/100 chance, now that they are the only ones remaining, why arent they both 1/2?
it takes a more careful approach to explain why this is not true, ie examining the probabilities as they evolve.
>>6658344 It's a good explanation because the issue here is false intuition.
Perhaps it's really easier if you write it all down with a probability tree. You have your preliminary choice and the rest. Yours has the probability n and the rest has the probability 1-n. This is true no matter how 1-n looks. The elimination of all doors but one is nothing more than the guy saying "Look, instead of opening that door, you can open AAAAALL of these, which have a 1-n chance of having the goat in it." The reason the chances don't change is because you already knew that at least all but one of them had to be losses. Opening them does not change this. It also does not eliminate the choice with n you made. What it does change is the way 1-n looks.
Apart from all of this: It is demonstrably true. So whatever.
>>6658347 what? the chances of you randomly picking correctly are much less than you randomly picking incorrectly on your initial choice. when he throws away 98 doors, you are left with the obvious difference of
a) i was insanely lucky in the beginning, and if i switch i will lose
b) it was overwhelmingly likely that i was wrong and if i switch i WILL win.
its an obvious choice and the explanation is a good one for pointing it out.
>>6658466 >http://pastebin.com/mKMhfMJx line 31. The problem is that the host does not randomly open a door. He KNOWS where the prize is, so he will always open a door that does not have the prize. Great that you are taking the experimental approach!
lol, but it's demonstrably true that switching results in a win 2/3 of the time. So whatever argument you're using (which I can barely read, so screw that) is necessarily flawed. What are you trying to accomplish?
Well, the only time you lose if you switch is if you get the door with the prize at the start. Since there are 3 doors, that's a 1/3 probability. Since there is a 1/3 probability of losing, there must be a 2/3 probability of winning.
>>6658344 You're a fucking idiot if you actually believe this. The problem people have with Monty Hall is conceptual, not mathematical. When you show them conceptually why you have a greater chance at winning when wrong doors are revealed, they can grasp the problem better.
Showing people like this a probability tree is as useless and showing a young earth creationist a skeleton of Lucy.
Also the Monty Hall problem is identical with any n number of doors > 2 when n-2 doors are revealed. If you don't understand that you're just as fucking dumb as OP.
>>6658757 No you failed to take into account that you were retarded. This is like 101 trolling post here, literally every day. Like there is no argument in this but the fact that you thought there was is hilarious.
>>6658754 Never judge a fish based on its ability to climb trees, otherwise it'll spend the rest of its life believing it's stupid. People like you that don't give other people a chance because you believe you're above everyone for as long as you know something they don't, need to die. I hope you have children and they watch you die in a fire. Or at least get your head caved in by someone you attempt to belittle.
>>6659056 Or maybe he expects that people lurk for at least a day before posting, as it stands this troll gets posted pretty much every day. Oh the people these days, literally expecting posters to know the board instead of just starting instantly to spam their retarded opinions.
Replace the 3 doors with the same set up except there are 1 million doors. The host removes 999,998 of them and asks if you want to switch. The odds you picked the right door in the first place are 1 in 1 million. You switch with absolute certainty. The probability that switching is successful is 999999/1000000.
Then you repeat it each time with 100000, 1000, 100, 10, etc. until you get to 3.
>>6659009 That's the point you stupid fuck. They say "one" because there is only "one" that is not chosen. The 100 doors explanation is good because it makes it very clear that "one" is just the stand-in for "All not chosen", when intuitively it feels like it's important that it's just one. It isn't; it's important that it's "all the doors not chosen", and the fact that it's "one" serves to obfuscate that.
I just looked up the rules and you are, in fact, wrong. If there are only two briefcases remaining and the top prize hasn't been revealed yet, it is equally likely to be in either case. The probability distribution is very different from the MH problem.
>>6658344 The point is, that it's more likely that you'll select the wrong door to begin with. If you're more likely to pick the wrong door and only wrong doors are eliminated, then it's more likely that the uneliminated door is the correct one.
>>6658254 Because there is a 2/3 chance you were wrong the first time, knowing what was behind one of the 2/3 doesn't increase you probability to 1/2. That door is still there whether or not you acknowledge it, and it still factors in the probability. Another way to look at it, is if you could choose either two doors, or one, you'd probably pick the one. But if I told you that door one was had a goat, you still have both doors.
>>6658446 It's to show that just because there is only two doors left, doesn't mean it's 50/50. Our minds have a hard time seeing that, but in this case it's obviously true, for the same reason it's not so obviously true in the three door example.
>>6663670 >knowledge does not increase probability Er... I don't think your argument works.
It's better to look at it as groups. You have your chosen group (one door) and the unchosen group (two doors). The probability that the car is in the unchosen group is 2/3. Showing you which of the doors does not change this, since you already knew at least one of them contained a goat. This does not change the fact that the probability of the defined groups. But it does change the probability within the unchosen group.
I don't know why we are still discussing this. It's a proven fact. Both mathematically and demonstrably. It has been explained from every angle. If you still don't get it, give up or read a book.
There is only one scenario where win the prize after switching: having guessed a goat in the first place. Guessing either door with a goat and then switching results in a win. The choices are independent, so the additive rule says that the probability of making either choice is the sum of their individual probablityes: 1/3 + 1/3. You have a 2/3 probability of winning if you make the switch.
The host flips a coin. You are asked to select the result. The host reveals that the coin has NOT landed on its side and asks whether or not you'd like to switch your answer. What is the probability you would win the toss if you switch your answer?
Three ( or N ) eggs are in the bowl. Only one of them is hard boiled. You select one egg, put it on the table. The chance that the table egg is hard boiled is 1/3 ( or 1/N ). The chance that the bowl has the hard boiled egg is 2/3 ( or (N-1)/N ). The show host breaks one egg ( or N-2 eggs ) from the bowl. The chance that the bowl has the hard boiled egg is STILL 2/3 ( or (N-1)/N ).
1) The host does NOT open a random door, the opened door always reveals a goat.
2) If the problem had 10 doors instead of 3, after you pick a door, the host would open 8 doors instead of only 1 door.
3) Question: How would the problem work (still with 3 doors) if the host randomly opened 1 of the 2 remaining doors, and it HAPPENED to be the one of the empty doors (or the one with the goat, etc.)?
Answer: The state of the host's mind doesn't matter when the goat is revealed. So for that one time, nothing new - it still pays off to switch.
For the times (50/50 chance) that the host randomly reveals the car, the show doesn't make sense.
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