I understand it, but I have trouble explaining it. Would be cool to see a discussion here.
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Don't know if this has been said yet, but the best way of explaining in my opinion, is to take a deck of cards. I lay all 52 cards on the table face down. If you select the Ace of Spades, you get a car, everything else you get nothing. So you choose a card randomly, in which case you have a 1/52 chance of it being the Ace of Spades. Now I look at the table and take away 50 of the remaining 51 cards, none of them are the Ace of Spades. So either you have the Ace of Spades, or it is the remaining card. Would you switch?
Of course. Your card had a 1/52 chance at the start, and that probability doesnt change. Therefore the other card must have a chance of 51/52.
Basically, the more choices there are, the easier it is to explain. If I have you choose one out of a million numbers, one of which will be a win (which is set at the start of the game), it's pretty obvious that if you choose a number you have a 1/million chance. But if I remove 999,998 of the incorrect numbers, the one left over obviously has a 99.9999% chance of being correct.
The other way of explaining is to literally list out all of the possibilities.
A) You choose goat 1, revealed door is goat 2, other door is car. ->switch you win
B) You choose goat 2, revealed door is goat 1, other door is car. ->switch you win
C) You choose the car, revealed door is a goat, other door is goat. ->switch you lose
So you can see that you have a 2/3 chance of winning if you switch
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This by far the best answer I ever came across. Thank you.
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So there is no advantage to switch? If so, you are incorrect. In fact, you are twice as more likely to win a car instead of a goat if you switch choices. Think about it.
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It is irrelevant. You first choose and you got a 33% chances to win. When he opens the other door and there are 2 left well then you have 50% chacnes in any of the doors. Changing your option would be irrelevant since your 33% change to 50$ chance.
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Nope, it's not. You actually have an advantage on switching your choice. I found this paradox extremely interesting because almost everyone would say it's 50/50 at the end which is incorrect. It's actually 66.6% to win a car and 33.3% to win a goat at the end if you switch your choice. I'd want to explain it but I'm not good at explaining things. If you don't believe me you can look it up on Wikipedia which brilliantly explains it.
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care to link the wikipedia article? I'm not sure how far I'm going to get searching wikipedia for "game shows with 3 doors the give you goats or cars". And as far as I can reason, after the goat is revealed behind door #3, there are two prizes left and two doors left. One door hides a goat and one door hides a car. Since there can be no way of knowing which is behind which, you have, at that point a 50% chance of the car being behind door #1 or door #2. I don't see how this can be any different.
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At first you have 66.6% to win a goat, and 33.3% to win a car. If you choose a goat door, and the host opens a goat door, switching is a better idea, because by switching, you change the prize. So, by switching, you'll have 66.6% of winning the car.
Took me a while to find this, it was very interesting, thanks :)
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It's very interesting once you understand it, isn't it? :)
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Sure, if you wish to win a goat just don't switch, then you're more likely to win a goat.
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I'd stick with my original choice of door #1.
It's a 50/50 chance either way and I'm not a flip flopper! >_<
I would actually be happier with a goat, I already have a car that I don't use but a goat would be awesome. I could milk it (if it were female) to make cheese (goat cheese is delicious) and I could train it to butt people I didn't like. ᕦ(ò_óˇ)ᕤ
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If you don't 'flip flop' it will give you a disadvantage if you wish to win a car.
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Not really, I originally picked door #1 (33% chance), the host opened door #3 revealing a goat (upping my chances to 50%).
The way I see it, sticking with my original choice gives me the same odds as changing to the second door, I've always had pretty decent instincts so I go with my "gut" and stick with door #1.
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You have 66% chance at the beginning to choose a goat, and 33% to choose a car. So 66% chances that if you switch you get the car, 33% you get a goat.
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Cant you just hear the goats through the curtains? O_o
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Haven't seen that movie. I can't actually remember how I came upon this.
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A curtain with a goat behind would be much smaller than one with a car. Just pick the bigger one for a 100% chance of success.
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Every time i see this explained, i shake my head because the person explains it terribly. Let me give it a shot.
Alright, when you first pick a door, THE CHANCES OF THERE BEING A GOAT BEHIND THE DOOR IS LARGER THAN THE CHANCES OF THE CAR BEING BEHIND THE DOOR.
So, look at it this way. The car is behind door 3.
You pick door 1.
The host opens door 2.
Door 2 is revealed to have a goat.
This is the crucial part -
Since you were ORIGINALLY more likely to have a goat behind the first door you picked, that means that there was a greater chance of the car being in one of the other TWO doors that you DIDNT PICK.
Since the host opened one of the doors you DIDNT pick and revealed the goat, then taking to account your original chances, it would benefit you to switch.
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Yeah. The important part of this problem is that the host KNOWS what is behind each door, and knowingly opened one with a goat.
If the host didn't know and opened a door that just happened to have a goat, then your chance would not increase by switching doors.
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I saw this on Mythbusters. When you chose the first door, you have 1/3 chance to get the prize and 2/3 chance to get a goat, if you stick you have 1/3 chance to pick the right one. Since the probability of chosing the wrong door is bigger than chosing the right one,when you pick a door you have 1/3 chance to win the car and there's 2/3 chance in the other two doors, when the host opens a door there's still 2/3 chance to win in the two doors that you didn't chose(and one of them is open), so you have 2/3 chance to win if you switch.
P.S. Sorry for the bad english
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Hm, I'll be honest and say I didn't have anything like this at school or read about that, but I guess my idea would be:
You select A, and the host who knows where the answer is picks C, so that eliminates C as one of the three answers. However, between the B and C he didn't choose B, which increases the chances of it being there as if the car is actually in B, he would never choose it, which adds up to the initial 33% chance and makes it a total of 66%, whereas your choice is 33%. Also, if your answer was a right one from the start, the host would probably not even begin the whole acting thing, which could actually become a 100% on the right answer of yours. I hope you understand what I wrote, but yeah, I agree with this theory.
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...and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Answer: Yes, there is an advantage.
Q: Why?
Don't Google this because it's no fun that way.
If you're thinking it's 50/50 at the end you're incorrect. If you give up and can't prove it's not look up 'Monty Hall problem'.
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