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Message: [QUOTE="SWC75, post: 5083197, member: 289"] I shuffle the three cards multiple times- when I first put them in groups of three. When I take a group of three for the next round and when I put them down, which is not in the order they were stacked in at that point. I look at the third card first. if it's black, it's not the Monte Hall scenario. I put those cards aside. If it's red, then the goat is behind the third door. That's the Monte Hall scenario. I then overturn the second card, then the first and record how many times the second card is black and how many times the first card is black. Through 53 times so far, the first card has been black 29 times. In the 1/3 vs. 2/3 scenario I should count the times the third card was black as if the third card was the second card. The third card has been black 25 times. That makes it 29-49, the 2/3 chance that you are looking for. But in 25 of those 49 times, the black card was the third one. When I first dealt with this, (it was 2008), I came up with three arguments for why the odds would change: The supposed answer to this is that when Monte opens the third door the odds that the car is behind door #2 go up but the odds that the car is behind door #1 don’t. What if Monte then opened door #2 and there was a goat there? Would the odds that the car was behind door #1 remain the same? Of course not- they would go to 100%. Therefore the opening of a door to reveal a goat changes the odds of the car being behind door #2. The opening of door #3 must have also changed the odds that the car was behind door #1. What if there were two doors? One car and one goat. The odds are obviously 50-50 that the car is behind door #1. Suppose then that Monte rolls up a third door, opens it and there’s a goat. What difference would that make? What’s different between this and there being three doors initially and Monte opens door #3 and there’s a goat there? Suppose the producer of the show, in a production meeting before the start of the season said “We’re going to have 99 contestants on this year’s show. To be fair to them, we’re going to put the car behind door #1 33 times, behind door #2 33 times and behind door #3 33 times. So when Monte opens door #3 to reveal the goat, what does that tell us? It tells us that this isn’t one of the 33 times the car is behind door #3. Instead, it’s one of the 33 times it’s behind door #2…or one of the 33 times it was behind door #1. It can’t be one of the 66 times it’s behind door #2 because the producer didn’t decide to put it behind door #2 66 times. I abandoned those arguments because I did the card game twice, with a single deck of cards and the results pointed to the 1/3 vs. 2/3 argument. I've now done it three times with two decks of cards and I'm at 29 for card #1, 24 for card #2 and 25 for card #3. Toss out card #3, (open that door), and it's 29-24. I'll do it a third time: 26 hands, 10 times the black card was #1, 9 times it was #2 and 7 times it was #3. It was #1 10 times and 16 times it wasn't - but 7 of those times it was #3. So Monte would not have asked for door #3 to be opened those seven times. Would he have asked for door #2 to be opened the other 9 times? Maybe. But the opening of a door doesn't add 7 cars to door #2 or 9 cars to door #3. Anyway, through 72 Monte Hall situations, (a red card, meaning a goat) was #3), #1 has had the black card, (a car 39 times. #2 has had the black card 33 times. Sherlock Holmes said “Eliminate all that is impossible, whatever remains is the explanation, however improbable.” I was willing to accept the 1/3 vs. 2/3 argument in 2008 because I did this with a deck of cards twice, (26 'hands'). I've now done it with two decks of card four times, (104 hands) and it doesn't hold up. With three doors, you are going to get a card behind each door about 1/3 of the time. Of the times, If it's not behind the third door, the odds it's behind either doors #1 or #2 both increase. Try it yourself. [/QUOTE]

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