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I have no doubt that the probabilities are $2/3$ by switching; my problem is trying to explain it to someone. His first point is the two choices are unrelated events. My argument is:

$1$) If at first you chose a door with a goat ($2$ cases of $3$), then for the second choice you will have a goat in your door and the car in the other. You win by switching.

$2$) If at first you chose the car door ($1$ case of $3$), then for the second choice you will have the car in your door and a goat in the other. You win by staying.

Textually, what he answered was: (He calls $E1$ to the first event, ie to the first selection, and $E3$ to the second selection)

"You could not have picked a Goat or Car in E1, it's impossible, the Game assigns no value to Your E1 pick, it's an unknown by Game definition. What is known is the Value of each Door in E3.

The difference in definition between E1 as a Constant or E1 as a Variable is the crux of the issue. All previous definitions of The Monty Hall define the Result of E1 as Variable. The Value of E1 is irrelevant. E1 as a Constant is why E3 is an independent event. With Odds calculated as Roulette, not as BlackJack."

The important thing about his argument, which I had not seen from other people, is that he thinks that the fact that you can not see what is behind the first door picked makes it completely irrelevant, and so the two choices are separated events.

Ronald Becerra
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    Ask him to clarify. His response doesn't make any sense, even as an incorrect idea. (I'm assuming it's his response that is inarticulate, rather than your description of his response, because the rest of your post is quite articulate.) – Wildcard Jan 06 '17 at 05:55
  • You might try saying, "Is a closed door a 'constant' just because you don't know what's behind it? The host knows." – Wildcard Jan 06 '17 at 05:56
  • If there is any doubt, I will copy his answer exactly as he put it. – Ronald Becerra Jan 06 '17 at 05:58
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    The whole issue with Monty Hall lies in the fact that Monty Hall never opens a door with a car behind it. If he always chose one of the remaining two doors at random, and sometimes chose a door with a car (and then forbade you to choose that door, of course), then switching and staying would be equivalent. – Ian Jan 06 '17 at 06:00
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    I like this answer that uses the technique of imagining the game with more than 3 doors. – pjs36 Jan 06 '17 at 06:02
  • I'm sorry but his answer is utterly incoherent. Not even wrong. Incoherent. You have three possibilities: E1) You picked the goat called martha. The host shows you the goat called Bill. You should switch. E2) You picked the goat called Bill. The host shows you the goat called Martha. You should switch. E3) You chose the car. The host randomly either shows you Bill or Martha. Doesn't matter which; You shouldn't switch. – fleablood Jan 06 '17 at 06:39

3 Answers3

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The rules are that Monty knows where the car is. And Monty always reveals the location of a goat. And that is important. If Monty doesn't play this way then the game may be different. And I have tried to explain this to people with good math skills and they still didn't get it.

When Marilyn Vos Savant first explained the answer she received many angry letters from math and statistics professors telling her she was wrong. The problem is not intuitive.

Take a pack of playing cards. Tell your friend that he wins if he can find the Ace of Spades. Spread the cards on the table and ask him to take one. Look at the rest. Separate out the Ace of spades. Turn the rest face up. Show that the ace is not in the pile of face up cards. Then ask if he want to keep his card or he wants to switch. If he still thinks there is only a 50% chance of winning, and nothing to be gained by switching you should be able to work out a wager.

Doug M
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  • Great practical illustration, no imagination involved. Incidentally a link for the Marilyn Vos Savant story would be nice. – Wildcard Jan 06 '17 at 06:23
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    @Wildcard here you go http://marilynvossavant.com/game-show-problem/ – Doug M Jan 06 '17 at 06:28
  • Ok, his response is that example is "so far from the Monty Hall problem as to be laughable". I will not continue on this. – Ronald Becerra Jan 06 '17 at 06:33
  • @RonaldBecerra If he thinks that then he doesn't understand the Monty Hall problem. What has been written here (assuming the other player doesn't look at what the dealer is doing) is exactly the same scenario as the Monty Hall problem except that there are 52 doors, 1 of which has a car, 51 of which have goats, and Monty Hall opens 50 doors all of which are required by the rules to show goats. – Ian Jan 06 '17 at 07:00
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Take it to the big limits.

Suppose there were 100 doors (and remember that Monty knows which door has the car behind it).

The contestant guesses door 100. And Monty opens door 1, door 2, door 3, ... but skips over door 38, and goes all the way to door 99 which is opened. And all open doors have a goat behind it.

Only doors 38 and 100 are left unopened.

Behind what door do you think the car is?

robert bristow-johnson
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One of the three closed doors has a car behind it. The probability that I point at the correct door is $1/3$. This means the probability that Monty's door is the correct one is $2/3$.

Austin Mohr
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