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Post by faustus5 on Jan 15, 2019 22:25:15 GMT
So you know that behind three doors there is either a goat or some other bogus "reward", or a sports car--two goats total, one sports car. The MC knows which is behind which door. You pick which one to open--you'll get what is behind it.
But before the MC opens the door of your choice, he chooses to open one door you had not chosen--and behind it is a goat.
He offers you a chance to change your mind--should he open the door of your original choice, or do you want to pick the other closed door?
What should you do? Does it make a difference?
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Post by CoolJGS☺ on Jan 15, 2019 22:30:39 GMT
It doesn’t make a difference beyond our risk threshold.
The odds did not improve just because he opened a door, our hopes just went up.
It reminds me of the Deal or No Deal Show where the final prize was decided at the onset and the rest of the showings uploaded to stir confidence in that first pick.
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Post by phludowin on Jan 15, 2019 22:46:34 GMT
The Monty Hall problem is not dead, I see. In Germany it's called the "Ziegenproblem" (Goat Problem).
It makes sense to change to the other closed door. But only if we know that the moderator (Monty Hall) always plays by the same rule. Meaning: He always opens a door, always revealing a goat. And he knows where the goats are. If he opens the door at random, and sometimes reveals a car, then it doesn't matter if you change or not; because then the odds are 50/50. And if he is known to reveal a goat door when you yourself picked the car; but the car door when you picked a goat, then you should stay.
But I doubt that such a dull game show would ever be aired.
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Post by Aj_June on Jan 15, 2019 22:57:40 GMT
So you know that behind three doors there is either a goat or some other bogus "reward", or a sports car--two goats total, one sports car. The MC knows which is behind which door. You pick which one to open--you'll get what is behind it.
But before the MC opens the door of your choice, he chooses to open one door you had not chosen--and behind it is a goat.
He offers you a chance to change your mind--should he open the door of your original choice, or do you want to pick the other closed door?
What should you do? Does it make a difference?
Your question may not be very clear to people who are not familiar with it. It could be stated this way better:
Once you have selected one of the three doors then the game show host will choose 1 door. Had you originally chosen the door with the car then the host will choose to open 1 of the two doors with goats behind them with equal probability. Had you originally chosen one of two doors with goats behind them then the host will not choose to open one of the two doors with equal probability. He will know which one has car and and which one has goat behind it and deliberately choose the door with the goat behind it (with a 100% probability).
Here you go now!
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Post by Catman on Jan 15, 2019 23:00:59 GMT
On Mythbusters, they confirmed that changing doors was the better option.
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Post by Aj_June on Jan 15, 2019 23:04:19 GMT
On Mythbusters, they confirmed that changing doors was the better option. The thing to understand is that the selection of the door by the host depends upon your original selection. If you originally selected the door with the car then the host selects one of the two doors with a different probability than he would if you originally selected one of two doors with goats.
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Post by Aj_June on Jan 15, 2019 23:05:12 GMT
Arlon10 sir. I do love to see your answer.
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Post by CoolJGS☺ on Jan 15, 2019 23:39:16 GMT
So you know that behind three doors there is either a goat or some other bogus "reward", or a sports car--two goats total, one sports car. The MC knows which is behind which door. You pick which one to open--you'll get what is behind it.
But before the MC opens the door of your choice, he chooses to open one door you had not chosen--and behind it is a goat.
He offers you a chance to change your mind--should he open the door of your original choice, or do you want to pick the other closed door?
What should you do? Does it make a difference?
Your question may not be very clear to people who are not familiar with it. It could be stated this way better:
Once you have selected one of the three doors then the game show host will choose 1 door. Had you originally chosen the door with the car then the host will choose to open 1 of the two doors with goats behind them with equal probability. Had you originally chosen one of two doors with goats behind them then the host will not choose to open one of the two doors with equal probability. He will know which one has car and and which one has goat behind it and deliberately choose the door with the goat behind it (with a 100% probability).
Here you go now!
Where am I going?
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Post by Aj_June on Jan 15, 2019 23:42:54 GMT
Your question may not be very clear to people who are not familiar with it. It could be stated this way better:
Once you have selected one of the three doors then the game show host will choose 1 door. Had you originally chosen the door with the car then the host will choose to open 1 of the two doors with goats behind them with equal probability. Had you originally chosen one of two doors with goats behind them then the host will not choose to open one of the two doors with equal probability. He will know which one has car and and which one has goat behind it and deliberately choose the door with the goat behind it (with a 100% probability).
Here you go now!
Where am I going? LOl...
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Post by Vegas on Jan 15, 2019 23:49:14 GMT
The answer to your question is: YES. You change doors. When you picked your first door the odds were 1 in 3. They are now 1 in 2... Your pick: 1 in 3... The other door: 1 in 2.. Your odds are better if you switch. Thank you, Monty Hall.
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Post by CoolJGS☺ on Jan 15, 2019 23:52:45 GMT
The answer to your question is: YES. You change doors. When you picked your first door the odds were 1 in 3. They are now 1 in 2... Your pick: 1 in 3... The other door: 1 in 2.. Your odds are better if you switch. Thank you, Monty Hall. That makes sense. You reset the risk. I still wouldn't change though. It just wouldn't be that big of a deal for me. I prefer Jeopardy to chance shows.
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Post by Aj_June on Jan 16, 2019 0:02:10 GMT
The answer to your question is: YES. You change doors. When you picked your first door the odds were 1 in 3. They are now 1 in 2... Your pick: 1 in 3... The other door: 1 in 2.. Your odds are better if you switch. Thank you, Monty Hall. That's a good way of explaining it. Much different than I am used to. Well done!
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Post by Arlon10 on Jan 16, 2019 0:11:19 GMT
Arlon10 sir. I do love to see your answer. The chance the first door has the prize is 1/3. That part is easy. The next part is tricky. It makes a difference whether the host knows where the prize is. In the given problem he does know. The chance the prize is behind one of the two remaining doors is 1 - 1/3 = 2/3. Because the host knows which door has the prize his opening of a wrong door has no effect on the 2/3 probability, it is still 2/3. You benefit from his certain information. The contestant should then choose the door he did not at first since its probability is 2/3 compared to 1/3 for his first choice. If the host chose a remaining door at random without knowing which has the prize it would be different. The probability the door he chose has no prize is 1/2 (one of two doors) times 2/3 (the total probability for the two) or 1/3. The probability the other of the two has the prize is also 1/3. If the door the host opens has no prize you now have two choices which had an equal probability of 1/3 which in present conditions would be 1/2. This is the reason people often assume it's an even chance in the problem as given. That can be confusing and difficult to understand so I wrote another way to look at it. Suppose there were 100 doors. The chance the contestant's first choice has the prize is 1/100. The chance the prize is behind one of the remaining doors is 99/100. Again it matters whether the host has information about where the prize is and uses that to open doors where the prize is not. If he knows where the prize is he can open 98 doors without changing the probability the remaining door has the prize is 99/100. The contestant should pick the remaining door then since the probability is 99/100 it has the prize. Here too, it would be different if the host randomly opens doors.
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Post by Rodney Farber on Jan 16, 2019 0:46:56 GMT
Switch ...
If the door you originally pick had a booby prize and you switch, you win.
If the door you originally pick had the grand prize and you switch, you lose.
What's the probability that you originally pick a door with a booby-prize?
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Post by goz on Jan 16, 2019 1:13:49 GMT
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Post by Arlon10 on Jan 16, 2019 1:27:27 GMT
The answer to your question is: YES. You change doors. When you picked your first door the odds were 1 in 3. They are now 1 in 2... Your pick: 1 in 3... The other door: 1 in 2.. Your odds are better if you switch. Thank you, Monty Hall. That's a good way of explaining it. Much different than I am used to. Well done! Except the original door has not been opened yet so the 1 in 2 has not been justified.
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Post by Aj_June on Jan 16, 2019 1:29:22 GMT
That's a good way of explaining it. Much different than I am used to. Well done! Except the original door has not been opened yet so the 1 in 2 has not been justified. It has been opened. They found a man suffering from Dunning-Kruger inside it instead of a goat.
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Post by general313 on Jan 16, 2019 1:31:35 GMT
That's a good way of explaining it. Much different than I am used to. Well done! Except the original door has not been opened yet so the 1 in 2 has not been justified. Actually the odds of winning if you switch doors is 2/3.
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Post by goz on Jan 16, 2019 1:35:26 GMT
Except the original door has not been opened yet so the 1 in 2 has not been justified. Actually the odds of winning if you switch doors is 2/3. On second thoughts and consulting my statistical evidence, can I please have the two goats, as one would need a friend?!?! That is the immutable theory of goat statistics help by polling goats.
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Post by Aj_June on Jan 16, 2019 1:39:39 GMT
Except the original door has not been opened yet so the 1 in 2 has not been justified. Actually the odds of winning if you switch doors is 2/3. Indeed. But Vegas's non-mathetmatical explanation could be be better understood by people not knowledgeable in probability theory. This type of tabular explanation could make it even easier to understand.
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