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larryshoat

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Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

This was a question I found .

Larry
 
And it's a trick question, isn't it! At least for someone as slow as I am with stuff like this...

Alice
 
Not a trick , but very tricky . This was a question asked of Marilyn Vos Savant and became known as the Monty Hall dilema and sparked a huge controversy and it rages on still today .

Larry
 
buckmaster33":1qjcn0x9 said:
No....because I want the goat!! I have enough cars and trucks.

:lol2: :lol2: :lol2: That may be the berst answer I've ever heard .

Larry
 
I would change my door cause now I can skew the odds.

edit - to clarify I had a 1:3 chance the first go now I got a 1:2 on the second. Sounds odd but changing the door will increase my odds of winning. Kinda like T values.
 
Jogee I think you're pretty close . The way I look at it is :if we choose door #1 we would all agree that the probability of winning the car is 1/3 and if the host always reveals a goat from behind the remaining 2 doors then by switching our odds would be 2/3 . Another way of looking at it would be if we choose door #1 and we agree that it has a 1/3 chance, which in turn means that there is a 2/3 chance that the car is not behind door #1 and if the host removes an incorrect answer from the remaining 2 doors the odds are 2/3 by switching .

We can set up a small game to prove this . Take 2 jokers and an ace from a deck of cards, let the ace represent the car . Have a partner lay out the three cards, then you pick 1 of the three, record whether you won (got the ace) or lost (got the joker), do this like 30 times . Next have your partner lay out the cards in the same way, make your pick, but this time have your partner reveal a joker from the remaining 2 cards, then switch from your original choice to the card that is left, do this 30 times and record the results .

Larry
 
larryshoat":3snvct2z said:
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

This was a question I found .

Larry

Depends, was the goat behind door #3 made by Pontiac?
 
cfpinz":20iuh0r2 said:
larryshoat":20iuh0r2 said:
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

This was a question I found .

Larry

Depends, was the goat behind door #3 made by Pontiac?

:lol2: That's never been my luck, for me when they say goat they don't mean GTO, they mean goat .

Larry
 
larryshoat":2uskyvsk said:
Jogee I think you're pretty close . The way I look at it is :if we choose door #1 we would all agree that the probability of winning the car is 1/3 and if the host always reveals a goat from behind the remaining 2 doors then by switching our odds would be 2/3 . Another way of looking at it would be if we choose door #1 and we agree that it has a 1/3 chance, which in turn means that there is a 2/3 chance that the car is not behind door #1 and if the host removes an incorrect answer from the remaining 2 doors the odds are 2/3 by switching .

We can set up a small game to prove this . Take 2 jokers and an ace from a deck of cards, let the ace represent the car . Have a partner lay out the three cards, then you pick 1 of the three, record whether you won (got the ace) or lost (got the joker), do this like 30 times . Next have your partner lay out the cards in the same way, make your pick, but this time have your partner reveal a joker from the remaining 2 cards, then switch from your original choice to the card that is left, do this 30 times and record the results .

Larry


I know the point of this was to spark debate, but I'm going to play along anyway. If you choose door #1, we all agree the odds are 1:3 you chose the car. Where I disagree is that the host is always going to take away one of the wrong answers (the goat). This leaves you with a new choice between only 2 doors (stay with what you got or take the other). The odds are 50/50 and always will be.
 
Well, I would encourage you to set up the card game and see what the outcome is .

Larry
 
cfpinz":35jkdgk0 said:
larryshoat":35jkdgk0 said:
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

This was a question I found .

Larry

Depends, was the goat behind door #3 made by Pontiac?
:lol: :lol: I had one of those! 1969 and that "olive green" color. She was a runner...
 
Well if the goat would be a 72 GTO judge like I had one time I'd deffinatly take it.
Saw one on the muscle car auction go for a whopping $126,000.

Sorry Larry don't mean to derail your thread or make anyone loose their train of thought.

Cal
 
Larry, this is similar to something I've used in statistics. For instance, when calculating statistical values you will end up with a bell curve. If your coefficient of variation is high, you are apt to come up with some negative numbers in the curve. Common sense would tell you this is impossible since there is no such thing as a negative valued widgit. Therefore you can throw these numbers out and skew the curve. This will in fact move the mean value to the right or - in theory - closer to the real answer. This is similar to opening a door and tossing it out. Some people call this massaging the numbers and this can sometimes get a little scary but surprisingly it works pretty darn good if you keep your wits about you when you do this.
 
Calman":2fwe4g1w said:
Well if the goat would be a 72 GTO judge like I had one time I'd deffinatly take it.
Saw one on the muscle car auction go for a whopping $126,000.

Sorry Larry don't mean to derail your thread or make anyone loose their train of thought.
Cal

No problem Cal, I think the GTO thread is more interesting anyway :D .

Larry
 
Have you ever looked at the Marilyn Is Wrong website? It's crazy. I prefer to go through life believing Marilyn is never wrong.
 
Seems to me just because you made a right choice the first time doesn't increase your odds of making a correct choice the second time since you have to make 2 correct choices.In my experience, I made a correct choice the first time,  now making a second correct choice is unlikely.  :lol: But in the scheme of the game where one wrong answer is going to be removed, are your odds not 50/50 to begin with.Deal or No Deal for example:  26 cases and you chose one.  You make it all the way to the end and they give you a chance to switch cases.  New choice new odds.  50/50 you are going to pick the correct case.  (Not a big deal because I would be choosing between $1 and $10.  :( )
 
ga. prime":170cdai4 said:
Have you ever looked at the Marilyn Is Wrong website? It's crazy. I prefer to go through life believing Marilyn is never wrong.

That's about the way I feel about it too . That website is about people that have something to prove . Marilyn has the highest recorded IQ, it's not like I'm second . I think Marilyn is the greatest and when she gives an answer I spend my time trying to figure out why I came up with the wrong answer .

Larry
 
1982vett":qf89igxl said:
Seems to me just because you made a right choice the first time doesn't increase your odds of making a correct choice the second time since you have to make 2 correct choices.In my experience, I made a correct choice the first time,  now making a second correct choice is unlikely.  :lol: But in the scheme of the game where one wrong answer is going to be removed, are your odds not 50/50 to begin with.Deal or No Deal for example:  26 cases and you chose one.  You make it all the way to the end and they give you a chance to switch cases.  New choice new odds.  50/50 you are going to pick the correct case.  (Not a big deal because I would be choosing between $1 and $10.  :( )

I know vett, that is the exact same way I approached it too . What convinced me I was wrong was the little card game I talked about in the earlier post, after I decided that switching worked, I thought about it for 2 years before I came up with an answer as to why it worked and that is that if I choose #1 and there is a 1/3 chance that the car is there, it also means that there is a 2/3 chance that it is not there, in other words there is a 2/3 chance that the car is behind one of the 2 doors that I didn't choose and if the host always opens a door with a goat behind it then the chance of the remaining door is 2/3 . If when we made a choice on the first door and then just took that door out of play, I think we would all agree that the odds of the car being behind the remaining 2 doors is 2/3, therefore if we eliminate a wrong answer from those remaining 2 doors the chance of the door that is left is 2/3 .

Very confusing I know, I think the trick to understanding this is that the host will always eliminate 1 wrong answer from the remaining 2 doors .

Larry
 

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