Sheep or a Car

chrisms

Diamond Member
Mar 9, 2003
6,615
0
0
Let's say you're on a game show, and you're given the choice of 3 doors: Behind one door is a car, and behind the others, sheep. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a sheep. He then says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice?
 

PhattyB

Member
Oct 20, 2004
160
0
0
There is no advantage. All it does is gaurentee you have a 50% chance of winning some sheep love.
 

chrisms

Diamond Member
Mar 9, 2003
6,615
0
0
Since it was already posted..

http://en.wikipedia.org/wiki/Monty_Hall_problem

The player's chances of winning the car double by switching to the door the host offers. The probability of initially choosing the car is one in three, which is the probability of winning the car by sticking with this choice. By contrast, the probability of initially choosing a door with a goat is two in three, and a player originally choosing a door with a goat wins by switching since after the host shows the other goat the only remaining door must be the door with the car.

When the player is asked whether to switch, there are three possible situations corresponding to the player's initial choice, each with probability ?:

* The player originally picked the door hiding goat number 1. The game host has shown the other goat.
* The player originally picked the door hiding goat number 2. The game host has shown the other goat.
* The player originally picked the door hiding the car. The game host has shown either of the two goats.

If the player chooses to switch, the player wins the car in the first two cases. A player choosing to stay with the initial choice wins in only the third case. Since in two out of three equally likely cases switching wins, the probability of winning by switching is ?. In other words, players who switch will win the car on average two times out of three.

The solution would be different if the host did not know what was behind each door, or if the host sometimes had the option of not offering the player the chance to switch. Some statements of the problem, notably the one in Parade Magazine, do not explicitly exclude these possibilities. For example, if the game host only offers the opportunity to switch if the contestant originally chooses the car, the probability of winning by switching is 0%. In the problem as stated by Mueser and Granberg, it is because the host must reveal a goat and must make the offer to switch that the player has a ? chance of winning by switching.
 

chuckywang

Lifer
Jan 12, 2004
20,139
1
0
Originally posted by: chrisms
Let's say you're on a game show, and you're given the choice of 3 doors: Behind one door is a car, and behind the others, sheep. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a sheep. He then says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice?

This is the Monty Hall problem. You ALWAYS switch.
 

Malak

Lifer
Dec 4, 2004
14,696
2
0
The reasoning behind the explanation is stupid and I will continue to dispute it. You are left with only 2 choices, the third choice is now moot and cannot be even considered in your decision. It is a 50/50 chance and there is no reason to switch nor to not switch. It comes down to chance. You have zero advantage.
 

chrisms

Diamond Member
Mar 9, 2003
6,615
0
0
It is not 50/50. I had trouble too until reading this part...

"It may be easier to appreciate the result by considering a 100 doors instead of just three. In this case there are 99 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 98 of the other doors revealing 98 goats ? imagine the host starting with the first door and going down a line of 100 doors, opening each one but skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 99 out of 100 times the other door will contain the prize, as 99 out of 100 times the player first picked a door with a goat. A rational player should switch."
 

Malak

Lifer
Dec 4, 2004
14,696
2
0
Chrisms, there is no 3rd door to choose once it is revealed, so the odds are 50/50. You choose one or the other, and you have no advantage. Their reasoning is flawed.
 

chrisms

Diamond Member
Mar 9, 2003
6,615
0
0
Originally posted by: Malak
Chrisms, there is no 3rd door to choose once it is revealed, so the odds are 50/50. You choose one or the other, and you have no advantage. Their reasoning is flawed.

On Marilyn vos Savant, record holder for the highest IQ:

"Perhaps the most famous event involving Marilyn vos Savant began with the following question in her September 9, 1990, column.

"Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, 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?" ?Craig F. Whitaker, Columbia, Maryland

This question, named "the Monty Hall problem" due to its resemblance to situations on the game show Let's Make a Deal, existed before Marilyn addressed it, but was brought to nationwide attention by her column. Marilyn's answer, that you should switch because door #2 has a 2/3 chance of winning whereas door #1 has only a 1/3 chance, provoked thousands of letters in response, nearly all arguing that she was wrong and that the doors are equally likely to win. A follow-up column affirming her answer only intensified the debate, which soon spread through the media, even reaching the front page of The New York Times. Among the ranks of her opponents were hundreds of academics with Ph.D.s, some of them professional mathematicians scolding her for propagating innumeracy.

Despite the criticism, Marilyn's answer was correct under the most common interpretation of the question; see Monty Hall problem for details. After a second follow-up in which she explained in more depth her reasoning and the conditions on which it was based, many readers, including academics who had previously argued against her, wrote to admit that she was right. Marilyn also called on school teachers across America to simulate the problem in their math classes. In a final column, she announced the results: out of more than a thousand schools which had performed the experiment, nearly 100% had found that it pays to switch. A majority of readers now agreed with her answer, and half of those whose letters had been published wrote to retract their arguments."
 

animalia

Banned
Dec 15, 2006
792
0
0
Originally posted by: chrisms
It is not 50/50. I had trouble too until reading this part...

"It may be easier to appreciate the result by considering a 100 doors instead of just three. In this case there are 99 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 98 of the other doors revealing 98 goats ? imagine the host starting with the first door and going down a line of 100 doors, opening each one but skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 99 out of 100 times the other door will contain the prize, as 99 out of 100 times the player first picked a door with a goat. A rational player should switch."

edit: ok I get it now....when choosing door 1 you had a 1/3 chance, but as soon as it was revealed that door 3 had a goat...this makes door 2 more probable to win?
 

chrisms

Diamond Member
Mar 9, 2003
6,615
0
0
Originally posted by: animalia
Originally posted by: chrisms
It is not 50/50. I had trouble too until reading this part...

"It may be easier to appreciate the result by considering a 100 doors instead of just three. In this case there are 99 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 98 of the other doors revealing 98 goats ? imagine the host starting with the first door and going down a line of 100 doors, opening each one but skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 99 out of 100 times the other door will contain the prize, as 99 out of 100 times the player first picked a door with a goat. A rational player should switch."

99/100 times is a bit more reassuring than 2/3. 99% vs 66%. In case 1 there are 66% GOATS and 33% CARS behind the doors. in case 2 there are 99% goats and 1% CARS. it is much easier to switch choices in case 2.

As the Wikipedia article states, the showing of the doors does not change the probability. It is merely a distraction. The 99 to 1 example is extreme to better prove the point.
 

iamaelephant

Diamond Member
Jul 25, 2004
3,816
1
81
Originally posted by: Malak
The reasoning behind the explanation is stupid and I will continue to dispute it. You are left with only 2 choices, the third choice is now moot and cannot be even considered in your decision. It is a 50/50 chance and there is no reason to switch nor to not switch. It comes down to chance. You have zero advantage.

Wow. You need your head examined. Go and read the article again. And again and again and again until your understand it.

Wait, you're a Christian right? Hhmm, explains a lot.
 

animalia

Banned
Dec 15, 2006
792
0
0
Originally posted by: chrisms
Originally posted by: animalia
Originally posted by: chrisms
It is not 50/50. I had trouble too until reading this part...

"It may be easier to appreciate the result by considering a 100 doors instead of just three. In this case there are 99 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 98 of the other doors revealing 98 goats ? imagine the host starting with the first door and going down a line of 100 doors, opening each one but skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 99 out of 100 times the other door will contain the prize, as 99 out of 100 times the player first picked a door with a goat. A rational player should switch."

99/100 times is a bit more reassuring than 2/3. 99% vs 66%. In case 1 there are 66% GOATS and 33% CARS behind the doors. in case 2 there are 99% goats and 1% CARS. it is much easier to switch choices in case 2.

As the Wikipedia article states, the showing of the doors does not change the probability. It is merely a distraction. The 99 to 1 example is extreme to better prove the point.

see edit above....

 

dafatha00

Diamond Member
Oct 19, 2000
3,871
0
76
I had to think about this for awhile because I was pretty sure that switching would do nothing. However, switching indeed doubles your chances. Think about it using a small sample size.

Example: Say doors 1 and 2 held a goat and door 3 held a car. Now say you open each door 12 times in which you are going to switch doors each time. For doors one and two, because you picked a sheep and you are now switching, you will get a car 24/36 times.

Now with the same example where you open each door 12 times, you decide to stay with your original choice. Now you will only get a car 12/36 times because each time you stick with door 1 or 2, you will wind up with a sheep.
 

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
49,606
166
111
www.slatebrookfarm.com
Originally posted by: chrisms
Originally posted by: Malak
Chrisms, there is no 3rd door to choose once it is revealed, so the odds are 50/50. You choose one or the other, and you have no advantage. Their reasoning is flawed.

On Marilyn vos Savant, record holder for the highest IQ:

"Perhaps the most famous event involving Marilyn vos Savant began with the following question in her September 9, 1990, column.

"Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, 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?" ?Craig F. Whitaker, Columbia, Maryland

This question, named "the Monty Hall problem" due to its resemblance to situations on the game show Let's Make a Deal, existed before Marilyn addressed it, but was brought to nationwide attention by her column. Marilyn's answer, that you should switch because door #2 has a 2/3 chance of winning whereas door #1 has only a 1/3 chance, provoked thousands of letters in response, nearly all arguing that she was wrong and that the doors are equally likely to win. A follow-up column affirming her answer only intensified the debate, which soon spread through the media, even reaching the front page of The New York Times. Among the ranks of her opponents were hundreds of academics with Ph.D.s, some of them professional mathematicians scolding her for propagating innumeracy.

Despite the criticism, Marilyn's answer was correct under the most common interpretation of the question; see Monty Hall problem for details. After a second follow-up in which she explained in more depth her reasoning and the conditions on which it was based, many readers, including academics who had previously argued against her, wrote to admit that she was right. Marilyn also called on school teachers across America to simulate the problem in their math classes. In a final column, she announced the results: out of more than a thousand schools which had performed the experiment, nearly 100% had found that it pays to switch. A majority of readers now agreed with her answer, and half of those whose letters had been published wrote to retract their arguments."

A bit of an exaggeration to make her look better or something??

It seems that a few people here have read the solution and remembered it. However, they are looking at an over-simplified case. You cannot argue that the odds are 50/50 or 2/3 / 1/3 unless you know the motivation of the person revealing the 3rd door. For example, suppose the person always revealed the door when the person had originally chosen the winning door, but didn't reveal a 3rd door when the person had a losing door. In this case, switching would carry a 0% probability of winning, not 2/3.

In order to answer the question, certain conditions or assumptions must be stated, such as the host always reveals one of the doors not picked.
 

mercanucaribe

Banned
Oct 20, 2004
9,763
1
0
As soon as I read that the game show host knows which door has one, I realized that that is the key.. Either you picked the car and the host picked the door to open randomly, or you picked a sheep and he had to pick the sheep instead of the car. I don't know the math, but it feels to me like your odds are better if you switch.
 

ElFenix

Elite Member
Super Moderator
Mar 20, 2000
102,425
8,388
126
the plane doesn't take off and .999... does not equal 1, goddamnit!
 

Malak

Lifer
Dec 4, 2004
14,696
2
0
Originally posted by: iamaelephant
Originally posted by: Malak
The reasoning behind the explanation is stupid and I will continue to dispute it. You are left with only 2 choices, the third choice is now moot and cannot be even considered in your decision. It is a 50/50 chance and there is no reason to switch nor to not switch. It comes down to chance. You have zero advantage.

Wow. You need your head examined. Go and read the article again. And again and again and again until your understand it.

Wait, you're a Christian right? Hhmm, explains a lot.

This is not the first time it has been brought up and I will despute it regadless of any article. You are given a choice between 2 doors. The third door is not a choice and whatever it reveals only means that your chances are either 50/50 or none at all. The reasoning is completely wrong. If this so-called genius cares to argue it with me, he/she can try. I will bury her/him.
 

Mo0o

Lifer
Jul 31, 2001
24,227
3
76
Originally posted by: Malak
Originally posted by: iamaelephant
Originally posted by: Malak
The reasoning behind the explanation is stupid and I will continue to dispute it. You are left with only 2 choices, the third choice is now moot and cannot be even considered in your decision. It is a 50/50 chance and there is no reason to switch nor to not switch. It comes down to chance. You have zero advantage.

Wow. You need your head examined. Go and read the article again. And again and again and again until your understand it.

Wait, you're a Christian right? Hhmm, explains a lot.

This is not the first time it has been brought up and I will despute it regadless of any article. You are given a choice between 2 doors. The third door is not a choice and whatever it reveals only means that your chances are either 50/50 or none at all. The reasoning is completely wrong. If this so-called genius cares to argue it with me, he/she can try. I will bury her/him.
But in the explanation it pays to swtich 2/3 of the time. and the trials the schools have run have shown that it pays to switch.


Think about it this way. Under your assumption, it does not pay to take ANY action since you say the door in front of you now has a 50/50 chance of having a car there. So under your assumption, you should be winning 50% of the time. OK. So if we are to run the game multiple times, and you always stick with choice 1, arey ou saying you'll win 1/2 the time or would you acutally win 1/3 of the time?
 

iamaelephant

Diamond Member
Jul 25, 2004
3,816
1
81
Originally posted by: Malak
Originally posted by: iamaelephant
Originally posted by: Malak
The reasoning behind the explanation is stupid and I will continue to dispute it. You are left with only 2 choices, the third choice is now moot and cannot be even considered in your decision. It is a 50/50 chance and there is no reason to switch nor to not switch. It comes down to chance. You have zero advantage.

Wow. You need your head examined. Go and read the article again. And again and again and again until your understand it.

Wait, you're a Christian right? Hhmm, explains a lot.

This is not the first time it has been brought up and I will despute it regadless of any article. You are given a choice between 2 doors. The third door is not a choice and whatever it reveals only means that your chances are either 50/50 or none at all. The reasoning is completely wrong. If this so-called genius cares to argue it with me, he/she can try. I will bury her/him.

I'm not sure what you don't get about this. I'm going to explain it as simply as I can. Please read and attempt to comprehend this, because I am telling you now that you are flat-out wrong.

A host offers you the choice of three doors. Two of them have sheep behind them, and one a car. I take it you understand that part. When you choose a door, the host, knowing what is behind each of the doors, opens one of the doors that you have not chosen. This door reveals a sheep.

Initially when you chose a door, you essentially divided the complete set of 3 doors into two parts, the one door you chose and the 2 remaining doors. So you know that there is a 2/3 chance that the car is behind one of the two doors, and a 1/3 chance that the car is behind the door you chose. Are you following? We now have two sets of doors. One with a 2/3 chance of finding a car, and one with a 1/3 chance of finding a car.

For your second choice, of course you will want to choose the side that has a 2/3 chance of finding a car. Well one of those doors is already open, so you now have one door with a 2/3 chance of a car and one door with a 1/3 chance. Get it? Do you understand now why one door has a higher chance of a car? I may not have explained it well but if you don't understand this still then you are beyond help - you are nothing but a brain dead moron.
 
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