Demon-Xanth
Lifer
- Feb 15, 2000
- 20,551
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Brown: ok, show me in ONE CASE of the possible situations where the chance is anything other than 50/50. The flaw in the logic is the knowledge. I do believe I listed every possible situation.
There is no reason to switch.
There are 2 choices left
Choice A = the car.
Choice B = the goat.
As someone pointed out, at this point the chances are 50/50.
You might switch, but your chances to pick the car will remain the same.
There are 2 choices left
Choice A = the car.
Choice B = the goat.
As someone pointed out, at this point the chances are 50/50.
You might switch, but your chances to pick the car will remain the same.
Duckers, look at it this way:
By switching, the first door you pick is the one you DON'T want. That way you get to open two of them. Your chances of accidentally discarding the car by switching are only 1/3, while you have a 2/3 chance of getting the car if you switch.
If you DON'T switch, you're relying on your ability to pick the correct door the first time, which unless you're psychic you're not going to do much better than 1/3 attempts.
Viper GTS
By switching, the first door you pick is the one you DON'T want. That way you get to open two of them. Your chances of accidentally discarding the car by switching are only 1/3, while you have a 2/3 chance of getting the car if you switch.
If you DON'T switch, you're relying on your ability to pick the correct door the first time, which unless you're psychic you're not going to do much better than 1/3 attempts.
Viper GTS
There is no reason to switch, if you don't want to take advantage of 2:3 odds vs. 1:3 odds.
Before first pick:
123
CGG
GCG
GGC
Before second pick (x denotes Monty the door Monty opened):
123
CGx
GCx
GxC
Demon-Xanth's logic example shows only two options after a door has been opened, when in reality there are three possibilities.
Assume you pick door #1. There is only a one in three chance you picked the right door. After Monty opens a door, the original three possibilities still hold, and as you can see, your chance of door #1 being correct is still only 1:3.
It really helped me to imagine a million doors, and I would have used that example but that would take a bit too long.
Before first pick:
123
CGG
GCG
GGC
Before second pick (x denotes Monty the door Monty opened):
123
CGx
GCx
GxC
Demon-Xanth's logic example shows only two options after a door has been opened, when in reality there are three possibilities.
Assume you pick door #1. There is only a one in three chance you picked the right door. After Monty opens a door, the original three possibilities still hold, and as you can see, your chance of door #1 being correct is still only 1:3.
It really helped me to imagine a million doors, and I would have used that example but that would take a bit too long.
Demon-Xanth, I thought I was clear in my explanation before.
Anyway, let me give you an example :
Suppose we have 3 doors, A, B, and C.
Behind A and B are goats, and behind C is car, but
the contestant doesn't know this.
Lets now have 2 contestants, one that will always change his
mind after his first pick (lets call him Mr. Change),
and another one that will always stay with her pick (lets call her Mrs. Stay).
Lets consider 3 situations for Mr. Change :
1. Mr. Change choose door A (which has goats), then Monty has to open
door B to show the door that has goats. Now Mr. Change will choose door C
which has car. He will win a car by choosing door A initially.
2. Mr. Change choose door B (which has goats), then Monty has to open
door A to show the door that has goats. Now Mr. Change will choose door C
which has car. He will win a car by choosing door B initially.
3. Mr. Change choose door C (which has car), then Monty can open door A or
B. Mr. Change then has to choose the door other than the one Monty picked.
Mr. Change will win a goat by choosing door C.
What does this tell you ?
By choosing the door with goats at first (2 out of 3 choices since there are
2 doors with goats), Mr. Change will win a car.
By choosing the door with car at first (1 out of 3 choices), Mr. Change will
win goats.
Now lets see 3 situations with Mrs. Stay
1. Mrs. Stay choose door A (contains goats), Monty will open door B (contains
goats). Mrs. Stay will stay with door A. She will win goats by choosing door A.
2. Mrs. Stay choose door B (contains goats), Monty will open door A (contains
goats). Mrs. Stay will stay with door B. She will win goats by choosing door B.
3. Mrs. Stay choose door C (contains car), Monty can open door A or B.
Mrs. Stay will stay with door C. She will win car by choosing door C.
Now we see that in order to win car, Mrs. Stay has to choose door which contains
car at first (which is 1 out of 3 choices).
Is it clear now ?
Anyway, let me give you an example :
Suppose we have 3 doors, A, B, and C.
Behind A and B are goats, and behind C is car, but
the contestant doesn't know this.
Lets now have 2 contestants, one that will always change his
mind after his first pick (lets call him Mr. Change),
and another one that will always stay with her pick (lets call her Mrs. Stay).
Lets consider 3 situations for Mr. Change :
1. Mr. Change choose door A (which has goats), then Monty has to open
door B to show the door that has goats. Now Mr. Change will choose door C
which has car. He will win a car by choosing door A initially.
2. Mr. Change choose door B (which has goats), then Monty has to open
door A to show the door that has goats. Now Mr. Change will choose door C
which has car. He will win a car by choosing door B initially.
3. Mr. Change choose door C (which has car), then Monty can open door A or
B. Mr. Change then has to choose the door other than the one Monty picked.
Mr. Change will win a goat by choosing door C.
What does this tell you ?
By choosing the door with goats at first (2 out of 3 choices since there are
2 doors with goats), Mr. Change will win a car.
By choosing the door with car at first (1 out of 3 choices), Mr. Change will
win goats.
Now lets see 3 situations with Mrs. Stay
1. Mrs. Stay choose door A (contains goats), Monty will open door B (contains
goats). Mrs. Stay will stay with door A. She will win goats by choosing door A.
2. Mrs. Stay choose door B (contains goats), Monty will open door A (contains
goats). Mrs. Stay will stay with door B. She will win goats by choosing door B.
3. Mrs. Stay choose door C (contains car), Monty can open door A or B.
Mrs. Stay will stay with door C. She will win car by choosing door C.
Now we see that in order to win car, Mrs. Stay has to choose door which contains
car at first (which is 1 out of 3 choices).
Is it clear now ?
Demon-Xanth
Lifer
- Feb 15, 2000
- 20,551
- 2
- 81
Viper GTS: why are your chances of discarding the car 1/3? There is two decisions being made, the opening of one wrong door makes the first decision a "no lose" condition making it irrelevant.
Wow! I laughed so hard at the thought of getting better odds by "switching"... of course that was before I realized that this was a serious discussion.
I hate to say it, but your odds are 50/50 once you are limited to two doors. Yes, 1/3 chance when there was three doors. Once the event is over, a new event has begun.... unless, of course, you believe that it is better to guess tails after heads have been tossed for the last 100 times....
Viper GTS
Well, I think I understand your point of view.
Why try to pick the car (1/3) when you can try to pick a goat (2/3)
That makes sense, but the question says:
---------
The contestant then chooses a door. Monty then opens a different door, revealing a goat.
Should the contestant change their mind? What are the odds of winning the car before and after switching?
---------
One goat is out, there are two options left: the car and the second goat.
AT THIS POINT the car could be in any of the 2 doors.
50/50 chances to pick the car.
Well, I think I understand your point of view.
Why try to pick the car (1/3) when you can try to pick a goat (2/3)
That makes sense, but the question says:
---------
The contestant then chooses a door. Monty then opens a different door, revealing a goat.
Should the contestant change their mind? What are the odds of winning the car before and after switching?
---------
One goat is out, there are two options left: the car and the second goat.
AT THIS POINT the car could be in any of the 2 doors.
50/50 chances to pick the car.
Demon-Xanth
Lifer
- Feb 15, 2000
- 20,551
- 2
- 81
allan: you are missing one possibility: CXG, nowhere is it stated WHICH goat is going to be shown.
Demon-Xanth:
It doesn't matter which goat is shown.
After pick two:
CGx - car behind first door
CxG
xCG - car behind second door
GCx
xGC - car behind third door
GxC
2:6 = 1:3.
It doesn't matter which goat is shown.
After pick two:
CGx - car behind first door
CxG
xCG - car behind second door
GCx
xGC - car behind third door
GxC
2:6 = 1:3.
Demon-Xanth
Lifer
- Feb 15, 2000
- 20,551
- 2
- 81
Ok, lets look at this a different way. Since everyone is saying "look at this a funky way". Prove that my logic has a fault. There is two events, not one. Since the first event cannot end in a loss, that event is irrelevant.
kranky
Elite Member
- Oct 9, 1999
- 21,019
- 156
- 106
For those who don't think switching doubles your chance of winning, do what hatboy did and test it yourself. You don't have to write code, just use pieces of paper to simulate it.
The logic isn't as complicated as some make it out to be. Keep in mind that just because Monty shows you a goat behind one of the doors you DIDN'T pick, that doesn't change the odds. Everybody knows that one of those other two doors had a goat behind it. Your original possibility was 1/3. The other two doors must total 2/3.
If you claim the odds are 1/2 if you switch, you're also saying the odds were 1/2 when you first chose, and they are obviously 1/3. The fact that he opens a door and shows a goat does not change your odds.
The logic isn't as complicated as some make it out to be. Keep in mind that just because Monty shows you a goat behind one of the doors you DIDN'T pick, that doesn't change the odds. Everybody knows that one of those other two doors had a goat behind it. Your original possibility was 1/3. The other two doors must total 2/3.
If you claim the odds are 1/2 if you switch, you're also saying the odds were 1/2 when you first chose, and they are obviously 1/3. The fact that he opens a door and shows a goat does not change your odds.
This discussion is turning stupid!
Some people are using complicated analogies to explain a very simple question, lol
Some people are using complicated analogies to explain a very simple question, lol
Another way I've seen this explained is to take it to the extreme.
Let's say there a MILLION doors, and 999,999 goats. You choose one door, & then Monty opens 999,998 doors. Thus leaving two doors left. Knowing the true identity of the other 999,998 doors does not change the fact that when you initially chose your door you had 1/1,000,000 odds of picking the car. The odds are much greater that the car lies behind the other door than they are that you chose correctly with those odds.
Many people here are overlooking the fact that revealing the contents of a door does not change the original probability that you chose the correct door. If you were to re-distribute the doors' contents randomly & then chose a door again, you're correct... The odds would be 50/50. But because the contents have remained constant the whole time, your odds are greatly increased by switching.
Go test this yourself. Go get three paper cups, & three marbles (two of one color, one of another). If you don't believe it, try it.
Viper GTS
Let's say there a MILLION doors, and 999,999 goats. You choose one door, & then Monty opens 999,998 doors. Thus leaving two doors left. Knowing the true identity of the other 999,998 doors does not change the fact that when you initially chose your door you had 1/1,000,000 odds of picking the car. The odds are much greater that the car lies behind the other door than they are that you chose correctly with those odds.
Many people here are overlooking the fact that revealing the contents of a door does not change the original probability that you chose the correct door. If you were to re-distribute the doors' contents randomly & then chose a door again, you're correct... The odds would be 50/50. But because the contents have remained constant the whole time, your odds are greatly increased by switching.
Go test this yourself. Go get three paper cups, & three marbles (two of one color, one of another). If you don't believe it, try it.
Viper GTS
The fact that he opens a door and shows a goat does not change your odds.
Obviously it does, because you would immediately elimate that door.
Obviously it does, because you would immediately elimate that door.
Demon:
You said that you don't know which goat he'll show, and your logic diagrams don't represent that.
After "You pick door one," you're missing:
XCG
XGC
Your other two diagrams are missing two possibilities each.
Looking at your first diagram, you have the car behind door #1 listed twice, but only once for behind doors #2 and #3.
You said that you don't know which goat he'll show, and your logic diagrams don't represent that.
After "You pick door one," you're missing:
XCG
XGC
Your other two diagrams are missing two possibilities each.
Looking at your first diagram, you have the car behind door #1 listed twice, but only once for behind doors #2 and #3.
Demon-Xanth, you don't give up, do you ?The key is Monty will open DIFFERENT door that has goats
(there are two keys there : one is different door, another
one is contain goats).
The first event (choosing the first time) is related
to the second event.
You can't say that since Monty will open a door with goat,
then we have only 2 doors to choose from the beginning.
As you can see from my example above (PLEASE READ THOSE),
even Monty has to change his pick based on the contestant first
pick. This is what I mean that the second event is related
to first event.
Here :
Choosing door with goats at first, and with ALWAYS switch attitude,
you have 100% of winning the car after you first pick.
Why ?
You pick the door with goat, Monty will open ANOTHER different door
with goat. What is left ? Door with car. Since you always switch
you have 100% of winning the car after you pick the door with goat at first.
kranky
Elite Member
- Oct 9, 1999
- 21,019
- 156
- 106
Duckers, try it this way. Would you agree that if you NEVER switch, that is the same as not being ALLOWED to switch? The results would be the same either way, right?
The chance of being right the first time is 1/3, right? So if you NEVER switch, your odds will always and forever be 1/3.
And therefore the odds of switching must be 2/3, since all possibilities must add up to 1.
The chance of being right the first time is 1/3, right? So if you NEVER switch, your odds will always and forever be 1/3.
And therefore the odds of switching must be 2/3, since all possibilities must add up to 1.
Demon-Xanth
Lifer
- Feb 15, 2000
- 20,551
- 2
- 81
Ok fine, I give up trying to convince people (even thought I stated 100% of the 12 possibilities) the same way I gave up trying to convince people that Shakespear was not the greatest author. I'll simply walk away and let Al Gore run thier school
What are the odds of winning the car before and after switching?
kranky,
What you are saying makes sense, but it all depends on how you want to analyze the question.
Part of the problem says:
Monty then opens a different door, revealing a goat.
I would say both arguments are right, but since one goat is out, the possibilities to pick the car at this point would be 50/50.
I agree with you on Shakespeare though.I give up also, either I have to take courses on "Teaching
and Explaining Math" or you have to learn or relearn
your "Probability" course.
Okay, I guess I give up too, since you stated 100% of only 12 of the 18 possibilities. Viper did well explaining the million doors, and kranky and br0wn both contributed well.
You really are missing six possibilities, since you're ignoring your own statement that you don't know which goat is shown.
Proper diagrams after Monty opens a door:
Pick the first door:
CGX
CXG
XCG
GCX
XGC
GXC
Pick the second:
CGX
CXG
XCG
GCX
XGC
GXC
Pick the third:
CGX
CXG
XCG
GCX
XGC
GXC
Note that the six possibilities are identical for each pick.
Oh well.
You really are missing six possibilities, since you're ignoring your own statement that you don't know which goat is shown.
Proper diagrams after Monty opens a door:
Pick the first door:
CGX
CXG
XCG
GCX
XGC
GXC
Pick the second:
CGX
CXG
XCG
GCX
XGC
GXC
Pick the third:
CGX
CXG
XCG
GCX
XGC
GXC
Note that the six possibilities are identical for each pick.
Oh well.
allan120...
It doesn't matter which goat you get. The fact is it's not the car. That's the only thing that matters. You can simplify things greatly by assuming the goats are identical. Or maybe they're sheep, & they're all named Dolly.
Viper GTS
It doesn't matter which goat you get. The fact is it's not the car. That's the only thing that matters. You can simplify things greatly by assuming the goats are identical. Or maybe they're sheep, & they're all named Dolly.
Viper GTS
For anybody who cares and/or still can't see it, you can find the program that I convinced myself with here: http://ajr.dhs.org/math. I wrote it in Java. TestWithSwitch switches doors after being shown the goat. TestWoSwitch does not. Both play the game 1 million times, but you can easily modify it so that it does less iterations. This made it really clear for me why I was wrong. Hopefully the code should be relatively easy to follow, even if you don't know much Java.
kranky
Elite Member
- Oct 9, 1999
- 21,019
- 156
- 106
Demon-Xanth, trying to convince someone that Shakespeare is the greatest author is still subjective. In this problem, switching doubles your chance of winning, and that's a absolute fact. Try it yourself and see.
duckers, In rereading your posts, you seem to be saying that the odds are 50/50 if you pick a door AFTER he shows a goat. I agree, but the question says you pick a door BEFORE Monty shows you a goat.
The flaw in the logic is that - once you pick a door - your odds do NOT change just because Monty shows you a goat behind a door you didn't pick. The odds would change ONLY if you haven't picked yet.
Try it yourself and see.duckers, In rereading your posts, you seem to be saying that the odds are 50/50 if you pick a door AFTER he shows a goat. I agree, but the question says you pick a door BEFORE Monty shows you a goat.
The flaw in the logic is that - once you pick a door - your odds do NOT change just because Monty shows you a goat behind a door you didn't pick. The odds would change ONLY if you haven't picked yet.
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