Sheep or a Car

[BigJ]

Originally posted by: 6000SUX

Originally posted by: sieistganzfett
6000sux, why you got to get over logical on the logic problem? you're acting like a lawyer!

Sorry. I don't like people telling me I'm wrong when I'm right. I'll stop now.

What about people telling you you're cute?

 

[NuclearNed]

I never actually get the answers to these riddles, so its amazing to me that the answer to this one seems so clear.

Always switch FTW.

 

[RapidSnail]

I haven't seen this posted before so here's my (hopefully unique) view.

Think of the doors as being divided into two groups: the door chosen (Group 1), and the doors not chosen (Group 2). Most of you will be capable enough to determine that each door has a 1:3 chance of containing the car. With that being said we can say that Group 1 has 1:3 chance of containing the car, while Group 2 has 2:3 chance of containing the car.

Extending the previous logic, wouldn't it be to your advantage if you were able to open both doors in Group 2 instead of just the one in Group 1? Of course it would! If you could open both doors in Group 2, you winning odds would increase from 1:3 to 2:3! Makes perfect sense, right?

Well that is exactly the chance the host is giving you! By switching doors, you are effectively opening both doors in Group 2, thereby increasing your odds to 2:3. What is behind the opened door does not matter, nor does it change the odds of winning. The host's move is only a ploy to skew the contestant's logic and make him complacent with his choice, meanwhile giving the host the advantage in odds!

 

[PlatinumGold]

Originally posted by: RapidSnail
I haven't seen this posted before so here's my (hopefully unique) view.

Think of the doors as being divided into two groups: the door chosen (Group 1), and the doors not chosen (Group 2). Most of you will be capable enough to determine that each door has a 1:3 chance of containing the car. With that being said we can say that Group 1 has 1:3 chance of containing the car, while Group 2 has 2:3 chance of containing the car.

Extending the previous logic, wouldn't it be to your advantage if you were able to open both doors in Group 2 instead of just the one in Group 1? Of course it would! If you could open both doors in Group 2, you winning odds would increase from 1:3 to 2:3! Makes perfect sense, right?

Well that is exactly the chance the host is giving you! By switching doors, you are effectively opening both doors in Group 2, thereby increasing your odds to 2:3. What is behind the opened door does not matter, nor does it change the odds of winning. The host's move is only a ploy to skew the contestant's logic and make him complacent with his choice, meanwhile giving the host the advantage in odds!

actually, it does matter what is behind the door the host opens. but we know from the premise of the game show that the host is not going to open a door with the car behind it.

hence, the odds of the door not chosen initially (by either the host or the contestant) has 66% chance of having the car behind it.

 

[Schfifty Five]

Originally posted by: PlatinumGold

Originally posted by: RapidSnail
I haven't seen this posted before so here's my (hopefully unique) view.

Think of the doors as being divided into two groups: the door chosen (Group 1), and the doors not chosen (Group 2). Most of you will be capable enough to determine that each door has a 1:3 chance of containing the car. With that being said we can say that Group 1 has 1:3 chance of containing the car, while Group 2 has 2:3 chance of containing the car.

Extending the previous logic, wouldn't it be to your advantage if you were able to open both doors in Group 2 instead of just the one in Group 1? Of course it would! If you could open both doors in Group 2, you winning odds would increase from 1:3 to 2:3! Makes perfect sense, right?

Well that is exactly the chance the host is giving you! By switching doors, you are effectively opening both doors in Group 2, thereby increasing your odds to 2:3. What is behind the opened door does not matter, nor does it change the odds of winning. The host's move is only a ploy to skew the contestant's logic and make him complacent with his choice, meanwhile giving the host the advantage in odds!

actually, it does matter what is behind the door the host opens. but we know from the premise of the game show that the host is not going to open a door with the car behind it.

hence, the odds of the door not chosen initially (by either the host or the contestant) has 66% chance of having the car behind it.

I think the point is that it doesn't matter that the host opens a door...

what it boils down to is: "do you want trade your single door for all the hosts' doors?"

 

[James Bond]

If you stick with door one, you stick with your original odds: 33%.
If you switch, you are affectively opening 2 doors (since he negates one): 66%.

 

[Maluno]

Sheep.

 

[BillyBatson]

i do not get this, if i picked #1 and he opened #3 did he not even open #1 because there was a goat in which case why didn't he just tell me and why did he open door #3 which the game is still going on? I picked #1 and he whould open #1! if he opens any other door to me it means the car was behind the door i chose and he is trying to avoid that so i would pick the same door

 

[PlatinumGold]

Originally posted by: BillyBatson
i do not get this, if i picked #1 and he opened #3 did he not even open #1 because there was a goat in which case why didn't he just tell me and why did he open door #3 which the game is still going on? I picked #1 and he whould open #1! if he opens any other door to me it means the car was behind the door i chose and he is trying to avoid that so i would pick the same door

no, it's not about game host playing mind games, the way it is set out,

the host MUST open a door regardless of whether the door you've chosen (#1) has the car behind it or not. so motivation of host is not an issue.

so, you choose #1 which as a 33% chance of having the car behind it, by opening door number 3, the host is saying in effect, by switching you get the odds of choosing doors 2 and 3 but he will just open one of the doors which he knows doesn't have the car behind it.

hence, we know that door #2 has a 66% chance of having the car behind it.

 

[kevinthenerd]

To everyone who said "50/50": here's semi-proof that you're wrong. (The odds of you being right are extremely low given the nature of the randomness I incorporated.)

The major thing everybody has a hard time getting their head around is that, if only one door has a goat when the host makes the decision, the host must pick the other, constraining the probability.

I wrote some code to check this. First here's the output:

Stay Win Count
33480

Stay Lose Count
66520

Stay Winning Percentage
.334



Switch Win Count
66520

Switch Lose Count
33480

Switch Winning Percentage
.665


Advantage factor for choosing to switch than to stay:
1.986



Now here's the code, written in good-ole trust QuickBASIC 4.5. If you don't trust the results then find the flaw in my code that justifies the error.



DECLARE FUNCTION Coin! ()
DECLARE FUNCTION GetDoor% ()
'Code to verify the solution to the Monty Hall problem

CONST Goat = -1
CONST Car = 1
DIM Door(1 TO 3) AS INTEGER
RANDOMIZE TIMER
DIM DoorNumber AS INTEGER
DIM PlayerChoice AS INTEGER
DIM HostOption1 AS INTEGER
DIM HostOption2 AS INTEGER
DIM HostChoice AS INTEGER
DIM HostCoin AS INTEGER
DIM PlayerDecision AS INTEGER

DIM SwitchLoseCount AS LONG
DIM SwitchWinCount AS LONG
DIM StayLoseCount AS LONG
DIM StayWinCount AS LONG

DIM X AS LONG


SwitchWinCount = 0
SwitchLoseCount = 0

StayWinCount = 0
StayLoseCount = 0

CLS

FOR X = 1 TO 100000

'Set up the game

Door(1) = Goat
Door(2) = Goat
Door(3) = Goat
DoorNumber = GetDoor
Door(DoorNumber) = Car


'The player makes the choice
PlayerChoice = GetDoor

'The host recommends another door

'First find out what doors he can choose
HostOption1 = 0
HostOption2 = 0
IF PlayerChoice = 1 THEN
HostOption1 = 2
HostOption2 = 3
END IF
IF PlayerChoice = 2 THEN
HostOption1 = 1
HostOption2 = 3
END IF
IF PlayerChoice = 3 THEN
HostOption1 = 1
HostOption2 = 2
END IF

'If only one door has a goat, the host must pick it
IF Door(HostOption1) = Car THEN HostChoice = HostOption2
IF Door(HostOption2) = Car THEN HostChoice = HostOption1

'If both doors have goats, pick one randomly as the host choice
IF Door(HostOption1) = Goat AND Door(HostOption2) = Goat THEN
HostCoin = Coin
IF HostCoin = 0 THEN
HostChoice = HostOption1
ELSE
HostChoice = HostOption2
END IF
END IF

'What if the player switches?
IF PlayerChoice <> 1 AND HostChoice <> 1 THEN PlayerDecision = 1
IF PlayerChoice <> 2 AND HostChoice <> 2 THEN PlayerDecision = 2
IF PlayerChoice <> 3 AND HostChoice <> 3 THEN PlayerDecision = 3
IF Door(PlayerDecision) = Car THEN SwitchWinCount = SwitchWinCount + 1
IF Door(PlayerDecision) = Goat THEN SwitchLoseCount = SwitchLoseCount + 1


'What if the player stays?
PlayerDecision = PlayerChoice
IF Door(PlayerDecision) = Car THEN StayWinCount = StayWinCount + 1
IF Door(PlayerDecision) = Goat THEN StayLoseCount = StayLoseCount + 1


NEXT X


PRINT "Stay Win Count"
PRINT StayWinCount
PRINT
PRINT "Stay Lose Count"
PRINT StayLoseCount
PRINT
PRINT "Stay Winning Percentage"
PRINT INT((StayWinCount / (StayWinCount + StayLoseCount)) * 1000) / 1000
PRINT
PRINT
PRINT
PRINT "Switch Win Count"
PRINT SwitchWinCount
PRINT
PRINT "Switch Lose Count"
PRINT SwitchLoseCount
PRINT
PRINT "Switch Winning Percentage"
PRINT INT((SwitchWinCount / (SwitchWinCount + SwitchLoseCount)) * 1000) / 1000
PRINT
PRINT
PRINT "Advantage factor for choosing to switch than to stay:"
PRINT INT((SwitchWinCount / StayWinCount) * 1000) / 1000

FUNCTION Coin
'Randomly returns 0 or 1
Coin = INT(RND * 2)
END FUNCTION

FUNCTION GetDoor%
GetDoor = INT(RND * 3!) + 1
END FUNCTION

 

[JujuFish]

Originally posted by: 6000SUX

Originally posted by: JujuFish

Originally posted by: 6000SUX
Don't try to school your betters.

I don't have to do so. You're doing a good job making an idiot of yourself for me. :thumbsup:

ROFLMAO You are probably feeling pretty poorly about yourself right now. You did it to yourself. Read the Wikipedia article and weep, dumbass.

Uh, no. If the host knows what's behind each door, it is impossible for him to open a door with disregard to what's behind that door (without tossing a coin or using some other sort of random generator). You're just mad because you read it as the host randomly opening a door, and you're trying to save face by acting like some sort of arrogant badass on an anonymous internet forum.

 

[kevinthenerd]

Ok, let me try to explain this again.


You have three initial choices, Goat1, Goat2, or Car. There are four possible scenarios.

Scenario 1 has 1/3 chance of happening:
You have 1/3 chance to pick Goat1. The host picks Goat2. You win by switching.

Scenario 2 has 1/3 chance of happening:
You have 1/3 chance to pick Goat2. The host picks Goat1. You win by switching.

Scenario 3 has 1/6 chance of happening:
You have 1/3 chance to pick the Car. The host picks Goat1 on a (1/2) chance. You lose by switching. (1/2)*(1/3)=(1/6)

Scenario 4 has 1/6 chance of happening:
You have 1/3 chance to pick the Car as in Scenario 3, but this time the host has 1/2 chance to pick Goat2. You lose by switching. (1/2)*(1/3)=(1/6)

Adding it up, you have 2/3 chance to win by switching and 1/3 chance to lose by switching.

 

[6000SUX]

Originally posted by: JujuFish

Originally posted by: 6000SUX

Originally posted by: JujuFish

Originally posted by: 6000SUX
Don't try to school your betters.

I don't have to do so. You're doing a good job making an idiot of yourself for me. :thumbsup:

ROFLMAO You are probably feeling pretty poorly about yourself right now. You did it to yourself. Read the Wikipedia article and weep, dumbass.

Uh, no. If the host knows what's behind each door, it is impossible for him to open a door with disregard to what's behind that door (without tossing a coin or using some other sort of random generator). You're just mad because you read it as the host randomly opening a door, and you're trying to save face by acting like some sort of arrogant badass on an anonymous internet forum.

PWNED. (by Wikipedia, no less) Pleased to have participated in your self-pwnage. Have a nice day, and please play again.

 

[dcdarren]

I'll throw my opinion in. I believe it is to your advantage to switch. Forgive me if someone already posted this.

Forget about Monty Hall opening the door at first. Since there are 2 sheep and 1 car, you have a greater chance of picking a sheep door. OK? You pick a door and must assume it is one of the sheep. So, going forward, you HAVE to assume that you picked one of the sheep. Now, Monty Hall opens the other sheep door. Since you've assumed that you picked a sheep door, then Monty actually opens a sheep door, you have to assume that the car is behind the third remaining door. So you switch.

OK, now go ahead and flame my logic...

 

[6000SUX]

Originally posted by: dcdarren
I'll throw my opinion in. I believe it is to your advantage to switch. Forgive me if someone already posted this.

Forget about Monty Hall opening the door at first. Since there are 2 sheep and 1 car, you have a greater chance of picking a sheep door. OK? You pick a door and must assume it is one of the sheep. So, going forward, you HAVE to assume that you picked one of the sheep. Now, Monty Hall opens the other sheep door. Since you've assumed that you picked a sheep door, then Monty actually opens a sheep door, you have to assume that the car is behind the third remaining door. So you switch.

OK, now go ahead and flame my logic...

No flames, but the logic is wrong. You don't have to assume that you picked a sheep, etc.

You have 1/3 chance of initially picking the door with the car. The rest of the time (2/3) you're better off switching because the car will be behind the other door.

 

[sieistganzfett]

Originally posted by: 6000SUX

Originally posted by: dcdarren
I'll throw my opinion in. I believe it is to your advantage to switch. Forgive me if someone already posted this.

Forget about Monty Hall opening the door at first. Since there are 2 sheep and 1 car, you have a greater chance of picking a sheep door. OK? You pick a door and must assume it is one of the sheep. So, going forward, you HAVE to assume that you picked one of the sheep. Now, Monty Hall opens the other sheep door. Since you've assumed that you picked a sheep door, then Monty actually opens a sheep door, you have to assume that the car is behind the third remaining door. So you switch.

OK, now go ahead and flame my logic...

No flames, but the logic is wrong. You don't have to assume that you picked a sheep, etc.

You have 1/3 chance of initially picking the door with the car. The rest of the time (2/3) you're better off switching because the car will be behind the other door.

you both seem to say the same thing to me, just worded very different. if there is a 66% chance i will find a dead horse behind a door, i'll assume i picked a door with a dead horse behind it, and switch once monty shows another door with a dead horse, since there was only the 33% chance of having a car behind what ever door i chose, i'll just assume the rest of the way going off the probability.

 

[FeuerFrei]

I've totally ignored this thread until now.

Who cares if the probability of winning with a switch goes up? The probability of winning by sticking to your original choice has also gone up ... thanks to the elimination of a third choice.

The fact that you chose already is irrelevant as it wasn't a final choice ... the host didn't open the door you asked for, so you choose again with a 50/50 chance. Yes your second choice is more likely to be correct, now that the options have narrowed, but who cares? You have one choice now - two options.

 

[Praxis1452]

Originally posted by: FeuerFrei
I've totally ignored this thread until now.

Who cares if the probability of winning with a switch goes up? The probability of winning by sticking to your original choice has also gone up ... thanks to the elimination of a third choice.

The fact that you chose already is irrelevant as it wasn't a final choice ... the host didn't open the door you asked for, so you choose again with a 50/50 chance. Yes your second choice is more likely to be correct, now that the options have narrowed, but who cares? You have one choice now - two options.

your wrong. It's not 50/50. It's still 1/3 chance of picking the car and 2/3 chance of picking a goat so switching means you win 2/3 times. Not 50/50.

 

[rikadik]

I have had this argument so many times with people... it is to your advantage to swap.

The easiest way to explain is that your chance of initially picking the car is 1/3 and your chance of initially picking the sheep is 2/3. After opening a door to reveal one of the sheep the game show host will always leave you with a choice between (a) a car and (b) a sheep. Therefore, if you initially pick the car and then swap you will always get a sheep. If you initially pick the sheep you will always get the car. Since you are TWICE as likely to have initially picked a sheep, you are TWICE as likely to end up with the car if you swap.

(I realise some people will have already said this but... I wanted to say it too!)

 

[Fayd]

Originally posted by: TheChort

Originally posted by: Fayd

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.

not really. if you switch, it's a 66% chance of success. the chance of the item being behind your current door is 33%.
reason being, with switching, you're getting doors 2 and 3 with 1 shot.
example to make it clearer. there is 1000 doors.
you choose door 1. the game show host, opens all doors except door 1, and door 687. do you switch to door 687?
of course. reason being? it's a 999/1000 chance that it's behind 687.

Do you realize that this explanation has been posted exactly the same way at least 3 times already.
And why are you quoting someone that hasn't posted in 2 days

to all:
PLEASE READ ALL THE POSTS BEFORE POSTING -- SOMEONE ELSE HAS PROBABLY SAID WHAT YOU ARE THINKING OF

maybe you've realized by now, i dont care.

and i didnt see anyone expand the problem in my quick perusal of the thread, aside from 1 guy say something about 4 doors.

 

[jjones]

Originally posted by: jjones
The only thing I know at this point is that behind every door is a dead horse that keeps getting flogged for no apparent reason.

Never had to quote myself before.

 

[FeuerFrei]

Originally posted by: hellokeith
Car--------Goat--------Goat
initial-------shown------switched-------LOSE

Car--------Goat--------Goat
switched---shown------initial-----------WIN

Car--------Goat--------Goat
switched---initial-------shown----------WIN

Switching won 2/3 of the time


Car----------Goat--------Goat
initial--------shown-------no switch-------WIN

Car----------Goat--------Goat
no switch----shown------initial------------LOSE

Car----------Goat--------Goat
no switch----initial-------shown-----------LOSE

not switching won 1/3 of the time


The reason why switching improves your odds is because you know you will be initially wrong 2/3 of the time. Since you know you will be initially wrong more often than you are right, and you are shown one of the wrong choices after your initial choice, you improve your odds by switching. It is not necessarily intuitive, but it is mathematically correct.

Now, if you are only given this scenario one time, there is of course no guarantee you will win even if you do switch, because you still have a 1/3 chance of losing.

Ok, I'm starting to agree.
I added the missing 4th scenario which makes it look like it's dead even.
But the first two scenarios in each case below can be condensed to one, like you did above, because if you picked the car initially the goat (sheep) shown by the host doesn't matter. He could show either. Either way you lose if you switch and win if you don't.

Car--------Goat--------Goat
initial-------shown------switched-------LOSE

Car--------Goat--------Goat
initial-----switched----shown-----------LOSE

Car--------Goat--------Goat
switched---shown------initial-----------WIN

Car--------Goat--------Goat
switched---initial-------shown----------WIN

Switching won 2/4 of the time?


Car----------Goat--------Goat
initial--------shown-----no switch---------WIN

Car----------Goat--------Goat
initial------no switch-----shown-----------WIN

Car----------Goat--------Goat
no switch----shown------initial------------LOSE

Car----------Goat--------Goat
no switch----initial-------shown-----------LOSE

not switching won 2/4 of the time?

Notice the added scenario (in italics) isn't an added outcome and therefore doesn't the likelihood of winning.

So yeah I agree with the guy I quoted. Possible outcome is 1/3 vs. 2/3. Man, took a bit of thinking.

 

[FeuerFrei]

Originally posted by: Praxis1452

Originally posted by: FeuerFrei
I've totally ignored this thread until now.

Who cares if the probability of winning with a switch goes up? The probability of winning by sticking to your original choice has also gone up ... thanks to the elimination of a third choice.

The fact that you chose already is irrelevant as it wasn't a final choice ... the host didn't open the door you asked for, so you choose again with a 50/50 chance. Yes your second choice is more likely to be correct, now that the options have narrowed, but who cares? You have one choice now - two options.

your wrong. It's not 50/50. It's still 1/3 chance of picking the car and 2/3 chance of picking a goat so switching means you win 2/3 times. Not 50/50.

Yeah, I'm wrong.

 

[kevinthenerd]

Did anyone verify my code? (Please?)

 

[GundamSonicZeroX]

Originally posted by: us3rnotfound
42