Whatchya Gonna Do (2 goats and a car)

[Guest]

Jennifer was selected to be on the popular TV game show "Whatchya Gonna Do?". As she jumped up and down in excitement, the host, Monty Barker, showed her three doors.

"Now, Jennifer, behind one of these doors I have personally placed a brand new Jaguar XJS"

"Eeeeeeee!", Jennifer squealed in delight.

"But", continued MB, "behind the other two doors, there are goats! Select a door and the prize behind it is yours. Which shall it be: Door #1?".

"Oh-oh-oh-oh", Jenny jumped.

"Door #2?"

"Ah-ah-ah-ah"

or "Door #3?"

"Um-um-um-um....well...ah...okay - TWO! I choose Door #2". The audience cheers.

"Door #2. Okay Jennifer", says the host, "but before we show you what's behind Door #2, let me tell you that I am glad you did not pick Door #1, because behind Door #1 is..." The first door swings open to reveal a shabby looking goat. Jenny bounces around the stage in joy because she was thinking of picking that door.

"Ah, but the name of the game is 'Watchya Gonna Do?'. You picked Door #2, but there is still Door #3. I'm going to offer you a choice: either you can keep the door you picked, Door #2, and if the car is behind it, I'll throw in a dinner for two at the Hoi Poloi Restaurant..." The audience gives a collective "oooooo" (who wouldn't want to eat at the exclusive Hoi Poloi?)

"... or you can trade it for what's behind Door #3 - AND if it is the car, I'll throw in a year's supply of Platypus Wax! Tell us about the Platypus Wax, Joe"

Joe's voiceover proceeds to tell us all about the wonders of using Platypus Wax on a Jaguar. As he drones on, Jennifer is nervously trying to decide. Until, MB spins around, points to her, and says (with the audience chorusing behind him)

"What - Chya - Gon - Na - Do?"

Should Jennifer stay with her original choice (Door #2), switch to Door #3, or does it make any difference?

Assume the dinner at the High Poloi and the Platypus Wax are worth about the same dollar amount.

Aaaaaaand, before any of you wise-acres start, Jennifer lives in Toronto - the Jaguar is of far more use to her than the goat.

Jennifer should switch. Contrary to what may seem intuitive, switching actually doubles her chances of winning the car.

This problem is just a re-wording of what is known as the Monty Hall Problem. The key to understanding it is that the host knows the locations of the car and goats. His knowledge changes his actions and thus affects the odds.

Here is a breakdown of all the possible scenarios that Jennifer faces and why Jennifer should switch:

• Door #2 has goat B (probability 1:3) - MB shows goat A behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances (1:3 x 1:1 x 1:1 = 1:3)[/*:m:6a353]
  
• Door #2 has the car (probability 1:3) - MB shows goat A behind Door #1 (1:2) - goat B is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)[/*:m:6a353]
  
• Door #2 has the car (probability 1:3) - MB shows goat B behind Door #1 (1:2) - goat A is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)[/*:m:6a353]
  

There are (1:3 + 1:3 = 2:3) chances that switching will get Jennifer the car, and only (1:6 + 1:6 = 1:3) chances she would get the car by not switching. She should switch.

A more general presentation of the reasoning is this:

At the start of the game, there is a 2:3 chance that Jennifer will pick a door with a goat behind it. If she does, the host will reveal the other other goat, and switching doors will get Jennifer the car.

There is a 1:3 chance she will pick the car. The host will then reveal a goat. Switching would win Jennifer a good supply of Ch?vre (and the disdain of her neighbours).

So, 2 out of 3 times switching gets the car. Simple - unintuitive, but simple.

Why does the host's knowledge change the odds. Because he does not randomly select a door to open - he always opens a door with a goat. By doing this he reduced the possible scenarios for Jennifer to the four listed above. If he randomly picked, then Jenny's chances, if the show progressed as presented, would be 50/50. However, there would also be a 1:3 chance that MB would open the wrong door and reveal the car's location (followed by a 1:1 chance that MB would be sacked and re-runs of McGyver would fill out the remainder of the season!)

[Guest]

I just figured it out.

She SHOULD switch doors. In the beginning, She has a 1/3, or 33% chance of getting the car. Once she picks a door and finds out that a different one had the goat, then 33% is switched from the goats, or negative, to the car, or positive, because she couldnt pick it. her probability of the other door being correct is 66%, so yes, she should switch.

[Guest]

okay this is going to sound crazy, but the answer IS one half. I don't understand quantum mechanics well enough to prove it mathematically, but according to quantum, until the act of measurement (i.e. opening the door) is performed, there is a double state in which both a car and a goat exist behind each door. It may seem like a cheap answer, but when I get back to college I'll ask my professor about it ;-)

[Guest]

I have to disagree that switching improves the chances of winning... when you look at it this way.

Let's say I play this game 100 times, and each time I pick door #2, and each time door #1 has a goat and is opened. If I stay with #2, 50 times I'll win the car, 50 times I'll win a goat. If I switch, the results should be the same.

If you're walking in a line with 2 friends through a minefield that you know has 2 randomly placed mines, and the friend to your left gets a leg blown off, do you change your course or keep walking straight?

[Guest]

I have to disagree that switching improves the chances of winning... when you look at it this way.

Let's say I play this game 100 times, and each time I pick door #2, and each time door #1 has a goat and is opened. If I stay with #2, 50 times I'll win the car, 50 times I'll win a goat. If I switch, the results should be the same.

If you're walking in a line with 2 friends through a minefield that you know has 2 randomly placed mines, and the friend to your left gets a leg blown off, do you change your course or keep walking straight?

To elaborate... if there is ALWAYS a goat behind door #1, then the car will be 50/50 behind doors 2 and 3. But the car CAN be behind door number 1, so switching does indeed improve your chances.

[Guest]

I haven't read all the posts yet, but wanted to say what has come to mind....

Mathematically, I can't see where being shown one door changes your odds of winning. NO matter which door you choose, there's a goat behind at least one of the remaining doors.

Personally, in this exact case, I would switch. WHY??? You ask.... It's simple.

If I don't switch (and win) he'll throw in dinner. If I do switch (and win) he'll thow in car wax.

Why offer me car wax with the door that hides a goat?

[Guest]

1a: [OPEN] [goat] (1/6)

2b: [goat] [OPEN] (1/6)

2: [OPEN] [car] (1/3)

3: [car] [OPEN] (1/3)

=> it is here you make your mistake with the chances that situation 1a and 1b occur

this means we have 1/6 + 1/6 = 1/3 chance on a goat behind the last door, and a 1/3 + 1/3 = 2/3 chance on a car.

OK, I take it back.... I can see the math now. Makes sence to me. The odds are also with you if you switch.

[Guest]

Another way of looking at it:

You pick a door.

You then have the option of either taking whatever is behind the door you have picked; or opening both of the other two doors and picking the best prize*.

Clearly you should switch.

*there may, of course be two goats...

[Guest]

I kind of barely get the idea of the answer here. I've heard this puzzle before. However it seems to me that in the beginning, there are 3 possible setups of how the doors/prizes are arranged

g g c

g c g

c g g

After revealing the first door to contain a goat, we can eliminate the last possiblity. And since the first door has been revealed, we can eliminate it from the choices altogether. So now our possibilities look like

g c

c g

In which case there is a 50% chance that door 2 contains a car, and a 50% chance that door 3 contains a car. In essence, the host is asking the lady to make a new choice between door 2 and door 3. On paper, the idea that she has a 66% chance to win the car might work, but in practice, it's still 50/50 no matter what she chooses. My reasoning is that the first round of choosing is made obsolete by revealing one of the unchosen doors and asking her to make a new choice. It's basically as if the first round never happened.

In other words, the only choice that actually decides whether or not she gets a car is her second choice. No matter her first choice, there would have been one goat left to be revealed. So this first choice is completely irrelevant to the second. She is now being asked to make a second choice that is completely independent of the first. The only connection to the first choice is a perceived one.

[Guest]

Your problem is actually the same as those who bring it down to 4 choices:

You select goat 1 and goat 2 is revealed

You select goat 2 and goat 1 is revealed

You select the car and goat 1 is revealed

You select the car and goat 2 is revealed

The problem with this situation is, if you followed the logic, your first choice always has a 50% chance of getting a car (25% plus 25%). Actually, the last two choices are part of one choice, so if you broke it down, there is a 33.33% chance of the first, 33.33% chance of the second, 16.66% of the third, and 16.66% chance of the fourth (because you only have a 33.33% chance to pick the car on first selection).

So, in your case, yes, every time we pick a goat it eliminates one of the selections, but you have to understand that each selection is not equally as likely after we have chosen that goat. I’ll try to explain.

Lets take your scenario:

1)G G C

2)G C G

3)C G G

And let’s say we choose door 2. If goat 1 is revealed, we can say it is either:

G G C or

G C G

So, you’re right so far. BUT (and the big but is) what is the likelihood of each scenario? If it were scenario 1, he will pick goal 1 every time. If it were scenario 2, he will pick goal 1 only half the time. This means, given that we know that number one is a goat, and that the host will always choose a goat (random in the case of two goats) option 1 is twice as likely as option 2, which is 66.66% instead of 33.3%.

[Guest]

This is actually used in a scene from 21.

Change doors because the original 33% chance has now gone up to 66% by switching doors.

[Guest]

This thread is the Michael Meyers of Brainden.

[Guest]

Your problem is actually the same as those who bring it down to 4 choices:

<snip>

So, you’re right so far. BUT (and the big but is) what is the likelihood of each scenario? If it were scenario 1, he will pick goal 1 every time. If it were scenario 2, he will pick goal 1 only half the time. This means, given that we know that number one is a goat, and that the host will always choose a goat (random in the case of two goats) option 1 is twice as likely as option 2, which is 66.66% instead of 33.3%.

Ok, I get it now.

if you look at it like this. You start with three possible setups.

g g c

g c g

c g g

Let's say you pick door 1 (your pick is colored in red). In the first example, the host would have to show you door number 2. What's left is

g . c

In example 2, the host would have to show you door number 3. What's left is

g c .

In example 3, he could show you either of the doors, either way, you're left with

c g . or c . g

So your possible second choice setups are

g c

g c

c g

Your first round pick is listed first, so 66% of the time you'll win if you switch.

Now let's say you pick door 2, your possible second choice setups are

g c

c g

g c

Your first round pick is listed first. Again, 2 out of 3 win if you switch

Now let's say you picked door 3, your possible second choice setups are

c g

g c

g c

Listing your first round pick first, still 66% of the possibilities are winners if you switch.

[Guest]

or more simply

On average, you will pick the car on your first try 1/3 of the time.

So 2/3 of the time you will pick one goat, and the remaining goat will be shown to you and taken out of the mix, leaving only the car.

So 2/3 of the time, changing will get you the car.

[Guest]

I disagree with most of you (believe the probability is equal) for two reasons

1) Probability is based on givens. e.g. given a 6 sided die what are the odds youll roll a one. So think about what youre given in this problem:

a. There is a car behind one of the three doors

b. There are goats behind the other two doors

c. There is not a goat behind door number one

given c. ALL you can conclude the goat must be behind door number either door 2 or door 3 therefore 50%

2)It makes a difference to know which door is revealed

if you say beforehand "youre gonna pick a door, i am going to open a door that isnt what you picked, and doesnt have a car behind it. would you like to change your pick after that?" then the answer would be yes and the probability of winning would be 2/3

most of your logic states that

a. car goat goat pick door two and get shown three; switch then win

b. goat car goat pick door two and get shown door one or three; switch and lose

c. goat goat car pick door two and get shown door one; switch and win

but you guy forget that when they pick door two and are shown door one option a. would be non applicable and it shouldnt be included in the final calculations.

you are left with just

b. goat car goat pick door two and get shown door one or three; switch and lose

c. goat goat car pick door two and get shown door one; switch and win

so

50-50!

[Guest]

I disagree with most of you (believe the probability is equal) for two reasons

1) Probability is based on givens. e.g. given a 6 sided die what are the odds youll roll a one. So think about what youre given in this problem:

a. There is a car behind one of the three doors

b. There are goats behind the other two doors

c. There is not a goat behind door number one

given c. ALL you can conclude the goat must be behind door number either door 2 or door 3 therefore 50%

That is not the riddle. If it were, we'd know where the car and goats are 100% of the time. I don't know where you're getting 50% from.

most of your logic states that

a. car goat goat pick door two and get shown three; switch then win

b. goat car goat pick door two and get shown door one or three; switch and lose

c. goat goat car pick door two and get shown door one; switch and win

but you guy forget that when they pick door two and are shown door one option a. would be non applicable and it shouldnt be included in the final calculations.

you are left with just

b. goat car goat pick door two and get shown door one or three; switch and lose

c. goat goat car pick door two and get shown door one; switch and win

so

50-50!

Wrong. No one has forgotten anything. The three possible scenarios you posted above are the possibilities before we know what door the host revealed, not after.

You are correct that if the contestant picks door two, and the host chooses to show a goat behind door 1 (there first must of course be a goat there) that we must have scenario b. or c. However, you are forgetting that having two choices does not mean both choices are equal. Since the host doesn't randomly reveal a door, he only reveals what's behind a door if it doesn't have a car behind it, switching gives the contestant a 2/3 probability of picking the door with the car behind it.

BTW, your logic that there are only two scenarios, therefore the probability of ending up with the car is 1/2 if the contestant switches contradicts what you wrote after 2) which you gave an answer of 2/3.

This is a simple way of looking at the problem which may help a lot:

You pick a door. The probability that it is the correct choice is 1/3 and the probability that the Jaguar is behind one of the other two is 2/3. If given the choice to switch your one door for the other two, would you do it? This is essentially the choice Monty has given the contestant, only he revealed a goat behind one of those doors first (which we already knew he could do).

Play with this simulator for a bit if you don't believe switching gives a 2/3 probability of winning. Switch every time and you'll find you win more than you lose.

[Prof. Templeton]

Great simulator. I played three times and won twice by switching as did almost 68% percent of other players. Go figure. You can always count on someone posting a 50/50 answer on this thread every few days. Just when you think the horse is about to be buried, someone comes along to beat on it a little more.

[Guest]

Theres an easy way to understand this problem atleast intuitively

In the absence of any information the probability of occurance of three equally likely events is 1/3 or 33.3333%.

In this puzzle when the person selects onf of the 3 doors, the probabilty of the car being behind the door is 1/3. Now the host already knows what are behind the 3 doors. By opening a door with a goat behind it the host is actually revealing information about whats behind the door the person selected.

For the host to have a choice in selecting a door with a goat behind it the person must select the door with car behind it - the probability of which is only 33%.

However for the host to have no choice in selection of a door that has a goat behind it the person must select a door with a goat behind it - the probability of which is 66%

So when the host reveals the door behind which is a goat, The probabilty that theres a car behind the door the person selected becomes 33% and the probabiilty that theres a goat behind the door the person selected becomes 66%. So the person has to switch to increase his probabilty of getting the car.

For the scenario where the probabilty is 50-50 whether the person swithces or not to occur, the host must also not know whats behind the doors. In this case the host cannot reveal any information and would have to open one of the remaining 2 doors at random. So in this scenario there is a non zero probabilty that the host opens the door with a car behind it whereas in the scenario above the probabilty that the host opens a door with a car behind it is always zero.

Edited September 5, 2008 by vimil

[Guest]

GET OVER IT! There is no 50/50 probability, EVER! research conditional probability.

Hopefully this can convince the non believers. EVEN IF the host picks a door at random. I will represent the doors as binary, but I will use x for both the door picked and the door opened by the host. There are 3 possible set ups, and 3 scenarios.

110

101

011

Now you can eleminte either door 1,2, or 3 leaving one of 9 scenarios

110 X10

101 X01

011 X11

110 1X0

101 1X1

011 0X1

110 11X

101 10X

011 01X

The host can then pick a door AT RANDOM! Leaving 18 scenarios

110 X10 XX0 Remaining door is 0

110 X10 X1X Remaining door is 1

101 X01 XX0 Remaining door is 0

101 X10 X1X Remaining door is 1

011 X11 XX1 Remaining door is 1

011 X11 X1X Remaining door is 1

110 1X0 XX0 Remaining door is 0

110 1X0 1XX Remaining door is 1

101 1X1 XX1 Remaining door is 1

101 1X1 1XX Remaining door is 1

011 0X1 XX1 Remaining door is 1

011 0X1 0XX Remaining door is 0

110 11X X1X Remaining door is 1

110 11X 1XX Remaining door is 1

101 10X X0X Remaining door is 0

101 10X 1XX Remaining door is 1

011 01X X1X Remaining door is 1

011 01X 0XX Remaining door is 0

Now if you see out of the 18 possible scenarios with a random selection by the host, 12 are 1 and 6 are 0.

2/3 porabality no matter what! Let it rest!

[Guest]

GET OVER IT! There is no 50/50 probability, EVER! research conditional probability.

Hopefully this can convince the non believers. EVEN IF the host picks a door at random. I will represent the doors as binary, but I will use x for both the door picked and the door opened by the host. There are 3 possible set ups, and 3 scenarios.

110

101

011

Now you can eleminte either door 1,2, or 3 leaving one of 9 scenarios

110 X10

101 X01

011 X11

110 1X0

101 1X1

011 0X1

110 11X

101 10X

011 01X

The host can then pick a door AT RANDOM! Leaving 18 scenarios

110 X10 XX0 Remaining door is 0

110 X10 X1X Remaining door is 1

101 X01 XX0 Remaining door is 0

101 X10 X1X Remaining door is 1

011 X11 XX1 Remaining door is 1

011 X11 X1X Remaining door is 1

110 1X0 XX0 Remaining door is 0

110 1X0 1XX Remaining door is 1

101 1X1 XX1 Remaining door is 1

101 1X1 1XX Remaining door is 1

011 0X1 XX1 Remaining door is 1

011 0X1 0XX Remaining door is 0

110 11X X1X Remaining door is 1

110 11X 1XX Remaining door is 1

101 10X X0X Remaining door is 0

101 10X 1XX Remaining door is 1

011 01X X1X Remaining door is 1

011 01X 0XX Remaining door is 0

Now if you see out of the 18 possible scenarios with a random selection by the host, 12 are 1 and 6 are 0.

2/3 porabality no matter what! Let it rest!

Hey Eyes I did not understand ur explanation. Anyway this is the famous monty hall problem and you can google about it for more information. Heres a link to enlighten you

http://brainyplanet.com/index.php/Monty%20...deca25f0f615f91

[Guest]

GET OVER IT! There is no 50/50 probability, EVER! research conditional probability.

<Snip>

Now if you see out of the 18 possible scenarios with a random selection by the host, 12 are 1 and 6 are 0.

2/3 porabality no matter what! Let it rest!

vimil is correct. If the host randomly reveals what's behind one of the two remaining doors and it happens to be a goat, then the contestant's probability of winning whether he switches or not is 1/2.

Just for fun, instead of two goats, let's consider a car, a goat and a pig. The only one of course which is considered a winning prize is the car.

Let's say the contestant decides in advance he will pick door #1 and the host will only reveal a losing prize (just as in the original riddle).

These are the possibilities and the results of what would happen if the contestant chooses to switch:

car goat pig...lose

car pig goat...lose

goat car pig...win

goat pig car...win

pig goat car...win

pig car goat...win

The contestant has a 2/3 probability of winning.

Now, what would happen if the contestant decides in advance he will pick door #1 and the host then randomly chooses to reveal what's behind door #2? Well, 1/3 of the time the host will reveal a car, but since we're only concerned with what would happen if he randomly opens a door and it reveals an animal, the following are the possibilities and the results of what would happen if the contestant chooses to switch:

car goat pig...lose

car pig goat...lose

goat pig car...win

pig goat car...win

The contestant has a 1/2 probability of winning. Likewise, if he chooses not to switch, he still has a 1/2 probability of winning.

[Guest]

<snip>

Now, what would happen if the contestant decides in advance he will pick door #1 and the host then randomly chooses to reveal what's behind door #2? Well, 1/3 of the time the host will reveal a car, but since we're only concerned with what would happen if he randomly opens a door and it reveals an animal, the following are the possibilities and the results of what would happen if the contestant chooses to switch:

car goat pig...lose

car pig goat...lose

goat pig car...win

pig goat car...win

The contestant has a 1/2 probability of winning. Likewise, if he chooses not to switch, he still has a 1/2 probability of winning.

Ok, let's say the prizes are car, goat and pig. If you pick the goat in the first round, the host will reveal the pig. If you pick the pig in the first round, the host will reveal the goat. If you pick the car in the first round, he will reveal a random animal. He can do this because he knows what is where. He will never reveal the car.

So...

Do you agree that in the first round there is a 2/3 chance that you will pick a loser? (you should because there are three doors, behind two of which are losers)

Do you agree that if you pick one of the losers in the first round, the other loser will be revealed, leaving only the car behind the other door? (you should because that's the way it works)

If you agree to these conditions (which are the way the "riddle" is set up), then you'll see that 2/3 of the time you'll start with a loser. If you start with a loser and decide to switch, you have no option but to win the car. So 2/3 of the time, switching wins the car.

Edited September 5, 2008 by pyrobryan

[Guest]

pyrobryan, was I not clear in my last post that I agree that the answer to the riddle is that switching will give a 2/3 probability of winning? If it wasn't clear to you, read some of my other posts in this thread. The section of my post that you quoted was regarding what the probability would be if the host chose to randomly open a door and an animal was revealed.

[Prof. Templeton]

@ Pyrobryan

I don't think you understand the scenario that Martini was describing. If the Host doesn't know where the prize is your odds are 50/50. Martini knows this topic better than anybody else I've seen try to explain it. 2/3 if the host knows where the prizes are and you switch and 50/50 if the host doesn't know. I don't even know why I'm speaking on Martini's behalf he's quite capable (and I was just looking for something to do).

[Guest]

The support is still appreciated, Professor.

[Prof. Templeton]

The support is still appreciated, Professor.

De Nada

"I've got a fever. And the only cure, is more cowbell" love it.