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]

change her mind and go for 3 because she now has a 50/50 chance vs a 33% chance.

[Guest]

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.

Let's look at the problem again. 1 equals prize 0 equals goat

100

If you nor the host knows the doors, you are essentialy picking 2 doors at random. In this scenario, 100

you pick d1 host picks d2, remaining is 0

you pick d1 host picks d3, remaining is 0

you pick d2 host picks d3, remaining is 1

you pick d3 host picks d2, remaining is 1

you pick d2 host picks d1 remaining is 0, he opened the prize

you pick d3 host picks d1, remaining is 0, he opened the prize

Out of 6 possibilities 4 of them leave the unpicked door at 0. 2 of those you have already seen the prize. I didn't see anywhere that if the host opens the prize you can't switch because of his mistake. If he opens the door to reveal the prize and the game is over then those 2 scenarios don't count, so only 4 remain, 2 winners 2 losers. So if the host opens the door with the car at random yes it's a 50/ 50 but only because you are eleminating the chance to switch with those scenarios.

[Guest]

If he opens the door to reveal the prize and the game is over then those 2 scenarios don't count, so only 4 remain, 2 winners 2 losers. So if the host opens the door with the car at random yes it's a 50/ 50 but only because you are eleminating the chance to switch with those scenarios.

In my scenario, the host opens a door at random and we see that a car is not revealed, so that possibility of him revealing the car is already out of the question. I was pretty clear in laying out the scenario. This is the scenario I believed vimil was describing. Reading what he wrote again, I'm not so sure:

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.

vimil, you say there is a non-zero probability of the host revealing a door with a goat behind it, so my scenario is not analogous to yours as mine supposes that the host chose to reveal what's behind a door at random and he ends up not revealing a car. Your scenario differs in that that possibility still exists. Can you explain what happens if the host reveals the car and why you believe the probability changes to 1/2?

[Guest]

In my scenario, the host opens a door at random and we see that a car is not revealed, so that possibility of him revealing the car is already out of the question. I was pretty clear in laying out the scenario. This is the scenario I believed vimil was describing. Reading what he wrote again, I'm not so sure:

vimil, you say there is a non-zero probability of the host revealing a door with a goat behind it, so my scenario is not analogous to yours as mine supposes that the host chose to reveal what's behind a door at random and he ends up not revealing a car. Your scenario differs in that that possibility still exists. Can you explain what happens if the host reveals the car and why you believe the probability changes to 1/2?

Exactly, if anything, if the host picks a door at random, you only have 1/3 overall chance whether you stay or switch

[Guest]

I remember reading an entire webpage on this exact question! I believe it was a wikipedia page. Logic questions always make me so agitated because you have to use a different part of your brain to think it out.

[Guest]

In my scenario, the host opens a door at random and we see that a car is not revealed, so that possibility of him revealing the car is already out of the question. I was pretty clear in laying out the scenario. This is the scenario I believed vimil was describing. Reading what he wrote again, I'm not so sure:

vimil, you say there is a non-zero probability of the host revealing a door with a goat behind it, so my scenario is not analogous to yours as mine supposes that the host chose to reveal what's behind a door at random and he ends up not revealing a car. Your scenario differs in that that possibility still exists. Can you explain what happens if the host reveals the car and why you believe the probability changes to 1/2?

If the host reveals the car then the question of switching never arises. It has become 100% certain that the car is behind the door the host opened and so there isnt any car behind the other 2 doors. What I wanted to say was that when the host himself does not know whats behind the doors and he opens a door revealing a goat then the probabilty does not favour switching. Its equally likely that the car is behind the door that person selected or the remaining unopened door.

Many people wonder how can opening of a door(unless theres a car behind the door opened) affect the probability of the presence of a car behind the other 2 doors, but they forget that the problem states that the host already knows whats behind the 3 doors. By making his decision on which door to open based on the door that the contestant has opened he is actually skewing the probabilities of the presence of the car behind the other 2 doors such that there is a lesser probability that the car is behind the door the contestant selected.

[Guest]

What I wanted to say was that when the host himself does not know whats behind the doors and he opens a door revealing a goat then the probabilty does not favour switching. Its equally likely that the car is behind the door that person selected or the remaining unopened door.

I agree with this, which is why I defended your conclusion. However, I can understand EyesOfTheDead's confusion as you stated "So in this scenario there is a non zero probabilty that the host opens the door with a car behind it". If there is a non-zero probability of opening a door with the car behind it, it seems you're describing a situation in which the host can still open a door and reveal the car.

No big deal. I understand you were referring to a scenario similar to the one I posted in post #121 in which we're calculating odds after the fact that the host opened a door and it happened that the car wasn't revealed. But I can understand the confusion the wording caused.

[Guest]

Mathematically the removal of one of the choices in the ORIGINAL odds is of nill value in the CURRENT state of probability. Although Thinking in those terms makes for an interesting mathematical equation exercise

The host knowing that the opened door had a goat did NOT remove 1/3 of the possible choices, the contestant DID by NOT picking it. Now we have a brand new randmoness exercise and your odds are now 50/50 .. Going back to the original scenario, to qualify the current odds by using the odds in the first set of values taints the current percentages. In fact this is now a new set of odds.

Based on this what are the odds that in a coin toss the con will land heads ?

Actually it is 1/3 because as unlikely as it is the coin could land on its side, but we remove the side possibility having previous knowledge (time nwledge) that the odds of that happening are astronomical , so the host just revaled a possibilty that was no longer part of the odds.

This is akin to the Three Card Monty "cheat"

You begin with three cards and the dealer , tricks you into choosing a card he knows is not the right card , so all you have to do to make the odds 50/50 is to absolubtly NOT bet on the card you are POSITIVE is the right card in fact removing one element of the odds. in the original premise , you already discarded 1/3 of the odds by NOT picking A goat) you do the same here by not picking the card that is "certainly the winning card"

And we all know that casinos will never play on 50/50 odds becuase the odds are they will loose

[Guest]

The host knowing that the opened door had a goat did NOT remove 1/3 of the possible choices, the contestant DID by NOT picking it. Now we have a brand new randmoness exercise and your odds are now 50/50 ..

Incorrect. The host revealing a goat does not change the probability of the contestants choice. Since we already knew the host could reveal a goat since he knows which door would reveal the car, the probability that the contestant will win the car remains 1/3 if he chooses not to switch. Did you read the many posts in this thread? It's been explained many times.

Going back to the original scenario, to qualify the current odds by using the odds in the first set of values taints the current percentages. In fact this is now a new set of odds.

Nope. Again, since the host does not randomly reveal what's behind a door (he purposely reveals a goat), there is no new set of odds if the contestant does not switch. There is however an increased probability that switching will win the contestant the car.

Based on this what are the odds that in a coin toss the con will land heads ?

Actually it is 1/3 because as unlikely as it is the coin could land on its side

Really? I'll bet you anything you like that I can correctly guess what side a fair coin will land on after 100 flips with much better accuracy than 33% (and I'm not even psychic).

This is akin to the Three Card Monty "cheat"

You begin with three cards and the dealer , tricks you into choosing a card he knows is not the right card , so all you have to do to make the odds 50/50 is to absolubtly NOT bet on the card you are POSITIVE is the right card in fact removing one element of the odds. in the original premise , you already discarded 1/3 of the odds by NOT picking A goat) you do the same here by not picking the card that is "certainly the winning card"

Different scenario. If given the choice of three cards where one is a winner, and you are sure a specific one is not a winner, then you have a 1/2 probability of picking the winner (assuming the three card monty dealer will not deceive you in another way).

In this scenario, the host reveals what's behind one door which is definitely not a winner after you choose a door. By sticking to the original choice, the probability remains 1/3. If the contestant chooses to switch, the probability increases to 2/3. See the many posts in this thread that provide the rationale for the probability increase.

And we all know that casinos will never play on 50/50 odds becuase the odds are they will loose

Hmm. So, the odds are that the gambler will win?

[Guest]

This problem has already been solved by others, yet allow me to use a mathematical model to prove this answer. (Which was probably already done, but hey)

You have three possibilities of where the car is 1, 2, or 3. Each with probability of 1/3.

Choose any door and the probability that you choose the correct door is 1/3, and 2/3 that you are wrong. Meaning, there is a 2/3 probability that of the two doors left not chosen, one contains a car, the other a donkey. And if you choose the car then there is a 1/3 probability that both are donkeys.

Here is the part where most people find contradicts itself, so follow carefully:

The host reveals one of the doors you did not choose to reveal a donkey. Now, there is 1/3 chance that you already picked out the car and the host is revealing doors to throw you off. The host can reveal the other doors with equal probability with probability 1/2. So, the odds of you being correct and the host revealing any particular door is 1/2 * 1/3 or 1/6. Revealing the other donkey also happens with probability 1/6. So the odds for this scenario is 1/3....the part where you choose the car to begin with.

However, there is a 2/3 probability that you did not choose the car. The host will reveal a donkey with 100% probability, yet, in both cases, there is only one option for him to choose the donkey. So 2/3 of the time you selected the donkey, and the host revealed the other donkey, therefore 2/3 of the time, if you switch you will have found the car, but 1/3 of the time you already choose the car.

The way to view this is by remembering you already selected a door, so the host cannot choose that door, and the host will not choose the door with the car behind it. However, if the host revealed a door before you choose, then initial logic is correct the odds are 1/2.

Reply to this if you are not convinced and I can go further into it, or you can suggest why it does not make sense to you.

[Guest]

Sorry I'm late to the party, but I've seen this problem here and there and ventured in for a look. My rationale was always "You have a 50/50 chance" and I was satisfied with that--I thought the rest was all mumbo-jumbo. But then I decided to think about it today and made a diagram showing the three scenarios--it's pretty easy to follow without going into advanced mathematical principles...

|-No Switch = Goat A

|-Goat A---

| |_Switch = Car

|

|

| |-No Switch = Goat B

|-Goat B---

| |_Switch = Car

|

|

| |-No Switch = Car

|-Car------

|_Switch = Goat (A or B--irrelevant)

The underlined choices comprise your starting position after picking a door.

The bolded choices (no switches) give you 2 Goats out of 3 Options. Basic probability of 67% Goat and 33% Car

The italicized choices (switches) give you 2 Cars out of 3 Options. Basic probability of 67% Car and 33% Goat

EDIT: Make diagram look neater

Edited October 9, 2008 by Mumbles140

[Guest]

The people who think it is 2/3 are wrong.

I think this has been pointed out many times but I didn't read all 14 pages to see if it was explained this way.

Since we know a goat will be eliminated no matter what, the odds of picking the car at the beginning are 50/50. You can ignore one of the goats totally. You know that you will either pick a goat or car.

Here is what the tree really looks like

Your pick Other box

Car Goat Switch you lose

Car Goat Don't switch you win

Goat Car Switch you win

Goat Car Don't switch you lose

So no matter what you KNOW you will end up with a goat and a car behind the last 2 boxes. The 3rd box is just to throw you off. But in the end, you have a 50/50 chance of winning a car. You have those odds right from the beginning. Basically all you are choosing is which goat will be eliminated. If you pick the car, then MB picks which goat is eliminated.

The 2/3 scenario doesn't work even if a box is eliminated at random.

Trying to make sure I cover all the bases here. The original problem of 3 boxes is just a setup for the 2 box choice. Thus the odds of getting the car are 50/50 from the beginning.

[unreality]

Mumbles, this is the diagram:

You picked Goat A originally

* a door is opened, Goat B is revealed

** Switch: You get a Car

** Don't Switch: You get Goat A

You picked Goat B originally

* a door is opened, Goat A is revealed

** Switch: You get a Car

** Don't Switch: You get Goat B

You picked the Car originally

* a door is opened, and a goat is revealed

** Switch: You get the other goat

** Don't Switch: You get the Car

This is the same diagram you made, I think, but you looked at it incorrectly after you made it...

Look at the three "Don't Switch" options. Two result in "Goat" and one in "Car"

Look at the three "Switch" options. Two result in "Car" and one in "Goat"

If you don't switch, you have a 1/3 chance of getting a car

If you do switch, you have a 2/3 chance of getting a car

[bonanova]

The people who think it is 2/3 are wrong.

I think this has been pointed out many times but I didn't read all 14 pages to see if it was explained this way.

Since we know a goat will be eliminated no matter what, the odds of picking the car at the beginning are 50/50. You can ignore one of the goats totally. You know that you will either pick a goat or car.

Here is what the tree really looks like

Your pick Other box

Car Goat Switch you lose

Car Goat Don't switch you win

Goat Car Switch you win

Goat Car Don't switch you lose

So no matter what you KNOW you will end up with a goat and a car behind the last 2 boxes. The 3rd box is just to throw you off. But in the end, you have a 50/50 chance of winning a car. You have those odds right from the beginning. Basically all you are choosing is which goat will be eliminated. If you pick the car, then MB picks which goat is eliminated.

The 2/3 scenario doesn't work even if a box is eliminated at random.

Trying to make sure I cover all the bases here. The original problem of 3 boxes is just a setup for the 2 box choice. Thus the odds of getting the car are 50/50 from the beginning.

Take your analysis to the limit, and see if you still go with it.

There are now 1 million doors: one car and nine hundred ninety-nine thousand nine hundred and ninety-nine goats.

You will pick a door, then you will be shown 999,998 goats.

Finally you will be given a chance to swap your choice with the 999,999th door.

You reason that you know that 999,998 goats will be eliminated.

There are only two doors where the car can be - the one you choose and the 999,999th door.

So they both have 50% chance of winning.

So you choose a door, stick with your choice and expect to win 50% of the time.

Right?

[Guest]

You are confused, Grasshopper. Your logic has fallen short. Let me show you the error of your ways.

IF I choose 1 out of a million then my chances of winning are 1 in a million. BUT now you eliminate 1. That means my chances of having the correct choice are 1 in 999,999. As you keep eliminating them my odds go up that I have the right one until I am at 50-50 chance of having the right one. But that is not quite correct either. What IS the right answer?

Look at it if boxes are eliminated at random.

Each choice has the same odds of being right. None of them will change just because some are eliminated. Each box has the same odds of winning. 1 in however many. BUT the odds do not shift to any of the remaining boxes no matter how many you remove. Only the odds for the eliminated box have changed. Each box still have 1 chance in however many of being the correct one. Finally you are down to 2 boxes. Each one has odds of 1 in a millon. So that equates to 50% since you now know that between these 2 you have a 100% chance of finding the winner.

Now think about it. IF the person doing the eliminating knows which box is correct and eliminates ALL other boxes except yours and 1 other (winner or not) then you have started with a 50/50 chance because you KNOW that you will come down to 1 winner and 1 loser box no matter what. It doesn't matter how many you start out with. It could be 50 gabillion boxes. In essence what happens is you choose and then the other person removes ALL but 2 boxes, yours and 1 other. You may as well have started out with 2 boxes.

You are trying to confuse odds of picking the winner from x number of boxes with what the actual choice is. The actual choice is 1 of 2 boxes. The rest of the boxes are irrelevant because you know they will be eliminated.

[Guest]

The underlined choices comprise your starting position after picking a door.

The bolded choices (no switches) give you 2 Goats out of 3 Options. Basic probability of 67% Goat and 33% Car

The italicized choices (switches) give you 2 Cars out of 3 Options. Basic probability of 67% Car and 33% Goat

This is the same diagram you made, I think, but you looked at it incorrectly after you made it...

Look at the three "Don't Switch" options. Two result in "Goat" and one in "Car"

Look at the three "Switch" options. Two result in "Car" and one in "Goat"

If you don't switch, you have a 1/3 chance of getting a car

If you do switch, you have a 2/3 chance of getting a car

Am I misreading here, or is that the same thing I said?

[bonanova]

You are confused, Grasshopper.

Your logic has fallen short.

Let me show you the error of your ways.

IF I choose 1 out of a million then my chances of winning are 1 in a million.

Absolutely correct.

See? It's not really hard to figure that out.

It isn't your logic that has fallen short [did you say grasshopper?] it's your reading, comprehension or

retention abilities. You've simply forgotten what you read and understood from the OP. Or you

remembered what you didn't understand from the OP. Or, you simply did not read the OP.

Whatever.

The problem that you are analyzing is this:

There are a million doors - one car, the rest goats.

1. Monty opens 999,998 goat doors, leaving two unopened.
   
2. You choose from the two remaining doors.
   

You have a 50% chance of getting the car.

Did you notice the subtle but important difference in the order of events?

By the way, if you still cling to your one-in-a-million-even-odds bets, that really fine with me,

I have no horse in this race - you can analyze these things any way you like.

But understand that the real test of correct analysis is to play the

game and come out with more money than you started with.

And anytime you want to play this game, just give me a call.

[Guest]

Now think about it. IF the person doing the eliminating knows which box is correct and eliminates ALL other boxes except yours and 1 other (winner or not) then you have started with a 50/50 chance because you KNOW that you will come down to 1 winner and 1 loser box no matter what. It doesn't matter how many you start out with. It could be 50 gabillion boxes. In essence what happens is you choose and then the other person removes ALL but 2 boxes, yours and 1 other. You may as well have started out with 2 boxes.

This is wrong. Since the person doing the eliminating didn't do so randomly, but only eliminated those which are not winners, you have the same probability of having a winning box as you did before he started- 1/1,000,000. The probability that the other box is a winner is 999,999/1,000,000.

You made the same mistake regarding the riddle. You said:

Since we know a goat will be eliminated no matter what, the odds of picking the car at the beginning are 50/50. You can ignore one of the goats totally. You know that you will either pick a goat or car.

This is wrong for the same reason as above. Since Monty Barker did not eliminate doors randomly, he only eliminated one door that definitely had a goat behind it, the contestant's probability of having originally chosen the door with a car behind it remains 1/3, it was never 50/50 and never becomes 50/50. The probability of the remaining door having a car behind it is 2/3 and the rationale for this has been explained many times in this thread.

Here is what the tree really looks like

Your pick Other box

Car Goat Switch you lose

Car Goat Don't switch you win

Goat Car Switch you win

Goat Car Don't switch you lose

No, there are three possibilities for each of the two scenarios (switching or sticking) as unreality has shown in his last post.

If you choose to switch, these are the three scenarios. Let's say the car is behind A.

You pick A. You switch. You lose.

You pick B. You switch. You win.

You pick C. You switch. You win.

Switching gives you a 2/3 probability of winning.

If you choose to stick, these are the three scenarios. Let's say the car is behind A.

You pick A. You stick. You win.

You pick B. You stick. You lose.

You pick C. You stick. You lose.

Sticking gives you a 1/3 probability of winning and the fact that Monty decides to show you where a goat is after you pick a door doesn't change this.

Since you admitted to not reading the entire thread, maybe you missed this simulator where you can experiment with switching and tallying what percentage of the time you win a car.

[Guest]

Hate to keep telling people this, but you are totally wrong. You are looking at the problem wrong.

Okay, start over.

You have 3 boxes. 2 goats, 1 car. You choose 1. Now your odds of picking the car are 1/3. Your odds of picking goat A = 1/3. Your odds of picking goat B = 1/3. So all of them are 1/3. So if MB just reveals what is behind your box, you always had a 1/3 chance. BUT that is not what happens.

Suppose he eliminates 1 but does not tell you what is behind it. Then you STILL have 1/3 chance of being right. So if he eliminates one and shows you what is behind it you STILL have a 1/3 chance of being right. The remaining box STILL has a 1/3 chance of being right. So you have 2 boxes each with 1/3 chance of being the car. The odds do not change simply because you eliminated 1 of the choices. OR you could divide the odds of the 1 between the other 2. Thus you have 1/3 + 1/6 for each one. Which is 2/6 + 1/6 = 3/6 = 1/2. No matter how you slice it you have a 50/50 chance of getting the car now.

The ONLY way to get a 2/3 chance of winning is if you choose and then he says "Pick a 2nd box".

Now let's look at what you said here :

"This is wrong. Since the person doing the eliminating didn't do so randomly, but only eliminated those which are not winners, you have the same probability of having a winning box as you did before he started- 1/1,000,000. The probability that the other box is a winner is 999,999/1,000,000."

Not at all. Totally wrong. Why do you think all the odds pass to the other box? There is nothing that says this anywhere. In reality the boxes ALL keep their odds of being the right box. Thus you come down to 2 boxes both with 1/1,000,000 odds of being the car. I mean I could twist it like you are doing and say that the box you have chosen has a 999,999/1,000,000 chance of being right. It is the exact same flawed logic you are using.

The fact remains, once you are down to 2 boxes, since you KNOW he did not eliminate the car, you have a 50/50 chance of being right.

Let me also add that the million scenario doesn't apply UNLESS you are going to have a choice to switch after each box is revealed. But it still does not matter, since each box has the same chance of being a winner all the way through.

"You pick A. You stick. You win.

You pick B. You stick. You lose.

You pick C. You stick. You lose."

Ooooops you missed 1. You pick A and you switch you lose. Why did you forget this one? Intentional?

Once again here are your REAL choices at the end.

Car is in A Box C is revealed to have a goat now ---

You picked A and stick you win

You picked A and switch you lose

You picked B and switch you win

You picked B and stick you lose

2 chances of getting the car and 2 chances of getting the goat. 50/50.

It doesn't matter if you picked C because it was eliminated. Notice how if I take YOUR scenario and pick C and it is eliminated then I lose with no chance of switching. You are forgetting to eliminate one box up there. BUT let's start from there

Car is in A you pick C

MB eliminates box B as having a goat in it.

Do you switch?

Here is what you have:

You picked C and stick you lose

You picked C and switch you win

Okay, now the car is in B and you pick B

MB eliminates a goat.

Do you swtich?

You stick you win

You swtich you lose

No matter how many combos I do, it all comes out to 50/50.

It doesn't matter what box you pick, a goat will be eliminated. That means you have a 50/50 chance of being right. All you are doing on the first pick is eliminating a goat.

Here is what is at stake, people believing a twisted logic solution that is wrong, or understanding the actual answer based on solid probabilities. Hate seeing people run around with the wrong information.

[Guest]

Of course in real life, Monty Hall answered this question and said odds have nothing to do with it. Since he knows which one is the winner, he will try to influence the person depending on whether or not they are supposed to win and for the drama of it all.

Like this:

You have picked box A and have the car but don't know it. Monty reveals box B ... a goat.

NOW he starts with "I'll give you blender for your box. How about $1000 to change to box B? How about ....."

He builds the tension and the game show gets great ratings. But he told everyone that he can read people and know who will stick and who will switch and how to play them. They have to have enough big winners to keep people interested but not enough that they go broke. Same as Vegas. Those slot machines are rigged big time. That came out in a court case where they (the casino owner) had to prove that the guy who supposedly won a jackpot had not because the machine was not set to have a winner at that time. The casino won the case.

[Guest]

Hate to keep telling people this, but you are totally wrong. You are looking at the problem wrong.

I linked to a site where you can experiment for yourself and I assume you didn't try it out. I'll post rebuttals to your arguments again.

Okay, start over.

You have 3 boxes. 2 goats, 1 car. You choose 1. Now your odds of picking the car are 1/3. Your odds of picking goat A = 1/3. Your odds of picking goat B = 1/3. So all of them are 1/3. So if MB just reveals what is behind your box, you always had a 1/3 chance. BUT that is not what happens.

Suppose he eliminates 1 but does not tell you what is behind it. Then you STILL have 1/3 chance of being right. So if he eliminates one and shows you what is behind it you STILL have a 1/3 chance of being right.

The remaining box STILL has a 1/3 chance of being right.

Wrong. Monty does not reveal what's behind one of the doors randomly; he only reveals a door that has a goat behind it. The remaining door has a 2/3 chance of having a car behind it and here's why:

Let's say Door A has the car behind it. Do the remaining doors have a car behind it 1/3 of the time or 2/3 of the time?:

Three possible scenarios- You pick A, B or C.

You pick A- Monty reveals what's behind one of the other doors. Does the remaining door have a car behind it? No

You pick B- Monty reveals what's behind one of the other doors. Does the remaining door have a car behind it? Yes

You pick C- Monty reveals what's behind one of the other doors. Does the remaining door have a car behind it? Yes

Now let's look at what you said here :

"This is wrong. Since the person doing the eliminating didn't do so randomly, but only eliminated those which are not winners, you have the same probability of having a winning box as you did before he started- 1/1,000,000. The probability that the other box is a winner is 999,999/1,000,000."

Not at all. Totally wrong. Why do you think all the odds pass to the other box?

Simple. The person doing the eliminating did not do so randomly; he only eliminated boxes which are losers. You already know that he has this knowledge and can choose to reveal losing boxes if he chooses to. His doing so does not change your probability of having the winning box one bit. A simpler analogy:

I have a bag with four marbles in it. One is white and three are black. You stick your hand in the bag and remove one without looking at it. There is a 1/4 probability that you picked the white one. Now I look in the bag and purposely remove two black marbles. Do you think you now have a 50/50 chance of having the white marble concealed in your hand? You must, as this is an exact analogy to the million boxes problem you posted and the OP. But of course the probability that you have a white marble remains 1/4 and did not get reduced to 1/2 (50/50 chance). What about the marble left in the bag? It has a 3/4 probability of being white. Don't believe it? Try it several times for yourself and record the results. Same thing with the riddle in the OP. You pick a door and have a 1/3 chance of it having a car behind it. Monty purposely reveals a goat. The probability of your door having the car behind it remains 1/3. The probability that the remaining door has a car behind it is 2/3.

"You pick A. You stick. You win.

You pick B. You stick. You lose.

You pick C. You stick. You lose."

Ooooops you missed 1. You pick A and you switch you lose. Why did you forget this one? Intentional?

What are you talking about? I didn't forget anything. I included that scenario as being one of the three that exist if the choice is to switch, and of those three scenarios, switching leaves a winning probability of 2/3. Are you actually accusing me of "forgetting" something intentionally, as if I am purposely arguing for the wrong answer?

Once again here are your REAL choices at the end.

Car is in A Box C is revealed to have a goat now ---

You picked A and stick you win

You picked A and switch you lose

You picked B and switch you win

You picked B and stick you lose

2 chances of getting the car and 2 chances of getting the goat. 50/50.

You are assuming that Box C is revealed without first looking at that being a possible choice of yours. You can't just look at A and B as being a choice, C must also be included. Tell me what you find wrong with these scenarios for switching and sticking:

If you choose to switch, these are the three scenarios. Let's say the car is behind A.

You pick A. You switch. You lose.

You pick B. You switch. You win.

You pick C. You switch. You win.

Switching gives you a 2/3 probability of winning.

If you choose to stick, these are the three scenarios. Let's say the car is behind A.

You pick A. You stick. You win.

You pick B. You stick. You lose.

You pick C. You stick. You lose.

Sticking gives you a 1/3 probability of winning and the fact that Monty decides to show you where a goat is after you pick a door doesn't change this.

It doesn't matter if you picked C because it was eliminated.

Of course it matters; if you chose C, Monty could not have chosen to reveal it. He only reveals what's behind one of the doors that you didn't pick along with it having a goat behind it.

Notice how if I take YOUR scenario and pick C and it is eliminated then I lose with no chance of switching. You are forgetting to eliminate one box up there. BUT let's start from there

I'm getting the feeling you don't understand the rules. Once you pick a door, Monty will reveal a goat behind one of the other doors.

Car is in A you pick C

MB eliminates box B as having a goat in it.

Do you switch?

Here is what you have:

You picked C and stick you lose

You picked C and switch you win

Okay, now the car is in B and you pick B

MB eliminates a goat.

Do you swtich?

You stick you win

You swtich you lose

No matter how many combos I do, it all comes out to 50/50.

You're not looking at all of the possible scenarios. First you looked at what happens if I pick C and the car is in A and then if I pick B and the car is in B. There are three scenarios for switching and three for sticking (the same ones I posted above).

It doesn't matter what box you pick, a goat will be eliminated. That means you have a 50/50 chance of being right. All you are doing on the first pick is eliminating a goat.

That solution does not follow that premise. There is a 1/3 probability that the contestant chose a car. The host choosing to reveal one goat when he knows what doors are concealing what does not change this. This is also true in the one million box and four marbles scenario.

Here is what is at stake, people believing a twisted logic solution that is wrong, or understanding the actual answer based on solid probabilities. Hate seeing people run around with the wrong information.

Yeah, I know how you feel. I guess you believe if you pull a marble out of my bag adhering to the conditions I stated above, after I remove two black marbles, you will have a 1/2 probability of having the white one. It would be pretty cool that you will be able to pull out one white marble out 1/2 the time simply because I purposely remove black marbles after you pull yours out randomly. Almost like you're psychic.

Now since you don't want others to subscribe to "twisted logic", what do you think of the following reliable and popular sites getting it all wrong?

http://mathforum.org/dr.math/faq/faq.monty.hall.html

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

http://montyhallproblem.com/

[Guest]

Let me try one more time. You are going for 3 false scenarios. There are only 2.

Car is behind Door A

You pick Door A he shows you Door B you stick you win

You pick Door A he shows you Door B you change you lose.

You Pick Door B he shows you Door C you stick you lose.

You pick Door B he shows you Door C you change you win.

You pick Door C he shows you Door B you stick you lose

You pick Door C he shows you Door B you change you win

3 wins 3 losses but picking the wrong door has the same results. It doesn't matter which wrong door you pick if you stick you lose if you change you win.

So there are 3 scenarios but it comes down to just 2.

You pick Door with Car and stick - Win

You pick Door with Car and change - Loss - you can only change to ONE Door, the one he did not reveal.

You pick Door with Goat and stick - Loss

You pick Door with Goat and change - Win because other goat has been revealed

You pick Door with Goat and stick - Loss

You pick Door with Goat and change - Win because other goat has been revealed

Why is this so hard to understand? Well, I guess because you don't want to.

3 wins 3 losses. But actually picking door with goat doesn't matter if you pick goat a or goat b.

Notice how this works compared to what you LEFT OUT

You pick A- Monty reveals what's behind one of the other doors. Does the remaining door have a car behind it? No

You pick B- Monty reveals what's behind one of the other doors. Does the remaining door have a car behind it? Yes

You pick C- Monty reveals what's behind one of the other doors. Does the remaining door have a car behind it? Yes

This is based on not changing. You are giving the odds of NOT picking the car. Review the correct answer until it sinks in.

about the marble left in the bag? It has a 3/4 probability of being white.

No, it does not. You need to get a grip on probabilities. Why would the marble in the bag have a greater probability now than the one in your hand now that you are down to 2? It won't. You are ASSUMING the one in the bag is the white one but it could be black.

4 marbles - Here are the scenarios

(adding elements doesn't change the results btw. you can go to a zillion and it won't change)

I pick the white so there are 3 black left. I keep the one I have - I win I swap I lose

I pick a black 1 he removes the 2 other black marbles - I swap I win I hold I lose.

That simple. 50/50 chance.

BECAUSE he is not removing marbles or doors or whatever at random the odds will go down to 50/50 with 2 left because you know 1 is the winner and 1 is a loser. IF they were being removed randomly then you would have a 1/4 chance of winning no matter what you did.

That bolded is what it comes down to no matter what. 1 is a winner and 1 is a loser. For the car/goats each actually has a 1/3 chance of being a winner. So you have 2 choices each with a 1/3 chance of being a winner. Even odds. You can go up to as many as you want, when it gets down to 2 you have even odds. That is such basic probability that I'm amazed this discussion is even happening.

The confusion is coming in because you are applying the initial odds to the final odds. You are saying correctly that there is a 2/3 chance that you will pick a goat. BUT each goat and the car have a 1/3 chance that you picked it. 1/3 Goat A 1/3 Goat B 1/3 the car. One goat eliminated means you are down to 1/3 Goat (remaining) and 1/3 Car. The odds do not suddenly shift to 2/3 Goat Remaining and 1/3 Car. You go into the final choice with a 1/3 chance you have picked the car and a 1/3 chance you have picked the goat.

[Guest]

Let me try one more time. You are going for 3 false scenarios. There are only 2.

Car is behind Door A

You pick Door A he shows you Door B you stick you win

You pick Door A he shows you Door B you change you lose.

You Pick Door B he shows you Door C you stick you lose.

You pick Door B he shows you Door C you change you win.

You pick Door C he shows you Door B you stick you lose

You pick Door C he shows you Door B you change you win

3 wins 3 losses but picking the wrong door has the same results. It doesn't matter which wrong door you pick if you stick you lose if you change you win.

See what you did there? You are looking at what happens when you add "what happens if you stick" to "what happens if you switch" instead of looking at them separately, which is what we're discussing- what happens if you switch vs. if you stick. We can't compare those two scenarios if we don't separate them, can we? I'll copy and paste what you wrote above, but I'll separate them so we can see what happens when looked at separately.

First what happens if you decide to stick:

You pick Door A he shows you Door B you stick you win

You Pick Door B he shows you Door C you stick you lose.

You pick Door C he shows you Door B you stick you lose

See that? You stick to your original choice and the probability of winning is 1/3. Monty purposely showing you a goat after the fact does not change this. Now I'll copy and paste what you wrote for what happens if the choice is to switch.

You pick Door A he shows you Door B you change you lose.

You pick Door B he shows you Door C you change you win.

You pick Door C he shows you Door B you change you win

The probability of winning is 2/3 when you switch.

Now, I'm guessing you may argue that they can't be separated for some reason or another. Then how about we suppose sticking isn't an option? If we're forced to switch, using the three possible scenarios for switching that you posted, we are left with this:

You pick Door A he shows you Door B you change you lose.

You pick Door B he shows you Door C you change you win.

You pick Door C he shows you Door B you change you win

Changing your mind yet?

BECAUSE he is not removing marbles or doors or whatever at random the odds will go down to 50/50 with 2 left because you know 1 is the winner and 1 is a loser. IF they were being removed randomly then you would have a 1/4 chance of winning no matter what you did.

You have it backwards. Because the marbles are not being removed randomly, the probability that the white marble is in your hand remains the same (1/4) and the probability that the white marble remains in the bag remains at 3/4 (even though it now only contains one marble).

If the marbles were removed randomly and they happened to be black, the probability that the white marble is in your hand is 1/2 and the probability that the white marble is in the bag is 1/2.

Same thing with the Monty Barker problem. Since the door that was revealed was not revealed randomly, the probability that the contestant picked the door with the car behind it remains at 1/3 even after Monty purposely reveals a goat behind one of the other doors. The probability that the remaining doors has the car behind it is still 2/3 (even though that now there is only one remaining door from the others).

If one of the doors you didn't choose were opened randomly and it happened to reveal a goat, the probability that the car is behind your door is 1/2 and the probability that the car is behind the remaining door is 1/2.

No, it does not. You need to get a grip on probabilities. Why would the marble in the bag have a greater probability now than the one in your hand now that you are down to 2? It won't. You are ASSUMING the one in the bag is the white one but it could be black.

I didn't make any assumptions. Yes, it could be black. But that probability is only 1/4.

There are three black marbles and one white one in a bag. You pick one out of the bag and hold it in your closed hand (never having saw its color). You agree that the probability that it is white is 1/4, correct? You agree that the probability that the white marble remains in the bag is 3/4, correct? If there is a 3/4 chance that the white marble is in the bag, why would this change if I look in the bag and purposely remove two black marbles?

If there is a 1/2 probability that the white marble is in your hand after I purposely pulled two black marbles out, then if we performed this experiment 100 times, about 50% of the time you will have picked the white marble blindly out of a bag of four marbles containing only one of which is white. Don't you see the obvious problem with your conclusion?

Are you going to bother with the simulator I linked to? You haven't even mentioned it. It would be nice if you gave it a chance. How about the other links I posted, one being to a website that specializes in math problems? Not helping to have you reevaluate your conclusion?

[bonanova]

You are confused, Grasshopper.

Call me that one more time and ...

The rest of the boxes are irrelevant because you know they will be eliminated.

Sure, if you only knew which ones they were... Doh.

Hello, Mc Fly .... *knocks you gently on the head*

[Guest]

Ooooh Grasshopper gets cranky.

Anyway, let me now really confuse you.

Let's look at it from the OTHER SIDE.

You have 3 choices. You don't know what you picked BUT what about MB?

Here are ALL of his choices for which one he KEEPS :

You pick CAR he picks GOAT A

You pick CAR he picks GOAT B

You pick GOAT A he picks CAR

You pick GOAT A he picks GOAT B

You pick GOAT B he picks CAR

You pick GOAT B he picks GOAT A

Hold on hold on. Don't go off the deep end again.

What we have here is called a CONDITIONAL. In other words the odds are not what they seem because Action B is conditional on Action A. There is a condition involved stating the CAR must be in the final 2. Thus we eliminate any result that does not have the CAR in the final 2.

So that leaves us with

You pick CAR he picks GOAT A

You pick CAR he picks GOAT B

You pick GOAT A he picks CAR

You pick GOAT B he picks CAR

Now what YOU have done is combine the first 2 into 1. Can't do it. They are seperate probabilities. What you CAN do is say this.

IF you pick the car he picks A or B.

IF you pick A or B he picks the Car

Can you do math? You end up with CAR = A+B and A+B = CAR

What you are trying to do is say that it makes sense that based on the initial odds, he will have the car to choose from and thus MUST choose the car. But the math doesn't work that way. Only knee jerk reaction works that way. The math says that he has a 2/4 chance of picking EITHER goat. And a 2/4 chance of picking the car.

Gotta love conditionals.

Anyway, this is just going to back and forth until the goats go home and the new model year rolls around. Take some classes in probability and see what you come up with.

Have fun!