The Monty Hall 3 Door Probability Problem (1 Viewer)

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The way this finally clicked for me was this:

START: You have a 1 in 3 chance of picking the car.

IF YOU DIDN'T PICK THE CAR (66% of the time):

There are 2 doors remaining.... 1 goat and 1 car...

monte gets the goat out of the way is basically saying "THERE'S THE CAR!"...GUARANTEED.

IF YOU DID PICK THE CAR (33% of the time):

There are 2 doors remaining.... Both goats... Monty is offering no help....

SO....

66% of the time... monty points you at the car (by not picking that door)

33% of the time.... monty offers no help because you've already picked the car

 

[Maurile Tremblay]

Let me throw this out there...Suppose you're playing the 3 doors game outside. The goats & car are placed behind the doors ahead of time. (you can't see them, the doors are huge)You walk out to play the game. You choose door one. Before door #1 is opened, a gust of wind blows door #3 open revealing a goat. Do you now have any reason to change to door #2?

No. In this situation the odds are 50%.

Are we sure about this?When you choose Door#1, you have a 33% chance of having the car. There is a 66% chance the car is behind one of the other 2 doors. If wind blows door#3 open revealing a goat, shouldn't you still switch? Or does that door being blown open at random change each to 50/50?I'm serious. I totally understand the whole Monty Hall problem, but this is puzzling.Door's 2 and 3 have a combined 66% chance before the wind, but only 50% after?ETA: I think it does change after thinking about it, if this scenario was played over and over, 33% of the time the wind would blow open the door with the car.

If the wind blows door #3 open revealing a goat, the chance that door #3 contains the prize goes from 33% to 0%.The question is where that original 33% goes. Does it get distributed evenly between door #1 and door #2, giving them each 50% total? Or does it all go to door #2, giving door #1 33% and door #2 67%?If Monty purposely opens door #3 knowing that door #3 is empty and you chose door #1, that means all of door #3's percentage goes to door #2.If the wind randomly opens door #3 not knowing anything, that distributes door #3's percentage between door #1 and door #2 evenly.

 

[Fennis]

Think of 100 doors. 1 car, 99 goats.

The contestant makes a pick. Has a 1% chance of being right, right? Say he picks door #4.

Monty opens 98 doors and shows goats. He opens door 1, 2, 3, skips the contestant's #4, 5, 6...etc.... skips #38, then 39-100. Now, he asks, "Switch?"

OK, so the contestant had a 1% chance that his door was right. Monty opened every door except the contestants, and, suspiciously, #38.

Still think it's 50-50, or is it more likely that the car was behind #38?

Odds are, the contestant will get it wrong 99% of the time. So, 99% of the time, Monty skips the car when he's opening doors.

Therefore, switching to the skipped door would be smart, right?

In the Monty Hall scenario the contestant only loses when he picks the winner first. Every other time, he wins.

If someone still doesnt get it, rather than 3 doors, or 100 doors, picture 10,609,444,162,190 doors. Make our debt work for you.

 

[[icon]]

Here's another one.....

Let's assume the money is behind door 1.

Let's go through scenarios in order (UNDERLINE = DOOR YOU PICKED)

Lets look at outcomes:

IF YOU DON'T SWITCH

$ 2 3 = WIN (Monte opens 2 or 3... you stay with $)

$ 2 3 = LOSE (Monte opens door 3.. you stay with 2)

$ 2 3 = LOSE (Monte opens door 2... you stay with 3)

OUTCOME: You win 33% of the time

IF YOU SWITCH:

$ 2 3 = LOSE (Monte opens 2 or 3...you switch to remaining goat)

$ 2 3 = WIN (Monte opens 3.... you switch to $)

$ 2 3 = WIN (Monte opens 2... you switch to $)

OUTCOME: You win 66% of the time

The outcome is the same if the money is behind doors 2 or 3.

You make the switch.. you win 2/3 of the time...

 

[satch]

As I said earlier, I think I now understand the concept, tell me if I'm correct:

You begin with 3 doors hiding 2 goats and one car. There is a 66% chance that the door you pick first will be a goat. If you base your gameplay on the assumption that you most likely picked a goat with your first choice door #1, then Monty revealing the other goat behind door#3 indicates that the car is most likely behind door#2. Therefore you would choose door#2.

 

[fatguyinalittlecoat]

As I said earlier, I think I now understand the concept, tell me if I'm correct:

You begin with 3 doors hiding 2 goats and one car. There is a 66% chance that the door you pick first will be a goat. If you base your gameplay on the assumption that you most likely picked a goat with your first choice door #1, then Monty revealing the other goat behind door#3 indicates that the car is most likely behind door#2. Therefore you would choose door#2.

Yes. This is a good explanation.

 

[Leroy Hoard]

Only took 4 pages this time to resolve.

 

[rick6668]

Let me throw this out there...Suppose you're playing the 3 doors game outside. The goats & car are placed behind the doors ahead of time. (you can't see them, the doors are huge)You walk out to play the game. You choose door one. Before door #1 is opened, a gust of wind blows door #3 open revealing a goat. Do you now have any reason to change to door #2?

No. In this situation the odds are 50%.

Are we sure about this?When you choose Door#1, you have a 33% chance of having the car. There is a 66% chance the car is behind one of the other 2 doors. If wind blows door#3 open revealing a goat, shouldn't you still switch? Or does that door being blown open at random change each to 50/50?I'm serious. I totally understand the whole Monty Hall problem, but this is puzzling.Door's 2 and 3 have a combined 66% chance before the wind, but only 50% after?ETA: I think it does change after thinking about it, if this scenario was played over and over, 33% of the time the wind would blow open the door with the car.

If the wind blows door #3 open revealing a goat, the chance that door #3 contains the prize goes from 33% to 0%.The question is where that original 33% goes. Does it get distributed evenly between door #1 and door #2, giving them each 50% total? Or does it all go to door #2, giving door #1 33% and door #2 67%?If Monty purposely opens door #3 knowing that door #3 is empty and you chose door #1, that means all of door #3's percentage goes to door #2.If the wind randomly opens door #3 not knowing anything, that distributes door #3's percentage between door #1 and door #2 evenly.

Yup. This is what I thought after thinking it over.

 

[satch]

Let me throw this out there...

Suppose you're playing the 3 doors game outside. The goats & car are placed behind the doors ahead of time. (you can't see them, the doors are huge)

You walk out to play the game. You choose door one. Before door #1 is opened, a gust of wind blows door #3 open revealing a goat. Do you now have any reason to change to door #2?

No. In this situation the odds are 50%.

Are we sure about this?When you choose Door#1, you have a 33% chance of having the car. There is a 66% chance the car is behind one of the other 2 doors. If wind blows door#3 open revealing a goat, shouldn't you still switch? Or does that door being blown open at random change each to 50/50?

I'm serious. I totally understand the whole Monty Hall problem, but this is puzzling.

Door's 2 and 3 have a combined 66% chance before the wind, but only 50% after?

ETA: I think it does change after thinking about it, if this scenario was played over and over, 33% of the time the wind would blow open the door with the car.

If the wind blows door #3 open revealing a goat, the chance that door #3 contains the prize goes from 33% to 0%.The question is where that original 33% goes. Does it get distributed evenly between door #1 and door #2, giving them each 50% total? Or does it all go to door #2, giving door #1 33% and door #2 67%?

If Monty purposely opens door #3 knowing that door #3 is empty and you chose door #1, that means all of door #3's percentage goes to door #2.

If the wind randomly opens door #3 not knowing anything, that distributes door #3's percentage between door #1 and door #2 evenly.

Yup. This is what I thought after thinking it over.

See, that doesn't make sense to me the way I've come to understand this concept. From the contestant's point of view, why does it matter if door#3 is opened revealing a goat by the wind or by Monty?Either way I knew going in that my 1st choice, door #1, had a 66% chance of being a goat.

When door #3 is opened revealing a goat, either by the wind or by Monty, I can reasonably say that if my 1st choice had a 66% chance of being a goat, and I know the other goat is behind door #3, than it's most likely that the car is behind door #2 and I should switch.

 

[David Yudkin]

Let me throw this out there...

Suppose you're playing the 3 doors game outside. The goats & car are placed behind the doors ahead of time. (you can't see them, the doors are huge)

You walk out to play the game. You choose door one. Before door #1 is opened, a gust of wind blows door #3 open revealing a goat. Do you now have any reason to change to door #2?

No. In this situation the odds are 50%.

Are we sure about this?When you choose Door#1, you have a 33% chance of having the car. There is a 66% chance the car is behind one of the other 2 doors. If wind blows door#3 open revealing a goat, shouldn't you still switch? Or does that door being blown open at random change each to 50/50?

I'm serious. I totally understand the whole Monty Hall problem, but this is puzzling.

Door's 2 and 3 have a combined 66% chance before the wind, but only 50% after?

ETA: I think it does change after thinking about it, if this scenario was played over and over, 33% of the time the wind would blow open the door with the car.

If the wind blows door #3 open revealing a goat, the chance that door #3 contains the prize goes from 33% to 0%.The question is where that original 33% goes. Does it get distributed evenly between door #1 and door #2, giving them each 50% total? Or does it all go to door #2, giving door #1 33% and door #2 67%?

If Monty purposely opens door #3 knowing that door #3 is empty and you chose door #1, that means all of door #3's percentage goes to door #2.

If the wind randomly opens door #3 not knowing anything, that distributes door #3's percentage between door #1 and door #2 evenly.

Yup. This is what I thought after thinking it over.

See, that doesn't make sense to me the way I've come to understand this concept. From the contestant's point of view, why does it matter if door#3 is opened revealing a goat by the wind or by Monty?Either way I knew going in that my 1st choice, door #1, had a 66% chance of being a goat.

When door #3 is opened revealing a goat, either by the wind or by Monty, I can reasonably say that if my 1st choice had a 66% chance of being a goat, and I know the other goat is behind door #3, than it's most likely that the car is behind door #2 and I should switch.

There is a difference between Door #3 being opened and having a goat (ie the wind outcome) vs. Door #3 being opened BECAUSE it's a goat.

 

[satch]

Let me throw this out there...

Suppose you're playing the 3 doors game outside. The goats & car are placed behind the doors ahead of time. (you can't see them, the doors are huge)

You walk out to play the game. You choose door one. Before door #1 is opened, a gust of wind blows door #3 open revealing a goat. Do you now have any reason to change to door #2?

No. In this situation the odds are 50%.

Are we sure about this?When you choose Door#1, you have a 33% chance of having the car. There is a 66% chance the car is behind one of the other 2 doors. If wind blows door#3 open revealing a goat, shouldn't you still switch? Or does that door being blown open at random change each to 50/50?

I'm serious. I totally understand the whole Monty Hall problem, but this is puzzling.

Door's 2 and 3 have a combined 66% chance before the wind, but only 50% after?

ETA: I think it does change after thinking about it, if this scenario was played over and over, 33% of the time the wind would blow open the door with the car.

If the wind blows door #3 open revealing a goat, the chance that door #3 contains the prize goes from 33% to 0%.The question is where that original 33% goes. Does it get distributed evenly between door #1 and door #2, giving them each 50% total? Or does it all go to door #2, giving door #1 33% and door #2 67%?

If Monty purposely opens door #3 knowing that door #3 is empty and you chose door #1, that means all of door #3's percentage goes to door #2.

If the wind randomly opens door #3 not knowing anything, that distributes door #3's percentage between door #1 and door #2 evenly.

Yup. This is what I thought after thinking it over.

See, that doesn't make sense to me the way I've come to understand this concept. From the contestant's point of view, why does it matter if door#3 is opened revealing a goat by the wind or by Monty?Either way I knew going in that my 1st choice, door #1, had a 66% chance of being a goat.

When door #3 is opened revealing a goat, either by the wind or by Monty, I can reasonably say that if my 1st choice had a 66% chance of being a goat, and I know the other goat is behind door #3, than it's most likely that the car is behind door #2 and I should switch.

There is a difference between Door #3 being opened and having a goat (ie the wind outcome) vs. Door #3 being opened BECAUSE it's a goat.

Can you explain the difference? Explain the thought process from the contestant's point of view in each scenario. Scenario A being where the contestant picks door #1 and the wind reveals a goat behind door #3. Scenario B being where the contestant picks door#1 and Monty reveals a goat behind door #3. If you can, please step through the contestant's reasoning in both scenario's,

 

[Disco Stu]

Let me throw this out there...

Suppose you're playing the 3 doors game outside. The goats & car are placed behind the doors ahead of time. (you can't see them, the doors are huge)

You walk out to play the game. You choose door one. Before door #1 is opened, a gust of wind blows door #3 open revealing a goat. Do you now have any reason to change to door #2?

No. In this situation the odds are 50%.

Are we sure about this?When you choose Door#1, you have a 33% chance of having the car. There is a 66% chance the car is behind one of the other 2 doors. If wind blows door#3 open revealing a goat, shouldn't you still switch? Or does that door being blown open at random change each to 50/50?

I'm serious. I totally understand the whole Monty Hall problem, but this is puzzling.

Door's 2 and 3 have a combined 66% chance before the wind, but only 50% after?

ETA: I think it does change after thinking about it, if this scenario was played over and over, 33% of the time the wind would blow open the door with the car.

If the wind blows door #3 open revealing a goat, the chance that door #3 contains the prize goes from 33% to 0%.The question is where that original 33% goes. Does it get distributed evenly between door #1 and door #2, giving them each 50% total? Or does it all go to door #2, giving door #1 33% and door #2 67%?

If Monty purposely opens door #3 knowing that door #3 is empty and you chose door #1, that means all of door #3's percentage goes to door #2.

If the wind randomly opens door #3 not knowing anything, that distributes door #3's percentage between door #1 and door #2 evenly.

Yup. This is what I thought after thinking it over.

See, that doesn't make sense to me the way I've come to understand this concept. From the contestant's point of view, why does it matter if door#3 is opened revealing a goat by the wind or by Monty?Either way I knew going in that my 1st choice, door #1, had a 66% chance of being a goat.

When door #3 is opened revealing a goat, either by the wind or by Monty, I can reasonably say that if my 1st choice had a 66% chance of being a goat, and I know the other goat is behind door #3, than it's most likely that the car is behind door #2 and I should switch.

There is a difference between Door #3 being opened and having a goat (ie the wind outcome) vs. Door #3 being opened BECAUSE it's a goat.

Can you explain the difference? Explain the thought process from the contestant's point of view in each scenario. Scenario A being where the contestant picks door #1 and the wind reveals a goat behind door #3. Scenario B being where the contestant picks door#1 and Monty reveals a goat behind door #3. If you can, please step through the contestant's reasoning in both scenario's,

The difference is that the wind had a 33% chance of revealing a car. Monty has a 0% chance of revealing a car.

 

[satch]

Let me throw this out there...

Suppose you're playing the 3 doors game outside. The goats & car are placed behind the doors ahead of time. (you can't see them, the doors are huge)

You walk out to play the game. You choose door one. Before door #1 is opened, a gust of wind blows door #3 open revealing a goat. Do you now have any reason to change to door #2?

No. In this situation the odds are 50%.

Are we sure about this?When you choose Door#1, you have a 33% chance of having the car. There is a 66% chance the car is behind one of the other 2 doors. If wind blows door#3 open revealing a goat, shouldn't you still switch? Or does that door being blown open at random change each to 50/50?

I'm serious. I totally understand the whole Monty Hall problem, but this is puzzling.

Door's 2 and 3 have a combined 66% chance before the wind, but only 50% after?

ETA: I think it does change after thinking about it, if this scenario was played over and over, 33% of the time the wind would blow open the door with the car.

If the wind blows door #3 open revealing a goat, the chance that door #3 contains the prize goes from 33% to 0%.The question is where that original 33% goes. Does it get distributed evenly between door #1 and door #2, giving them each 50% total? Or does it all go to door #2, giving door #1 33% and door #2 67%?

If Monty purposely opens door #3 knowing that door #3 is empty and you chose door #1, that means all of door #3's percentage goes to door #2.

If the wind randomly opens door #3 not knowing anything, that distributes door #3's percentage between door #1 and door #2 evenly.

Yup. This is what I thought after thinking it over.

See, that doesn't make sense to me the way I've come to understand this concept. From the contestant's point of view, why does it matter if door#3 is opened revealing a goat by the wind or by Monty?Either way I knew going in that my 1st choice, door #1, had a 66% chance of being a goat.

When door #3 is opened revealing a goat, either by the wind or by Monty, I can reasonably say that if my 1st choice had a 66% chance of being a goat, and I know the other goat is behind door #3, than it's most likely that the car is behind door #2 and I should switch.

There is a difference between Door #3 being opened and having a goat (ie the wind outcome) vs. Door #3 being opened BECAUSE it's a goat.

Can you explain the difference? Explain the thought process from the contestant's point of view in each scenario. Scenario A being where the contestant picks door #1 and the wind reveals a goat behind door #3. Scenario B being where the contestant picks door#1 and Monty reveals a goat behind door #3. If you can, please step through the contestant's reasoning in both scenario's,

The difference is that the wind had a 33% chance of revealing a car. Monty has a 0% chance of revealing a car.

Can you explain how that difference factors into the contestants line of reasoning when deciding whether or not to stay with door #1 or switch to door #2?

 

[videoguy505]

Let me throw this out there...

Suppose you're playing the 3 doors game outside. The goats & car are placed behind the doors ahead of time. (you can't see them, the doors are huge)

You walk out to play the game. You choose door one. Before door #1 is opened, a gust of wind blows door #3 open revealing a goat. Do you now have any reason to change to door #2?

No. In this situation the odds are 50%.

Are we sure about this?When you choose Door#1, you have a 33% chance of having the car. There is a 66% chance the car is behind one of the other 2 doors. If wind blows door#3 open revealing a goat, shouldn't you still switch? Or does that door being blown open at random change each to 50/50?

I'm serious. I totally understand the whole Monty Hall problem, but this is puzzling.

Door's 2 and 3 have a combined 66% chance before the wind, but only 50% after?

ETA: I think it does change after thinking about it, if this scenario was played over and over, 33% of the time the wind would blow open the door with the car.

If the wind blows door #3 open revealing a goat, the chance that door #3 contains the prize goes from 33% to 0%.The question is where that original 33% goes. Does it get distributed evenly between door #1 and door #2, giving them each 50% total? Or does it all go to door #2, giving door #1 33% and door #2 67%?

If Monty purposely opens door #3 knowing that door #3 is empty and you chose door #1, that means all of door #3's percentage goes to door #2.

If the wind randomly opens door #3 not knowing anything, that distributes door #3's percentage between door #1 and door #2 evenly.

Yup. This is what I thought after thinking it over.

See, that doesn't make sense to me the way I've come to understand this concept. From the contestant's point of view, why does it matter if door#3 is opened revealing a goat by the wind or by Monty?Either way I knew going in that my 1st choice, door #1, had a 66% chance of being a goat.

When door #3 is opened revealing a goat, either by the wind or by Monty, I can reasonably say that if my 1st choice had a 66% chance of being a goat, and I know the other goat is behind door #3, than it's most likely that the car is behind door #2 and I should switch.

There is a difference between Door #3 being opened and having a goat (ie the wind outcome) vs. Door #3 being opened BECAUSE it's a goat.

Can you explain the difference? Explain the thought process from the contestant's point of view in each scenario. Scenario A being where the contestant picks door #1 and the wind reveals a goat behind door #3. Scenario B being where the contestant picks door#1 and Monty reveals a goat behind door #3. If you can, please step through the contestant's reasoning in both scenario's,

The difference is that the wind had a 33% chance of revealing a car. Monty has a 0% chance of revealing a car.

Can you explain how that difference factors into the contestants line of reasoning when deciding whether or not to stay with door #1 or switch to door #2?

From the contestants POV:"Before this game starts, I know that no matter which door I pick, if Monty knowingly reveals a goat, then: by switching, I will always win the car in the situations in which I picked a goat first. If I start with a goat, and Monty reveals the other goat, then I know by switching I pick the car. Since I start with a goat 2 out of 3 times, I know that always switching will result in a win in the long run."

"However, if a door is opened by accident, my 'always switching' strategy doesn't work. I could start with the car and switch to the goat. Or I could have started with the other goat and switched to the car. All I know is that the one goat I do see, I didn't start with.

Since the wind is random, played many times, there will be occasions where it randomly blows open the door containing the car. In those situations, I could start with a goat, decide to switch, and lose."

In the "Monty knows what's going on" scenario, it's not possible to start with a goat, switch, and lose. However, in the "random chance opened the door" scenario, it is. So your starting selection doesn't help you out at all.

Imagine many doors, or, if you want, Deal or No Deal. When the contestant makes it to the final 2 cases, one may have a million and the other a dollar. Right then, Howie offers the chance to switch, and it's 50-50. Each case is equally likely to hold the million, because the contestant could have revealed the million at any time leading up to that point. However, if the contestant picked a case at the start of the game, Case #1, Howie purposely opened 23 known loser cases in a row to just leave one containing a million dollars in it, obviously the odds are against the contestant having the winning case from the beginning and it's smart to switch. But when there's no knowledge behind it, and just dumb luck, it isn't. Many, many contestants play Deal or No Deal and reveal the million dollars before reaching the final switch opportunity.

 

[bryhamm]

Anyone watch Mythbusters? They proved this out in a recent episode.

 

[SacramentoBob]

Anyone watch Mythbusters? They proved this out in a recent episode.

Thought it was a waste of an episode since I wouldn't call it a "myth". Also could've just written a computer simulation, but that's not exciting TV.

 

[St. Louis Bob]

Anyone watch Mythbusters? They proved this out in a recent episode.

Thought it was a waste of an episode since I wouldn't call it a "myth". Also could've just written a computer simulation, but that's not exciting TV.

But informative none-teh-less.

 

[jon_mx]

And that is why you go for 2 early.

 

[kupcho1]

I missed the bit about the wind knocking the door down the first time around

 

[FlapJacks]

Frick: I'm thinking of a number between on and million...go ahead and guess, after your guess, I will reduce the choices to two numbers, including your guess

Freak: 249,372

Frick: I can tell you with certainty that the number is 249,372 or 12. Do you want to stay with 249.372 or switch to 12?

 

[cstu]

This website explains it very simply.

 

[belljr]

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