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The Monty Hall 3 Door Probability Problem (1 Viewer)

satch

satch

Footballguy
Monty Hall Problem

Here's a quick summary:

There are 3 doors. Behind one of the doors is a car, behind the other 2 doors are goats. The contestant must try to pick the door that has the car behind it. Let's say the contestant chooses door #1. Before the host reveals what's behind door #1, he opens door #3 revealing a goat. This leaves doors #1 and #2 still unopened, one of them must have a car behind it. The host then gives the contestant the opportunity to change his pick from door #1 to door #2. What should the contestant do? Stay with his original choice of door #1, or switch to door #2? Does it matter?

Supposedly, the correct answer is: The contestant should choose door #2, because door #2 has a higher probability of having the car behind it than door #1. This has been "proven" by computer simulations.

:lmao: :lmao: :lmao: :lmao: :confused: :confused:

There are many web pages, articles written about this. I've read a few of them and I just can't understand how this could be true. The contestant has no idea which door the car is behind. When given 3 choices, each door has a 33% chance of being right. When one of the three options is eliminated, each the 2 remaining doors now have a 50% chance of being right. It should make absolutely no difference which door the contestant picked originally. They're just doors. His picking one doesn't have any effect on what's actually behind it, and should have no effect on probability.

Can anyone explain this in a way that makes sense? Specifically, explain how after door #3 has been eliminated, the 2 remaining doors do not have a 50/50 chance of being right. 2 doors, one car, one choice.

 
Last edited by a moderator:
The best way to visualize the logic is to pose the exact same scenario, but increase the number of doors.

edit - or so I've been told.

 
Last edited by a moderator:
fatguyinalittlecoat said:
Oh boy.The important thing to know is that Monty knows where the prize is and he will always open one that doesn't have a prize.
Click to expand...
How does Monty's knowledge of which door hides the car have anything to do with it from the contestants point of view? The contestant still doesn't know which door hides the car, so when given 2 doors to choose from it's 50/50.
 
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.

 
Last edited by a moderator:
When he originally picked door #1 he had a 33% chance of being correct. Adding additional information doesn't change that fact.

 
Here's the standard for 3 doors.

You pick one of the doors, he opens another, then asks "Switch?"

1) You pick the car, he opens a goat door, you switch, LOSE.

2) You pick goat #1, he opens the door for goat #2, you switch, WIN.

3) You pick goat #3, he opens the door for goat #1, you switch, WIN.

Meanwhile, not switching means you had to have been right on the first choice, 1/3. By switching, you win 2/3.

 
Have a friend pick a card out of a deck (without you looking) and then have him arrange the 52 cards on the table (therefore he'll know where the card he picked is). Then, turn back around and pick a card. At this point, your friend will turn over 50 cards that are NOT his selected card from the deck.

At this point he offers you a choice - keep your card or switch. What do you think would be the best option now?

-QG

 
When you chose your door you had a 1/3 chance of being right.

Now Monte opens a door that he knows has a goat.

Your odds of having the correct box are still 1/3 since nothing has changed (Since Monte knew the door he opened was the goat).

Therefore the odds the other door has the car are 2/3.

 
videoguy505 said:
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.
Click to expand...
Bingo. 2 out of 3 chance to win with 3 doors.
 
shining path said:
The best way to visualize the logic is to pose the exact same scenario, but increase the number of doors.edit - or so I've been told.
Click to expand...
Isn't this strictly about percentages? When given 4 choices, each choice has a 25% chance of being correct. When given 3 choices, each choice has a 33% chance of being correct. When given 2 choices, each choice has a 50% chance of being correct. I don't understand the concept that "choosing" a door to start somehow magically changes the law of percentages. The contestant's "original choice" of doors has zero effect on anything.
 
satch said:
shining path said:
The best way to visualize the logic is to pose the exact same scenario, but increase the number of doors.edit - or so I've been told.
Click to expand...
Isn't this strictly about percentages? When given 4 choices, each choice has a 25% chance of being correct. When given 3 choices, each choice has a 33% chance of being correct. When given 2 choices, each choice has a 50% chance of being correct.
Click to expand...
This is only true if you're assuming randomness.
 
videoguy505 said:
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.
Click to expand...
This actually makes sense to me.Thank you.

 
Disco Stu said:
When he originally picked door #1 he had a 33% chance of being correct. Adding additional information doesn't change that fact.
Click to expand...
Of course it does. He originally had a 33% chance of being correct because there were 3 choices. Now there are 2. Now he has a 50% chance of being correct.Using your logic, what happens if you eliminate door#2 as well? Now there's only one door remaining which must have the car behind it. Adding this additional information, does he still only have a 33% chance? Of course not, he has a 100% chance of being right because as each door is eliminated, his odds of being right increase. Doesn't matter if you start wth 100 or 3.
 
satch said:
shining path said:
The best way to visualize the logic is to pose the exact same scenario, but increase the number of doors.

edit - or so I've been told.
Click to expand...
Isn't this strictly about percentages? When given 4 choices, each choice has a 25% chance of being correct. When given 3 choices, each choice has a 33% chance of being correct. When given 2 choices, each choice has a 50% chance of being correct. I don't understand the concept that "choosing" a door to start somehow magically changes the law of percentages. The contestant's "original choice" of doors has zero effect on anything.
Click to expand...
Post 6.Imagine you're playing Deal or No Deal. Howie has you pick a briefcase out of the 25. So you have a 1-in-25 chance of picking the one with a million dollars in it, right? Then he says, you can have that one case, or, you can have ALL of the remaining 24 cases? What would you do? You'd switch to the remaining 24, figuring there's a 96% chance it's there and only 4% it's in the one you picked.

After you switch, Howie starts opening your 24 briefcases, one at a time... and on purpose starts with low dollar values until there's just one case left, say, $1,000,000 and $100. You really think you picked the $1,000,000 first half the time? Or is Howie screwing with you and purposely skipped the mil and you picked the $100 first?

 
satch said:
Disco Stu said:
When he originally picked door #1 he had a 33% chance of being correct. Adding additional information doesn't change that fact.
Click to expand...
Of course it does. He originally had a 33% chance of being correct because there were 3 choices. Now there are 2. Now he has a 50% chance of being correct.Using your logic, what happens if you eliminate door#2 as well? Now there's only one door remaining which must have the car behind it. Adding this additional information, does he still only have a 33% chance? Of course not, he has a 100% chance of being right because as each door is eliminated, his odds of being right increase. Doesn't matter if you start wth 100 or 3.
Click to expand...
You're still not taking into account that Monty is selectively choosing which door to reveal.
 
satch said:
Disco Stu said:
When he originally picked door #1 he had a 33% chance of being correct. Adding additional information doesn't change that fact.
Click to expand...
Of course it does. He originally had a 33% chance of being correct because there were 3 choices. Now there are 2. Now he has a 50% chance of being correct.Using your logic, what happens if you eliminate door#2 as well? Now there's only one door remaining which must have the car behind it. Adding this additional information, does he still only have a 33% chance? Of course not, he has a 100% chance of being right because as each door is eliminated, his odds of being right increase. Doesn't matter if you start wth 100 or 3.
Click to expand...
Monte always opens a door with a goat behind it. Always. So by being given the chance to switch, you are essentially being asked if you want one door, or the other two.
 
satch said:
shining path said:
The best way to visualize the logic is to pose the exact same scenario, but increase the number of doors.

edit - or so I've been told.
Click to expand...
Isn't this strictly about percentages? When given 4 choices, each choice has a 25% chance of being correct. When given 3 choices, each choice has a 33% chance of being correct. When given 2 choices, each choice has a 50% chance of being correct. I don't understand the concept that "choosing" a door to start somehow magically changes the law of percentages. The contestant's "original choice" of doors has zero effect on anything.
Click to expand...
It is about percentages, the original percentages. Viewing it as a new choice, 50/50, discards information. Take the original percentages into account to include all the information.

In the 100 doors scenario, for each door opened the original percentage "moves" to the skipped one. They start out at 1% each, but, when #38 is skipped, each goat "adds" 1% to the skipped door. At the end, 99%.

It's basically taking "the field" in a sports wager, betting on everyone else not included in the original.

 
satch said:
Disco Stu said:
When he originally picked door #1 he had a 33% chance of being correct. Adding additional information doesn't change that fact.
Click to expand...
Of course it does. He originally had a 33% chance of being correct because there were 3 choices. Now there are 2. Now he has a 50% chance of being correct.Using your logic, what happens if you eliminate door#2 as well? Now there's only one door remaining which must have the car behind it. Adding this additional information, does he still only have a 33% chance? Of course not, he has a 100% chance of being right because as each door is eliminated, his odds of being right increase. Doesn't matter if you start wth 100 or 3.
Click to expand...
It has already been said in many different ways but here you go, You are answering your own question, just put it into a negative. He has a 2/3 chance of being WRONG!!! When the goat behind #3 is gone, switching increases chances of being correct dramatically. Yeah you might lose because your initial odds are not bad and a 1/3 chance is better than 1/100, but you will always INCREASE odds of winning by switching. Numb3rs ran this scenario on an episode about a year or two ago.When multiplied by more opportunities it makes perfect sense.
 
satch said:
Disco Stu said:
When he originally picked door #1 he had a 33% chance of being correct. Adding additional information doesn't change that fact.
Click to expand...
Of course it does. He originally had a 33% chance of being correct because there were 3 choices. Now there are 2. Now he has a 50% chance of being correct.Using your logic, what happens if you eliminate door#2 as well? Now there's only one door remaining which must have the car behind it. Adding this additional information, does he still only have a 33% chance? Of course not, he has a 100% chance of being right because as each door is eliminated, his odds of being right increase. Doesn't matter if you start wth 100 or 3.
Click to expand...
No. Your door not being revealed as wrong doesn't improve the odds of it being right.And what I meant by "additional information" is the predetermined additional information that Monty gives you in this scenario... not any and all additional information available in the universe.
 
Monty Hall Simulator. It'll keep track for you, go ahead and run 20 or more times not switching, then 20 or more times switching, and see what the numbers are. Or, use the thing at the bottom to have it run 1000 times for you and compare.
 
Last edited by a moderator:
Do you think you would have better chances by choosing one door or two doors?

Now pick any two doors(say 1 and 2) but you tell Monty you want door number 3. Monty opens up the loser of 1 and two and then you switch back, effectively picking two doors and doubling your chances.

 
I am really dumb. I dont get it. I have read all posts.

If you pick the winner, why do you lose. Do you have to switch. Why not stay

 
AcerFC said:
I am really dumb. I dont get it. I have read all posts. If you pick the winner, why do you lose. Do you have to switch. Why not stay
Click to expand...
You don't know what's behind the doors, but Monty does. I should have specified that I was talking about always switching.If you are running the "Always Switch" strategy, you only lose if you pick the winner first.If you are running the "Always Keep" strategy, you only win if you pick the winner first.
 
videoguy505 said:
AcerFC said:
I am really dumb. I dont get it. I have read all posts.

If you pick the winner, why do you lose. Do you have to switch. Why not stay
Click to expand...
You don't know what's behind the doors, but Monty does. I should have specified that I was talking about always switching.If you are running the "Always Switch" strategy, you only lose if you pick the winner first.

If you are running the "Always Keep" strategy, you only win if you pick the winner first.
Click to expand...
Thanks, this cleared it up
 
fatguyinalittlecoat said:
satch said:
Disco Stu said:
When he originally picked door #1 he had a 33% chance of being correct. Adding additional information doesn't change that fact.
Click to expand...
Of course it does. He originally had a 33% chance of being correct because there were 3 choices. Now there are 2. Now he has a 50% chance of being correct.Using your logic, what happens if you eliminate door#2 as well? Now there's only one door remaining which must have the car behind it. Adding this additional information, does he still only have a 33% chance? Of course not, he has a 100% chance of being right because as each door is eliminated, his odds of being right increase. Doesn't matter if you start wth 100 or 3.
Click to expand...
You're still not taking into account that Monty is selectively choosing which door to reveal.
Click to expand...
I want to understand how that matters, but I don't. I pick door #1. If he shows a goat behind door #3, then the car is either behind door#1 or door #2. If he shows a goat behind door #2, then the car is either behind door #1 or door #3. Either way I'm left with a 50% chance of being right.
 
What if the outcome is not known ahead of time in terms of knowing where the car is? For example, in Deal or No Deal you have the option of switching your case for the last remaining case but that's an outstanding offer for all players and no one is picking another case to open like with Monty and the goats.

 
FatMax said:
satch said:
Disco Stu said:
When he originally picked door #1 he had a 33% chance of being correct. Adding additional information doesn't change that fact.
Click to expand...
Of course it does. He originally had a 33% chance of being correct because there were 3 choices. Now there are 2. Now he has a 50% chance of being correct.Using your logic, what happens if you eliminate door#2 as well? Now there's only one door remaining which must have the car behind it. Adding this additional information, does he still only have a 33% chance? Of course not, he has a 100% chance of being right because as each door is eliminated, his odds of being right increase. Doesn't matter if you start wth 100 or 3.
Click to expand...
Monte always opens a door with a goat behind it. Always. So by being given the chance to switch, you are essentially being asked if you want one door, or the other two.
Click to expand...
But you already know that one of "the other two" doors has a goat behind it. So you're left with 2 unknown doors. 50/50 chance of being right with no way for the contestant to know.
 
What are the odds that Monty switches where the car is to reflect what the players usually do, swicth or keep. Serious question

 
David Yudkin said:
What if the outcome is not known ahead of time in terms of knowing where the car is? For example, in Deal or No Deal you have the option of switching your case for the last remaining case but that's an outstanding offer for all players and no one is picking another case to open like with Monty and the goats.
Click to expand...
This is where the confusion is coming from. It sounds like some are confusing Monty's thought process with the contestants. The only thing the contestant can possibly base his decision on is percentages. That's it. Whether or not Monty is trying to confuse him, or screw with him can never change the law of percentages. It doesn't matter what Monty knows, the contestant is still clueless.
 
satch said:
FatMax said:
satch said:
Disco Stu said:
When he originally picked door #1 he had a 33% chance of being correct. Adding additional information doesn't change that fact.
Click to expand...
Of course it does. He originally had a 33% chance of being correct because there were 3 choices. Now there are 2. Now he has a 50% chance of being correct.Using your logic, what happens if you eliminate door#2 as well? Now there's only one door remaining which must have the car behind it. Adding this additional information, does he still only have a 33% chance? Of course not, he has a 100% chance of being right because as each door is eliminated, his odds of being right increase. Doesn't matter if you start wth 100 or 3.
Click to expand...
Monte always opens a door with a goat behind it. Always. So by being given the chance to switch, you are essentially being asked if you want one door, or the other two.
Click to expand...
But you already know that one of "the other two" doors has a goat behind it. So you're left with 2 unknown doors. 50/50 chance of being right with no way for the contestant to know.
Click to expand...
Do it with cards. Tell you wife/kid/mistress/coworker/whoever deal out a deck of cards in stacks of three. Don't use any leftover cards. For the sake of this problem, let's assume that the highest card is the car, and the other two are goats. You'll have to assign a rank of suits too, just in case you get two cards that are the same within a stack of three. With one stack, pick one card. Either card one, two or three. Don't look at the card. Then have whoever show the lowest card of the two left over. Then either stay or switch, but do it the same throughout all the groups. Either stay all the time, or switch all the time. Record how many times you were right and how many times you were wrong. Switching will be somewhere around 66%
 
satch said:
fatguyinalittlecoat said:
satch said:
Disco Stu said:
When he originally picked door #1 he had a 33% chance of being correct. Adding additional information doesn't change that fact.
Click to expand...
Of course it does. He originally had a 33% chance of being correct because there were 3 choices. Now there are 2. Now he has a 50% chance of being correct.Using your logic, what happens if you eliminate door#2 as well? Now there's only one door remaining which must have the car behind it. Adding this additional information, does he still only have a 33% chance? Of course not, he has a 100% chance of being right because as each door is eliminated, his odds of being right increase. Doesn't matter if you start wth 100 or 3.
Click to expand...
You're still not taking into account that Monty is selectively choosing which door to reveal.
Click to expand...
I want to understand how that matters, but I don't. I pick door #1. If he shows a goat behind door #3, then the car is either behind door#1 or door #2. If he shows a goat behind door #2, then the car is either behind door #1 or door #3. Either way I'm left with a 50% chance of being right.
Click to expand...
Here's another way to try explaining:You pick a door and have a 33% chance of being correct.Monty is always going to open up another door, and it is always going to be a goat. That's how he rolls.Since you had a 33% chance of being right, Monty has 2 goat doors to pick from 33% of the time and 1 goat door/1 car door 67% of the time.Odds favor that Monty finds himself in the second situation, with 1 goat door/1 car door left to open.Therefore, odds favor that the door Monty does not open has the car.Therefore, you should switch to whatever door Monty does not open. This switch will be correct 67% of the time, and wrong 33% of the time.
 
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