Monty Hall Problem - Numberphile
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- Entertaining and deceptive PROBABILITY: Each door has 1/3 probability of having the prize. Now, that's an entertaining deception, as only one door has the prize; the others have no prize. You just don't -know- which is which. So, the 1/3 probability, while entertaining and plausible in the abstract, fails. I will explain.
- INSIGHTFUL OPINION: Once another door 'with no prize' is open, those who magically lump 1/3+1/3 and come up with a probability of 2/3 are just confusing themselves and others. 'with no prize' is key here.
- FACT: After you choose, and once another door is open, the only way you could win by staying (not switching) is if the door you choose initially has the prize. IF is a keyword here.
- FACT: If the prize is behind one of the two other doors, you will lose if you stay (if you don't switch. IF is a keyword here.
- PROBABILITY: You have a better chance (theoretically twice the probability, in the abstract) assuming that you initially picked the wrong door and switching rather than assuming you picked the right door initially and staying. ASSUMING is a keyword here.
- DECEPTIVE PROBAILITY ADVANTAGE: It's better to switch. Very clear!
- REALITY CHECK: Since you don't 'know' (you're only assuming that you picked the wrong door initially, the deceptive probability advantage of switching fails as it offers no GUARANTEE that the -other- unopened door has the prize (as opposed to the door you initially picked, because even though probability favors switching, and you have a better abstract chance if you switch, you don't know where the prize is after all (you're ultimately left with a choice of two unopened doors, and the prize may be behind the unopened door you initially picked or the other unopened door, and it all boils down to a 50-50 luck of the draw.
MY FINAL RULING: Yes, Probability advantage favors switching but offers no GUARANTEES of winning whatsoever. IN REALITY, probability is entertaining in the abstract for simple minds. But, after all that is said and done, you're left with a choice of two unopened doors, either one can have the prize.
Date: 2022-04-08
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Comments and reviews: 9
Jim
After you have picked a door, what if Monty gave you an opportunity to give up your door for an opportunity to win if the prize is behind either of the 2 remaining doors? Assuming that each door has a 1/3 probability of having the car behind it, you would be exchanging a 1/3 probability of winning for a 2/3 probability, doubling your chance of winning.
We know that only 1 door has a prize behind it, so, for 2 doors, at least 1 must conceal a zonk. Now, before Monty gives you the opportunity to give up your door for 2 doors, he has the staff open 1 of those 2 doors that they know conceals a zonk. The probabilities haven't changed because we already knew that at least 1 door of the 2 concealed a zonk. If you had chosen door #1 and there was a car behind door #2, they would have opened door #3. If there were a car behind door #3, they would have opened door #2. If the car were behind your door, they would have opened either door #2 or #3, but not told you that the other door also concealed a zonk. So, the probability that 1 of the 2 doors conceals the prize is unchanged at 2/3, but we now can narrow it down to 1 door having 2/3 probability and the other 0, instead of each 1/3. Monty gives us the opportunity to give up our door for the other, unopened, door. This really is the same as the first case of giving up 1 door for 2 doors.
It would be a different story if Monty had the staff open a door randomly, so they could open the door with the car behind it. In that case, when a zonk appears, the remaining doors each have equal probability (1/2) of concealing the car.
When you switched doors, you have doubled your chances of winning, but that doesn't mean that you couldn't still lose. In the long run, about 2/3 of the contestants who switched doors would go home with cars. The remaining 1/3 will wish that they had stayed with their hunches and not switched doors!
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After you have picked a door, what if Monty gave you an opportunity to give up your door for an opportunity to win if the prize is behind either of the 2 remaining doors? Assuming that each door has a 1/3 probability of having the car behind it, you would be exchanging a 1/3 probability of winning for a 2/3 probability, doubling your chance of winning.
We know that only 1 door has a prize behind it, so, for 2 doors, at least 1 must conceal a zonk. Now, before Monty gives you the opportunity to give up your door for 2 doors, he has the staff open 1 of those 2 doors that they know conceals a zonk. The probabilities haven't changed because we already knew that at least 1 door of the 2 concealed a zonk. If you had chosen door #1 and there was a car behind door #2, they would have opened door #3. If there were a car behind door #3, they would have opened door #2. If the car were behind your door, they would have opened either door #2 or #3, but not told you that the other door also concealed a zonk. So, the probability that 1 of the 2 doors conceals the prize is unchanged at 2/3, but we now can narrow it down to 1 door having 2/3 probability and the other 0, instead of each 1/3. Monty gives us the opportunity to give up our door for the other, unopened, door. This really is the same as the first case of giving up 1 door for 2 doors.
It would be a different story if Monty had the staff open a door randomly, so they could open the door with the car behind it. In that case, when a zonk appears, the remaining doors each have equal probability (1/2) of concealing the car.
When you switched doors, you have doubled your chances of winning, but that doesn't mean that you couldn't still lose. In the long run, about 2/3 of the contestants who switched doors would go home with cars. The remaining 1/3 will wish that they had stayed with their hunches and not switched doors!
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kegginstructure
Actually, there are two ways to interpret the problem. It depends on whether you count the -swap/stay- question as a continuation of the problem or as a separate question. IF you count it as a separate question it IS a 50/50 case, and I have simulated it to prove that point. However, if you take it as a continuation then probabilities are in the 1/3 vs 2/3 pattern.
The thing that you CANNOT do (and I challenge all of the math wizards to answer this) is -How do you justify the asymmetric distribution of probability when you see the zonk and have two doors left. - After the initial pick but before the first reveal, each door has equal probability, 1/3 each. Now after the first reveal (of a zonk) you know that there are TWO places where the car can be. But to say it is 2/3 odds on the unpicked door means you assigned probabilities unevenly. You took all of the 1/3 probability from the zonk door and assigned it to the unchosen door, when the part I cannot justify is why you didn't assign 1/6 probability to each of the two remaining doors.
When I did this by a simulation, I had two cases. In the first case, I simulated the swap/stay two ways. In the case where I randomly chose to swap or stay, it was 50/50, but in the case where I followed the 1/3-2/3 rule, I saw the 1/3 - 2/3 pattern. Casey L. points out correctly that when you make the second choice dependent on the first one, the rule holds - but if you flip a coin to stay or swap, the 50/50 pattern holds. So it depends on whether you see this as one problem with two parts or as two problems. IF you see it as two problems, it is 50/50 for the 2nd choice.
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Actually, there are two ways to interpret the problem. It depends on whether you count the -swap/stay- question as a continuation of the problem or as a separate question. IF you count it as a separate question it IS a 50/50 case, and I have simulated it to prove that point. However, if you take it as a continuation then probabilities are in the 1/3 vs 2/3 pattern.
The thing that you CANNOT do (and I challenge all of the math wizards to answer this) is -How do you justify the asymmetric distribution of probability when you see the zonk and have two doors left. - After the initial pick but before the first reveal, each door has equal probability, 1/3 each. Now after the first reveal (of a zonk) you know that there are TWO places where the car can be. But to say it is 2/3 odds on the unpicked door means you assigned probabilities unevenly. You took all of the 1/3 probability from the zonk door and assigned it to the unchosen door, when the part I cannot justify is why you didn't assign 1/6 probability to each of the two remaining doors.
When I did this by a simulation, I had two cases. In the first case, I simulated the swap/stay two ways. In the case where I randomly chose to swap or stay, it was 50/50, but in the case where I followed the 1/3-2/3 rule, I saw the 1/3 - 2/3 pattern. Casey L. points out correctly that when you make the second choice dependent on the first one, the rule holds - but if you flip a coin to stay or swap, the 50/50 pattern holds. So it depends on whether you see this as one problem with two parts or as two problems. IF you see it as two problems, it is 50/50 for the 2nd choice.
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Roger
Actually, your chance of winning is 50% from the start. The mathmatics would only show that if the door was picked at random. If the doors were picked at random, you would have a 33% chance of losing from the 1st pick. Since they aren't going to pick your door, you have 0% chance of winning after the 1st pick since they are not going to pick your door, so that gives you a 2/3 chance of NOT losing after 1st pick. Now, there are 2 doors left. 2 doors, 1 winner is always going to be 50% because you have no information on either of those doors. 50% of 2/3 is still 1/3.
Since they are not going to pick your door, and they are not going to pick the winning door, you have a 100% chance of NOT losing after 1s pick. Two doors 1 winner is still 50% chance because you have no information on either door. That gives you a 50% chance of NOT losing.
The actual chance of winning is determined by how many doors are left at your final decision. Look at it this way. There are 100 doors and 2 people playing. You then elliminate 98 doors and all are losers. Which one has a better chance of winning?
That is horrible math to assume the other one gets the entire 1/3 chance of the other number going to the other door. How do you justify that?
edit: My mistake. I didn't factor in the part where they won't pick the door you chose to open 1st. That actually does switch the odds into the other door's favor.
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Actually, your chance of winning is 50% from the start. The mathmatics would only show that if the door was picked at random. If the doors were picked at random, you would have a 33% chance of losing from the 1st pick. Since they aren't going to pick your door, you have 0% chance of winning after the 1st pick since they are not going to pick your door, so that gives you a 2/3 chance of NOT losing after 1st pick. Now, there are 2 doors left. 2 doors, 1 winner is always going to be 50% because you have no information on either of those doors. 50% of 2/3 is still 1/3.
Since they are not going to pick your door, and they are not going to pick the winning door, you have a 100% chance of NOT losing after 1s pick. Two doors 1 winner is still 50% chance because you have no information on either door. That gives you a 50% chance of NOT losing.
The actual chance of winning is determined by how many doors are left at your final decision. Look at it this way. There are 100 doors and 2 people playing. You then elliminate 98 doors and all are losers. Which one has a better chance of winning?
That is horrible math to assume the other one gets the entire 1/3 chance of the other number going to the other door. How do you justify that?
edit: My mistake. I didn't factor in the part where they won't pick the door you chose to open 1st. That actually does switch the odds into the other door's favor.
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the
If player elects to always stay with first choice:
Player selects door: 1 - 2 - 3
Car is at door 1: .W - L - L
Car is at door 2: .L - W - L
Car is at door 3: .L - L - W
Any argument that says there is no advantage to switching first has to clarify whether they really mean the odds are 1/2 vs 1/2, or if they mean that switching, like staying, stands to win 1/3.
If they mean that the probabilities are 1/2 vs 1/2, then they have to show how the strategy of always staying stands to win 1/2 of all possible games. To accomplish this, they would have to explain how the player could pick the goat, elect to stay, and then somehow stand to win that game part of the time. And, they would have to explain how the player could pick the car, stay, and then somehow stand to lose that game part of the time. And this process would have to end up with equal probabilities of winning and losing. It's going to be a very challenging thing to prove! Hopefully their proof will set them up to also prove the complementary argument that switching stands to win half of all possible games as well, because that too is going to be a challenge. Really, the Clay Mathematics Institute ought to award a prize to any person who can come up with that proof.
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If player elects to always stay with first choice:
Player selects door: 1 - 2 - 3
Car is at door 1: .W - L - L
Car is at door 2: .L - W - L
Car is at door 3: .L - L - W
Any argument that says there is no advantage to switching first has to clarify whether they really mean the odds are 1/2 vs 1/2, or if they mean that switching, like staying, stands to win 1/3.
If they mean that the probabilities are 1/2 vs 1/2, then they have to show how the strategy of always staying stands to win 1/2 of all possible games. To accomplish this, they would have to explain how the player could pick the goat, elect to stay, and then somehow stand to win that game part of the time. And, they would have to explain how the player could pick the car, stay, and then somehow stand to lose that game part of the time. And this process would have to end up with equal probabilities of winning and losing. It's going to be a very challenging thing to prove! Hopefully their proof will set them up to also prove the complementary argument that switching stands to win half of all possible games as well, because that too is going to be a challenge. Really, the Clay Mathematics Institute ought to award a prize to any person who can come up with that proof.
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Armastat
Here is a problem with the rationale shown in the video. The three door problem never existed, its an illusion.
2/3 chance? No, even tho there are 2 doors other than the one u select, the third door will NEVER be offered to you. it will always be removed from the choices before you get to make it. And then you are now given a NEW choice between 2 doors. At that moment you have a 50% chance.
No matter if you select the winning door or a losing door - as a first choice - you will never get any other result other than the two doors which will always contain the winning door and a losing door. The first choice is NOT a choice at all.
EDIT: Okay i just realized I was focused too hard on the second step. Yes you can select the winning door with your first choice (1 in 3 chance to win. But THAT is not the Monty Hall Problem, which is what happens when you don't select the winning door. The two situations are mutually exclusive. either you win with a 1/3 probability or you dont and end up with a 50/50. you NEVER have a 2/3 chance of losing.
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Here is a problem with the rationale shown in the video. The three door problem never existed, its an illusion.
2/3 chance? No, even tho there are 2 doors other than the one u select, the third door will NEVER be offered to you. it will always be removed from the choices before you get to make it. And then you are now given a NEW choice between 2 doors. At that moment you have a 50% chance.
No matter if you select the winning door or a losing door - as a first choice - you will never get any other result other than the two doors which will always contain the winning door and a losing door. The first choice is NOT a choice at all.
EDIT: Okay i just realized I was focused too hard on the second step. Yes you can select the winning door with your first choice (1 in 3 chance to win. But THAT is not the Monty Hall Problem, which is what happens when you don't select the winning door. The two situations are mutually exclusive. either you win with a 1/3 probability or you dont and end up with a 50/50. you NEVER have a 2/3 chance of losing.
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Emergency
I disagree, but I'm also an idiot. however. I believe that the probability is 50/50, and is so from the beginning of the game. Why? The game is manipulated that way. It does not matter what door you pick, (1, 2, or 3, Monty will always open a door you did not choose and you will always be shown a zonk. Your initial choice is completely irrelevant, and has absolutely zero impact on the final outcome. Monty knows where the car is so he always has the option to open a door with a zonk, and always will.
If you choose a door with a car you will be shown a zonk from a door you didn't pick. If you choose a door with a zonk you will be shown a zonk from a door you didn't pick. No matter what choices you made with the first door, you will ALWAYS end up with two closed doors, one with a car, one with a zonk.
So the real game only begins when you are down to the two doors. Therefore 50/50.
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I disagree, but I'm also an idiot. however. I believe that the probability is 50/50, and is so from the beginning of the game. Why? The game is manipulated that way. It does not matter what door you pick, (1, 2, or 3, Monty will always open a door you did not choose and you will always be shown a zonk. Your initial choice is completely irrelevant, and has absolutely zero impact on the final outcome. Monty knows where the car is so he always has the option to open a door with a zonk, and always will.
If you choose a door with a car you will be shown a zonk from a door you didn't pick. If you choose a door with a zonk you will be shown a zonk from a door you didn't pick. No matter what choices you made with the first door, you will ALWAYS end up with two closed doors, one with a car, one with a zonk.
So the real game only begins when you are down to the two doors. Therefore 50/50.
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Sabeeh
Great video however I don't follow mathematically. Your explanation assumes the probabilities remain as they were before the new information is known. But surely the probability changes given the new information. We are told the 98 doors are wrong so now the only relevant information is that the car is either behind your door or the other one. How can the probability remain at 1/100 for your door, you are ignoring the new information. It must be 50 / 50 chance now and no advantage of switching.
However it makes sense if we ignore most of the maths ie initially you pick a door with 1/100 probability of being a car. The host will always eliminate all other doors such that there must be a car behind one of the last 2 doors. Chances are that you picked the wrong door and this means the host MUST pick the door with the car, hence you should switch.
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Great video however I don't follow mathematically. Your explanation assumes the probabilities remain as they were before the new information is known. But surely the probability changes given the new information. We are told the 98 doors are wrong so now the only relevant information is that the car is either behind your door or the other one. How can the probability remain at 1/100 for your door, you are ignoring the new information. It must be 50 / 50 chance now and no advantage of switching.
However it makes sense if we ignore most of the maths ie initially you pick a door with 1/100 probability of being a car. The host will always eliminate all other doors such that there must be a car behind one of the last 2 doors. Chances are that you picked the wrong door and this means the host MUST pick the door with the car, hence you should switch.
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ian
For anyone still struggling - maybe this will help. -
Here is the same maths problem described differently: -
Imagine there are two contestants who get to the final game. -
Contestant A gets to pick ONE of the 3 doors. -
- If his door has the car he wins and the game stops. -
- If his door has a goat then we give Contestant B a chance. -
On those occassions where Contestant A loses and Contestant B gets his chance we know that the car must be behind one of the last 2 doors. -
Contestant B gets to pick ONE of the 2 remaining doors. -
But before he chooses WE TELL HIM WHICH DOOR THE GOAT IS BEHIND. -
Who would you rather be - Contestant A or Contestant B? -
- Being Contestant A is the same as not switching. -
- Being Contestant B is the same as switching.
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For anyone still struggling - maybe this will help. -
Here is the same maths problem described differently: -
Imagine there are two contestants who get to the final game. -
Contestant A gets to pick ONE of the 3 doors. -
- If his door has the car he wins and the game stops. -
- If his door has a goat then we give Contestant B a chance. -
On those occassions where Contestant A loses and Contestant B gets his chance we know that the car must be behind one of the last 2 doors. -
Contestant B gets to pick ONE of the 2 remaining doors. -
But before he chooses WE TELL HIM WHICH DOOR THE GOAT IS BEHIND. -
Who would you rather be - Contestant A or Contestant B? -
- Being Contestant A is the same as not switching. -
- Being Contestant B is the same as switching.
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Daniel
I have agonized over this problem trying to understand the logic/mathematics and now I have a follow up question which might help me clarify. If someone smarter than I can kindly answer.
In my hypothetical scenario. As you play the game (with all the previously stated rules, you get disqualified (from the game) for some reason or other at the point when you had already picked door 1 and the second door (door 2) had already been opened by the host.
And then some random guy (James) is called off the street (was not previously present in the room when you were playing) and he finds door 2 open but because he was not in the room at that point, doesn't know you picked door 1.
Does door 3 still become a better option for him (James) or is it now a 50/50 (1/2) chance?
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I have agonized over this problem trying to understand the logic/mathematics and now I have a follow up question which might help me clarify. If someone smarter than I can kindly answer.
In my hypothetical scenario. As you play the game (with all the previously stated rules, you get disqualified (from the game) for some reason or other at the point when you had already picked door 1 and the second door (door 2) had already been opened by the host.
And then some random guy (James) is called off the street (was not previously present in the room when you were playing) and he finds door 2 open but because he was not in the room at that point, doesn't know you picked door 1.
Does door 3 still become a better option for him (James) or is it now a 50/50 (1/2) chance?
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