Monty Hall Problem (best explanation) - Numberphile
video description
Now we just have 2 doors left. One door has to have a car behind it according to the game. The 1st door chosen represents the 1 out of 3 chance probability and stays with that door. The other door left that Monty did not open represents or offers a 1 out of 2 chance of getting the car so it would be most likely to benefit of choosing that door because it represents a 1 out of 2 chance of getting the car compared to the first door chosen that represents a 1 out of 3 chance of getting the car. me think so
Date: 2022-04-08
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Comments and reviews: 9
The
All eliminating the third door does is give both the remaining doors a 16. 5% extra chance of being the right door. The 100 door scenario though 100% factual, is irrelevant because it isn't even close to the same scenario. The way they are presenting this 3 door scenario is essentially giving away a car. In the 100 door scenario you guess 1 door, and 98 other doors are closed off. The odd of you guessing the right door out of a 100 is HIGHLY improbable, so you would ask yourself -Out of the 100 doors what are the odds I picked the right one? - And you would pick the other door surely. In the 3 door scenario, picking 1 door and closing another actually just gives both door a 16. 5% higher chance of being the right door essentially making it a 50/50. Though the 100 door scenario is something I had never thought about before and both scenarios were interesting to think about.
It is a 50/50 in this 3 door scenario because in the final equation there are only 2 doors to choose from.
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All eliminating the third door does is give both the remaining doors a 16. 5% extra chance of being the right door. The 100 door scenario though 100% factual, is irrelevant because it isn't even close to the same scenario. The way they are presenting this 3 door scenario is essentially giving away a car. In the 100 door scenario you guess 1 door, and 98 other doors are closed off. The odd of you guessing the right door out of a 100 is HIGHLY improbable, so you would ask yourself -Out of the 100 doors what are the odds I picked the right one? - And you would pick the other door surely. In the 3 door scenario, picking 1 door and closing another actually just gives both door a 16. 5% higher chance of being the right door essentially making it a 50/50. Though the 100 door scenario is something I had never thought about before and both scenarios were interesting to think about.
It is a 50/50 in this 3 door scenario because in the final equation there are only 2 doors to choose from.
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Austin
There is a 1/3 chance that the door the player selects has the car behind it, whereas there is a 2/3 chance that the player picks a door that has a goat behind it. If the player picks the door with the car behind it, the host has to randomly select one of the remaining two doors to reveal a goat. In this situation, it is not advantageous to switch. However, if the player picks a door that has a goat behind it, the host only has one option to choose, because the host can't reveal the door with the car behind it, and the host also can't reveal the door that the player selected. Therefore, it is advantageous for the player to change their selection in this situation because it guarantees that they will select the car. But there is only a 2/3 chance of this situation happening. It is more likely to win if the player changes their selection simply because it is more likely that there is a goat behind the original door that they selected.
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There is a 1/3 chance that the door the player selects has the car behind it, whereas there is a 2/3 chance that the player picks a door that has a goat behind it. If the player picks the door with the car behind it, the host has to randomly select one of the remaining two doors to reveal a goat. In this situation, it is not advantageous to switch. However, if the player picks a door that has a goat behind it, the host only has one option to choose, because the host can't reveal the door with the car behind it, and the host also can't reveal the door that the player selected. Therefore, it is advantageous for the player to change their selection in this situation because it guarantees that they will select the car. But there is only a 2/3 chance of this situation happening. It is more likely to win if the player changes their selection simply because it is more likely that there is a goat behind the original door that they selected.
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Jason
one can do this concretely without abstractions. Just get three cups, and put, let's say, a peanut under one, and a raisin under the other two. Have someone do the hiding for you, plus removing a cup with a raisin, you will come out even picking the peanut on the switches and stand pats. I don't see what Monty knowing has to do with it. He is just cutting your odds from 1/3 to 1/2 or even. I have done this and would trust this hands on approach more than a computer. I think the error is thinking there is probability when each of these actions is discreet. There are not three doors in any single instance we are talking about. One has been eliminated. And that would be true for each discreet time it is done. So there are never three doors or three possibilities. Only two.
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one can do this concretely without abstractions. Just get three cups, and put, let's say, a peanut under one, and a raisin under the other two. Have someone do the hiding for you, plus removing a cup with a raisin, you will come out even picking the peanut on the switches and stand pats. I don't see what Monty knowing has to do with it. He is just cutting your odds from 1/3 to 1/2 or even. I have done this and would trust this hands on approach more than a computer. I think the error is thinking there is probability when each of these actions is discreet. There are not three doors in any single instance we are talking about. One has been eliminated. And that would be true for each discreet time it is done. So there are never three doors or three possibilities. Only two.
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subductionzone
It does not even matter if Monty knows. Think of it as two populations. When you make your first choice you separate the doors into two populations. The odds are set at the point that you make that choice. The odds of your original choice do not change if Monty did not know. The problem is that people forget in the trivial case of Monty revealing a car we would know that it would be pointless to switch. You could run a simulation. You would find that the trivial case arose one time out of three, and that would not count as a -game- since we know that Monty revealed a goat. Of the remaining times the goat would still be behind the door that Monty did not open two times out of three and one time out of thee the contestant would be the winner.
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It does not even matter if Monty knows. Think of it as two populations. When you make your first choice you separate the doors into two populations. The odds are set at the point that you make that choice. The odds of your original choice do not change if Monty did not know. The problem is that people forget in the trivial case of Monty revealing a car we would know that it would be pointless to switch. You could run a simulation. You would find that the trivial case arose one time out of three, and that would not count as a -game- since we know that Monty revealed a goat. Of the remaining times the goat would still be behind the door that Monty did not open two times out of three and one time out of thee the contestant would be the winner.
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Sebastian
And again, just to prove that this programme and the professor are confusing the influences of 'human behaviour' with -mathematics. -
Well, I simply stated this in the prior explanation when put the same problem in card form:
Another example of how this example above is wrong: -
If you have 3 playing cards. Ace, King, Queen. -
And you are trying to find the Ace, and you pick the Queen first, and there's 2 more cards left. -
Then there of course 2 cards left, you are only have one of these to choose from so the chances are 1/2. 50/50. -
In which care it doesn't make a blind bit of difference whether you switch or not, 'unless' it is Monty who is turning the cards. who knows where the Ace is!
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And again, just to prove that this programme and the professor are confusing the influences of 'human behaviour' with -mathematics. -
Well, I simply stated this in the prior explanation when put the same problem in card form:
Another example of how this example above is wrong: -
If you have 3 playing cards. Ace, King, Queen. -
And you are trying to find the Ace, and you pick the Queen first, and there's 2 more cards left. -
Then there of course 2 cards left, you are only have one of these to choose from so the chances are 1/2. 50/50. -
In which care it doesn't make a blind bit of difference whether you switch or not, 'unless' it is Monty who is turning the cards. who knows where the Ace is!
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Sanjay
%matplotlib inline
import numpy as np
import matplotlib. pyplot as plt
def monte_hall(num_rooms=10, iterations=100, switch=True):
hits = 0
rooms = ['goat'] - num_rooms
rooms[0] = 'car'
for k in range(iterations):
pick_room = np. random. choice(num_rooms, 1)[0]
if (pick_room == 0) and (switch): continue
elif (pick_room == 0) and (not switch): hits += 1
elif (pick_room! = 0) and (switch): hits += 1
else: continue
hits = hits/iterations
miss = 1 - hits
return hits, miss
hits_miss = monte_hall(num_rooms=3, iterations=1000, switch=True)
print(f'hits, miss -hits_miss-')
plt. pie(hits_miss, labels = ['hits', 'miss'])
plt. show)
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%matplotlib inline
import numpy as np
import matplotlib. pyplot as plt
def monte_hall(num_rooms=10, iterations=100, switch=True):
hits = 0
rooms = ['goat'] - num_rooms
rooms[0] = 'car'
for k in range(iterations):
pick_room = np. random. choice(num_rooms, 1)[0]
if (pick_room == 0) and (switch): continue
elif (pick_room == 0) and (not switch): hits += 1
elif (pick_room! = 0) and (switch): hits += 1
else: continue
hits = hits/iterations
miss = 1 - hits
return hits, miss
hits_miss = monte_hall(num_rooms=3, iterations=1000, switch=True)
print(f'hits, miss -hits_miss-')
plt. pie(hits_miss, labels = ['hits', 'miss'])
plt. show)
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GetMeThere1
You can NEVER go wrong when you use an exhaustive solution. Fortunately, in this case, it's ridiculously easy to use an exhaustive solution. There are 3 conditions: oxx, xox, xxo, where the valued pick is the -o- door. And there are three strategies: pick door 1, 2, or 3. If you apply all three strategies to all three conditions, and always take the switching choice, you will see that you will always win 2 out of 3 times under all three conditions. Therefore you KNOW that switching gives you 2/3 probability since you've tried ALL combinations. If you don't switch you will win only 1/3 of the time with each strategy under all three conditions.
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You can NEVER go wrong when you use an exhaustive solution. Fortunately, in this case, it's ridiculously easy to use an exhaustive solution. There are 3 conditions: oxx, xox, xxo, where the valued pick is the -o- door. And there are three strategies: pick door 1, 2, or 3. If you apply all three strategies to all three conditions, and always take the switching choice, you will see that you will always win 2 out of 3 times under all three conditions. Therefore you KNOW that switching gives you 2/3 probability since you've tried ALL combinations. If you don't switch you will win only 1/3 of the time with each strategy under all three conditions.
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Andy
This is how I understand it.
I would only have a 1/3 chance of correctly choosing door with the car. So therefore, I have a 2/3 chance of being wrong and choosing the goat.
So assuming I pick door A, there is a 2/3 chance of the car being behind doors B or C.
As there is more chance of me initially choosing a goat then
My choice (1/3) Alternative 1 - 1/3 = 2/3
goat goat and car
In this case the goat is removed from the alternative, but I still only have a 1/3 chance of my choice being correct and, by logic, the alternative is still 1 - 1/3 = 2/3. Logic dictates that you go with the option that has a higher chance.
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This is how I understand it.
I would only have a 1/3 chance of correctly choosing door with the car. So therefore, I have a 2/3 chance of being wrong and choosing the goat.
So assuming I pick door A, there is a 2/3 chance of the car being behind doors B or C.
As there is more chance of me initially choosing a goat then
My choice (1/3) Alternative 1 - 1/3 = 2/3
goat goat and car
In this case the goat is removed from the alternative, but I still only have a 1/3 chance of my choice being correct and, by logic, the alternative is still 1 - 1/3 = 2/3. Logic dictates that you go with the option that has a higher chance.
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Amir
What if Monty only sometimes reveals a remaining door and offer if you want to switch? He may the choose to do so only when he wants to compel you to switch (based on the assumption that you are already familiar with the odds. In such a case you are better off staying with your original selection. But then if everyone realizes all of that, then you don't know which angle he may be manipulating by. Generally, from the host's perspective, giving the contestant the option to change their selection when you know they have already made the right selection would heavily benefit the game show's interests in not giving away a car.
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What if Monty only sometimes reveals a remaining door and offer if you want to switch? He may the choose to do so only when he wants to compel you to switch (based on the assumption that you are already familiar with the odds. In such a case you are better off staying with your original selection. But then if everyone realizes all of that, then you don't know which angle he may be manipulating by. Generally, from the host's perspective, giving the contestant the option to change their selection when you know they have already made the right selection would heavily benefit the game show's interests in not giving away a car.
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