What's wrong with this equal probability solution for Monty Hall Problem?
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I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective gives me equal probability.
Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door.
Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)?
Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely.
Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car.
What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks.
probability monty-hall
edited Feb 3 '16 at 4:10
Greek - Area 51 Proposal
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asked Jan 14 '16 at 23:13
OneZeroOneZero
13112
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add a comment |
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I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective gives me equal probability.
Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door.
Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)?
Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely.
Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car.
What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks.
probability monty-hall
edited Feb 3 '16 at 4:10
Greek - Area 51 Proposal
3,206769105
asked Jan 14 '16 at 23:13
OneZeroOneZero
13112
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What's wrong with the process (and what's really behind the whole Monty Hall paradox) is the fact that revealing a goat doesn't change the fact that there's a $1/3$ chance of $A$ containing a car. No matter what door you pick, there will always be at least one goat door you didn't pick, so by just having that fact confirmed (by revealing a goat), you haven't really recieved any information that changes anything about the door you picked.
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– Arthur
Jan 14 '16 at 23:19
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"If we have two doors, and know a car is behind one, then we have a 50/50 chance of choosing the right door." True, but only if that is all you know. If you have two doors, and one door is on a tiny doll house, and one is on a full size garage, you know the chance is not 50/50. If both doors were clear glass, again the chance is not 50/50. There was information used in the Monty Hall problem to reduce down to the set of two doors, and although now you only see two, one piece of information you have is that Monty had full knowledge of what was happening, and he would always open a goat door.
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– piCookie
Nov 7 '16 at 20:42
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I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective gives me equal probability.
Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door.
Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)?
Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely.
Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car.
What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks.
probability monty-hall
edited Feb 3 '16 at 4:10
Greek - Area 51 Proposal
3,206769105
asked Jan 14 '16 at 23:13
OneZeroOneZero
13112
$endgroup$
I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective gives me equal probability.
Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door.
Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)?
Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely.
Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car.
What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks.
probability monty-hall
probability monty-hall
edited Feb 3 '16 at 4:10
Greek - Area 51 Proposal
3,206769105
asked Jan 14 '16 at 23:13
OneZeroOneZero
13112
edited Feb 3 '16 at 4:10
Greek - Area 51 Proposal
3,206769105
asked Jan 14 '16 at 23:13
OneZeroOneZero
13112
edited Feb 3 '16 at 4:10
Greek - Area 51 Proposal
3,206769105
edited Feb 3 '16 at 4:10
Greek - Area 51 Proposal
3,206769105
edited Feb 3 '16 at 4:10
Greek - Area 51 Proposal
3,206769105
3,206769105
asked Jan 14 '16 at 23:13
OneZeroOneZero
13112
asked Jan 14 '16 at 23:13
OneZeroOneZero
13112
asked Jan 14 '16 at 23:13
OneZeroOneZero
13112
13112
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What's wrong with the process (and what's really behind the whole Monty Hall paradox) is the fact that revealing a goat doesn't change the fact that there's a $1/3$ chance of $A$ containing a car. No matter what door you pick, there will always be at least one goat door you didn't pick, so by just having that fact confirmed (by revealing a goat), you haven't really recieved any information that changes anything about the door you picked.
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– Arthur
Jan 14 '16 at 23:19
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"If we have two doors, and know a car is behind one, then we have a 50/50 chance of choosing the right door." True, but only if that is all you know. If you have two doors, and one door is on a tiny doll house, and one is on a full size garage, you know the chance is not 50/50. If both doors were clear glass, again the chance is not 50/50. There was information used in the Monty Hall problem to reduce down to the set of two doors, and although now you only see two, one piece of information you have is that Monty had full knowledge of what was happening, and he would always open a goat door.
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– piCookie
Nov 7 '16 at 20:42
add a comment |
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What's wrong with the process (and what's really behind the whole Monty Hall paradox) is the fact that revealing a goat doesn't change the fact that there's a $1/3$ chance of $A$ containing a car. No matter what door you pick, there will always be at least one goat door you didn't pick, so by just having that fact confirmed (by revealing a goat), you haven't really recieved any information that changes anything about the door you picked.
$endgroup$
– Arthur
Jan 14 '16 at 23:19
$begingroup$
"If we have two doors, and know a car is behind one, then we have a 50/50 chance of choosing the right door." True, but only if that is all you know. If you have two doors, and one door is on a tiny doll house, and one is on a full size garage, you know the chance is not 50/50. If both doors were clear glass, again the chance is not 50/50. There was information used in the Monty Hall problem to reduce down to the set of two doors, and although now you only see two, one piece of information you have is that Monty had full knowledge of what was happening, and he would always open a goat door.
$endgroup$
– piCookie
Nov 7 '16 at 20:42
$begingroup$
What's wrong with the process (and what's really behind the whole Monty Hall paradox) is the fact that revealing a goat doesn't change the fact that there's a $1/3$ chance of $A$ containing a car. No matter what door you pick, there will always be at least one goat door you didn't pick, so by just having that fact confirmed (by revealing a goat), you haven't really recieved any information that changes anything about the door you picked.
$endgroup$
– Arthur
Jan 14 '16 at 23:19
$begingroup$
What's wrong with the process (and what's really behind the whole Monty Hall paradox) is the fact that revealing a goat doesn't change the fact that there's a $1/3$ chance of $A$ containing a car. No matter what door you pick, there will always be at least one goat door you didn't pick, so by just having that fact confirmed (by revealing a goat), you haven't really recieved any information that changes anything about the door you picked.
$endgroup$
– Arthur
Jan 14 '16 at 23:19
$begingroup$
"If we have two doors, and know a car is behind one, then we have a 50/50 chance of choosing the right door." True, but only if that is all you know. If you have two doors, and one door is on a tiny doll house, and one is on a full size garage, you know the chance is not 50/50. If both doors were clear glass, again the chance is not 50/50. There was information used in the Monty Hall problem to reduce down to the set of two doors, and although now you only see two, one piece of information you have is that Monty had full knowledge of what was happening, and he would always open a goat door.
$endgroup$
– piCookie
Nov 7 '16 at 20:42
$begingroup$
"If we have two doors, and know a car is behind one, then we have a 50/50 chance of choosing the right door." True, but only if that is all you know. If you have two doors, and one door is on a tiny doll house, and one is on a full size garage, you know the chance is not 50/50. If both doors were clear glass, again the chance is not 50/50. There was information used in the Monty Hall problem to reduce down to the set of two doors, and although now you only see two, one piece of information you have is that Monty had full knowledge of what was happening, and he would always open a goat door.
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– piCookie
Nov 7 '16 at 20:42
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8 Answers
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Therefore, showing an open door with a goat reveals nothing new about which door has the car.
Two-thirds of the time, our initial pick will be wrong. In that case, Monty cannot randomly pick a door out of the remaining two to reveal. He must pick the door that has a goat. This gives the additional insight into which door has a car.
Conversely, one-third of the time, our initial pick will be correct so Monty can then open any of the two remaining two doors. Switching in that case will then lead us to the wrong door. The good news is, that only happens one-third of the time!
edited Feb 3 '16 at 4:11
Greek - Area 51 Proposal
3,206769105
answered Jan 14 '16 at 23:28
Kelvin SohKelvin Soh
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Your error is a common one: You have confused the event that the host reveals door $C$ with the event that there is a goat behind door $C$. This is distinction enough to see: if the car is, in fact, behind door $A$, then the host will only reveal door $C$ with a probability of $50%$, whereas there is definitely a goat behind door $C$.
In particular, let $C_{text{goat}}$ be the event that there's a goat behind door $C$, and $C_{text{revealed}}$ be the event that door $C$ is revealed, and $A_{text{car}}$ be the event that there is a car behind door $A$. Then, one can see that $A_{text{car}}$ and $C_{text{revealed}}$ are independent of each other, since the host reveals door $B$ or $C$ randomly with $50%$ probability if there's a car behind door $A$, and if there's not a car behind door $A$, the host is forced to pick the door which hasn't got a car behind it - which is $B$ or $C$ with probability $50%$. Thus, one concludes that
$$P(A_{text{car}}|C_{text{revealed}})=P(A_{text{car}})=frac{1}3.$$
The calculation you've done (annotated with a $neq$ sign where things go wrong) is:
$$P(A_{text{car}}|C_{text{revealed}})neq P(A_{text{car}}|C_{text{goat}})=frac{1}2$$
answered Jan 14 '16 at 23:33
Milo BrandtMilo Brandt
40.1k476140
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Well, this was one one the most convincing argument that I have seen so far. Thank you.
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– onurcanbektas
Jul 20 '18 at 2:48
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Let me try to guess your birthday. My guess is April $2$nd.
Ignoring leap years, remove $363$ dates that are not your birthday and that are not April $2$nd (because I chose that), so that we are now left with only two options. You offer me to switch to the date that I didn't pick, which I do.
That I was right in my initial guess is unaffected by the dates that you subsequently remove.
answered Jan 14 '16 at 23:23
MankindMankind
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What is we see "switching door" as a "second guess"? If I give you two dates, and one is my bday, then you have a 50/50 chance, right? Then how is it different? Not switching = guessing 4/2, and switching = guess the other one.
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– OneZero
Jan 14 '16 at 23:25
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@OneZero Yes, switching doors corresponds to a second guess. If you give me only two dates, with no prior choices, then it's $50/50$. But you give me $365$ dates, and tell me to choose one of them. The probability that I'm right is $1/365$. You then eliminate $363$ possibilities, leaving me with $2$ choices, one of which is my original pick. Just because I now have only two choices left doesn't mean it's $50/50$, and the reason is that I made my choice back when there were $365$ options.
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– Mankind
Jan 14 '16 at 23:32
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Very nice way to put it. If someone isn't convinced, just try it 10 times with random friends and see how much you'll be right :-D
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– Ant
Apr 8 '16 at 7:04
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The flaw in your thought process is that in the $50:50$ case with only two doors where one is a goat and one is a car, that is the only information you have. In the other case where you have two goat doors and one car door, when you open a goat door, you gain some information by doing this. Thus you appear to have the same $50:50$ situation, but you have some added information.
answered Jan 14 '16 at 23:25
flabby99flabby99
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Can you elaborate on this? What's added?
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– OneZero
Jan 14 '16 at 23:26
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Have you seen an actual proof/ working out of the probabilities of the monty hall problem using bayes law and conditional probability?
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– flabby99
Jan 14 '16 at 23:29
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Also I think it might potentially be helpful to imagine a situation where there is 100 doors, 99 of them have goats and 1 of them has a car. You choose a door. Monty opens 98 that have goats, and asks should you switch?
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– flabby99
Jan 14 '16 at 23:32
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The idea of assigning $50/50$ probabilities to two alternatives
makes sense only when the two alternatives are truly symmetric.
If you just start with two doors, you know there is a car behind
exactly one, and nothing else has happened concerning those doors,
then $50/50$ probabilities make sense.
But once you have chosen one of three doors and Monty has opened
another door (having been forced by the rules of the game to open
a door that wasn't yours and didn't have the car),
you no longer have a symmetric situation. One of the remaining
doors is an available choice because you already chose it once;
the other is an available choice either (A) because you already
chose the car and Monty got to take one of the "goat" doors away at random,
or (B) because you chose a goat initially and Monty had to take away
the remaining "goat" door, leaving the door with the car.
Case (B) is twice as likely to occur as case (A), not $50/50$,
and which door has the car is now completely a function of whether
you're in case (A) or case (B), so it's also not $50/50$.
answered Apr 8 '16 at 6:59
David KDavid K
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The error in your argument is that you don't start with two doors, you start with three.
If you pick one door out of three, you have a one in three chance of having picked the right one ($tfrac{1}{3}$), and a two in three chance of having picked the wrong one ($tfrac{2}{3}$).
If you have picked the wrong door, Monty shows you which door has the car behind it, by opening the other door. You win a car by switching.
Conversely, if you have picked the right door, switching makes you lose the car.
Since there's a $tfrac{2}{3}$ chance of picking the wrong door at first, there's a $tfrac{2}{3}$ of winning the car after switching. The chance of picking the right door is $tfrac{1}{3}$, so there's a $tfrac{1}{3}$ chance of losing the car after switching.
answered Jul 22 '16 at 8:43
SQBSQB
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example 1 which shows there would be no advantage in shifting: lets say there are one million doors and I pick No 1. Monty reveals 999998 doors which are empty so now we have only 2 doors, one that I picked and another one. According to the paradox the possibility would be 1/999999 = almost 100% certain that the shift would have advantage. Obviously wrong.
Convinced?
example 2 which shows there would be no advantage in shifting:
Lets say instead of one player we have 2 players and one of them choose door one and another choose door 2 and Monty opens door three. If the argument is true shifting would increase the chance for both players. Obviously not possible.
Convinced?
answered Jun 28 '17 at 12:27
user7432810user7432810
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I'm amazed people still argue this.
Example 1 yes you would have an extremely high chance of winning if you selected the door that was filtered out 999,999 doors. If not, you are basically arguing that Door 1 had a 50/50 shot at winning a contest that had 1 out of 1,000,000 odds. The fact the game show host had to filter the losing 999,998 doors out of the set makes switching advantageous since that door is weighted by the 999,998 losing doors that were filtered out.
Example 2. This example isn't applicable to the stated problem. When you introduce a 2nd player the host no longer can filter 2 doors down to 1 door unless both players choose the same door. If they choose the same door then it's the same classic Monty Hall problem just 2 iterations happening simultaneously. If 2 players select different doors then Monty has to open the 3rd unchosen door. According to you, that is 50/50 right? This is your premise but you're forgetting another option.
What you're forgetting is if neither player selected the correct door. In this case, Monty would have to Open player 1 or 2 door. This would then force that particular player to move as he now knows for a fact he will lose without switching. The player who's door was opened would switch obviously but he now has a 50/50 shot instead of 1/3. The player who's door wasn't opened should switch as this had no impact to him other than the typical Monty hall problem, i.e he now has 2/3 chance.
answered Jun 19 '18 at 1:14
BrianBrian
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Therefore, showing an open door with a goat reveals nothing new about which door has the car.
Two-thirds of the time, our initial pick will be wrong. In that case, Monty cannot randomly pick a door out of the remaining two to reveal. He must pick the door that has a goat. This gives the additional insight into which door has a car.
Conversely, one-third of the time, our initial pick will be correct so Monty can then open any of the two remaining two doors. Switching in that case will then lead us to the wrong door. The good news is, that only happens one-third of the time!
edited Feb 3 '16 at 4:11
Greek - Area 51 Proposal
3,206769105
answered Jan 14 '16 at 23:28
Kelvin SohKelvin Soh
1,645714
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Therefore, showing an open door with a goat reveals nothing new about which door has the car.
Two-thirds of the time, our initial pick will be wrong. In that case, Monty cannot randomly pick a door out of the remaining two to reveal. He must pick the door that has a goat. This gives the additional insight into which door has a car.
Conversely, one-third of the time, our initial pick will be correct so Monty can then open any of the two remaining two doors. Switching in that case will then lead us to the wrong door. The good news is, that only happens one-third of the time!
edited Feb 3 '16 at 4:11
Greek - Area 51 Proposal
3,206769105
answered Jan 14 '16 at 23:28
Kelvin SohKelvin Soh
1,645714
$endgroup$
add a comment |
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Therefore, showing an open door with a goat reveals nothing new about which door has the car.
Two-thirds of the time, our initial pick will be wrong. In that case, Monty cannot randomly pick a door out of the remaining two to reveal. He must pick the door that has a goat. This gives the additional insight into which door has a car.
Conversely, one-third of the time, our initial pick will be correct so Monty can then open any of the two remaining two doors. Switching in that case will then lead us to the wrong door. The good news is, that only happens one-third of the time!
edited Feb 3 '16 at 4:11
Greek - Area 51 Proposal
3,206769105
answered Jan 14 '16 at 23:28
Kelvin SohKelvin Soh
1,645714
$endgroup$
Therefore, showing an open door with a goat reveals nothing new about which door has the car.
Two-thirds of the time, our initial pick will be wrong. In that case, Monty cannot randomly pick a door out of the remaining two to reveal. He must pick the door that has a goat. This gives the additional insight into which door has a car.
Conversely, one-third of the time, our initial pick will be correct so Monty can then open any of the two remaining two doors. Switching in that case will then lead us to the wrong door. The good news is, that only happens one-third of the time!
edited Feb 3 '16 at 4:11
Greek - Area 51 Proposal
3,206769105
answered Jan 14 '16 at 23:28
Kelvin SohKelvin Soh
1,645714
edited Feb 3 '16 at 4:11
Greek - Area 51 Proposal
3,206769105
edited Feb 3 '16 at 4:11
Greek - Area 51 Proposal
3,206769105
edited Feb 3 '16 at 4:11
Greek - Area 51 Proposal
3,206769105
3,206769105
answered Jan 14 '16 at 23:28
Kelvin SohKelvin Soh
1,645714
answered Jan 14 '16 at 23:28
Kelvin SohKelvin Soh
1,645714
answered Jan 14 '16 at 23:28
Kelvin SohKelvin Soh
1,645714
1,645714
add a comment |
add a comment |
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Your error is a common one: You have confused the event that the host reveals door $C$ with the event that there is a goat behind door $C$. This is distinction enough to see: if the car is, in fact, behind door $A$, then the host will only reveal door $C$ with a probability of $50%$, whereas there is definitely a goat behind door $C$.
In particular, let $C_{text{goat}}$ be the event that there's a goat behind door $C$, and $C_{text{revealed}}$ be the event that door $C$ is revealed, and $A_{text{car}}$ be the event that there is a car behind door $A$. Then, one can see that $A_{text{car}}$ and $C_{text{revealed}}$ are independent of each other, since the host reveals door $B$ or $C$ randomly with $50%$ probability if there's a car behind door $A$, and if there's not a car behind door $A$, the host is forced to pick the door which hasn't got a car behind it - which is $B$ or $C$ with probability $50%$. Thus, one concludes that
$$P(A_{text{car}}|C_{text{revealed}})=P(A_{text{car}})=frac{1}3.$$
The calculation you've done (annotated with a $neq$ sign where things go wrong) is:
$$P(A_{text{car}}|C_{text{revealed}})neq P(A_{text{car}}|C_{text{goat}})=frac{1}2$$
answered Jan 14 '16 at 23:33
Milo BrandtMilo Brandt
40.1k476140
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Well, this was one one the most convincing argument that I have seen so far. Thank you.
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– onurcanbektas
Jul 20 '18 at 2:48
add a comment |
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Your error is a common one: You have confused the event that the host reveals door $C$ with the event that there is a goat behind door $C$. This is distinction enough to see: if the car is, in fact, behind door $A$, then the host will only reveal door $C$ with a probability of $50%$, whereas there is definitely a goat behind door $C$.
In particular, let $C_{text{goat}}$ be the event that there's a goat behind door $C$, and $C_{text{revealed}}$ be the event that door $C$ is revealed, and $A_{text{car}}$ be the event that there is a car behind door $A$. Then, one can see that $A_{text{car}}$ and $C_{text{revealed}}$ are independent of each other, since the host reveals door $B$ or $C$ randomly with $50%$ probability if there's a car behind door $A$, and if there's not a car behind door $A$, the host is forced to pick the door which hasn't got a car behind it - which is $B$ or $C$ with probability $50%$. Thus, one concludes that
$$P(A_{text{car}}|C_{text{revealed}})=P(A_{text{car}})=frac{1}3.$$
The calculation you've done (annotated with a $neq$ sign where things go wrong) is:
$$P(A_{text{car}}|C_{text{revealed}})neq P(A_{text{car}}|C_{text{goat}})=frac{1}2$$
answered Jan 14 '16 at 23:33
Milo BrandtMilo Brandt
40.1k476140
$endgroup$
$begingroup$
Well, this was one one the most convincing argument that I have seen so far. Thank you.
$endgroup$
– onurcanbektas
Jul 20 '18 at 2:48
add a comment |
$begingroup$
Your error is a common one: You have confused the event that the host reveals door $C$ with the event that there is a goat behind door $C$. This is distinction enough to see: if the car is, in fact, behind door $A$, then the host will only reveal door $C$ with a probability of $50%$, whereas there is definitely a goat behind door $C$.
In particular, let $C_{text{goat}}$ be the event that there's a goat behind door $C$, and $C_{text{revealed}}$ be the event that door $C$ is revealed, and $A_{text{car}}$ be the event that there is a car behind door $A$. Then, one can see that $A_{text{car}}$ and $C_{text{revealed}}$ are independent of each other, since the host reveals door $B$ or $C$ randomly with $50%$ probability if there's a car behind door $A$, and if there's not a car behind door $A$, the host is forced to pick the door which hasn't got a car behind it - which is $B$ or $C$ with probability $50%$. Thus, one concludes that
$$P(A_{text{car}}|C_{text{revealed}})=P(A_{text{car}})=frac{1}3.$$
The calculation you've done (annotated with a $neq$ sign where things go wrong) is:
$$P(A_{text{car}}|C_{text{revealed}})neq P(A_{text{car}}|C_{text{goat}})=frac{1}2$$
answered Jan 14 '16 at 23:33
Milo BrandtMilo Brandt
40.1k476140
$endgroup$
Your error is a common one: You have confused the event that the host reveals door $C$ with the event that there is a goat behind door $C$. This is distinction enough to see: if the car is, in fact, behind door $A$, then the host will only reveal door $C$ with a probability of $50%$, whereas there is definitely a goat behind door $C$.
In particular, let $C_{text{goat}}$ be the event that there's a goat behind door $C$, and $C_{text{revealed}}$ be the event that door $C$ is revealed, and $A_{text{car}}$ be the event that there is a car behind door $A$. Then, one can see that $A_{text{car}}$ and $C_{text{revealed}}$ are independent of each other, since the host reveals door $B$ or $C$ randomly with $50%$ probability if there's a car behind door $A$, and if there's not a car behind door $A$, the host is forced to pick the door which hasn't got a car behind it - which is $B$ or $C$ with probability $50%$. Thus, one concludes that
$$P(A_{text{car}}|C_{text{revealed}})=P(A_{text{car}})=frac{1}3.$$
The calculation you've done (annotated with a $neq$ sign where things go wrong) is:
$$P(A_{text{car}}|C_{text{revealed}})neq P(A_{text{car}}|C_{text{goat}})=frac{1}2$$
answered Jan 14 '16 at 23:33
Milo BrandtMilo Brandt
40.1k476140
answered Jan 14 '16 at 23:33
Milo BrandtMilo Brandt
40.1k476140
answered Jan 14 '16 at 23:33
Milo BrandtMilo Brandt
40.1k476140
answered Jan 14 '16 at 23:33
Milo BrandtMilo Brandt
40.1k476140
40.1k476140
$begingroup$
Well, this was one one the most convincing argument that I have seen so far. Thank you.
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– onurcanbektas
Jul 20 '18 at 2:48
add a comment |
$begingroup$
Well, this was one one the most convincing argument that I have seen so far. Thank you.
$endgroup$
– onurcanbektas
Jul 20 '18 at 2:48
$begingroup$
Well, this was one one the most convincing argument that I have seen so far. Thank you.
$endgroup$
– onurcanbektas
Jul 20 '18 at 2:48
$begingroup$
Well, this was one one the most convincing argument that I have seen so far. Thank you.
$endgroup$
– onurcanbektas
Jul 20 '18 at 2:48
add a comment |
$begingroup$
Let me try to guess your birthday. My guess is April $2$nd.
Ignoring leap years, remove $363$ dates that are not your birthday and that are not April $2$nd (because I chose that), so that we are now left with only two options. You offer me to switch to the date that I didn't pick, which I do.
That I was right in my initial guess is unaffected by the dates that you subsequently remove.
answered Jan 14 '16 at 23:23
MankindMankind
10.4k72444
$endgroup$
$begingroup$
What is we see "switching door" as a "second guess"? If I give you two dates, and one is my bday, then you have a 50/50 chance, right? Then how is it different? Not switching = guessing 4/2, and switching = guess the other one.
$endgroup$
– OneZero
Jan 14 '16 at 23:25
$begingroup$
@OneZero Yes, switching doors corresponds to a second guess. If you give me only two dates, with no prior choices, then it's $50/50$. But you give me $365$ dates, and tell me to choose one of them. The probability that I'm right is $1/365$. You then eliminate $363$ possibilities, leaving me with $2$ choices, one of which is my original pick. Just because I now have only two choices left doesn't mean it's $50/50$, and the reason is that I made my choice back when there were $365$ options.
$endgroup$
– Mankind
Jan 14 '16 at 23:32
$begingroup$
Very nice way to put it. If someone isn't convinced, just try it 10 times with random friends and see how much you'll be right :-D
$endgroup$
– Ant
Apr 8 '16 at 7:04
add a comment |
$begingroup$
Let me try to guess your birthday. My guess is April $2$nd.
Ignoring leap years, remove $363$ dates that are not your birthday and that are not April $2$nd (because I chose that), so that we are now left with only two options. You offer me to switch to the date that I didn't pick, which I do.
That I was right in my initial guess is unaffected by the dates that you subsequently remove.
answered Jan 14 '16 at 23:23
MankindMankind
10.4k72444
$endgroup$
$begingroup$
What is we see "switching door" as a "second guess"? If I give you two dates, and one is my bday, then you have a 50/50 chance, right? Then how is it different? Not switching = guessing 4/2, and switching = guess the other one.
$endgroup$
– OneZero
Jan 14 '16 at 23:25
$begingroup$
@OneZero Yes, switching doors corresponds to a second guess. If you give me only two dates, with no prior choices, then it's $50/50$. But you give me $365$ dates, and tell me to choose one of them. The probability that I'm right is $1/365$. You then eliminate $363$ possibilities, leaving me with $2$ choices, one of which is my original pick. Just because I now have only two choices left doesn't mean it's $50/50$, and the reason is that I made my choice back when there were $365$ options.
$endgroup$
– Mankind
Jan 14 '16 at 23:32
$begingroup$
Very nice way to put it. If someone isn't convinced, just try it 10 times with random friends and see how much you'll be right :-D
$endgroup$
– Ant
Apr 8 '16 at 7:04
add a comment |
$begingroup$
Let me try to guess your birthday. My guess is April $2$nd.
Ignoring leap years, remove $363$ dates that are not your birthday and that are not April $2$nd (because I chose that), so that we are now left with only two options. You offer me to switch to the date that I didn't pick, which I do.
That I was right in my initial guess is unaffected by the dates that you subsequently remove.
answered Jan 14 '16 at 23:23
MankindMankind
10.4k72444
$endgroup$
Let me try to guess your birthday. My guess is April $2$nd.
Ignoring leap years, remove $363$ dates that are not your birthday and that are not April $2$nd (because I chose that), so that we are now left with only two options. You offer me to switch to the date that I didn't pick, which I do.
That I was right in my initial guess is unaffected by the dates that you subsequently remove.
answered Jan 14 '16 at 23:23
MankindMankind
10.4k72444
answered Jan 14 '16 at 23:23
MankindMankind
10.4k72444
answered Jan 14 '16 at 23:23
MankindMankind
10.4k72444
answered Jan 14 '16 at 23:23
MankindMankind
10.4k72444
10.4k72444
$begingroup$
What is we see "switching door" as a "second guess"? If I give you two dates, and one is my bday, then you have a 50/50 chance, right? Then how is it different? Not switching = guessing 4/2, and switching = guess the other one.
$endgroup$
– OneZero
Jan 14 '16 at 23:25
$begingroup$
@OneZero Yes, switching doors corresponds to a second guess. If you give me only two dates, with no prior choices, then it's $50/50$. But you give me $365$ dates, and tell me to choose one of them. The probability that I'm right is $1/365$. You then eliminate $363$ possibilities, leaving me with $2$ choices, one of which is my original pick. Just because I now have only two choices left doesn't mean it's $50/50$, and the reason is that I made my choice back when there were $365$ options.
$endgroup$
– Mankind
Jan 14 '16 at 23:32
$begingroup$
Very nice way to put it. If someone isn't convinced, just try it 10 times with random friends and see how much you'll be right :-D
$endgroup$
– Ant
Apr 8 '16 at 7:04
add a comment |
$begingroup$
What is we see "switching door" as a "second guess"? If I give you two dates, and one is my bday, then you have a 50/50 chance, right? Then how is it different? Not switching = guessing 4/2, and switching = guess the other one.
$endgroup$
– OneZero
Jan 14 '16 at 23:25
$begingroup$
@OneZero Yes, switching doors corresponds to a second guess. If you give me only two dates, with no prior choices, then it's $50/50$. But you give me $365$ dates, and tell me to choose one of them. The probability that I'm right is $1/365$. You then eliminate $363$ possibilities, leaving me with $2$ choices, one of which is my original pick. Just because I now have only two choices left doesn't mean it's $50/50$, and the reason is that I made my choice back when there were $365$ options.
$endgroup$
– Mankind
Jan 14 '16 at 23:32
$begingroup$
Very nice way to put it. If someone isn't convinced, just try it 10 times with random friends and see how much you'll be right :-D
$endgroup$
– Ant
Apr 8 '16 at 7:04
$begingroup$
What is we see "switching door" as a "second guess"? If I give you two dates, and one is my bday, then you have a 50/50 chance, right? Then how is it different? Not switching = guessing 4/2, and switching = guess the other one.
$endgroup$
– OneZero
Jan 14 '16 at 23:25
$begingroup$
What is we see "switching door" as a "second guess"? If I give you two dates, and one is my bday, then you have a 50/50 chance, right? Then how is it different? Not switching = guessing 4/2, and switching = guess the other one.
$endgroup$
– OneZero
Jan 14 '16 at 23:25
$begingroup$
@OneZero Yes, switching doors corresponds to a second guess. If you give me only two dates, with no prior choices, then it's $50/50$. But you give me $365$ dates, and tell me to choose one of them. The probability that I'm right is $1/365$. You then eliminate $363$ possibilities, leaving me with $2$ choices, one of which is my original pick. Just because I now have only two choices left doesn't mean it's $50/50$, and the reason is that I made my choice back when there were $365$ options.
$endgroup$
– Mankind
Jan 14 '16 at 23:32
$begingroup$
@OneZero Yes, switching doors corresponds to a second guess. If you give me only two dates, with no prior choices, then it's $50/50$. But you give me $365$ dates, and tell me to choose one of them. The probability that I'm right is $1/365$. You then eliminate $363$ possibilities, leaving me with $2$ choices, one of which is my original pick. Just because I now have only two choices left doesn't mean it's $50/50$, and the reason is that I made my choice back when there were $365$ options.
$endgroup$
– Mankind
Jan 14 '16 at 23:32
$begingroup$
Very nice way to put it. If someone isn't convinced, just try it 10 times with random friends and see how much you'll be right :-D
$endgroup$
– Ant
Apr 8 '16 at 7:04
$begingroup$
Very nice way to put it. If someone isn't convinced, just try it 10 times with random friends and see how much you'll be right :-D
$endgroup$
– Ant
Apr 8 '16 at 7:04
add a comment |
$begingroup$
The flaw in your thought process is that in the $50:50$ case with only two doors where one is a goat and one is a car, that is the only information you have. In the other case where you have two goat doors and one car door, when you open a goat door, you gain some information by doing this. Thus you appear to have the same $50:50$ situation, but you have some added information.
answered Jan 14 '16 at 23:25
flabby99flabby99
14211
$endgroup$
$begingroup$
Can you elaborate on this? What's added?
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– OneZero
Jan 14 '16 at 23:26
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Have you seen an actual proof/ working out of the probabilities of the monty hall problem using bayes law and conditional probability?
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– flabby99
Jan 14 '16 at 23:29
$begingroup$
Also I think it might potentially be helpful to imagine a situation where there is 100 doors, 99 of them have goats and 1 of them has a car. You choose a door. Monty opens 98 that have goats, and asks should you switch?
$endgroup$
– flabby99
Jan 14 '16 at 23:32
add a comment |
$begingroup$
The flaw in your thought process is that in the $50:50$ case with only two doors where one is a goat and one is a car, that is the only information you have. In the other case where you have two goat doors and one car door, when you open a goat door, you gain some information by doing this. Thus you appear to have the same $50:50$ situation, but you have some added information.
answered Jan 14 '16 at 23:25
flabby99flabby99
14211
$endgroup$
$begingroup$
Can you elaborate on this? What's added?
$endgroup$
– OneZero
Jan 14 '16 at 23:26
$begingroup$
Have you seen an actual proof/ working out of the probabilities of the monty hall problem using bayes law and conditional probability?
$endgroup$
– flabby99
Jan 14 '16 at 23:29
$begingroup$
Also I think it might potentially be helpful to imagine a situation where there is 100 doors, 99 of them have goats and 1 of them has a car. You choose a door. Monty opens 98 that have goats, and asks should you switch?
$endgroup$
– flabby99
Jan 14 '16 at 23:32
add a comment |
$begingroup$
The flaw in your thought process is that in the $50:50$ case with only two doors where one is a goat and one is a car, that is the only information you have. In the other case where you have two goat doors and one car door, when you open a goat door, you gain some information by doing this. Thus you appear to have the same $50:50$ situation, but you have some added information.
answered Jan 14 '16 at 23:25
flabby99flabby99
14211
$endgroup$
The flaw in your thought process is that in the $50:50$ case with only two doors where one is a goat and one is a car, that is the only information you have. In the other case where you have two goat doors and one car door, when you open a goat door, you gain some information by doing this. Thus you appear to have the same $50:50$ situation, but you have some added information.
answered Jan 14 '16 at 23:25
flabby99flabby99
14211
answered Jan 14 '16 at 23:25
flabby99flabby99
14211
answered Jan 14 '16 at 23:25
flabby99flabby99
14211
answered Jan 14 '16 at 23:25
flabby99flabby99
14211
14211
$begingroup$
Can you elaborate on this? What's added?
$endgroup$
– OneZero
Jan 14 '16 at 23:26
$begingroup$
Have you seen an actual proof/ working out of the probabilities of the monty hall problem using bayes law and conditional probability?
$endgroup$
– flabby99
Jan 14 '16 at 23:29
$begingroup$
Also I think it might potentially be helpful to imagine a situation where there is 100 doors, 99 of them have goats and 1 of them has a car. You choose a door. Monty opens 98 that have goats, and asks should you switch?
$endgroup$
– flabby99
Jan 14 '16 at 23:32
add a comment |
$begingroup$
Can you elaborate on this? What's added?
$endgroup$
– OneZero
Jan 14 '16 at 23:26
$begingroup$
Have you seen an actual proof/ working out of the probabilities of the monty hall problem using bayes law and conditional probability?
$endgroup$
– flabby99
Jan 14 '16 at 23:29
$begingroup$
Also I think it might potentially be helpful to imagine a situation where there is 100 doors, 99 of them have goats and 1 of them has a car. You choose a door. Monty opens 98 that have goats, and asks should you switch?
$endgroup$
– flabby99
Jan 14 '16 at 23:32
$begingroup$
Can you elaborate on this? What's added?
$endgroup$
– OneZero
Jan 14 '16 at 23:26
$begingroup$
Can you elaborate on this? What's added?
$endgroup$
– OneZero
Jan 14 '16 at 23:26
$begingroup$
Have you seen an actual proof/ working out of the probabilities of the monty hall problem using bayes law and conditional probability?
$endgroup$
– flabby99
Jan 14 '16 at 23:29
$begingroup$
Have you seen an actual proof/ working out of the probabilities of the monty hall problem using bayes law and conditional probability?
$endgroup$
– flabby99
Jan 14 '16 at 23:29
$begingroup$
Also I think it might potentially be helpful to imagine a situation where there is 100 doors, 99 of them have goats and 1 of them has a car. You choose a door. Monty opens 98 that have goats, and asks should you switch?
$endgroup$
– flabby99
Jan 14 '16 at 23:32
$begingroup$
Also I think it might potentially be helpful to imagine a situation where there is 100 doors, 99 of them have goats and 1 of them has a car. You choose a door. Monty opens 98 that have goats, and asks should you switch?
$endgroup$
– flabby99
Jan 14 '16 at 23:32
add a comment |
$begingroup$
The idea of assigning $50/50$ probabilities to two alternatives
makes sense only when the two alternatives are truly symmetric.
If you just start with two doors, you know there is a car behind
exactly one, and nothing else has happened concerning those doors,
then $50/50$ probabilities make sense.
But once you have chosen one of three doors and Monty has opened
another door (having been forced by the rules of the game to open
a door that wasn't yours and didn't have the car),
you no longer have a symmetric situation. One of the remaining
doors is an available choice because you already chose it once;
the other is an available choice either (A) because you already
chose the car and Monty got to take one of the "goat" doors away at random,
or (B) because you chose a goat initially and Monty had to take away
the remaining "goat" door, leaving the door with the car.
Case (B) is twice as likely to occur as case (A), not $50/50$,
and which door has the car is now completely a function of whether
you're in case (A) or case (B), so it's also not $50/50$.
answered Apr 8 '16 at 6:59
David KDavid K
55.7k345121
$endgroup$
add a comment |
$begingroup$
The idea of assigning $50/50$ probabilities to two alternatives
makes sense only when the two alternatives are truly symmetric.
If you just start with two doors, you know there is a car behind
exactly one, and nothing else has happened concerning those doors,
then $50/50$ probabilities make sense.
But once you have chosen one of three doors and Monty has opened
another door (having been forced by the rules of the game to open
a door that wasn't yours and didn't have the car),
you no longer have a symmetric situation. One of the remaining
doors is an available choice because you already chose it once;
the other is an available choice either (A) because you already
chose the car and Monty got to take one of the "goat" doors away at random,
or (B) because you chose a goat initially and Monty had to take away
the remaining "goat" door, leaving the door with the car.
Case (B) is twice as likely to occur as case (A), not $50/50$,
and which door has the car is now completely a function of whether
you're in case (A) or case (B), so it's also not $50/50$.
answered Apr 8 '16 at 6:59
David KDavid K
55.7k345121
$endgroup$
add a comment |
$begingroup$
The idea of assigning $50/50$ probabilities to two alternatives
makes sense only when the two alternatives are truly symmetric.
If you just start with two doors, you know there is a car behind
exactly one, and nothing else has happened concerning those doors,
then $50/50$ probabilities make sense.
But once you have chosen one of three doors and Monty has opened
another door (having been forced by the rules of the game to open
a door that wasn't yours and didn't have the car),
you no longer have a symmetric situation. One of the remaining
doors is an available choice because you already chose it once;
the other is an available choice either (A) because you already
chose the car and Monty got to take one of the "goat" doors away at random,
or (B) because you chose a goat initially and Monty had to take away
the remaining "goat" door, leaving the door with the car.
Case (B) is twice as likely to occur as case (A), not $50/50$,
and which door has the car is now completely a function of whether
you're in case (A) or case (B), so it's also not $50/50$.
answered Apr 8 '16 at 6:59
David KDavid K
55.7k345121
$endgroup$
The idea of assigning $50/50$ probabilities to two alternatives
makes sense only when the two alternatives are truly symmetric.
If you just start with two doors, you know there is a car behind
exactly one, and nothing else has happened concerning those doors,
then $50/50$ probabilities make sense.
But once you have chosen one of three doors and Monty has opened
another door (having been forced by the rules of the game to open
a door that wasn't yours and didn't have the car),
you no longer have a symmetric situation. One of the remaining
doors is an available choice because you already chose it once;
the other is an available choice either (A) because you already
chose the car and Monty got to take one of the "goat" doors away at random,
or (B) because you chose a goat initially and Monty had to take away
the remaining "goat" door, leaving the door with the car.
Case (B) is twice as likely to occur as case (A), not $50/50$,
and which door has the car is now completely a function of whether
you're in case (A) or case (B), so it's also not $50/50$.
answered Apr 8 '16 at 6:59
David KDavid K
55.7k345121
answered Apr 8 '16 at 6:59
David KDavid K
55.7k345121
answered Apr 8 '16 at 6:59
David KDavid K
55.7k345121
answered Apr 8 '16 at 6:59
David KDavid K
55.7k345121
55.7k345121
add a comment |
add a comment |
$begingroup$
The error in your argument is that you don't start with two doors, you start with three.
If you pick one door out of three, you have a one in three chance of having picked the right one ($tfrac{1}{3}$), and a two in three chance of having picked the wrong one ($tfrac{2}{3}$).
If you have picked the wrong door, Monty shows you which door has the car behind it, by opening the other door. You win a car by switching.
Conversely, if you have picked the right door, switching makes you lose the car.
Since there's a $tfrac{2}{3}$ chance of picking the wrong door at first, there's a $tfrac{2}{3}$ of winning the car after switching. The chance of picking the right door is $tfrac{1}{3}$, so there's a $tfrac{1}{3}$ chance of losing the car after switching.
answered Jul 22 '16 at 8:43
SQBSQB
1,75611026
$endgroup$
add a comment |
$begingroup$
The error in your argument is that you don't start with two doors, you start with three.
If you pick one door out of three, you have a one in three chance of having picked the right one ($tfrac{1}{3}$), and a two in three chance of having picked the wrong one ($tfrac{2}{3}$).
If you have picked the wrong door, Monty shows you which door has the car behind it, by opening the other door. You win a car by switching.
Conversely, if you have picked the right door, switching makes you lose the car.
Since there's a $tfrac{2}{3}$ chance of picking the wrong door at first, there's a $tfrac{2}{3}$ of winning the car after switching. The chance of picking the right door is $tfrac{1}{3}$, so there's a $tfrac{1}{3}$ chance of losing the car after switching.
answered Jul 22 '16 at 8:43
SQBSQB
1,75611026
$endgroup$
add a comment |
$begingroup$
The error in your argument is that you don't start with two doors, you start with three.
If you pick one door out of three, you have a one in three chance of having picked the right one ($tfrac{1}{3}$), and a two in three chance of having picked the wrong one ($tfrac{2}{3}$).
If you have picked the wrong door, Monty shows you which door has the car behind it, by opening the other door. You win a car by switching.
Conversely, if you have picked the right door, switching makes you lose the car.
Since there's a $tfrac{2}{3}$ chance of picking the wrong door at first, there's a $tfrac{2}{3}$ of winning the car after switching. The chance of picking the right door is $tfrac{1}{3}$, so there's a $tfrac{1}{3}$ chance of losing the car after switching.
answered Jul 22 '16 at 8:43
SQBSQB
1,75611026
$endgroup$
The error in your argument is that you don't start with two doors, you start with three.
If you pick one door out of three, you have a one in three chance of having picked the right one ($tfrac{1}{3}$), and a two in three chance of having picked the wrong one ($tfrac{2}{3}$).
If you have picked the wrong door, Monty shows you which door has the car behind it, by opening the other door. You win a car by switching.
Conversely, if you have picked the right door, switching makes you lose the car.
Since there's a $tfrac{2}{3}$ chance of picking the wrong door at first, there's a $tfrac{2}{3}$ of winning the car after switching. The chance of picking the right door is $tfrac{1}{3}$, so there's a $tfrac{1}{3}$ chance of losing the car after switching.
answered Jul 22 '16 at 8:43
SQBSQB
1,75611026
answered Jul 22 '16 at 8:43
SQBSQB
1,75611026
answered Jul 22 '16 at 8:43
SQBSQB
1,75611026
answered Jul 22 '16 at 8:43
SQBSQB
1,75611026
1,75611026
add a comment |
add a comment |
$begingroup$
example 1 which shows there would be no advantage in shifting: lets say there are one million doors and I pick No 1. Monty reveals 999998 doors which are empty so now we have only 2 doors, one that I picked and another one. According to the paradox the possibility would be 1/999999 = almost 100% certain that the shift would have advantage. Obviously wrong.
Convinced?
example 2 which shows there would be no advantage in shifting:
Lets say instead of one player we have 2 players and one of them choose door one and another choose door 2 and Monty opens door three. If the argument is true shifting would increase the chance for both players. Obviously not possible.
Convinced?
answered Jun 28 '17 at 12:27
user7432810user7432810
112
$endgroup$
add a comment |
$begingroup$
example 1 which shows there would be no advantage in shifting: lets say there are one million doors and I pick No 1. Monty reveals 999998 doors which are empty so now we have only 2 doors, one that I picked and another one. According to the paradox the possibility would be 1/999999 = almost 100% certain that the shift would have advantage. Obviously wrong.
Convinced?
example 2 which shows there would be no advantage in shifting:
Lets say instead of one player we have 2 players and one of them choose door one and another choose door 2 and Monty opens door three. If the argument is true shifting would increase the chance for both players. Obviously not possible.
Convinced?
answered Jun 28 '17 at 12:27
user7432810user7432810
112
$endgroup$
add a comment |
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example 1 which shows there would be no advantage in shifting: lets say there are one million doors and I pick No 1. Monty reveals 999998 doors which are empty so now we have only 2 doors, one that I picked and another one. According to the paradox the possibility would be 1/999999 = almost 100% certain that the shift would have advantage. Obviously wrong.
Convinced?
example 2 which shows there would be no advantage in shifting:
Lets say instead of one player we have 2 players and one of them choose door one and another choose door 2 and Monty opens door three. If the argument is true shifting would increase the chance for both players. Obviously not possible.
Convinced?
answered Jun 28 '17 at 12:27
user7432810user7432810
112
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example 1 which shows there would be no advantage in shifting: lets say there are one million doors and I pick No 1. Monty reveals 999998 doors which are empty so now we have only 2 doors, one that I picked and another one. According to the paradox the possibility would be 1/999999 = almost 100% certain that the shift would have advantage. Obviously wrong.
Convinced?
example 2 which shows there would be no advantage in shifting:
Lets say instead of one player we have 2 players and one of them choose door one and another choose door 2 and Monty opens door three. If the argument is true shifting would increase the chance for both players. Obviously not possible.
Convinced?
answered Jun 28 '17 at 12:27
user7432810user7432810
112
edited Jun 29 '17 at 11:56
answered Jun 28 '17 at 12:27
user7432810user7432810
112
answered Jun 28 '17 at 12:27
user7432810user7432810
112
answered Jun 28 '17 at 12:27
user7432810user7432810
112
112
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I'm amazed people still argue this.
Example 1 yes you would have an extremely high chance of winning if you selected the door that was filtered out 999,999 doors. If not, you are basically arguing that Door 1 had a 50/50 shot at winning a contest that had 1 out of 1,000,000 odds. The fact the game show host had to filter the losing 999,998 doors out of the set makes switching advantageous since that door is weighted by the 999,998 losing doors that were filtered out.
Example 2. This example isn't applicable to the stated problem. When you introduce a 2nd player the host no longer can filter 2 doors down to 1 door unless both players choose the same door. If they choose the same door then it's the same classic Monty Hall problem just 2 iterations happening simultaneously. If 2 players select different doors then Monty has to open the 3rd unchosen door. According to you, that is 50/50 right? This is your premise but you're forgetting another option.
What you're forgetting is if neither player selected the correct door. In this case, Monty would have to Open player 1 or 2 door. This would then force that particular player to move as he now knows for a fact he will lose without switching. The player who's door was opened would switch obviously but he now has a 50/50 shot instead of 1/3. The player who's door wasn't opened should switch as this had no impact to him other than the typical Monty hall problem, i.e he now has 2/3 chance.
answered Jun 19 '18 at 1:14
BrianBrian
1
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add a comment |
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I'm amazed people still argue this.
Example 1 yes you would have an extremely high chance of winning if you selected the door that was filtered out 999,999 doors. If not, you are basically arguing that Door 1 had a 50/50 shot at winning a contest that had 1 out of 1,000,000 odds. The fact the game show host had to filter the losing 999,998 doors out of the set makes switching advantageous since that door is weighted by the 999,998 losing doors that were filtered out.
Example 2. This example isn't applicable to the stated problem. When you introduce a 2nd player the host no longer can filter 2 doors down to 1 door unless both players choose the same door. If they choose the same door then it's the same classic Monty Hall problem just 2 iterations happening simultaneously. If 2 players select different doors then Monty has to open the 3rd unchosen door. According to you, that is 50/50 right? This is your premise but you're forgetting another option.
What you're forgetting is if neither player selected the correct door. In this case, Monty would have to Open player 1 or 2 door. This would then force that particular player to move as he now knows for a fact he will lose without switching. The player who's door was opened would switch obviously but he now has a 50/50 shot instead of 1/3. The player who's door wasn't opened should switch as this had no impact to him other than the typical Monty hall problem, i.e he now has 2/3 chance.
answered Jun 19 '18 at 1:14
BrianBrian
1
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add a comment |
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I'm amazed people still argue this.
Example 1 yes you would have an extremely high chance of winning if you selected the door that was filtered out 999,999 doors. If not, you are basically arguing that Door 1 had a 50/50 shot at winning a contest that had 1 out of 1,000,000 odds. The fact the game show host had to filter the losing 999,998 doors out of the set makes switching advantageous since that door is weighted by the 999,998 losing doors that were filtered out.
Example 2. This example isn't applicable to the stated problem. When you introduce a 2nd player the host no longer can filter 2 doors down to 1 door unless both players choose the same door. If they choose the same door then it's the same classic Monty Hall problem just 2 iterations happening simultaneously. If 2 players select different doors then Monty has to open the 3rd unchosen door. According to you, that is 50/50 right? This is your premise but you're forgetting another option.
What you're forgetting is if neither player selected the correct door. In this case, Monty would have to Open player 1 or 2 door. This would then force that particular player to move as he now knows for a fact he will lose without switching. The player who's door was opened would switch obviously but he now has a 50/50 shot instead of 1/3. The player who's door wasn't opened should switch as this had no impact to him other than the typical Monty hall problem, i.e he now has 2/3 chance.
answered Jun 19 '18 at 1:14
BrianBrian
1
$endgroup$
I'm amazed people still argue this.
Example 1 yes you would have an extremely high chance of winning if you selected the door that was filtered out 999,999 doors. If not, you are basically arguing that Door 1 had a 50/50 shot at winning a contest that had 1 out of 1,000,000 odds. The fact the game show host had to filter the losing 999,998 doors out of the set makes switching advantageous since that door is weighted by the 999,998 losing doors that were filtered out.
Example 2. This example isn't applicable to the stated problem. When you introduce a 2nd player the host no longer can filter 2 doors down to 1 door unless both players choose the same door. If they choose the same door then it's the same classic Monty Hall problem just 2 iterations happening simultaneously. If 2 players select different doors then Monty has to open the 3rd unchosen door. According to you, that is 50/50 right? This is your premise but you're forgetting another option.
What you're forgetting is if neither player selected the correct door. In this case, Monty would have to Open player 1 or 2 door. This would then force that particular player to move as he now knows for a fact he will lose without switching. The player who's door was opened would switch obviously but he now has a 50/50 shot instead of 1/3. The player who's door wasn't opened should switch as this had no impact to him other than the typical Monty hall problem, i.e he now has 2/3 chance.
answered Jun 19 '18 at 1:14
BrianBrian
1
answered Jun 19 '18 at 1:14
BrianBrian
1
answered Jun 19 '18 at 1:14
BrianBrian
1
answered Jun 19 '18 at 1:14
BrianBrian
1
1
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What's wrong with the process (and what's really behind the whole Monty Hall paradox) is the fact that revealing a goat doesn't change the fact that there's a $1/3$ chance of $A$ containing a car. No matter what door you pick, there will always be at least one goat door you didn't pick, so by just having that fact confirmed (by revealing a goat), you haven't really recieved any information that changes anything about the door you picked.
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– Arthur
Jan 14 '16 at 23:19
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"If we have two doors, and know a car is behind one, then we have a 50/50 chance of choosing the right door." True, but only if that is all you know. If you have two doors, and one door is on a tiny doll house, and one is on a full size garage, you know the chance is not 50/50. If both doors were clear glass, again the chance is not 50/50. There was information used in the Monty Hall problem to reduce down to the set of two doors, and although now you only see two, one piece of information you have is that Monty had full knowledge of what was happening, and he would always open a goat door.
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– piCookie
Nov 7 '16 at 20:42