"Monty Hall" problem revisited
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bw
Idéfix
MaxEntropy_Man
Seva Lamberdar
8 posters
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Re: "Monty Hall" problem revisited
Thanks to all the guys who participated in this very interesting discussion and posted their comments here. Seva
Re: "Monty Hall" problem revisited
As Seva correctly stated there are 4 distinct possibilities, 2 leading to a win by switching and 2 leading to a loss by switching  but not all possibilities are necessarily equally likely.
In order to arrive at the correct solution you need to assign probabilities to each possibility. So, assuming the prize is behind Door C then:
(i) Contestant Picks Door A (Prob. = 1/3) and host has to open Door B (Prob. = 1). Probability of this result occurring is 1/3 * 1 = 1/3. Switching WINS
(ii) Contestant Picks Door B (Prob. = 1/3) and host has to open Door A (Prob. = 1). Probability of this result occurring is 1/3 * 1 = 1/3. Switching WINS
(iii) Contestant Picks Door C (Prob. = 1/3) and host opens Door A (Prob. = 1/2). Probability of this result occurring is 1/3 * 1/2 = 1/6. Switching LOSES
(iv) Contestant Picks Door C (Prob. = 1/3) and host opens Door B (Prob. = 1/2). Probability of this result occurring is 1/3 * 1/2 = 1/6. Switching LOSES
So adding them all up, switching wins with a total probability of 2/3, and loses with a total probability of 1/3. QED
And if you're wondering why the probability of the host opening Door A in case 3 (and Door B in case 4) is 1/2, it is because he he is equally likely to open either, and makes a random choice as to which one to open
In order to arrive at the correct solution you need to assign probabilities to each possibility. So, assuming the prize is behind Door C then:
(i) Contestant Picks Door A (Prob. = 1/3) and host has to open Door B (Prob. = 1). Probability of this result occurring is 1/3 * 1 = 1/3. Switching WINS
(ii) Contestant Picks Door B (Prob. = 1/3) and host has to open Door A (Prob. = 1). Probability of this result occurring is 1/3 * 1 = 1/3. Switching WINS
(iii) Contestant Picks Door C (Prob. = 1/3) and host opens Door A (Prob. = 1/2). Probability of this result occurring is 1/3 * 1/2 = 1/6. Switching LOSES
(iv) Contestant Picks Door C (Prob. = 1/3) and host opens Door B (Prob. = 1/2). Probability of this result occurring is 1/3 * 1/2 = 1/6. Switching LOSES
So adding them all up, switching wins with a total probability of 2/3, and loses with a total probability of 1/3. QED
And if you're wondering why the probability of the host opening Door A in case 3 (and Door B in case 4) is 1/2, it is because he he is equally likely to open either, and makes a random choice as to which one to open
Last edited by PalmerEldritch on Thu Apr 18, 2013 4:12 am; edited 1 time in total (Reason for editing : Formatting)
PalmerEldritch Posts : 13
Join date : 20130418
Re: "Monty Hall" problem revisited
Hello bw, wellspotted, I am a Phillip Dick fan.
PalmerEldritch Posts : 13
Join date : 20130418
Re: "Monty Hall" problem revisited
PalmerEldritch wrote:Hello bw, wellspotted, I am a Phillip Dick fan.
ah, you read that post before i deleted it.
nice  you new here or someone with a new handle? if you are new, welcome, hope you enjoy the stay etc.
bw Posts : 2922
Join date : 20121115
Re: "Monty Hall" problem revisited
PalmerEldritch wrote:As Seva correctly stated there are 4 distinct possibilities, 2 leading to a win by switching and 2 leading to a loss by switching  but not all possibilities are necessarily equally likely.
In order to arrive at the correct solution you need to assign probabilities to each possibility. So, assuming the prize is behind Door C then:
(i) Contestant Picks Door A (Prob. = 1/3) and host has to open Door B (Prob. = 1). Probability of this result occurring is 1/3 * 1 = 1/3. Switching WINS
(ii) Contestant Picks Door B (Prob. = 1/3) and host has to open Door A (Prob. = 1). Probability of this result occurring is 1/3 * 1 = 1/3. Switching WINS
(iii) Contestant Picks Door C (Prob. = 1/3) and host opens Door A (Prob. = 1/2). Probability of this result occurring is 1/3 * 1/2 = 1/6. Switching LOSES
(iv) Contestant Picks Door C (Prob. = 1/3) and host opens Door B (Prob. = 1/2). Probability of this result occurring is 1/3 * 1/2 = 1/6. Switching LOSES
So adding them all up, switching wins with a total probability of 2/3, and loses with a total probability of 1/3. QED
And if you're wondering why the probability of the host opening Door A in case 3 (and Door B in case 4) is 1/2, it is because he he is equally likely to open either, and makes a random choice as to which one to open
In the above cases '(iii)' and '(iv)', the probability of host opening an empty door (A or B) is 1, since the host knows that both doors are empty (having neither the player nor the prize). The probability for the host in cases '(iii)' and '(iv)' is with respect to picking an empty door and not door A or B.
Btw try the other method (Let's look at this problem in another way.) I mentioned on page 1 earlier.
Re: "Monty Hall" problem revisited
Seva Lamberdar wrote:PalmerEldritch wrote:As Seva correctly stated there are 4 distinct possibilities, 2 leading to a win by switching and 2 leading to a loss by switching  but not all possibilities are necessarily equally likely.
In order to arrive at the correct solution you need to assign probabilities to each possibility. So, assuming the prize is behind Door C then:
(i) Contestant Picks Door A (Prob. = 1/3) and host has to open Door B (Prob. = 1). Probability of this result occurring is 1/3 * 1 = 1/3. Switching WINS
(ii) Contestant Picks Door B (Prob. = 1/3) and host has to open Door A (Prob. = 1). Probability of this result occurring is 1/3 * 1 = 1/3. Switching WINS
(iii) Contestant Picks Door C (Prob. = 1/3) and host opens Door A (Prob. = 1/2). Probability of this result occurring is 1/3 * 1/2 = 1/6. Switching LOSES
(iv) Contestant Picks Door C (Prob. = 1/3) and host opens Door B (Prob. = 1/2). Probability of this result occurring is 1/3 * 1/2 = 1/6. Switching LOSES
So adding them all up, switching wins with a total probability of 2/3, and loses with a total probability of 1/3. QED
And if you're wondering why the probability of the host opening Door A in case 3 (and Door B in case 4) is 1/2, it is because he he is equally likely to open either, and makes a random choice as to which one to open
In the above cases '(iii)' and '(iv)', the probability of host opening an empty door (A or B) is 1, since the host knows that both doors are empty (having neither the player nor the prize). The probability for the host in cases '(iii)' and '(iv)' is with respect to picking an empty door and not door A or B.
Btw try the other method (Let's look at this problem in another way.) I mentioned on page 1 earlier.
If you want to consider cases (iii) and (iv) as 1 case then the probability of the host opening an empty door (either A or B) is indeed 1, and the total probability becomes :
Cases (iii) AND (iv). The contestant picks Door C (Prob=1/3) AND the host opens either Door A or Door B (Prob = 1). Switching LOSES.
You get the same result  2/3 in favour of switching. It was you who wanted to consider them as separate cases (Door A in one case and Door B in the second case), and if you do that the Probability of the host opening Door A = 1/2 and the Probability of the host opening Door B = 1/2 (so that the Probability of the host opening EITHER Door A OR Door B = 1/2 + 1/2 = 1, as you add the individual probabilities together)
Last edited by PalmerEldritch on Thu Apr 18, 2013 6:19 am; edited 1 time in total (Reason for editing : clarification)
PalmerEldritch Posts : 13
Join date : 20130418
Re: "Monty Hall" problem revisited
PalmerEldritch wrote:Seva Lamberdar wrote:PalmerEldritch wrote:As Seva correctly stated there are 4 distinct possibilities, 2 leading to a win by switching and 2 leading to a loss by switching  but not all possibilities are necessarily equally likely.
In order to arrive at the correct solution you need to assign probabilities to each possibility. So, assuming the prize is behind Door C then:
(i) Contestant Picks Door A (Prob. = 1/3) and host has to open Door B (Prob. = 1). Probability of this result occurring is 1/3 * 1 = 1/3. Switching WINS
(ii) Contestant Picks Door B (Prob. = 1/3) and host has to open Door A (Prob. = 1). Probability of this result occurring is 1/3 * 1 = 1/3. Switching WINS
(iii) Contestant Picks Door C (Prob. = 1/3) and host opens Door A (Prob. = 1/2). Probability of this result occurring is 1/3 * 1/2 = 1/6. Switching LOSES
(iv) Contestant Picks Door C (Prob. = 1/3) and host opens Door B (Prob. = 1/2). Probability of this result occurring is 1/3 * 1/2 = 1/6. Switching LOSES
So adding them all up, switching wins with a total probability of 2/3, and loses with a total probability of 1/3. QED
And if you're wondering why the probability of the host opening Door A in case 3 (and Door B in case 4) is 1/2, it is because he he is equally likely to open either, and makes a random choice as to which one to open
In the above cases '(iii)' and '(iv)', the probability of host opening an empty door (A or B) is 1, since the host knows that both doors are empty (having neither the player nor the prize). The probability for the host in cases '(iii)' and '(iv)' is with respect to picking an empty door and not door A or B.
Btw try the other method (Let's look at this problem in another way.) I mentioned on page 1 earlier.
If you want to consider cases (iii) and (iv) as 1 case then the probability of the host opening an empty door (either A or B) is indeed 1, and the total probability becomes :
Cases (iii) AND (iv). The contestant picks Door C (Prob=1/3) AND the host opens either Door A or Door B (Prob = 1). Switching LOSES.
You get the same result  2/3 in favour of switching. It was you who wanted to consider them as separate cases (Door A in one case and Door B in the second case), and if you do that the Probability of the host opening Door A = 1/2 and the Probability of the host opening Door B = 1/2 (so that the Probability of the host opening EITHER Door A OR Door B = 1/2 + 1/2 = 1, as you add the individual probabilities together)
try also the other method I mentioned on page 1 earlier, "Let's look at this problem in another way."
Btw since the host knows where the prize is and which door has been opened by the player, the probability of his opening an empty door in all the above cases 'i', 'ii', 'iii' and 'iv' is 1.
Re: "Monty Hall" problem revisited
I'd prefer to look at the 4 cases that you initially presented.
The probability of the host opening an empty door cannot = 1 in ALL 4 cases otherwise you'd get a total probability of 4/3  clearly impossible.
It is = 1 in cases (i) and (ii), but only = 1/2 in cases (iii) and (iv)
Unless what you're now saying is:
Case (iii) The contestant picks Door C and the host opens an empty door (either Door A or Door B)
Case (iv) The contestant picks Door C and the host opens an empty door (either Door A or Door B)
then clearly cases (iii) and (iv) are identical and can only be counted once. I'd be interested to know what YOU calculate the probabilities for the 4 cases you originally presented to be.
As to "Let's look at this problem in another way." what you are doing here is trying to conflate the probabilities of 2 independent random variables (The Door the Car is Behind, and The Door the Contestant Picks) by adding them together and getting an answer of 2/3. This equates to the statement: "What is the Probability the Contestant picks Door A OR the Probability the Car is behind Door A? In the context of this problem that question is meaningless.
What you can do is ask: What is the Probability the Contestant picks Door A AND the Probability the Car is also behind Door A? And the answer is 1/3 * 1/3 = 1/9 (which stands to reason as there are 3 doors to conceal the car and 3 picks for each door, giving a total of 9 equiprobable cases covering all possible options).
The probability of the host opening an empty door cannot = 1 in ALL 4 cases otherwise you'd get a total probability of 4/3  clearly impossible.
It is = 1 in cases (i) and (ii), but only = 1/2 in cases (iii) and (iv)
Unless what you're now saying is:
Case (iii) The contestant picks Door C and the host opens an empty door (either Door A or Door B)
Case (iv) The contestant picks Door C and the host opens an empty door (either Door A or Door B)
then clearly cases (iii) and (iv) are identical and can only be counted once. I'd be interested to know what YOU calculate the probabilities for the 4 cases you originally presented to be.
As to "Let's look at this problem in another way." what you are doing here is trying to conflate the probabilities of 2 independent random variables (The Door the Car is Behind, and The Door the Contestant Picks) by adding them together and getting an answer of 2/3. This equates to the statement: "What is the Probability the Contestant picks Door A OR the Probability the Car is behind Door A? In the context of this problem that question is meaningless.
What you can do is ask: What is the Probability the Contestant picks Door A AND the Probability the Car is also behind Door A? And the answer is 1/3 * 1/3 = 1/9 (which stands to reason as there are 3 doors to conceal the car and 3 picks for each door, giving a total of 9 equiprobable cases covering all possible options).
PalmerEldritch Posts : 13
Join date : 20130418
Re: "Monty Hall" problem revisited
PalmerEldritch wrote:What you can do is ask: What is the Probability the Contestant picks Door A AND the Probability the Car is also behind Door A? And the answer is 1/3 * 1/3 = 1/9 (which stands to reason as there are 3 doors to conceal the car and 3 picks for each door, giving a total of 9 equiprobable cases covering all possible options).
After the host throws open the empty door (let’s say C), which he always knows beforehand as empty (due to the placing of the prize behind a certain door and the player choosing a door for himself), the choice involving two remaining doors (A or B) for the prize and by the player in terms of two sides of coin can be considered as the head H (corresponding to door A) and the tail T (corresponding to door B). These choices are mutually exclusive in terms of H and T, implying that the player or the prize can’t be both at door A and door B.
Consider also that the prize and the player are represented now by two coins, coin 1 and coin 2, respectively.
We can then represent all the possible choices involving prize, player and the two remaining doors A and B (after the host has opened the empty door with the prior knowledge about the door being empty) in terms of coins 1 and 2 their heads and tails.
From the above, prize at door A will be designated as 1H (head in the case of coin 1) and the player at door A as 2H. Similarly, prize at door B will be 1T and the player at door B as 2T.
The conditions for switch to work (leading to a win), i.e. either the prize is at door A (1H) and the player at door B (2T), or the prize at door B (1T) and the player at door A (2H): in terms of probability, the probability for the switch to work,
P(s) = P(1H)*P(2T) + P(1T)*P(2H) ………….. (Eq. 1)
Similarly, the conditions for the switch not to work (failing to win), i.e. either the prize is at door A (1H) and the player at door A (2H), or the prize at door B (1T) and the player at door B (2T), which in terms of probability for switch not producing a win,
P(f) = P(1H)*P(2H) + P(1T)*P(2T) ……………(Eq. 2).
Since the probability for head or tail for any coin is ½, i.e. P(1H) = P(1T) = P(2H) = P(2T) = 1/2;
The probability of a switch leading to a prize from Eq. 1, P(s) = 1/2;
And the probability switch leading to no win (using Eq. 2), P(f) = 1/2
Re: "Monty Hall" problem revisited
Let's stick with the 4 cases you defined in your 1st post :
"In situation (v), the player picks door C. Host
opens door A, then the switching by the player to door B leads to
missing the prize (failure in switching).
In
situation (vi), the player picks door C. Host opens door B, then the
switching by the player to door A leads to missing the prize (failure in switching)."
You later go on to say, and I quote:
"To the contrary, cases (v) and (vi) are two separate possibilities of
switching, just like possible switchings in (i) and (iii). Therefore,
(v) and (vi) need to be treated and counted separately as 2 (instead of
1) in probability calculations." (my emphasis)
So I ask again, what are your probability calculations for cases (v) and (vi)? I have correctly calculated them as 1/6 and 1/6.
Are you saying:
a) they are 1/3 and 1/3 the same as cases (i) and (iii) OR
b) all 4 cases have an equal probability of 1/4
If a) then the total probability of all 4 cases is 4/3 which is impossible.
If b) then this implies the contestant has a 50% chance of picking Door C, again an impossibility (unless the contestant is psychic)
One of the above must be true if you disagree with my calculations. Which is it?
(I think it is best to stick to the original problem of 3 doors, 2 goats and 1 car, rather than muddying the waters by introducing coins. I'm sure you agree, you said as much in your earlier posts)
"In situation (v), the player picks door C. Host
opens door A, then the switching by the player to door B leads to
missing the prize (failure in switching).
In
situation (vi), the player picks door C. Host opens door B, then the
switching by the player to door A leads to missing the prize (failure in switching)."
You later go on to say, and I quote:
"To the contrary, cases (v) and (vi) are two separate possibilities of
switching, just like possible switchings in (i) and (iii). Therefore,
(v) and (vi) need to be treated and counted separately as 2 (instead of
1) in probability calculations." (my emphasis)
So I ask again, what are your probability calculations for cases (v) and (vi)? I have correctly calculated them as 1/6 and 1/6.
Are you saying:
a) they are 1/3 and 1/3 the same as cases (i) and (iii) OR
b) all 4 cases have an equal probability of 1/4
If a) then the total probability of all 4 cases is 4/3 which is impossible.
If b) then this implies the contestant has a 50% chance of picking Door C, again an impossibility (unless the contestant is psychic)
One of the above must be true if you disagree with my calculations. Which is it?
(I think it is best to stick to the original problem of 3 doors, 2 goats and 1 car, rather than muddying the waters by introducing coins. I'm sure you agree, you said as much in your earlier posts)
PalmerEldritch Posts : 13
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Re: "Monty Hall" problem revisited
PalmerEldritch wrote:Let's stick with the 4 cases you defined in your 1st post :
Good. Then you don't calculate the overall probability using some arbitrary values, like if the host opens a door which he already knows is empty ... the probability of the door being empty is 1/2, which is wrong (because in this case the probability is 1 since the host already knows the door is empty).
If there are 3 doors, 2 of which are empty and you know which 2 doors are empty, then the probability of you opening an empty door is 1, and not 1/3 or 1/2.
Btw the 4 cases in the above were given to show 4 possibilities (leading to wins in 2 cases and no wins in 2 cases) and not for calculating probability by using arbitrary values.
If you want to use and calculate probabilities, follow the method in my previous post.
Re: "Monty Hall" problem revisited
Good, then we are agreed, we will consider the 4 cases you defined in your 1st post as (to
quote you again) “solutions to the difficult and complex problems are pursued by using simple and easy to understand cases in the beginning” . And the easiest, simplest, and most rigorous method
to find the solution to this problem is calculate the probabilities of each of the 4 cases.
To recap the 4 cases are:
“ In situation (i), the player picks door A. Host opens door B, then the switching
by the player to door C leads to the prize (success in switching).
In situation (iii), the player picks door B. Host opens door A, then the
switching by the player to door C leads to the prize (success in switching).
In situation (v), the player picks door C. Host opens door A, then the
switching by the player to door B leads to missing the prize (failure in switching).
In situation (vi), the player picks door C. Host opens door B, then the
switching by the player to door A leads to missing the prize (failure in switching)”
Each case consists of 2 events: event 1) the player picks a door, and event 2) the host picks an empty door to open.
In order to calculate the probability for each case we need to calculate the probability of both events happening. (BTW, in event 2 we are NOT calculating the probability that the door the host opens is empty  as you said in your previous post  but the probability of the host opening a particular empty door, a subtle but very important difference. We already know that the door the host opens has to be empty, but we don't know which door he opens).
So we can make the following statements:
a) the probability the player picks any particular door = 1/3, and applies to all 4 cases.
b) In case (i) the probability the host opens door B = 1, as it is the only door he can open. Or to put it another way: the probability the host opens door A (the door picked) = 0, and the probability the host opens door C (the door hiding the car) =0
c) In case (iii) the probability the host opens door A = 1, as it is the only door he can open, similar in all aspects to case(i)
Now we come to cases (v) and (vi) where we appear to disagree. In these 2 cases the player has picked the door hiding the car, and the host therefore has a choice between 2 empty doors to open. Given the host makes a random choice between the 2 doors we assign a probability of ½ to him opening door A and a probability of ½ to
him opening door B. This is not, I can assure you, an arbitrary value as you appear to think, in fact it is the only value we can assign when making a random pick between 2 indistinguishable options.
We have now calculated the probability for each of the 4 cases
occurring, and these are:
case(i) probability = 1/3 * 1 =1/3
case (iii) probability = 1/3 * 1 = 1/3
case (v) probability = 1/3 * 1/2= 1/6
case (vi) probability = 1/3 * 1/2 = 1/6
Total probability for all 4 cases = 1/3 + 1/3 + 1/6 + 1/6 = 1
Total probability for successful switching cases (i) and (iii) = 1/3 +1/3 = 2/3
Total probability for switching failure cases (v) and (vi) = 1/6 + 1/6 = 1/3
If you don’t agree with my probability calculations for these 4 cases will you provide your own as I requested in my previous post?
quote you again) “solutions to the difficult and complex problems are pursued by using simple and easy to understand cases in the beginning” . And the easiest, simplest, and most rigorous method
to find the solution to this problem is calculate the probabilities of each of the 4 cases.
To recap the 4 cases are:
“ In situation (i), the player picks door A. Host opens door B, then the switching
by the player to door C leads to the prize (success in switching).
In situation (iii), the player picks door B. Host opens door A, then the
switching by the player to door C leads to the prize (success in switching).
In situation (v), the player picks door C. Host opens door A, then the
switching by the player to door B leads to missing the prize (failure in switching).
In situation (vi), the player picks door C. Host opens door B, then the
switching by the player to door A leads to missing the prize (failure in switching)”
Each case consists of 2 events: event 1) the player picks a door, and event 2) the host picks an empty door to open.
In order to calculate the probability for each case we need to calculate the probability of both events happening. (BTW, in event 2 we are NOT calculating the probability that the door the host opens is empty  as you said in your previous post  but the probability of the host opening a particular empty door, a subtle but very important difference. We already know that the door the host opens has to be empty, but we don't know which door he opens).
So we can make the following statements:
a) the probability the player picks any particular door = 1/3, and applies to all 4 cases.
b) In case (i) the probability the host opens door B = 1, as it is the only door he can open. Or to put it another way: the probability the host opens door A (the door picked) = 0, and the probability the host opens door C (the door hiding the car) =0
c) In case (iii) the probability the host opens door A = 1, as it is the only door he can open, similar in all aspects to case(i)
Now we come to cases (v) and (vi) where we appear to disagree. In these 2 cases the player has picked the door hiding the car, and the host therefore has a choice between 2 empty doors to open. Given the host makes a random choice between the 2 doors we assign a probability of ½ to him opening door A and a probability of ½ to
him opening door B. This is not, I can assure you, an arbitrary value as you appear to think, in fact it is the only value we can assign when making a random pick between 2 indistinguishable options.
We have now calculated the probability for each of the 4 cases
occurring, and these are:
case(i) probability = 1/3 * 1 =1/3
case (iii) probability = 1/3 * 1 = 1/3
case (v) probability = 1/3 * 1/2= 1/6
case (vi) probability = 1/3 * 1/2 = 1/6
Total probability for all 4 cases = 1/3 + 1/3 + 1/6 + 1/6 = 1
Total probability for successful switching cases (i) and (iii) = 1/3 +1/3 = 2/3
Total probability for switching failure cases (v) and (vi) = 1/6 + 1/6 = 1/3
If you don’t agree with my probability calculations for these 4 cases will you provide your own as I requested in my previous post?
Last edited by PalmerEldritch on Fri Apr 19, 2013 8:39 am; edited 1 time in total (Reason for editing : formatting)
PalmerEldritch Posts : 13
Join date : 20130418
Re: "Monty Hall" problem revisited
"If you don’t agree with my probability calculations for these 4 cases will you provide your own as I requested in my previous post?" PalmerEldritch
>>> As I said earlier the host opening the door is irrelevant in this problem.
The door opening by host amounts to nonconsequntial probablity (not to be used in probability calculations) and helps only in reducing physically the 3 door problem into a situation involving 2 doors and two selectors (player and the prize).
The probability calculation for 2 doors and 2 selectors (player and prize) by using the example of coins was already posted by me earlier, indicating that the probability the switch works and wins the prize, P(s) = 1/2, same as the probability switch doesn't win the prize, p(f) =1/2.
>>> As I said earlier the host opening the door is irrelevant in this problem.
The door opening by host amounts to nonconsequntial probablity (not to be used in probability calculations) and helps only in reducing physically the 3 door problem into a situation involving 2 doors and two selectors (player and the prize).
The probability calculation for 2 doors and 2 selectors (player and prize) by using the example of coins was already posted by me earlier, indicating that the probability the switch works and wins the prize, P(s) = 1/2, same as the probability switch doesn't win the prize, p(f) =1/2.
Re: "Monty Hall" problem revisited
I'm disappointed, but not surprised, that yet again you refused to answer my question. Nevermind, you did at least calculate a "probability that switching produces a win P(w)" at the end of your 1st post, and from that it is possible to derive probabilities for each of the 4 cases. Before I get to that, I'll address some of the comments you made in your last post:
"the host opening the door is irrelevant in this problem"
If that was the case (and it isn't) why is the door the host opens specifically described in each of the 4 cases?
"The door opening by host amounts to nonconsequntial probablity (not to be used in probability calculations)"
You are joking right? Why not apply the same 'reasoning' to "the door selection by the player" and "the initial placement of the car behind a door" and eliminate all the variables in the problem from the probability calculations?
"The probability calculation for 2 doors ....was already posted by me earlier,
indicating that the probability the switch works and wins the prize,
P(s) = 1/2, same as the probability switch doesn't win the prize, p(f)
=1/2"
All that post indicated was that if you flip a coin Pr(H) = Pr(T) = 1/2.
Now, back to the probability calculation in your 1st post.
"... probability that switching produces a win, P(w), number of prize winning switching cases (i and iii) / total number of real cases (i, iii, v and vi), P(w) = 2 / 4 = 1/2"
This statement is actually true according to probability theory, provided all cases are equally likely.
In other words provided Pr(case(i)) = Pr(case(iii)) = Pr(case(v)) = Pr(case(vi)).
If true, then the probability of each case = 1/4. So far so good.
Now consider case(i): "the player picks door A. Host opens door B"
This case consists of 2 events:
event1 "the player picks door A" and
event2 "the Host opens door B"
The laws of probability state that:
Pr(event1 and event2) = Pr(event1) * Pr(event2).
Pr(event2) = 1 (as door B is the only door the host can open).
Therefore Pr(event1), the Pr(player picks door A), must =1/4 since the total probability of this case happening = 1/4 as has already been determined.
Repeating the same calculation for case(iii) yields a similar result,
Pr(player picks door B) = 1/4.
If Pr(player picks door A) = 1/4 and Pr(player picks door B) = 1/4, then Pr(player picks door C) must = 1/2. And therein lies your problem.
Because clearly: Pr(player picks door A) = Pr(player picks door B) = Pr(player picks door C) = 1/3
Given the probability calculations above are accurate (which they are) then the only conclusion we can draw is that our original premise is invalid, i.e. the 4 cases are NOT equally likely.
And if the 4 cases are not equally likely then the conclusion you made at the end of your original post
"P(w) or P(nw) = 1/2 … which indicates that the switching has no effect probabilistically in winning the prize" is FALSE.
This should come as no surprise if you had read my last comment, as I conclusively proved that the 4 cases are indeed not equally likely and that P(w) = 2/3 and P(nw) = 1/3
"the host opening the door is irrelevant in this problem"
If that was the case (and it isn't) why is the door the host opens specifically described in each of the 4 cases?
"The door opening by host amounts to nonconsequntial probablity (not to be used in probability calculations)"
You are joking right? Why not apply the same 'reasoning' to "the door selection by the player" and "the initial placement of the car behind a door" and eliminate all the variables in the problem from the probability calculations?
"The probability calculation for 2 doors ....was already posted by me earlier,
indicating that the probability the switch works and wins the prize,
P(s) = 1/2, same as the probability switch doesn't win the prize, p(f)
=1/2"
All that post indicated was that if you flip a coin Pr(H) = Pr(T) = 1/2.
Now, back to the probability calculation in your 1st post.
"... probability that switching produces a win, P(w), number of prize winning switching cases (i and iii) / total number of real cases (i, iii, v and vi), P(w) = 2 / 4 = 1/2"
This statement is actually true according to probability theory, provided all cases are equally likely.
In other words provided Pr(case(i)) = Pr(case(iii)) = Pr(case(v)) = Pr(case(vi)).
If true, then the probability of each case = 1/4. So far so good.
Now consider case(i): "the player picks door A. Host opens door B"
This case consists of 2 events:
event1 "the player picks door A" and
event2 "the Host opens door B"
The laws of probability state that:
Pr(event1 and event2) = Pr(event1) * Pr(event2).
Pr(event2) = 1 (as door B is the only door the host can open).
Therefore Pr(event1), the Pr(player picks door A), must =1/4 since the total probability of this case happening = 1/4 as has already been determined.
Repeating the same calculation for case(iii) yields a similar result,
Pr(player picks door B) = 1/4.
If Pr(player picks door A) = 1/4 and Pr(player picks door B) = 1/4, then Pr(player picks door C) must = 1/2. And therein lies your problem.
Because clearly: Pr(player picks door A) = Pr(player picks door B) = Pr(player picks door C) = 1/3
Given the probability calculations above are accurate (which they are) then the only conclusion we can draw is that our original premise is invalid, i.e. the 4 cases are NOT equally likely.
And if the 4 cases are not equally likely then the conclusion you made at the end of your original post
"P(w) or P(nw) = 1/2 … which indicates that the switching has no effect probabilistically in winning the prize" is FALSE.
This should come as no surprise if you had read my last comment, as I conclusively proved that the 4 cases are indeed not equally likely and that P(w) = 2/3 and P(nw) = 1/3
PalmerEldritch Posts : 13
Join date : 20130418
Re: "Monty Hall" problem revisited
PalmerEldritch wrote:I'm disappointed, but not surprised, that yet again you refused to answer my question. Nevermind, you did at least calculate a "probability that switching produces a win P(w)"
You simply don't agree with my answer, so let's forget about the whole thing. You stick to your answer ("probability of winning after the switch is 2/3") and I will stick to mine ("probability after the switch is 1/2 and makes no difference."
Re: "Monty Hall" problem revisited
Seva Lamberdar wrote:PalmerEldritch wrote:I'm disappointed, but not surprised, that yet again you refused to answer my question. Nevermind, you did at least calculate a "probability that switching produces a win P(w)"
You simply don't agree with my answer, so let's forget about the whole thing. You stick to your answer ("probability of winning after the switch is 2/3") and I will stick to mine ("probability after the switch is 1/2 and makes no difference."
I don't agree with your answer because your answer is incorrect. I thought you might be interested in understanding why your answer is incorrect. Apparently not.
PalmerEldritch Posts : 13
Join date : 20130418
Re: "Monty Hall" problem revisited
Huzefa Kapasi wrote:so who won?
the borg.
MaxEntropy_Man Posts : 14702
Join date : 20110428
Re: "Monty Hall" problem revisited
Huzefa Kapasi wrote:so who won?
OR
Depends what you want to believe in.
Re: "Monty Hall" problem revisited
Huzefa Kapasi wrote:so who won?
OR
lol!
Captain Bhankas Posts : 676
Join date : 20130205
Re: "Monty Hall" problem revisited
When it comes to puzzles like the Monty Hall Problem you want to believe in mathematics.Seva Lamberdar wrote:
Depends what you want to believe in.
PalmerEldritch Posts : 13
Join date : 20130418
Re: "Monty Hall" problem revisited
PalmerEldritch wrote:When it comes to puzzles like the Monty Hall Problem you want to believe in mathematics.Seva Lamberdar wrote:
Depends what you want to believe in.
Surely, but only that much mathematics which is useful and necessary to solve the problem correctly.
Re: "Monty Hall" problem revisited
Seva Lamberdar wrote:PalmerEldritch wrote:When it comes to puzzles like the Monty Hall Problem you want to believe in mathematics.Seva Lamberdar wrote:
Depends what you want to believe in.
Surely, but only that much mathematics which is useful and necessary to solve the problem correctly.
That is all I used. I could have gone with Bayes Theory (which would have shown the same result) but conditional probability calculations are sometimes harder to follow.
PalmerEldritch Posts : 13
Join date : 20130418
Re: "Monty Hall" problem revisited
I have already said enough about this. There are basic differences the way we look and interpret this problem, which also decides as to what we consider the extent of useful and necessary math.PalmerEldritch wrote:Seva Lamberdar wrote:PalmerEldritch wrote:When it comes to puzzles like the Monty Hall Problem you want to believe in mathematics.Seva Lamberdar wrote:
Depends what you want to believe in.
Surely, but only that much mathematics which is useful and necessary to solve the problem correctly.
That is all I used. I could have gone with Bayes Theory (which would have shown the same result) but conditional probability calculations are sometimes harder to follow.
Re: "Monty Hall" problem revisited
There is only one correct solution to this problem and mathematics unequivocally proves it. Your interpretation of the problem and your solution is simply incorrect and is unsupported by any maths.
PalmerEldritch Posts : 13
Join date : 20130418
Re: "Monty Hall" problem revisited
You are right about only one correct solution to this problem but not about mathematics unequivocally proving it. Mathematics is just a tool, like computers etc., to get results quantitatively and not to justify or prove that your understanding and interpretation of the problem is right. So you can forget about the other half of your statement ("Your interpretation of the problem and your solution is simply incorrect and is unsupported by any maths").PalmerEldritch wrote:There is only one correct solution to this problem and mathematics unequivocally proves it. Your interpretation of the problem and your solution is simply incorrect and is unsupported by any maths.
Re: "Monty Hall" problem revisited
Seva Lamberdar wrote:You are right about only one correct solution to this problem but not about mathematics unequivocally proving it. Mathematics is just a tool, like computers etc., to get results quantitatively and not to justify or prove that your understanding and interpretation of the problem is right. So you can forget about the other half of your statement ("Your interpretation of the problem and your solution is simply incorrect and is unsupported by any maths").PalmerEldritch wrote:There is only one correct solution to this problem and mathematics unequivocally proves it. Your interpretation of the problem and your solution is simply incorrect and is unsupported by any maths.
Seva try this simulation :
http://www.grandillusions.com/simulator/montysim.htm
In the settings let the number of runs be 1000. Try this once for keeping your choice and once for changing your choice.
Guest Guest
Re: "Monty Hall" problem revisited
Rashmun, there is no need for it. Most of the simulations (basically the computer programs) related to the Monty Hall problem are set using considerations which I am not in agreement with.
Re: "Monty Hall" problem revisited
Seva Lamberdar wrote:Rashmun, there is no need for it. Most of the simulations (basically the computer programs) related to the Monty Hall problem are set using considerations which I am not in agreement with.
What considerations in the computer simulations are you not in agreement with?
Mathematics is NOT a tool, like computers etc., to get results quantitatively. Have you never heard of mathematical proofs? It is one of the few disciplines where you can conclusively prove a hypothesis. And your solution is still unsupported by any mathematics.
PalmerEldritch Posts : 13
Join date : 20130418
Re: "Monty Hall" problem revisited
PalmerEldritch wrote:Seva Lamberdar wrote:Rashmun, there is no need for it. Most of the simulations (basically the computer programs) related to the Monty Hall problem are set using considerations which I am not in agreement with.
What considerations in the computer simulations are you not in agreement with?
Mathematics is NOT a tool, like computers etc., to get results quantitatively. Have you never heard of mathematical proofs? It is one of the few disciplines where you can conclusively prove a hypothesis. And your solution is still unsupported by any mathematics.
What you call mathematical proof of a hypothesis is really the symbolic form / representation of hypothesis to get numerical and other types of information. When Newton says (based on Kepler's laws etc.) that the rate of change of momentum is directly proportional to the impressed force, its symbolic notation (F = m*a) is not a mathematical proof of the above statement. Note, F=m*a as a mathematical tool or expression based on the hypothesis yields numerical and other types of information which is found to be in agreement with the physical phenomenon and that is what validates the hypothesis.
Re: "Monty Hall" problem revisited
Nevermind Newton's 2nd Law of Motion (which is a law of physics), what considerations in the computer simulations are you not in agreement with?
In fact, why not explain clearly and concisely what your interpretation of the Monty Hall Problem is, so that the rest of us can see how it differs from the usual interpretation (as it must surely do for you get an answer of 1/2)?
In fact, why not explain clearly and concisely what your interpretation of the Monty Hall Problem is, so that the rest of us can see how it differs from the usual interpretation (as it must surely do for you get an answer of 1/2)?
PalmerEldritch Posts : 13
Join date : 20130418
Re: "Monty Hall" problem revisited
another writeup on Internet on this topic:
"A new approach to the Monty Hall problem"  http://www.reenigne.org/blog/montyhall/
"A new approach to the Monty Hall problem"  http://www.reenigne.org/blog/montyhall/
Re: "Monty Hall" problem revisited
Monty Hall passes away at 96. May he RIP!
http://www.cnn.com/2017/09/30/tvshows/montyhalldeadat96/index.html
http://www.cnn.com/2017/09/30/tvshows/montyhalldeadat96/index.html
Re: "Monty Hall" problem revisited
"The example of starting with 100 buckets (with one of them containing gold), a player selecting one of the buckets for himself (1/100 probability for gold initially) and then the host removing 98 empty buckets seems to be the best example in favor of not making a switch, because after 98 empty buckets are knowingly taken out of play by the host the new chance (probability) for the prize (gold) in the 99th bucket (let's say selected by the player initially) or in the 100th bucket (not selected by the player initially) is same or 1/2 (with or without switch), and neither 1/100 (without switching) nor 99/100 (after switching). "
Ref.: A comment to "The Monty Hall Problem, Explained
And why you should ALWAYS switch" Dec. 22, 2022, https://www.cantorsparadise.com/themontyhallproblemexplainedb09ab4ce369b
Ref.: A comment to "The Monty Hall Problem, Explained
And why you should ALWAYS switch" Dec. 22, 2022, https://www.cantorsparadise.com/themontyhallproblemexplainedb09ab4ce369b
Re: "Monty Hall" problem revisited
Seva Lamberdar wrote:"The example of starting with 100 buckets (with one of them containing gold), a player selecting one of the buckets for himself (1/100 probability for gold initially) and then the host removing 98 empty buckets seems to be the best example in favor of not making a switch, because after 98 empty buckets are knowingly taken out of play by the host the new chance (probability) for the prize (gold) in the 99th bucket (let's say selected by the player initially) or in the 100th bucket (not selected by the player initially) is same or 1/2 (with or without switch), and neither 1/100 (without switching) nor 99/100 (after switching). "
Ref.: A comment to "The Monty Hall Problem, Explained
And why you should ALWAYS switch" Dec. 22, 2022, https://www.cantorsparadise.com/themontyhallproblemexplainedb09ab4ce369b
Let's consider another example in the following, using a deck of 52 cards (with the ace of Spades as the prize / desired card), to show that switching strategy does not improve odds of winning (while the probability after switching in finding the ace of Spades does not improve / increase).
A player first picks a card from the deck of cards without knowing what it is exactly (whether the ace of Spades or some other card?
From the remaining 51 cards then, after the first card has already been taken and remains unknown, 50 nonace of Spades cards are exposed, thus leaving only one card hidden (unexposed) on the table.
What then is the probability that card in the hand of the player or the one left on the table is the ace of Spades? It is 1/2 (with 50 nonace of cards already taken out of the deck and exposed). Moreover, it is clear that the player holding the unknown card in his hand will not have his odds for ace of Spades improved (>1/2) even if he makes a switch in favor of the unexposed (52nd) card on the table.
It's worth noting that even though the player had chosen the card initially himself, he did not know what card it exactly was (e.g. ace of Spades, etc.).
In another hypothetical situation, if the player had precisely known / seen the card in hand (as ace of Spades, or not) then it would not be a problem of probability and choice, because player already knows (based on the knowledge about card in hand whether it is ace of Spades, or not) what to do (to make a switch, or not).
Specifically, if the player knows he is holding the ace of Spades in hand then he wouldn't make a switch, whereas if he knows he is not having the ace of Spades in hand then he will make a switch. But that is quite different from the Monty Hall problem where the door selected initially by the player remains unknown to him (whether, or not, the door leads to the prize?).
Note: use the following link to read the comments on Facebook post (Dec. 2022):
https://www.facebook.com/cantor.paradise/posts/pfbid02k34rsk8HzXEZDng1ctefZ9kdC2GkjuL5SpXifGfqDewWd7TEn52jNa3VFkR39y6Xl
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