"Monty Hall" problem revisited
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bw
Idéfix
MaxEntropy_Man
Seva Lamberdar
8 posters
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"Monty Hall" problem revisited
“… the question was what is the probability that after making a choice, and that the host is always going to throw open an empty door, switching produces a win.” (https://such.forumotion.com/t12261sevaji#94571)
Solution:
Consider the following six possibilities with three doors, A, B and C, and the prize behind door C.
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 (ii), the player picks door A. If host opens door C, the prize is exposed and the game ends. Since the host always would open the empty door this situation (opening the door C with prize).is avoided by the host. This is therefore a null case (neither involving opening the door by the host, nor switching by the player).
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 (iv), the player picks door B. If host opens door C, the prize is exposed and the game ends. Since the host always would open the empty door this situation (opening the door C with prize).is avoided by the host. This is therefore a null case (neither involving opening the door by the host, nor switching by the player).
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).
As indicated in the above, there are only four real situations involving opening the door by the host and switching by the player (after the initial door selection by the player), i.e. situations i, iii, v and vi. Two of these four situations involving switching are helpful in getting the prize (situations i and iii) and the other two not (situations v and vi).
Thus, 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
Similarly, probability that switching (after the player has selected a door and the host has opened an empty door) will produce no win, P(nw),
number of no prize winning cases while switching (v and vi) / total number of real cases (i, iii, v and vi),
P(nw) = 2 / 4 = 1/2
In other words, after the player has selected a door and then the host has opened an empty door, the probability of winning the prize by switching or not winning the prize by switching is same (1/2),
i.e. P(w) or P(nw) = 1/2 … which indicates that the switching has no effect probabilistically in winning the prize.
Solution:
Consider the following six possibilities with three doors, A, B and C, and the prize behind door C.
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 (ii), the player picks door A. If host opens door C, the prize is exposed and the game ends. Since the host always would open the empty door this situation (opening the door C with prize).is avoided by the host. This is therefore a null case (neither involving opening the door by the host, nor switching by the player).
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 (iv), the player picks door B. If host opens door C, the prize is exposed and the game ends. Since the host always would open the empty door this situation (opening the door C with prize).is avoided by the host. This is therefore a null case (neither involving opening the door by the host, nor switching by the player).
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).
As indicated in the above, there are only four real situations involving opening the door by the host and switching by the player (after the initial door selection by the player), i.e. situations i, iii, v and vi. Two of these four situations involving switching are helpful in getting the prize (situations i and iii) and the other two not (situations v and vi).
Thus, 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
Similarly, probability that switching (after the player has selected a door and the host has opened an empty door) will produce no win, P(nw),
number of no prize winning cases while switching (v and vi) / total number of real cases (i, iii, v and vi),
P(nw) = 2 / 4 = 1/2
In other words, after the player has selected a door and then the host has opened an empty door, the probability of winning the prize by switching or not winning the prize by switching is same (1/2),
i.e. P(w) or P(nw) = 1/2 … which indicates that the switching has no effect probabilistically in winning the prize.
Re: "Monty Hall" problem revisited
ah, i see your fallacy. read the wording of the problem carefully again. it only says "switching". cases (v) and (vi) are really one case. switching is just switching. it doesn't matter which of those two possibilities actually happens.
and from a practical standpoint, if someone was doing this experiment and keeping count to answer the question that was originally posed, they would count (v) and (vi) as one and the same. in other words the column headings under which the cases are being enumerated are "switch" and "no switch". there is no sub heading under "switch".
and from a practical standpoint, if someone was doing this experiment and keeping count to answer the question that was originally posed, they would count (v) and (vi) as one and the same. in other words the column headings under which the cases are being enumerated are "switch" and "no switch". there is no sub heading under "switch".
MaxEntropy_Man Posts : 14702
Join date : 20110428
Re: "Monty Hall" problem revisited
MaxEntropy_Man wrote:ah, i see your fallacy. read the wording of the problem carefully again. it only says "switching". cases (v) and (vi) are really one case. switching is just switching. it doesn't matter which of those two possibilities actually happens.
and from a practical standpoint, if someone was doing this experiment and keeping count to answer the question that was originally posed, they would count (v) and (vi) as one and the same. in other words the column headings under which the cases are being enumerated are "switch" and "no switch". there is no sub heading under "switch".
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.
Re: "Monty Hall" problem revisited
Incidentally, in the Sulekha CH discussion on Monty Hall problem several years ago I had indicated the same thing that switching, after the host opens an empty door, has no advantage in winning the prize, because the probability of landing at the ‘prize’ door after switching is only half (or 50%) just like the probability of landing at the empty door after switching is half (or 50%).
Re: "Monty Hall" problem revisited
Sevaji, you are confusing  and mixing up  states of nature and the actions of the player.
You can approach the problem either starting from the states of nature, or from the choices open to the player. In that old CH thread, I took the former approach and Max took the latter. Either way, the solution is the same: switching doubles your odds of winning.
Max already posted his solution here. I will post mine here. Study both carefully, and you shall find the error in yours.
At the beginning, there are three possibilities:
So picking any particular door gives you onetothree odds of winning. Now the player has chosen C and the host opens an "empty" door. He has a choice of which door to open only if scenario 3 is the reality. In the other two scenarios, he just has to open the one empty door that the player hasn't picked. This is how the scenarios look, with the opened door crossed out.
Now let us consider the two options that the player has.
Option 1: Stick to your choice of C. If the player does this, he wins in scenario 3, but loses in scenarios 1 and 2. Since each of the three scenarios had an equal likelihood of occuring to begin with, the odds of winning are onetothree.
Option 2: Switch your choice. In scenario 1 and 2, the player wins. In scenario 3, he loses. So the odds of winning are twotothree.
The odds of winning are twice as large with option 2 ("switching your choice.") So the player should switch.
You can approach the problem either starting from the states of nature, or from the choices open to the player. In that old CH thread, I took the former approach and Max took the latter. Either way, the solution is the same: switching doubles your odds of winning.
Max already posted his solution here. I will post mine here. Study both carefully, and you shall find the error in yours.
At the beginning, there are three possibilities:
Scenario  Door A  Door B  Door C 
1  car  empty  empty 
2  empty  car  empty 
3  empty  empty  car 
Scenario  Door A  Door B  Door C 
1  car  empty  
2  car  empty  
3  empty or  empty or  car 
Option 1: Stick to your choice of C. If the player does this, he wins in scenario 3, but loses in scenarios 1 and 2. Since each of the three scenarios had an equal likelihood of occuring to begin with, the odds of winning are onetothree.
Option 2: Switch your choice. In scenario 1 and 2, the player wins. In scenario 3, he loses. So the odds of winning are twotothree.
The odds of winning are twice as large with option 2 ("switching your choice.") So the player should switch.
Idéfix Posts : 8808
Join date : 20120426
Location : Berkeley, CA
Re: "Monty Hall" problem revisited
Idéfix wrote:Sevaji, you are confusing  and mixing up  states of nature and the actions of the player.
You can approach the problem either starting from the states of nature, or from the choices open to the player. In that old CH thread, I took the former approach and Max took the latter. Either way, the solution is the same: switching doubles your odds of winning.
Max already posted his solution here. I will post mine here. Study both carefully, and you shall find the error in yours.
At the beginning, there are three possibilities:So picking any particular door gives you onetothree odds of winning. Now the player has chosen C and the host opens an "empty" door. He has a choice of which door to open only if scenario 3 is the reality. In the other two scenarios, he just has to open the one empty door that the player hasn't picked. This is how the scenarios look, with the opened door crossed out.
Scenario Door A Door B Door C 1 car empty empty 2 empty car empty 3 empty empty car Now let us consider the two options that the player has.
Scenario Door A Door B Door C 1 car emptyempty 2 emptycar empty 3 empty or emptyempty or emptycar
Option 1: Stick to your choice of C. If the player does this, he wins in scenario 3, but loses in scenarios 1 and 2. Since each of the three scenarios had an equal likelihood of occuring to begin with, the odds of winning are onetothree.
Option 2: Switch your choice. In scenario 1 and 2, the player wins. In scenario 3, he loses. So the odds of winning are twotothree.
The odds of winning are twice as large with option 2 ("switching your choice.") So the player should switch.
+1 for the solution and,
+1 for putting your finger on seva's source of confusion.
good exposition. enjoyed reading that.
MaxEntropy_Man Posts : 14702
Join date : 20110428
Re: "Monty Hall" problem revisited
MaxEntropy_Man wrote:Idéfix wrote:Sevaji, you are confusing  and mixing up  states of nature and the actions of the player.
You can approach the problem either starting from the states of nature, or from the choices open to the player. In that old CH thread, I took the former approach and Max took the latter. Either way, the solution is the same: switching doubles your odds of winning.
Max already posted his solution here. I will post mine here. Study both carefully, and you shall find the error in yours.
At the beginning, there are three possibilities:So picking any particular door gives you onetothree odds of winning. Now the player has chosen C and the host opens an "empty" door. He has a choice of which door to open only if scenario 3 is the reality. In the other two scenarios, he just has to open the one empty door that the player hasn't picked. This is how the scenarios look, with the opened door crossed out.
Scenario Door A Door B Door C 1 car empty empty 2 empty car empty 3 empty empty car Now let us consider the two options that the player has.
Scenario Door A Door B Door C 1 car emptyempty 2 emptycar empty 3 empty or emptyempty or emptycar
Option 1: Stick to your choice of C. If the player does this, he wins in scenario 3, but loses in scenarios 1 and 2. Since each of the three scenarios had an equal likelihood of occuring to begin with, the odds of winning are onetothree.
Option 2: Switch your choice. In scenario 1 and 2, the player wins. In scenario 3, he loses. So the odds of winning are twotothree.
The odds of winning are twice as large with option 2 ("switching your choice.") So the player should switch.
+1 for the solution and,
+1 for putting your finger on seva's source of confusion.
good exposition. enjoyed reading that.
No confusion here.
Just count the number of random possibilities (total 4, corresponding to cases i, iii, v and vi) creating 4 choices for switch, resulting in 2 wins and 2 losses. The probability for a win in the switch is therefore 2 / 4 or 1/2.
Re: "Monty Hall" problem revisited
[sevaji, bw deleted her post. i am reproducing it from memory because it directly addresses the distribution you have made in post 1 of this thread. it will also help you understand why prof. max is right in saying that you have made a double addition. credit for this post goes to bw.]
imagine that you are given a choice to select 2 doors over 1. thus the host opens the 2 doors and the result is declared. in this event, there are 3 possibilities and 2 lead to success. so the probability of success is 2/3.
now in which order the 2 doors you have chosen are opened is not important. no matter which sequence you open them in or even if you open them simultaneously, the event is recorded as 1 event. [this is the double addition in your list of possibilities]
this is the meaning of "switching" that should be applied in monty hall. the above scenario is analogous to the "switching" scenario.
thus, intuitively, which would you choose?
1) to select 1 door (and sticking to it  prob. 1/3), or
2) select 2 doors (same as switching  prob 2/3... in this case, your first selection is the door you intentionally wish to eliminate to travel to the 2/3rd probability of success of the remaining 2 doors combined).
edit. you can think of this problem in this way: the guest says, "i choose door 1 but i already have decided to switch." in this case the remaining 2 doors are simultaneously flung open and the result declared.
imagine that you are given a choice to select 2 doors over 1. thus the host opens the 2 doors and the result is declared. in this event, there are 3 possibilities and 2 lead to success. so the probability of success is 2/3.
now in which order the 2 doors you have chosen are opened is not important. no matter which sequence you open them in or even if you open them simultaneously, the event is recorded as 1 event. [this is the double addition in your list of possibilities]
this is the meaning of "switching" that should be applied in monty hall. the above scenario is analogous to the "switching" scenario.
thus, intuitively, which would you choose?
1) to select 1 door (and sticking to it  prob. 1/3), or
2) select 2 doors (same as switching  prob 2/3... in this case, your first selection is the door you intentionally wish to eliminate to travel to the 2/3rd probability of success of the remaining 2 doors combined).
edit. you can think of this problem in this way: the guest says, "i choose door 1 but i already have decided to switch." in this case the remaining 2 doors are simultaneously flung open and the result declared.
Guest Guest
Re: "Monty Hall" problem revisited
seva,
if you were told that you may choose either one door or a set of two doors and you win if one of the doors in the set has the prize behind it, what will you choose? 1 door or 2 doors and why?
again, extend this to 10 doors  choose 1 or 9, 100 doors  choose 1 or 99 and so on. the probability of the "set" containing the prize remains unchanged no matter how many doors are opened.
if you were told that you may choose either one door or a set of two doors and you win if one of the doors in the set has the prize behind it, what will you choose? 1 door or 2 doors and why?
again, extend this to 10 doors  choose 1 or 9, 100 doors  choose 1 or 99 and so on. the probability of the "set" containing the prize remains unchanged no matter how many doors are opened.
bw Posts : 2922
Join date : 20121115
Re: "Monty Hall" problem revisited
bw wrote:seva,
if you were told that you may choose either one door or a set of two doors and you win if one of the doors in the set has the prize behind it, what will you choose? 1 door or 2 doors and why?
again, extend this to 10 doors  choose 1 or 9, 100 doors  choose 1 or 99 and so on. the probability of the "set" containing the prize remains unchanged no matter how many doors are opened.
BW and HK,
Let's not worry about the suppositions and hypothetical conditions. This problem is about only "Monty Hall" and as stated in the beginning (in Max's words).
Just think calmly about what is indicated in the above statement, i.e. the player opening randomly any door first, followed by the host always opening an empty door and then the player having the opportunity / choice to make a switch (which he might or might not make). The question finally is on the probability of winning the prize after a switch?
The answer of course, as indicated earlier, is half (or 50%), after taking all the possibilities into consideration.
Re: "Monty Hall" problem revisited
Seva Lamberdar wrote:bw wrote:seva,
if you were told that you may choose either one door or a set of two doors and you win if one of the doors in the set has the prize behind it, what will you choose? 1 door or 2 doors and why?
again, extend this to 10 doors  choose 1 or 9, 100 doors  choose 1 or 99 and so on. the probability of the "set" containing the prize remains unchanged no matter how many doors are opened.
BW and HK,
Let's not worry about the suppositions and hypothetical conditions. This problem is about only "Monty Hall" and as stated in the beginning (in Max's words).
could you please answer my question though? if you had to choose between 99 doors and 1 door, what will you pick?
bw Posts : 2922
Join date : 20121115
Re: "Monty Hall" problem revisited
bw wrote:Seva Lamberdar wrote:bw wrote:seva,
if you were told that you may choose either one door or a set of two doors and you win if one of the doors in the set has the prize behind it, what will you choose? 1 door or 2 doors and why?
again, extend this to 10 doors  choose 1 or 9, 100 doors  choose 1 or 99 and so on. the probability of the "set" containing the prize remains unchanged no matter how many doors are opened.
BW and HK,
Let's not worry about the suppositions and hypothetical conditions. This problem is about only "Monty Hall" and as stated in the beginning (in Max's words).
could you please answer my question though? if you had to choose between 99 doors and 1 door, what will you pick?
BW, we are only discussing the Monty Hall problem (as stated earlier in Max's words) in this thread. If you want to talk about another problem, please start another thread.
Re: "Monty Hall" problem revisited
woo woo! that's what's called a tiltawhirl mat slam in professional wrestling!Seva Lamberdar wrote:
BW, we are only discussing the Monty Hall problem (as stated earlier in Max's words) in this thread. If you want to talk about another problem, please start another thread.
Guest Guest
Re: "Monty Hall" problem revisited
Huzefa Kapasi wrote:woo woo! that's what's called a tiltawhirl mat slam in professional wrestling!Seva Lamberdar wrote:
BW, we are only discussing the Monty Hall problem (as stated earlier in Max's words) in this thread. If you want to talk about another problem, please start another thread.
but predictable.
a lot of reasoning in mathematics and science is to look at related simpler cases, or going the other way, i.e. generalize, and argue by analogy. i am sure it is not unknown to sevaji, but for reasons known only to himself he is stuck in his stubborn position. one can only try so much after which one has to conclude the obvious, it is not the inability to see the truth, but ego that prevents one from accepting it. sad.
MaxEntropy_Man Posts : 14702
Join date : 20110428
Re: "Monty Hall" problem revisited
Seva Lamberdar wrote:bw wrote:Seva Lamberdar wrote:bw wrote:seva,
if you were told that you may choose either one door or a set of two doors and you win if one of the doors in the set has the prize behind it, what will you choose? 1 door or 2 doors and why?
again, extend this to 10 doors  choose 1 or 9, 100 doors  choose 1 or 99 and so on. the probability of the "set" containing the prize remains unchanged no matter how many doors are opened.
BW and HK,
Let's not worry about the suppositions and hypothetical conditions. This problem is about only "Monty Hall" and as stated in the beginning (in Max's words).
could you please answer my question though? if you had to choose between 99 doors and 1 door, what will you pick?
BW, we are only discussing the Monty Hall problem (as stated earlier in Max's words) in this thread. If you want to talk about another problem, please start another thread.
dear seva, i was told i was wasting my time engaging you on this topic but i did not listen.
bw Posts : 2922
Join date : 20121115
Re: "Monty Hall" problem revisited
yes, it is clearly not the inability to see the truth but that the ego that is coming in the way. case closed as far as i am concerned.MaxEntropy_Man wrote:Huzefa Kapasi wrote:woo woo! that's what's called a tiltawhirl mat slam in professional wrestling!Seva Lamberdar wrote:
BW, we are only discussing the Monty Hall problem (as stated earlier in Max's words) in this thread. If you want to talk about another problem, please start another thread.
but predictable.
a lot of reasoning in mathematics and science is to look at related simpler cases, or going the other way, i.e. generalize, and argue by analogy. i am sure it is not unknown to sevaji, but for reasons known only to himself he is stuck in his stubborn position. one can only try so much after which one has to conclude the obvious, it is not the inability to see the truth, but ego that prevents one from accepting it. sad.
Guest Guest
Re: "Monty Hall" problem revisited
i am not singling out sevaji. i am sure we all suffer from the same malady to a larger or lesser extent in some other aspects of life. i think this is why children and their brains are marvelous.
MaxEntropy_Man Posts : 14702
Join date : 20110428
Re: "Monty Hall" problem revisited
Seva Lamberdar wrote:MaxEntropy_Man wrote:Idéfix wrote:Sevaji, you are confusing  and mixing up  states of nature and the actions of the player.
You can approach the problem either starting from the states of nature, or from the choices open to the player. In that old CH thread, I took the former approach and Max took the latter. Either way, the solution is the same: switching doubles your odds of winning.
Max already posted his solution here. I will post mine here. Study both carefully, and you shall find the error in yours.
At the beginning, there are three possibilities:So picking any particular door gives you onetothree odds of winning. Now the player has chosen C and the host opens an "empty" door. He has a choice of which door to open only if scenario 3 is the reality. In the other two scenarios, he just has to open the one empty door that the player hasn't picked. This is how the scenarios look, with the opened door crossed out.
Scenario Door A Door B Door C 1 car empty empty 2 empty car empty 3 empty empty car Now let us consider the two options that the player has.
Scenario Door A Door B Door C 1 car emptyempty 2 emptycar empty 3 empty or emptyempty or emptycar
Option 1: Stick to your choice of C. If the player does this, he wins in scenario 3, but loses in scenarios 1 and 2. Since each of the three scenarios had an equal likelihood of occuring to begin with, the odds of winning are onetothree.
Option 2: Switch your choice. In scenario 1 and 2, the player wins. In scenario 3, he loses. So the odds of winning are twotothree.
The odds of winning are twice as large with option 2 ("switching your choice.") So the player should switch.
+1 for the solution and,
+1 for putting your finger on seva's source of confusion.
good exposition. enjoyed reading that.
No confusion here.
Just count the number of random possibilities (total 4, corresponding to cases i, iii, v and vi) creating 4 choices for switch, resulting in 2 wins and 2 losses. The probability for a win in the switch is therefore 2 / 4 or 1/2.
Your logic works if the pick was random. It is NOT. The host chooses to open the door that is empty and not picked.
Mosquito Posts : 706
Join date : 20110428
Re: "Monty Hall" problem revisited
MaxEntropy_Man wrote:Huzefa Kapasi wrote:woo woo! that's what's called a tiltawhirl mat slam in professional wrestling!Seva Lamberdar wrote:
BW, we are only discussing the Monty Hall problem (as stated earlier in Max's words) in this thread. If you want to talk about another problem, please start another thread.
but predictable.
a lot of reasoning in mathematics and science is to look at related simpler cases, or going the other way, i.e. generalize, and argue by analogy. i am sure it is not unknown to sevaji, but for reasons known only to himself he is stuck in his stubborn position. one can only try so much after which one has to conclude the obvious, it is not the inability to see the truth, but ego that prevents one from accepting it. sad.
Yes, solutions to the difficult and complex problems are pursued by using simple and easy to understand cases in the beginning. I had even tested for that reason the huge computer programs I wrote for complex systems (aerospace and nuclear), by applying them first to simple and down to earth examples.
Unfortunately, the case involving BW and HK is just the opposite.
They already seem to have a problem with 3 doors in this discussion, yet they think things will become clear in the example using 100 doors. It makes no sense at all. I would agree and even consider it logical if they go down to the example of 3 doors to clarify things during a 100 door discussion, but not the other way around (trying to jump to a discussion on 100 doors because things are not clear at the 3 door level).
Re: "Monty Hall" problem revisited
Mosquito wrote:Seva Lamberdar wrote:MaxEntropy_Man wrote:Idéfix wrote:Sevaji, you are confusing  and mixing up  states of nature and the actions of the player.
You can approach the problem either starting from the states of nature, or from the choices open to the player. In that old CH thread, I took the former approach and Max took the latter. Either way, the solution is the same: switching doubles your odds of winning.
Max already posted his solution here. I will post mine here. Study both carefully, and you shall find the error in yours.
At the beginning, there are three possibilities:So picking any particular door gives you onetothree odds of winning. Now the player has chosen C and the host opens an "empty" door. He has a choice of which door to open only if scenario 3 is the reality. In the other two scenarios, he just has to open the one empty door that the player hasn't picked. This is how the scenarios look, with the opened door crossed out.
Scenario Door A Door B Door C 1 car empty empty 2 empty car empty 3 empty empty car
Scenario Door A Door B Door C 1 car emptyempty 2 emptycar empty 3 empty or emptyempty or emptycar
Now let us consider the two options that the player has.
Option 1: Stick to your choice of C. If the player does this, he wins in scenario 3, but loses in scenarios 1 and 2. Since each of the three scenarios had an equal likelihood of occuring to begin with, the odds of winning are onetothree.
Option 2: Switch your choice. In scenario 1 and 2, the player wins. In scenario 3, he loses. So the odds of winning are twotothree.
The odds of winning are twice as large with option 2 ("switching your choice.") So the player should switch.
+1 for the solution and,
+1 for putting your finger on seva's source of confusion.
good exposition. enjoyed reading that.
No confusion here.
Just count the number of random possibilities (total 4, corresponding to cases i, iii, v and vi) creating 4 choices for switch, resulting in 2 wins and 2 losses. The probability for a win in the switch is therefore 2 / 4 or 1/2.
Your logic works if the pick was random. It is NOT. The host chooses to open the door that is empty and not picked.
You need to read the explanation in the beginning carefully.
There are only 6 possibilities in this problem, 2 of them being "null".
Based on the remaining 4 cases / possibilities, switch works only in 2 cases (resulting in a prize) which indicates the probability of prize after the switch as 2/4 or 1/2.
Re: "Monty Hall" problem revisited
i've been watching the proceedings here with considerable interest. what has happened is that kapasi has discredited seva on behalf of The Main Bore (TMB).
He enticed seva initially with easy questions about tossing dice, into restating the incorrect answer to the monty hall problem that he had given some years ago when i had first posted it and a solution based on conditional probabilities, at sulekha.
seva is wrong. still, it is surprising and disappointing to watch others being drawn into this nastiness for questionable motives of their own. they should know better than to serve as kapasi's accomplices in performing his caste duty of seeking redemption from wifebeating by ingratiating himself to TMB.
and southern indians, for heaven's bloody sakes, could you put an end to the inanity of attaching a sarcastic "ji" to seva's name every time you mention or address him?
He enticed seva initially with easy questions about tossing dice, into restating the incorrect answer to the monty hall problem that he had given some years ago when i had first posted it and a solution based on conditional probabilities, at sulekha.
seva is wrong. still, it is surprising and disappointing to watch others being drawn into this nastiness for questionable motives of their own. they should know better than to serve as kapasi's accomplices in performing his caste duty of seeking redemption from wifebeating by ingratiating himself to TMB.
and southern indians, for heaven's bloody sakes, could you put an end to the inanity of attaching a sarcastic "ji" to seva's name every time you mention or address him?
Jeremiah Mburuburu Posts : 1251
Join date : 20110909
Re: "Monty Hall" problem revisited
ha ha JM. my probability questions were for the larger audience (i explained the reason  i play online yahtzee and ask myself these questions) but i got drawn to sevaji when rishi solicited his help for a mathematical problem and of course i was immediately reminded of sevaji and monty hall.Jeremiah Mburuburu wrote:i've been watching the proceedings here with considerable interest. what has happened is that kapasi has discredited seva on behalf of The Main Bore (TMB).
He enticed seva initially with easy questions about tossing dice, into restating the incorrect answer to the monty hall problem that he had given some years ago when i had first posted it and a solution based on conditional probabilities, at sulekha.
seva is wrong. still, it is surprising and disappointing to watch others being drawn into this nastiness for questionable motives of their own. they should know better than to serve as kapasi's accomplices in performing his caste duty of seeking redemption from wifebeating by ingratiating himself to TMB.
and southern indians, for heaven's bloody sakes, could you put an end to the inanity of attaching a sarcastic "ji" to seva's name every time you mention or address him?
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Re: "Monty Hall" problem revisited
Jeremiah Mburuburu wrote:i've been watching the proceedings here with considerable interest. what has happened is that kapasi has discredited seva on behalf of The Main Bore (TMB).
He enticed seva initially with easy questions about tossing dice, into restating the incorrect answer to the monty hall problem that he had given some years ago when i had first posted it and a solution based on conditional probabilities, at sulekha.
seva is wrong.
How? As indicated in the beginning, there are only 4 real independent situations, after 2 "null" cases are deleted from the total six. Out of the 4 real situations (cases) 2 can result in the prize and other 2 not, after a player makes the switch. This means the probability that player making the switch will win the prize is 2 / 4 or 1/2.
Re: "Monty Hall" problem revisited
Huzefa Kapasi wrote:ha ha JM. my probability questions were for the larger audience (i explained the reason  i play online yahtzee and ask myself these questions) but i got drawn to sevaji when rishi solicited his help for a mathematical problem and of course i was immediately reminded of sevaji and monty hall.Jeremiah Mburuburu wrote:i've been watching the proceedings here with considerable interest. what has happened is that kapasi has discredited seva on behalf of The Main Bore (TMB).
He enticed seva initially with easy questions about tossing dice, into restating the incorrect answer to the monty hall problem that he had given some years ago when i had first posted it and a solution based on conditional probabilities, at sulekha.
seva is wrong. still, it is surprising and disappointing to watch others being drawn into this nastiness for questionable motives of their own. they should know better than to serve as kapasi's accomplices in performing his caste duty of seeking redemption from wifebeating by ingratiating himself to TMB.
and southern indians, for heaven's bloody sakes, could you put an end to the inanity of attaching a sarcastic "ji" to seva's name every time you mention or address him?
I had forgotten about the entire dicussion on this topic HK ... didn't know it meant so much to you and Carvaka (he even has the entire discussion package with him from long ago).
Re: "Monty Hall" problem revisited
Jeremiah Mburuburu wrote:
and southern indians, for heaven's bloody sakes, could you put an end to the inanity of attaching a sarcastic "ji" to seva's name every time you mention or address him?
so when you use "na" at the end of a sentence, you expect us to infer that you are using it mockingly, but you won't extend the same courtesy to other southern indians who use other northindianisms?
MaxEntropy_Man Posts : 14702
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Re: "Monty Hall" problem revisited
Jeremiah Mburuburu wrote:i've been watching ... blah, blah and 100 more words of blah... to TMB.
Calm yourself down, Jerji. Such excitement cannot be good for you.Jeremiah Mburuburu wrote:...for heaven's bloody sakes...
Yes, we can. But we will not. We might, if you fixed the error in your request and reposted it. Or we might not.Jeremiah Mburuburu wrote:and southern indians, could you put an end to the inanity of attaching a sarcastic "ji" to seva's name every time you mention or address him?
Idéfix Posts : 8808
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Re: "Monty Hall" problem revisited
Sevaji, all those old blogs you carefully copypaste from one site to another with references to other older blogs... do you carry them around with you in a "package"? If so, I have good news for you. You don't need to. It is 2013, and we have had search technology for a few years now.Seva Lamberdar wrote:he even has the entire discussion package with him from long ago.
Idéfix Posts : 8808
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Re: "Monty Hall" problem revisited
paul erdős also took a long time to accept that it is better to switch  paul hoffman has an account of this in his delightful book "the man who loved only numbers" (p 237)
bw Posts : 2922
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Re: "Monty Hall" problem revisited
bw wrote:paul erdős also took a long time to accept that it is better to switch  paul hoffman has an account of this in his delightful book "the man who loved only numbers" (p 237)
i didn't know that!
but i got curious after you posted this info and found this from: http://www.codinghorror.com/blog/2009/06/montyhallmontyfallmontycrawl.html
The army PhD who wrote in may have been correct that if all those PhDs were wrong, it would be a sign of trouble. But Marilyn was correct. When told of this, Paul Erdos, one of the leading mathematicians of the 20th century, said, "That's impossible." Then, when presented with a formal mathematical proof of the correct answer, he still didn't believe it and grew angry. Only after a colleague arranged for a computer simulation in which Erdos watched hundreds of trials that came out 2to1 in favor of switching did Erdos concede that he was wrong.
MaxEntropy_Man Posts : 14702
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Re: "Monty Hall" problem revisited
apparently pigeons and young children are better than adult humans.
http://www.livescience.com/6150pigeonsbeathumanssolvingmontyhallproblem.html
http://www.livescience.com/6150pigeonsbeathumanssolvingmontyhallproblem.html
Pigeons know better
To shed light on why humans often fall short of the best strategy with this kind of problem, scientists investigated pigeons, which often perform quite impressively on tasks requiring them to estimate relative probabilities, in some cases eclipsing human performance. Other animals do not always share the same biases as people, and therefore might help provide explanations for our behavior.
Scientists tested six pigeons with an apparatus with three keys. The keys lit up white to show a prize was available. After the birds pecked a key, one of the keys the bird did not choose deactivated, showing it was a wrong choice, and the other two lit up green. The pigeons were rewarded with bird feed if they made the right choice.
In the experiments, the birds quickly reached the best strategy for the Monty Hall problem — going from switching roughly 36 percent of the time on day one to some 96 percent of the time on day 30.
On the other hand, 12 undergraduate student volunteers failed to adopt the best strategy with a similar apparatus, even after 200 trials of practice each.
Why people don't get it
One possible reason people are worse than pigeons at the Monty Hall problem might be due to how people learn.
Past research with university students found they almost universally believed that staying and switching were equally likely to win, while younger students believed this less. Only in the youngest group tested — a bunch of 8th graders — did a significant although small fraction of students figure out switching was the best strategy. It may be that education leads people to acquire ways of thinking that, while efficient, can interfere with certain kinds of performance.
"During 'education,' which I would take to encompass not just formal education, but also one's general life experience, we acquire heuristics — rules of thumb that, either consciously or unconsciously, allow us to respond to a complex world quickly," said researcher Walter Herbranson, a comparative psychologist at Whitman College in Walla Walla, Washington. "But while these heuristics are fast and generally accurate, they're not correct 100 percent of the time."
The scientists propose the curious difference between pigeon and human behavior might be rooted in the difference between classical and empirical probability. In classical probability, one tries to figure out every possible outcome and make predictions without collecting data. In empirical probability, one makes predictions after tracking outcomes over time.
Pigeons likely use empirical probability to solve the Monty Hall problem and appear to do so quite successfully.
"Different species often find very different solutions to the same problems," Herbranson said. "We humans have ways of tackling probabilitybased problems that generally work pretty well for us, the Monty Hall dilemma being one notable exception. Pigeons apparently have a different approach, one that just happens to be better suited to the Monty Hall dilemma."
Empirical probability is a slower, less elegant, bruteforce method that can be tricked by the kind of random fluctuations seen in real data, Herbranson said, but it doesn't employ any mental rules of thumb that can lead to traps such as the Monty Hall problem. In a similar way, the visual systems we depend on to quickly make sense of the world around us can lead to our susceptibility to visual illusions, he added.
Indeed, the aforementioned mathematician Paul Erdos demonstrated the power of empirical probability nicely as well. According to his biography, Erdos refused to accept the explanations of colleagues for the correct solution, and was eventually convinced only after he was shown a simple computer simulation than ran the problem hundreds of times. In other words, "after Erdos approached the problem like a pigeon, he was able to embrace the right answer," Herbranson said.
MaxEntropy_Man Posts : 14702
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Re: "Monty Hall" problem revisited
from hoffman's book:
on his PC, vazsonyi ran a monte carlo simulation of the monty hall dilemma. erdős, who never had much use for computers, watched the pc randomly choose whether to switch or stick. the outcome of hundreds of trials favoured switching two to one and erdős conceded that he was wrong. but the simulation was no more satisfying than the computer proof of the four colour map theorem. it wasn't the Book proof. it didn't reveal why it was better to switch. erdős, who found vazsonyi's explanations lacking, was ready to leave.
.
.
.
erdős did not forget the monty hall problem. he called graham and demanded the Book proof. "the key to the monty hall problem", graham said, "is knowing ahead of time that the host is always going to give you the chance to pick another door. that's part of the rules of the game which you have to figure into your thinking." erdos accepted graham's explanation.
on his PC, vazsonyi ran a monte carlo simulation of the monty hall dilemma. erdős, who never had much use for computers, watched the pc randomly choose whether to switch or stick. the outcome of hundreds of trials favoured switching two to one and erdős conceded that he was wrong. but the simulation was no more satisfying than the computer proof of the four colour map theorem. it wasn't the Book proof. it didn't reveal why it was better to switch. erdős, who found vazsonyi's explanations lacking, was ready to leave.
.
.
.
erdős did not forget the monty hall problem. he called graham and demanded the Book proof. "the key to the monty hall problem", graham said, "is knowing ahead of time that the host is always going to give you the chance to pick another door. that's part of the rules of the game which you have to figure into your thinking." erdos accepted graham's explanation.
bw Posts : 2922
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Re: "Monty Hall" problem revisited
i just gave my older d this problem. at first she gave me the expected wrong answer. i then said her answer was wrong and only told her only that the odds were better and asked her to think about it some more. about thirty seconds later to my amazement she shot back with the following.
she said to begin with the odds of having picked an empty door is 2/3. being shown an empty door later doesn't change that original fact. switching only makes sense if you had originally picked an empty door. so she reasoned that one needs to calculate the composite probability of two successive events a & b:
(a) pick an empty door
(b) given that you have picked an empty door, and after being shown an empty door, switch and get a car.
the probability of (a) is 2/3, and the probability of (b) given (a) has occurred, and that the host has shown you the other empty door, is 1. so the probability for the composite event is (2/3)(1) = 2/3.
in essence she was applying baye's theorem without ever having heard of it.
she said to begin with the odds of having picked an empty door is 2/3. being shown an empty door later doesn't change that original fact. switching only makes sense if you had originally picked an empty door. so she reasoned that one needs to calculate the composite probability of two successive events a & b:
(a) pick an empty door
(b) given that you have picked an empty door, and after being shown an empty door, switch and get a car.
the probability of (a) is 2/3, and the probability of (b) given (a) has occurred, and that the host has shown you the other empty door, is 1. so the probability for the composite event is (2/3)(1) = 2/3.
in essence she was applying baye's theorem without ever having heard of it.
MaxEntropy_Man Posts : 14702
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Re: "Monty Hall" problem revisited
Wow, that's fascinating. I usually have a personal bias towards empirical methods, particularly when the problem at hand is complex. It is interesting to see a renowned mathematician be convinced by empirical results when he was not convinced by theoretical reasoning.MaxEntropy_Man wrote:bw wrote:paul erdős also took a long time to accept that it is better to switch  paul hoffman has an account of this in his delightful book "the man who loved only numbers" (p 237)
i didn't know that!
but i got curious after you posted this info and found this from: http://www.codinghorror.com/blog/2009/06/montyhallmontyfallmontycrawl.htmlThe army PhD who wrote in may have been correct that if all those PhDs were wrong, it would be a sign of trouble. But Marilyn was correct. When told of this, Paul Erdos, one of the leading mathematicians of the 20th century, said, "That's impossible." Then, when presented with a formal mathematical proof of the correct answer, he still didn't believe it and grew angry. Only after a colleague arranged for a computer simulation in which Erdos watched hundreds of trials that came out 2to1 in favor of switching did Erdos concede that he was wrong.
Idéfix Posts : 8808
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Re: "Monty Hall" problem revisited
Idéfix wrote:Wow, that's fascinating. I usually have a personal bias towards empirical methods, particularly when the problem at hand is complex. It is interesting to see a renowned mathematician be convinced by empirical results when he was not convinced by theoretical reasoning.MaxEntropy_Man wrote:bw wrote:paul erdős also took a long time to accept that it is better to switch  paul hoffman has an account of this in his delightful book "the man who loved only numbers" (p 237)
i didn't know that!
but i got curious after you posted this info and found this from: http://www.codinghorror.com/blog/2009/06/montyhallmontyfallmontycrawl.htmlThe army PhD who wrote in may have been correct that if all those PhDs were wrong, it would be a sign of trouble. But Marilyn was correct. When told of this, Paul Erdos, one of the leading mathematicians of the 20th century, said, "That's impossible." Then, when presented with a formal mathematical proof of the correct answer, he still didn't believe it and grew angry. Only after a colleague arranged for a computer simulation in which Erdos watched hundreds of trials that came out 2to1 in favor of switching did Erdos concede that he was wrong.
i don't believe he was convinced by the simulation. he was still seeking a proof from The Book.
Last edited by bw on Fri Apr 12, 2013 12:06 am; edited 1 time in total
bw Posts : 2922
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Re: "Monty Hall" problem revisited
This is amazing! The birds learned from their own experience and adjusted their strategy, although they don't have the tools of probability theory. And human beings are blinded by their theoretical frameworks. Because they expect to see the result of the theory play out, they fail to notice a different result when they perform the task 200 times. All in all, this is a great illustration of the limitations of our heuristics.In the experiments, the birds quickly reached the best strategy for the Monty Hall problem — going from switching roughly 36 percent of the time on day one to some 96 percent of the time on day 30.
Idéfix Posts : 8808
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Re: "Monty Hall" problem revisited
That's a really elegant solution!MaxEntropy_Man wrote:i just gave my older d this problem. at first she gave me the expected wrong answer. i then said her answer was wrong and only told her only that the odds were better and asked her to think about it some more. about thirty seconds later to my amazement she shot back with the following.
she said to begin with the odds of having picked an empty door is 2/3. being shown an empty door later doesn't change that original fact. switching only makes sense if you had originally picked an empty door. so she reasoned that one needs to calculate the composite probability of two successive events a & b:
(a) pick an empty door
(b) given that you have picked an empty door, and after being shown an empty door, switch and get a car.
the probability of (a) is 2/3, and the probability of (b) given (a) has occurred, and that the host has shown you the other empty door, is 1. so the probability for the composite event is (2/3)(1) = 2/3.
in essence she was applying baye's theorem without ever having heard of it.
Idéfix Posts : 8808
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Re: "Monty Hall" problem revisited
Jeremiah Mburuburu wrote:and southern indians, for heaven's bloody sakes, could you put an end to the inanity of attaching a sarcastic "ji" to seva's name every time you mention or address him?
Captain Bhankas Posts : 676
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Re: "Monty Hall" problem revisited
Idéfix wrote:That's a really elegant solution!MaxEntropy_Man wrote:i just gave my older d this problem. at first she gave me the expected wrong answer. i then said her answer was wrong and only told her only that the odds were better and asked her to think about it some more. about thirty seconds later to my amazement she shot back with the following.
she said to begin with the odds of having picked an empty door is 2/3. being shown an empty door later doesn't change that original fact. switching only makes sense if you had originally picked an empty door. so she reasoned that one needs to calculate the composite probability of two successive events a & b:
(a) pick an empty door
(b) given that you have picked an empty door, and after being shown an empty door, switch and get a car.
the probability of (a) is 2/3, and the probability of (b) given (a) has occurred, and that the host has shown you the other empty door, is 1. so the probability for the composite event is (2/3)(1) = 2/3.
in essence she was applying baye's theorem without ever having heard of it.
this morning i told her though her answer is correct, it is incomplete because she has to cover the whole universe of possibilities. if we define the goal as achieving a win, the universe of possibilities include:
(I): choosing wrongly first and then switching to win.
(II): choosing rightly first, and then not switching to win.
she only calculated p(I) = (2/3)(1). to complete the calculation, i told her she needed to calculate p(II) = (1/3)(1). and that the required probability is p(I)/[p(I)+p(II)].
this is a great teaching tool for conditional probability.
if i may speculate a bit on the psychological reason for why people get tripped up, they are evaluating the probability at the moment they are asked the question, forgetting completely the first step, i.e. they forget that the odds of them having made the wrong choice to begin with are higher.
MaxEntropy_Man Posts : 14702
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Re: "Monty Hall" problem revisited
Let’s look at this problem in another way.
The probability that the player initially chooses a certain door, Pi(pl) = 1/3.
Similarly, the probability that the prize initially is behind a certain door, Pi(pr) = 1/3 (and not 2/3, by wrongly subtracting the value of Pi(pl) from 1).
Thus the probability (Pi(pl)=1/3) related to the choice of a door by the player initially is same / equal to the probability (Pi(pr)=1/3) for the prize behind a certain door.
After the host throws open the empty door and before the switch is made, the total for new probabilities involving the player’s chosen door (Pn(pl)) and the prized door (Pn(pr)) must add to 1, increasing in equal amount (proportion) from the initial identical value 1/3 in each case (Pi(pl) and Pi(pr)).
In other words, before the switch is called (after the host throws open the door), the new probability related to the player’s door (Pn(pl)) is 1/2 (instead of the initial Pi(pl) = 1/3) and the new probability related to the prize door (Pn(pr)) is also 1/2 (instead of the initial Pi(pr) = 1/3), such that Pn(pl) + Pn(pr) =1.
Moreover, the probability in each case (corresponding initially to Pi(pl) = 1/3 for player’s door and Pi(pr) = 1/3 for the prize door) gets transformed by the same / equal amount, increasing equally to the new value 1/2 for player’s door (Pn(pl)) and 1/2 for prize door (Pn(pr)).
Considering the new probability values (Pn(pl)) and Pn(pr)) corresponding to the player’s door and the prize door are 1/2 in each case, there is no advantage in terms of heightened probability for winning the prize by making a switch after the host throws open the empty door.
The probability that the player initially chooses a certain door, Pi(pl) = 1/3.
Similarly, the probability that the prize initially is behind a certain door, Pi(pr) = 1/3 (and not 2/3, by wrongly subtracting the value of Pi(pl) from 1).
Thus the probability (Pi(pl)=1/3) related to the choice of a door by the player initially is same / equal to the probability (Pi(pr)=1/3) for the prize behind a certain door.
After the host throws open the empty door and before the switch is made, the total for new probabilities involving the player’s chosen door (Pn(pl)) and the prized door (Pn(pr)) must add to 1, increasing in equal amount (proportion) from the initial identical value 1/3 in each case (Pi(pl) and Pi(pr)).
In other words, before the switch is called (after the host throws open the door), the new probability related to the player’s door (Pn(pl)) is 1/2 (instead of the initial Pi(pl) = 1/3) and the new probability related to the prize door (Pn(pr)) is also 1/2 (instead of the initial Pi(pr) = 1/3), such that Pn(pl) + Pn(pr) =1.
Moreover, the probability in each case (corresponding initially to Pi(pl) = 1/3 for player’s door and Pi(pr) = 1/3 for the prize door) gets transformed by the same / equal amount, increasing equally to the new value 1/2 for player’s door (Pn(pl)) and 1/2 for prize door (Pn(pr)).
Considering the new probability values (Pn(pl)) and Pn(pr)) corresponding to the player’s door and the prize door are 1/2 in each case, there is no advantage in terms of heightened probability for winning the prize by making a switch after the host throws open the empty door.
Re: "Monty Hall" problem revisited
True, I was thinking of it another way... if you want to calculate the probability of a win given a strategy ("switch"), then you need to calculate the sum of probabilities of:MaxEntropy_Man wrote:this morning i told her though her answer is correct, it is incomplete because she has to cover the whole universe of possibilities. if we define the goal as achieving a win, the universe of possibilities include:
(I): choosing wrongly first and then switching to win.
(II): choosing rightly first, and then not switching to win.
(a) choosing wrongly first and then switching, and
(b) choosing rightly first and then switching.
It is obvious that the probability of winning with scenario b is 0.
Idéfix Posts : 8808
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Re: "Monty Hall" problem revisited
Sevaji, this is the source of your confusion. There is one random variable here  the location of the prize. (That is what I called "states of nature" above, borrowing FF's term from the old thread. You are treating the actions of the player as a random variable. The probability that player chooses door A may not be 1/3. If the player is superstitious and believes that A is his lucky letter, the probability of him picking A may be close to 1.Seva Lamberdar wrote:The probability that the player initially chooses a certain door, Pi(pl) = 1/3.
Similarly, the probability that the prize initially is behind a certain door, Pi(pr) = 1/3
Idéfix Posts : 8808
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Re: "Monty Hall" problem revisited
Idéfix wrote:True, I was thinking of it another way... if you want to calculate the probability of a win given a strategy ("switch"), then you need to calculate the sum of probabilities of:MaxEntropy_Man wrote:this morning i told her though her answer is correct, it is incomplete because she has to cover the whole universe of possibilities. if we define the goal as achieving a win, the universe of possibilities include:
(I): choosing wrongly first and then switching to win.
(II): choosing rightly first, and then not switching to win.
(a) choosing wrongly first and then switching, and
(b) choosing rightly first and then switching.
It is obvious that the probability of winning with scenario b is 0.
yes true. so many different ways to skin the cat. maybe that's what she meant. will discuss with her later this evening.
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Re: "Monty Hall" problem revisited
Idéfix wrote:Seva Lamberdar wrote:The probability that the player initially chooses a certain door, Pi(pl) = 1/3.
Similarly, the probability that the prize initially is behind a certain door, Pi(pr) = 1/3
Sevaji, this is the source of your confusion. There is one random variable here  the location of the prize. (That is what I called "states of nature" above, borrowing FF's term from the old thread. You are treating the actions of the player as a random variable. The probability that player chooses door A may not be 1/3. If the player is superstitious and believes that A is his lucky letter, the probability of him picking A may be close to 1.
Similarly, the game show people may be superstitious and like to put the prize only behind door C, thus having the probability for prize close to 1 (behind door C).
Re: "Monty Hall" problem revisited
And if that happens, most players will adjust their strategy based on empirical results, but you won't.Seva Lamberdar wrote:Idéfix wrote:Seva Lamberdar wrote:The probability that the player initially chooses a certain door, Pi(pl) = 1/3.
Similarly, the probability that the prize initially is behind a certain door, Pi(pr) = 1/3
Sevaji, this is the source of your confusion. There is one random variable here  the location of the prize. (That is what I called "states of nature" above, borrowing FF's term from the old thread. You are treating the actions of the player as a random variable. The probability that player chooses door A may not be 1/3. If the player is superstitious and believes that A is his lucky letter, the probability of him picking A may be close to 1.
Similarly, the game show people may be superstitious and like to put the prize only behind door C, thus having the probability for prize close to 1 (behind door C).
Idéfix Posts : 8808
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Re: "Monty Hall" problem revisited
what you don't know: the word "bloody" is used quite unemotionally in nonindian english. on the other hand, dr. bhoshle, "the deputy dean" of k.e.m. hospital, if he ever uses it, would do so only in serious anger. i used it in mild, mostly feigned irritation,Captain Bhankas wrote:Jeremiah Mburuburu wrote:and southern indians, for heaven's bloody sakes, could you put an end to the inanity of attaching a sarcastic "ji" to seva's name every time you mention or address him?
Jeremiah Mburuburu Posts : 1251
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Re: "Monty Hall" problem revisited
and behind door #1 there is a flaw (an online version of the game):
http://www.nytimes.com/2008/04/08/science/08monty.html
related article:
http://www.nytimes.com/2008/04/08/science/08tier.html?_r=0
http://www.nytimes.com/2008/04/08/science/08monty.html
related article:
http://www.nytimes.com/2008/04/08/science/08tier.html?_r=0
MaxEntropy_Man Posts : 14702
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Re: "Monty Hall" problem revisited
Interesting article and very useful simulation.
Sevaji, please play the simulation 200 times. Play 100 times with "switch" and another 100 times with "don't switch." Please report your results here.
Sevaji, please play the simulation 200 times. Play 100 times with "switch" and another 100 times with "don't switch." Please report your results here.
Idéfix Posts : 8808
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Re: "Monty Hall" problem revisited
Seva Lamberdar wrote:Nothing new there ... the same old rehashing.
seva  will you try the NYT simulation? if not, why not?
MaxEntropy_Man Posts : 14702
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Re: "Monty Hall" problem revisited
MaxEntropy_Man wrote:Seva Lamberdar wrote:Nothing new there ... the same old rehashing.
seva  will you try the NYT simulation? if not, why not?
Max, I have already expressed my basic ideas about this problem. Therefore the simulation in this case has no use for me. Moreover, most of the activities for me at this stage, including the Internet discussions, are just hobbies and meant to pass time. Not much interest or use for me to get involved very deeply in anything anymore.
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