Master Angela, Dr. Pinheiro, and the Monty Hall Puzzle: Part 2, Discussing Dr. Pinheiro's Solution

Mrs. Angela Kotsiras Secondary School Mathematics Teacher Founder and editor of mathsteachhersonly.com Master of Education (Mathematics) University of Melbourne Melbourne, VIC, Australia E-mail mrskotsiras@gmail.com

Dr. Marcia Pinheiro Lecturer at IICSE University Certified Translator and Interpreter Portuguese & English NAATI  40296          Member: PROz, RGMIA, Ancient Philosophy PhD in Philosophy and Mathematics Master in Philosophy Certified TESOL/TEFL professional Licentiate in Mathematics PO Box 12396 A’Beckett St Melbourne, VIC, AU, 8006 Tel 0416915138 E-mail drmarciapinheiro@gmail.com Web Research Gate

It seems like we need to reach common ground first and then work from there.

Firstly do you agree that we need to prove that

Pr( winning by switching given the host opens a particular door)= 2/3 ?

No, Angela, I don’t. I will repeat here what I have written on the blog post recently:

In short, if we go Mathematics with the Month Hall Problem, we get ½ of chance of winning, since if the door we chose first is the door they chose, only sticking to our first choice will return win, but we have stick or swap (2 possible choices, only one wins). If the door we chose first is not the one they chose, we only win if we swap, what is one in two again (2 possible choices, only one wins). In this way, 2 wins/4 possible choices or 1/2. If you see things differently, you will have to explain to me how you get to your conclusions. Sorry.

This is how I originally thought too but now I wish to find some common ground and then see all possible outcomes on a tree diagram or a table with the corresponding probabilities.  I agree with the 12 outcomes listed from your first post that were all considered equally likely. The probability of these outcomes may be where I now see things differently.

Firstly I need to know if you agree with the following statements.

1.      The car is randomly allocated to a particular door.

2.      The contestant randomly chooses a door.

3.      The host knows which door will reveal the car.

4.      The door revealed by the host is always a donkey.

5.      The door revealed by the host depends on the door chosen by the contestant.

6.      If the contestant chooses the door that will reveal the car then the host will need to randomly choose to open one of the other two remaining doors.

Problem: What is the probability of winning by switching? 

Yes, Angela. I do agree with your premises, all of them, from 1 to 6, I reckon. I cannot immediately see what could be wrong with any of them, definitely not.

Great! We have a common ground.

So if I want to construct a tree diagram for this problem

I can begin by firstly saying that the car can be randomly allocated to either door 1, door 2 or door 3 each with a probability of 1/3.

Secondly, we can assume that the contestant randomly chooses a door, each with a probability of 1/3. So if the car was behind door 1 the contestant could choose either Door 1, 2 or 3. They could also choose door 1,2 or 3 if the car was behind door 2 or door 3. Hence the following tree diagram  illustrates this.

Thirdly since the host knows which door will reveal the car and the door revealed by the host will always be a donkey, then if the car was behind door 1 and the contestant chose door 1 then the host will need to randomly choose to open one of the two remaining doors with a donkey, each with a probability of 1/2.

This would also be the case if the car was behind door 2 or door 3 and the contestant chose door 2 or door 3 respectively.

The partially completed tree diagram illustrates this.

For the remaining cases the host has only one option to choose if the contestant has not chosen a door with the car.  So for example if the car was allocated to Door 1 and the contestant chose door 2, then the only door host Monty could open is Door 3. Hence the probability of Monty opening door 3 given car is allocated to Door 1 and the contestant chooses door 2 is 1.

Hence the completed third stage of the tree diagram is illustrated below

Lastly if the contestant chooses to win by switching after Monty opens a door to reveal a donkey then the final stage of the tree diagram can be completed. For example if the car is behind Door 1 and the contestant chooses Door 1 and Monty opens door 2 then the contestant will choose door 3 if he chooses to win by swapping. In this case however, the contestant would have lost.

The tree diagram below shows the final stage and corresponding probabilities of each of the 12 possible outcomes (win or loss).

Hence

Pr(Win by switching) = 1/9 x 6 = 2/3 that is, if a contestant decided to win by switching his original choice, they would have a higher chance of winning (2/3) than staying with their original choice.

Similarly if the contestant decided to not switch and stick to their original choice then the Pr (Win by not switching)= Pr (loss by switching) 1/18x6=1/3.  

Angela, I do think you guessed the reasoning of the fellows you have mentioned in your first text with me, no doubts about it: That must indeed have been how they got their own result. Here I refer to (Pinheiro and Kotsiras, 2016):

By definition, the conditional probability of winning by switching given that the contestant initially picks door 1 and the host opens door 3 is the probability for the event car is behind door 2 and host opens door 3 divided by the probability for host opens door 3. These probabilities can be determined referring to the conditional probability from the decision tree (Chun 1991; Carlton 2005; Grinstead and Snell 2006:137–138). The conditional probability of winning by switching is  1/3 /(1/3+1/6) which is 2/3. (Selvin 1975b).

The problem with their thinking is that, first of all, conditional probability has to be calculated in a slightly different manner, as you can see on the same blog post of the extract.

Second, we know, from very simple writing, what the problem amounts to, and you can see yourself:

In short, if we go Mathematics with the Monty Hall Problem, we get 1/2  of chance of winning, since if the door we chose first is the door they chose, only sticking to our first choice will return win, but we have stick or swap (2 possible choices, only one wins). If the door we chose first is not the one they chose, we only win if we swap, what is one in two again (2 possible choices, only one wins). In this way, 2 wins/4 possible choices or 1/2. 

(Pinheiro, 2016a).

I still tried to do as you did and find some path to get to the 2/3, as you can see on the same site (Pinheiro, 2016a), paragraphs that follow the above extract.

Notwithstanding, I ended up realizing the mistake that I had committed when I wrote (Pinheiro, 2016c) :

Notice that, in our optimisation table, we forgot to eliminate one door when considering sticking and swapping. If we take away one case from each, we will actually have six cases and we win in three, that is, in half, so that now it is all the same, as it should be. That is for the glory of Maths! 

(Pinheiro, 2016b)

I am sorry Marcia I cannot see why you would take away one case. I can see there are a total of 12 outcomes not six if we take all that we know into consideration. However the probabilities of these outcomes are not all equal and the doors opened by the host are dependent on the contestant’s first choice.

As for your diagram, Angela, we can only see things from one perspective when analysing a problem in Combinatorics, so that we either use the perspective of the presenter, and you then have your 1/3 on his first choice and then 1 on his second choice, or we use the perspective of the audience member, when we have 1/3 on the first choice, then 1/2 (swap or stick) for each possibility from the first set. If they chose the right door the first time and they stick, they win, so that we get 1/6 x 3 for this one. If they chose the wrong door the first time and they swap, they win, so that we get 1/6 x 3 for this one. If they chose the wrong door the first time and they stick, they lose, so that we have 1/6 x 3 for this one. If they chose the right door and they swap, they lose, so that we have 1/6 x 3 for this one. In this case, we have 1/2 chance in the end for swap and 1/2 chance in the end for stick in terms of strategy.

From the presenter’s perspective I agree with the 1/3 in his first choice but it is not always 1 on his second choice since if a contestant chooses a door with the car, the presenter will need to randomly choose one of the remaining 2 doors, each with a probability of 1/2. In this case the presenter’s second choice has a probability of ½ not 1.

Angela, we don’t know what probability you refer to here, I suppose. In Combinatorics, we must have an objective, so say the presenter will have the probability of choosing door number 2 in the end. Then you are probably right. You talked about the presenter choosing one door to conceal the car, if I am not mistaken. That would certainly give him 1/3 for number 2. If the audience member chooses number 2 and the car is there, the presenter cannot open number 2, so that it is 1/3 and 0, what will give us 0. If the audience member chooses number 3 or 1 and the car is on number 2, the presenter has a chance of 1/3 x 0 = 0 of choosing number 2. If the audience member swaps, the reasoning is yet another. If we talk about how many choices the presenter has on each step of the process, we then have 1 in 3 for the first round. In the second round, if the door chosen by the member of the audience contains the car, he has 2 in 2 or 1 chance of getting a door without a car when he opens the door, since the chosen door will never be opened.

I can also add this to the tree diagram by leaving out the last stage of the tree and replacing ‘loss’ with ‘staying’ and ‘win’ with ‘switching’. In that case the Pr (switching to win)=2/3 and Pr(staying to win)=1/3 which is again the same.

Marcia I think it is very important that we take into consideration what we know. If we don’t then I agree with you as in the end we really have 2 doors and the chances of choosing the right door will be 50-50.

That is, as Kalid Azad wrote in his post ‘Understanding the Monty Hall Problem’

‘Suppose your friend walks into the game after you’ve picked a door and Monty has revealed a goat — but he doesn’t know the reasoning that Monty used.

He sees two doors and is told to pick one: he has a 50-50 chance! He doesn’t know why one door or the other should be better (but you do). The main confusion is that we think we’re like our buddy — we forget (or don’t realize) the impact of Monty’s filtering.’

‘The fatal flaw of the Monty Hall paradox is not taking Monty’s filtering into account, thinking the chances are the same before and after he filters the other doors.’

This has also helped me understand that if there were 1000 doors and a contestant randomly chose a door, they would have a 1/1000 chance of winning the car if they stayed with their original choice. No matter what the host does, this probability will not change. Hence when the host eliminates 998 doors then the remaining door will have a 999/1000 chance of revealing the car as the probabilities have to add to 1. Hence switching will always give the best chance of winning.

Dear Angela, that cannot be right: If the presenter goes and opens 998 doors, you know that 998 doors do not have the car, so that your universe of choice has been reduced from 1000 to 2 and therefore your probability of winning is now much higher. Whilst in the first choice you had 1/1000, as you yourself said, in the second round you had 1/2, so that it is now tremendously higher, not only higher. 

Otherwise we can just ignore what has happened and see there are now 2 doors to choose from. In this case there will be a 50-50 chance of choosing the door with the car. 

So, that is better. That sounds right.

Thank you for your time Marcia. I appreciate your efforts to make me see things differently. I am grateful for the time I spent in understanding this problem and all the underlying information.

My students had a great time discussing this problem with their peers and their families and were able to see a number of ways this problem could be interpreted and solved.

I really don’t think how this problem could have any result that is different from 50%, Angela.

References

Pinheiro, M. R. (2016). Monty Hall: In Short. Retrieved 29 October 2016 from http://mathematicalcircle.blogspot.com.au/2016/10/monty-hall-in-short.html

Pinheiro, M. R. & Kotsiras, A. (2016). Master Angela, Dr. Pinheiro, and the Monty Hall Puzzle: Part 1, Discussing Dr. Pinheiro’s Solution

http://mathematicalcircle.blogspot.com.au/2016/10/master-angela-dr-pinheiro-and-monty.html

Pinheiro, M. R. (2016a). Monty Hall: In Short. http://mathematicalcircle.blogspot.com.au/2016/10/monty-hall-in-short.html

Pinheiro, M. R. (2016b). Now the Detail: Optimisation Finally Equals Mathematics. http://mathematicalcircle.blogspot.com.au/2016/10/now-detail-optimisation-finally-equals.html

Pinheiro, M. R. (2016c). Master Angela and Inspiration: Our Solution? http://mathematicalcircle.blogspot.com.au/2016/10/master-angela-and-inspiration-our.html

Kotsiras, A. (2016). Better Explained blog post.

https://betterexplained.com/articles/understanding-the-monty-hall-problem/

Foundations and Exponents

A fellow from Research Gate posted this:

What is wrong??

2/6 is equal to 1/3 and we have :

(-1)^(2/6)=((-1)^2)^ 1/6=1 and

(-1)^(1/3)=-1

what is wrong??

Answer:

You have wrongly applied a property, which is the commutative property for exponents. 

You originally had (-1) raised to something. 

You can pass the exponent to inside of the brackets, but, in this case, it goes to the top of the 1 you had. 

If you want to keep the brackets, then you would have 2*1/6, not 2^(1/6): They are different things. 

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Now the Detail: Optimisation Finally Equals Mathematics

Notice that, in our optimisation table, we forgot to eliminate one door when considering sticking and swapping. If we take away one case from each, we will actually have six cases and we win in three, that is, in half, so that now it is all the same, as it should be. That is for the glory of Maths! 

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Monty Hall: In Short

In short, if we go Mathematics with the Monty Hall Problem, we get 1/2  of chance of winning, since if the door we chose first is the door they chose, only sticking to our first choice will return win, but we have stick or swap (2 possible choices, only one wins). If the door we chose first is not the one they chose, we only win if we swap, what is one in two again (2 possible choices, only one wins). In this way, 2 wins/4 possible choices or 1/2. 

If we go Optimisation, we have 9 possible cases and 3/9 win and that means sticking and 6/9 win and that means swapping, so that 1/3 means sticking and 2/3 means swapping, and, from there, we recommend always swapping as a good strategy.  




Do you believe that Optimisation could return a different result? Then please read Optimum, since, first of all, not all is to be believed. 

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Management x Mathematics

This is looking a lot like a rule for me; this thing of telling one thing from another. 


We must tell the difference between purely human, mathematical, and computer language.


We must also tell what is management from what is Mathematics. 


This is all coming from discussing the so famous Monty Hall Problem. 


More specifically, it is coming from the post Monty

Imagine a ship that is sinking. 


The captain has choices to make, let's say who will get a boat, and therefore who will get saved.


There are 36 seats, and 42 passengers. 


The captain may go history and find out that, in an interval of 10 years, whenever they decided to save young people first, a person who mattered the most, let's say, a scientist of many publications, died. 


The captain decides to save everyone who is old first.


Because they are slow people, 10 kids have died.


The captain's decision was a managerial one. 

If people ask mathematicians what they think about it, they will probably answer that there should be enough boats and seats, so that everyone could be saved. 


An ex-fellow, from Senai, R., teacher of Physics, told me this joke: The guy was walking and got lost. He found a man in a tree and decided to ask: Where am I? The guy, from the tree answered: You are under a tree, this tree is where I am now, and you are on a road that is inside of a forest. What is the profession of the guy who was in the tree? I starred at R. and had no answer: I don't know, I said. He said: Mathematician (no, don't think like that, that Brazilian men are always hostile to women at work and frequently offend them: It is not true at least sometimes...). I go: Why? R. said: he gave a completely useless and obvious answer (notice that the answer may be useless and obvious, but it is still totally true, restrictions of purely human language considered).

That is indeed the case at least here. 


Management has its own ways, so that they may indeed decide to choose Swap all the time because they are lazy, because they have seen the historic results of the game and think they will have more chances of winning. 


A mathematician will say: at this stage, on the second question, if you swap, you have 50% of chance of winning. If you stick, you also  have 50%. Up to you. 

Notice that a larger chance, due to historic results, does not mean you will win if you swap. 


You  can (obviously) still be the unlucky person who swaps and loses. 


In this way, the mathematician would not have any recommendation for you, just like in the case involving the boats and the humans. 


If given more time to think, they would probably say that the kids could be a hope of getting a brilliant scientist, the old person could be the person to donate their kidney to somebody important and they would then be saved, their blood type is exotic, etc., like they would probably say that everyone is important in their own way, so that they would not give any decision on what to do in such a situation: each head, a sentence, basically. 


The optimiser, however, would say: Swap! They would also say: Old people first. I suppose in this way one can understand better why the confusion appears. 


Coming back to R., Mathematics will always tell the truth, that is, what is perhaps obvious, and, for some, interested in real-life situations, perhaps useless. 


Mathematics is worried about everything being correct, about all being perfect. 


It would be imperfect suggesting that a person always swapped in the game because they may lose. 


We can at most say what is said: historically, people who swapped won more times. 


Even so, it is possible that that is not a reality in the history of the game: we could easily have the least likely option always happening, all the way through, and that is actually what a mathematician would be obliged to tell you. 


It is quite possible that, in another show, that be not Monty Hall's, everyone who stuck to their first choice, absolutely everyone, won, and most of the people who swapped lost. 


If probability meant something, everyone who chooses it would win the Lotto. 


Yet, if the games were for that purpose, the organisers would break quite frequently, is it not? 


Games are always chancy things. 


All we can say, as mathematicians, is what we say: At that stage, you have 50-50. 


As another point, stick and swap is not what you are actually doing in the problem at that height: you are actually choosing one in two doors instead. 


Of course, this is when we are talking about Mathematics, and, as R. said, mathematicians are useless and all they say is obvious, so that everything is the most objective thing as possible. 


Sticking and swapping are not as objective as choosing, which is definitely what repeats. 


Combinatorics is about repetition, and sticking and swapping are novelties introduced there to confuse us. 


We are repeating action, not doing something new: what we are doing is choosing again, this time between two instead of three. 


Mathematics needs the right words to be used, and that is why we created Classical Logic. 


It does not work in any other universe. 


Please read Words for Science. 


In R.'s terms, useful would be someone who can tell others what to do: Swap or stick. 


In the Mathematicians' World, we give people choices: We only talk about things that are one hundred percent true. 


If we advised people, and then told them to always swap when they went to the Monty Hall Show, we would be taking away their freedom: We must make sure they choose and take responsibility for their choices. 


If they lose, they will blame the optimiser and R., not us. 


We are useless in the sense that we don't tell you what to do.


We are useful in the sense that we truly empower you.

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