Master Angela, Dr. Pinheiro, and the Monty Hall Puzzle: Part 1, 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

So, I am sorry, Angela, I do feel like calling you Professor Angela Kotsiras instead, but I will force myself to call you Angela, as you requested. 

I understand you have a strong interest in the Monty Hall Problem. I would like to know how this interest appeared and what exactly makes you think this is an interesting problem. 

This month I am teaching at a school where students are learning about counting methods and associated probability. I would like to present the Monty Hall problem to them but I want to be clear about its solution so I do not guide them the wrong way. The problem has always interested me as I intuitively said that the contestant would have a 50-50 chance of choosing the right door, but further reading suggests it is not the case, at least not until I read your article The Monty Hall Show and Murphy’s score (Pinheiro, 2016b).

I had a copy of Professor Posamentier’s book called Math Wonders (Pinheiro, 2016a) to Inspire Teachers and Students and I read his explanations  which suggested there is a higher chance of getting the car if the contestant swaps. He uses the example of 1000 doors. I also discussed the problem with a colleague who suggested I think about the problem as the contestant choosing a particular door 900 times. They would then expect to get the car, in the long run, 300 times based on the door they initially choose. If they switched they would expect to get the car 600 times. Since the host opened the door with the donkey, they would expect to get the car 600 times if they switched. This made sense even though intuitively I would have said there is a 50-50 chance.

I now have come across the following tree diagram, which uses Bayes’ Theorem, showing the probability of every possible outcome if the player initially chose Door 1, that is, the conditional probability of winning by switching given that the player initially chooses Door 1 and the host opens door 3 is probability of the event the car is behind door 2 and host opens Door 3 divided by the probability for the host opens door 3. This would then also apply if he chose Door 2 or Door 3.

(Wikipedia, 2016)

As such it is better to switch than to keep.

Angela, what you present is really interesting. I also understand very well what you say: You want to do what is right and obviously teach only right things to your students. That is what we usually call ethics in teaching, I suppose: We should only teach what we know to be right. I was once doing probation in Primary Teaching and I had the teacher who was supposed to give me the example on how to teach actually asking the students how to spell a certain word. I was at the back of the room and wanted to help her, so that I kept on moving, I moved quite frantically, but she would never look at me, as if determined to only accept her own ways. I obviously knew the answer for that one, is it not, Angela? She should have trusted her probationer. Oh, well, that was the sigmatoid (Pinheiro, 2015) choveu, a sigmatoid that belongs to the Portuguese language and means rained. She was feeling like writing chuveu, which is wrong spelling, and she felt the pain of the doubt right at the beginning of her writing. She then took the catastrophic path of going for voting. As she asked, the class ALSO went 50/50 on the topic, so that half the kids raised their hands to say it was chuveu, and the other half raised their hands to say it was choveu. It is a shame that she never ever wanted to know my opinion, Angela. I was blue of so much waving at her from the back, and, back then, I did not have any better idea on how to go. She then went and wrote chuveu, very unfortunately. Oh, well, I tried to make sense of your graph, but I sincerely could not understand it. Apart from the beginning, where we can clearly see it would be 1/3 of chance if we chose any of the doors, I could not make sense of much. Perhaps you mean, on the first lines, that if the car is hidden behind Door 1 and the presenter opens Door 2 or 3, you still have 1/2 chance for each remaining door, chance of finding the car. I don’t know how to make sense of the lines that follow that however: The car is then behind Door 2. Why would we not have the same reasoning? Opening Door 1 or 3, then 1/2. Same for the last group, when we would have opened Door 1 or 2, then 1/2. In this case, all would return the same 1/6 and we would have equal probability. I think that is clear enough. As a mathematician, you must also know that we can add all probabilities in the end to make sure we did the right thing. Only by having 1/6 in each option can we get 6 x 1/6 = 1, which should be our whole, is it not? At the moment, you have 2/6 + 2/3 = 2/6 + 4/6 = 6/6, which is equal to 6/6, which would be our 1, is it not? Notwithstanding, you don't have uniformity in reasoning, but the multiplicative reasoning bears uniformity. Perhaps you will now come back to your old decision, which was agreeing with me. Will you?

The probabilities of all the outcomes in the tree diagram add to 1. 

Angela: I went back to the diagram after you wrote that. In a sense, you are right. I do understand part of what you say in the above paragraph, I think. The first column does give you one, since it is 1/3+1/3+1/3=1. The second column gives you 1/2+1/2+1+1=3, so that this is not one and we already got something that opposes what you wrote. The third and last column gives you 1/6+1/6+1/3+1/3=2/6+2/3=2/6+4/6=6/6, so that you are not wrong about the sum of this column, but the problem is how they get there, you see. You must observe that Combinatorics comes from multiplicative reasoning, so that there is usually equal distribution for each option, what is missing there. If you have two options for the first case, you would expect two options for all of them, which is what I am pointing out: The missing cases, the cases that they did not count. You are right about the sum in the column, but not about how you get to it in that case.

The following may make this clearer which was the source from which the tree diagram was taken from.

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).

How is this wrong?

Angela, once more, Combinatorics is multiplicative reasoning in its purest, the how we multiply from having a sum, basically. I am not disagreeing with the concept of conditional probability, which is the probability of an event happening given that another event took place before that event, so say the probability that I pick a flower from inside of a box when there are two boxes in six with flowers inside and my fellow went before me and drew one. If he found a flower, I have one in five. If he didn’t, I have 2 in five, so that my probability of finding a flower in those boxes is conditioned to what he finds. Either you go for the tree or you go for formulae, I suppose. If you use the tree and conditional probability, you will then have 1/6 for each line, but remember we should have two branches for each case instead, for each line you see in the first column of your graph, so that instead of 1/2, 1/2, 1, and 1, we should get 1/2, 1/2, 1/2, 1/2, 1/2, and 1/2. I however would have a completely different diagram. That was my point, what I previously said. Now, if we go with the formulae, we have to be careful to, first of all, pick the right formula, is it not? In Brazil, it was very common using a book they have named Tabelão, which I learned how to use with Prof. Alice, a female professor from FURG, and I simply love whenever a person helps me add positively to my body of knowledge, information or insight, do you? Prof. Alice was really cool, to be sincere, as for her figure. I actually remember having mentioned her to the sinister Trevor in that end of 2001. You don’t believe, but this woman had the courage of teaching us wearing shorts and revealing tops, having a short revolutionary hair, and she used to go around on a motorcycle. I still cannot believe that woman did that in Brazil, South of Brazil. Wow! She was one that entered my book of idols, basically, but for several reasons, not only her daring looks and attitude: Once more, I learned how to use Tabelão (Big Table) with her. That was a red book, enormous, basically containing all mathematical formulae one can think of. Oh, well, I don’t have my Tabelão with me, Angela, since I suffer really heavy crime for more than 14 years, but, if I go through the effort of finding this formula, I think I am sure we will have to agree with disagreeing with your extract at some point. Let’s do that, Angela! You are not here as I write this paragraph, so here I go:

(UAH, 2016)

We want to choose between swapping and sticking given that, let’s say, the other person chose door number n. That is our actual problem, right? Event B is then winning when choosing n. They had three possible choices and chose n. Event A is winning (getting the car), regardless of the choice made (sticking or swapping). If you observe well, choosing, when hearing the second question, is equivalent to simply restarting the game, like which one now? If we stick, we choose one of three options. If we swap, we choose one of two remaining options. Our events are independent, it seems, since we have what is called replenishment or putting the item back to the bag. That is important when we analyse things in Combinatorics, this issue of independence. Our probability of the intersection, in this case, is just multiplying P(A) by P(B) because of the independence. In this case, we have 1/3 * 1/3 = 1/9 for the numerator of the fraction. P(B) is 1/3, so that we have 1/3 as a final result. Now, if we consider that one door disappears from the game, we have 1/3 and 1/2 in the tree of choices, as you see in your own diagram/quote. The symbol \B means given that B occurred, as you know. In our own example, involving the boxes and the flowers, the fellow going there and opening the box with the flower would be the event that had already occurred. 

References

Wikipedia. (2016). Monty Hall Problem. https://en.wikipedia.org/wiki/Monty_Hall_problem

UAH.(2016). Conditional Probability. http://www.math.uah.edu/stat/prob/Conditional.html

Pinheiro, M. R. (2015). Words for Science. Indian Journal of Applied Research, 5(5). http://www.academia.edu/12181924/Words_for_Science

Pinheiro, M. R. (2016). Marcia Ricci Pinheiro. https://www.researchgate.net/profile/Marcia_Pinheiro4/publication/308256028_The_Monty_Hall_Show_and_Murphy's_Score/links/57df294408ae72d72eac21ab.pdf?origin=publication_detail

Pinheiro, M. R. (2016a). Monty Hall, Prof. Posamentier, and us.  https://drmarciapinheiro.wordpress.com/2016/09/29/monty-hall-prof-posamentier-and-us/

Pinheiro, M. R. (2013). The Monty Hall Problem and a few moments of shame for Modern Science and scientists: Newcastle, 2000, Australia. http://mathematicalcircle.blogspot.com.au/2013/09/the-monty-hall-problem-and-few-moments.html

Pinheiro, M. R. (2016b). The Monty Hall Show and Murphy’s Score. https://www.researchgate.net/profile/Marcia_Pinheiro4/publication/308256028_The_Monty_Hall_Show_and_Murphy's_Score/links/57df294408ae72d72eac21ab.pdf?origin=publication_detail