Mrs.
Angela Kotsiras
Secondary
School Mathematics Teacher
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
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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
