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
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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
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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
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.
