Critical thinking is vital to being a good mathematician. You don't go with what seems to be true or what makes the most sense at first glance; you have to analyze. You have to take a deep look at the problem, see how all its part interrelate, and then, having identified everything, solve it using other solutions which have been previously proven. In this way, it works much like science.
There's a rather famous puzzle in mathematics called the Monty Hall problem which, sadly, not many get exposed to in their educations. You can solve it using complex probability formulas, or you can approach it using critical thinking. It came to mind regarding my Ray Comfort *Seems* to be Logical.
The Problem
You are on a game show where there are three doors presented to you. Behind one door is a car and behind the other two doors are goats. If you pick the door which has the car behind it, you win it; if you pick a door which has a goat behind it, you win nothing. The car and goats were randomly placed behind them before the show and the host, Monty Hall, knows which one has the car behind it.
You are asked to pick one of the three doors. Once you pick, it remains closed. Monty then opens one of the two remaining doors which you did not choose. The door he chooses must have a goat behind it. So, if the door you chose has a goat behind it, he will open the other door which has a goat behind it. If the door you chose has the car behind it, he will open one of the goat doors at random. Understand?
There are now two doors remaining: one with a goat and one with the car. You have already chosen one of the doors. Monty turns to you and asks: "Would you like to switch your choice to the other door?"
The problem: Is it (A) advantageous for you to switch your choice now, (B) disadvantageous for you to switch your choice now, or (C) neither advantageous nor disadvantageous to switch right now?
The Answer
(A) Yes, it is advantageous to switch. But why? The overwhelming majority of people answering this answer (C) as they think their odds are equal for all doors. It would seem that assessing the probability does not require considering the past.
The probability of you winning increases from 33% (1/3) to 67% (2/3). But why? The same doors are there, minus one. So, hasn't it just gone from 33% (one out of three) to 50% (one out of two)? Here's an explanation of it, using just critical thinking:
In the first stage (where you make your initial choice), the probability of you picking a car is 33% and the probability of you picking a goat is 67%. So, the probability of you guessing incorrectly is 67%.
In the intermediary stage, the host removes one of the losing doors.
In the second stage, the probability of you having picked the wrong door is still 67%. Therefore, since there are only two doors left, the probability of you being wrong decreases from 67% to 33%, if you switch.
Since the host knows where the goat is, he is guaranteed to eliminate the one with the goat behind it. As such, you have to take into consideration the past.
3 comments:
Let's not make a deal. I seem to remember reading that if the contestant kept his original choice he got $100 whether or not there was a car behind the door. (The probabilities are the same, of course. And $100 was worth more back then -- though not, officially, as much as the car.)
I first came across this problem in Mark Haddon's book "The Curious Incident of the Dog in the Night-time". O.K. it's a children's book, but I made a mistake when I was buying a book to read on a plane at the Copenhagen airport. Mark Haddon went out to point out that many professional mathematicians derided the correct answer when it was posed in "Parade" in September 1990. Not because the problem posed failed to state that Monty Hall would ALWAYS open a door, regardless of what you picked (which is true, if he were malicious he would open your door if you picked a goat, and open another door if you picked the car), but rather that the chance of getting a car always increases to 50%, regardless of whether you stick or change.
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