I'm pretty sure most everyone is wrong about the "Let's Make a Deal" riddle. Have you heard this one? Check it out if you haven't, and just think on it for a while. I maintain that there is no advantage to switching doors after the host opens one of the doors. It's abuse of probability mathematics. In reality, your choice before the host opens a door has no consequence. Your choice after the host opens a door is the only one that matters, and that choice is 50/50. Please enlighten me if I am being a complete fool.
UPDATE: I like this guy's explanation, if you insist on cohering the first and second choices. Since the host will always pick a goat, it's as though there were only 2 doors available in the first place: 1 with a goat, and 1 with a car. The 3rd choice is only an illusion. Let's try this with 4 doors, labeled @, #, *, and Q. 25%, 25%, 25%, and 25%. Then, Monty opens @ and *, revealing a cardboard cut-out of Michael Richards and a poorly-drawn portrait of Snoopy. Now, take into account that the desired prize is behind door #. Your chances of picking Q were no greater than your chances of picking # in the beginning. It was 25% and 25%. Therefore, at this stage of the game, you are no more likely to have chosen the losing door than the prize door (50/50), and your next choice is 50/50 as well. Lots of good discussion here, along with plenty of bogus conclusions.
UPDATE 2 (the last): Ah, here are some people with empirical evidence proving me wrong, and an interesting statement from someone as stubborn as I am. My head is still spinning....
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10 years ago
9 comments:
Here's my approach:
The contestant has two decisions to make: which door to choose, and whether or not to switch. She has three options for the first decision (choose the car, choose goat #1, choose goat #2), and two options for the second decision (to switch or not to switch). That means there are 6 possible combinations of decisions, given below:
A. car
a. switch (win a goat)
b. no switch (win the car)
B. goat 1
a. switch (win the car)
b. no switch (win goat 1)
C. goat 2
a. switch (win the car)
b. no switch (win goat 2)
The reason the problem is counterintuitive is that of the 6 potential combinations, 3 result in winning the car and 3 result in winning a goat. 50% chance, right? BUT, the riddle only asks about whether or not it is advantageous to switch. In other words, for the purpose of the riddle, only the second decision matters. And when you tally up the results of those 2 options (i.e. switch and no switch), you see that switching results in winning the car 2 times out of 3, and not switching results in winning the car only 1 time out of 3.
Convinced?
Here's how I think of it:
There's a 66.6% chance the car's behind a door you didn't choose, then Monty does you the favor of taking one of those doors out of the equation.
The 66.6% chance doesn't go away, it shifts behind the door you didn't choose and Monty didn't open.
Ok, just forget the first choice; it's unrelated to the second choice. I consider it outside the scope of the problem. The 6 possible combinations you've proposed, Adam, are not equally likely. I think that's the main problem with the math. Your first choice is 1 in 3 (33%), and your second choice is 1 in 2 (50%). The riddle asks about the second choice, so why include the first choice in the calculation?
I think it's a form of the gambler's fallacy, but an especially tricky one. People tend to think that if they flip a coin 100 times and it lands on heads every time, that their chances of getting tails on the 101st toss are greater than 50%, because you want to think of all of the tosses as one cohesive object/entity/problem and do a calculation (0.5x100=0.005, so the chances are 5/1000; good odds), but they're not. It's abuse of probability math. Similar problem here.
So all it took to convince you was a stubborn guy with empirical evidence? I could've done that!
For me, I'll take mathematical, deductive reasoning over "empirical," inductive reasoning any day. Math is not something that can be "abused," as you say; it can only be sound or unsound. If the math seems to support a conclusion that is clearly false, it is because the person who performed the mathematical operations used unsound math--not because she abused proper math. I still maintain that the math I used in my example above is unimpeachable.
As long as I'm procrastinating at work anyway, I might as well prove that you are wrong about the situation in which the contestant is presented with 4 doors...
I'll use the same logic I used earlier: Contestant has three options for the first decision (choose the car, choose goat 1, choose goat 2, choose goat 3), and two options for the second decision (to not switch, to switch to a remaining door, to switch to the other remaining door). 4x3=12 possible combinations.
A. car
a. no switch (win car)
b. switch to goat (win goat)
c. switch to goat (win goat)
B. goat 1
a. no switch (win goat)
b. switch to car (win car)
c. switch to goat (win goat)
C. goat 2
a. no switch (win goat)
b. switch to car (win car)
c. switch to goat (win goat)
D. goat 3
a. no switch (win goat)
b. switch to car (win car)
c. switch to goat (win goat)
Given the possible outcomes, the contestant has a 3/8 chance of winning the car if she switches. She only has a 1/4 (or 2/8) chance of winning if she doesn't switch. Therefore, she would put herself in an advantageous position by switching.
I would propose, in fact, that no matter how many doors there are, the contestant will ALWAYS benefit from switching. Of course, the statistical advantage shrinks slightly the more doors you add, but it will always exist.
Aww shit, Jeff-rey! You just got had, Home
Slice. Props to my man AG - bustin' a deductive-logic-cap inna "Let's Make a Deal" ass! Yo yo yo --- A-Rizzle in the @#(*ing HOUSE, dogggggg!!!!!
"Math is not something that can be abused..."? Ach! Semantics. When "unsound math" is used to prove a point, math is being used improperly (i.e. "abused"). Adam, you must define "abuse" differently from how Merriam-Webster does....
Theory is great stuff, but reality is the ultimate authority--at least, in an argument like this--don't you think? You can craft a most eloquent proof and convince your fellow philosophers that the world is flat, but as soon as someone orbits the Earth and sees that it is actually spherical, you must accept that you were wrong.
I could probably convince a 12-year-old that switching doors is not advantageous by just explaining my reasoning to him, but I'd still be entirely wrong. My advice for 12-year-olds is to do their own testing, no matter how convinced they are. Evidence rules. Long live empiricism!
Of course, like unsound math, there exists also unsound evidence. I (like many others) have not the ability to distinguish sound probability math from unsound probability math with much certainty, so when a question can be so easily answered by a simple experiment, why trust the math? I'll trust a 12-year-old's simple experiment before I'll trust anyone's probability math. Anyone's.
Dear Jeff,
My first thought is that you were confusing a math problem for a science problem. Generally speaking, math problems are solved by deductive reasoning and science problems are solved by inductive reasoning, and where it is possible, deductive reasoning is preferable. Sound deductive reasoning elicits proof, whereas sound inductive reasoning elicits evidence. Evidence and proof are not the same thing.
That does not mean that one cannot reasonably apply inductive reasoning (or empiricism, or science, or whatever you choose to call it) to a math problem. Suppose I tell a child to perform the mathematical operation 10x10, and then that child tells me he doesn't know what I mean. I explain to him that "ten times ten" means the total amount when you put together ten groups of ten. So he goes and collects a bunch of bugs, makes ten piles of ten, moves them all into one big pile, counts them, and then comes back in and tells me, "100." That child used inductive reasoning (i.e. "because I tried X and Y happened, X results in Y") to come up with a correct result to a math problem.
But then suppose I ask the child how he knows that making ten piles of ten bugs and counting them will always result in 100? Or alternately, how he knows that he might not get a different answer if he conducts the same test with 10 piles of rocks? Without the cognitive structure to generalize and deduce based on mathematical principles, the only thing the child can do is head back outside and try it again. After he tries it a bunch of times and always gets the same result, it will eventually become evident that 10 times 10 always equals 100. (Although in order to be right, his experiment would naturally need to test for all the potential variables.) Empiricism would work, but it's hard to see how that method is preferable to math. Even though the child clearly isn't familiar with multiplication, it would be better to eventually teach him, rather than have him going through the same tedious scientific process every time someone gives him a multiplication problem.
What's true of 10 x 10 is true of the "Let's Make a Deal" riddle. Tedious empiricism will eventually make it evident that it is advantageous for the contestant to switch, but it is hardly preferable when a simpler and more accurate technique is available. The advantage of deductive reasoning to inductive reasoning is even more pronounced in the case of probability problems like this one; while inductive reasoning will lead fairly quickly to the conclusion that it is advantageous for the contestant to switch, deductive reasoning will actually tell you exactly what the advantage is: 2 to 1. Empiricism produces no such exactness.
Love,
Adam
To my dearly beloved Smasher:
Generally, I agree with you: "...where it is possible, deductive reasoning is preferable." However, in this situation, I knew that sound deductive reasoning was possible, I just couldn't distinguish it from unsound reasoning. I didn't know who to trust--mathematician/scientist A, mathematician/scientist B, or kid with sticks. Sure, the kid could be counting wrong, but he counted a few times and he always came up with about 100.
I think it's similar to when you try to convince your old decrepit bio professor that you've got a really strong acid in your beaker by showing him the read-out on the digital pH-meter and he doesn't believe you, and says, "Gimme that," dips his finger in the beaker, touches it to his tongue, tastes how sour it is, and concedes. The digital pH-meter is too fancy. It's beyond his understanding, and that makes him uncomfortable.
The math is too tricky for me in this problem. There are different ways to look at the problem and different ways to "run the numbers," you know? I trust the simpler method. I guess I'm just a simpleton.
Forever yours,
J. Michael
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