
Electric Wizard |

Peter Wason (1966), paraphrased, wrote:
You are shown a set of four cards placed on a table, each of which has a number on one side and a colored patch on the other side. The visible faces of the cards show 3, 8, red and brown. Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on one face, then its opposite face is red?Only 10% of people get this question right. This number is more or less culture-independent.
Let's see:
Truth table for A -> B:
F -> T => T
F -> F => T
T -> T => T
T -> F => F
A := card shows even number
B := opposite face is red
Testing: card shows even number -> opposite face is red => ?
A is False for card showing 3.
A is True for card showing 8.
B is False for brown card.
B is True for the red card.
The only cards I would turn over are the 8 and the brown card.
Because they the ones that can test the fourth condition in my Truth Table.
.
Am I Right? (Or Am I Left?) I have my fingers crossed...

Orfamay Quest |

The Wason task wasn't really intended to be a challenge for the thread, but since people seem to be having fun with it, I will hold off for a while on providing the answer.
I think my basic point is made; we've had four answers submitted so far, and since they're all different, at most one can be right, suggesting that this is a deceptively difficult task. (I'd also point out that analysis and explanation of this task has swallowed Ph.D. students since, oh, about 1966.)

Freehold DM |

Electric Wizard wrote:i really am surprised the play-out in this tread is supporting the thesis.I'm going to politely disagree.
I've seen two people actually answer the question. Big Norse Wolf has the "correct" answer, and raven1272 performs the *very* common error (sorry, Raven! Don't like singling you out here, but no one else was brave enough to try!) of assuming that a 13'x10' section minus a 4'x4' section is a 9'x6' section.
So so far the results are 50/50.
And I would believe that more than 50% of ALL adults (nationality notwithstanding) would make Raven's error. It's one of the most common ones I've seen in teaching geometry. Square footage is far more convoluted than most people think.
The other couple of dozen posts (give or take) are just general comments on the "unreality" of such questions. Would you really buy a square yard of carpet and cut it into teensy pieces just to save money, or would you spend a bit extra to minimize the number of separate pieces?
And now, a true story because it was one of my favorite teaching moments: I was teaching Statistics at a community college, which is NEVER a fun task. After a particularly nasty example, a student raised his hand and said, "You're not Dr. Good, you're Dr. Evil!"
So since it was four days before Halloween, I rolled with it. After class on October 30 I went to San Francisco, shaved my head, got professional-quality scar makeup and costume, and arrived on the 31st in full Dr. Evil regalia, complete with leather-clad beautiful female enforcer (courtesy of my wife). I got a standing ovation from the class.
And the rest of the semester was, "I am trying to take over the world using this device. It works only x% of the time, or on only y% of the people. I need to control this many people for my scheme to work. What are my chances of success?"
The students loved it. I even had one student who was failing horribly who wrote on an exam, "I refuse to help you in your nefarious schemes, so in good conscience I cannot...
are you sure you're a math teacher? You seem to be against trick questions and not too into ridiculing others who get answers wrong.

NobodysHome |
1 person marked this as a favorite. |

are you sure you're a math teacher? You seem to be against trick questions and not too into ridiculing others who get answers wrong.
It gets worse. I used to tell my students, "I've never met anyone who's bad at math. I've only met people who have had a bad math teacher."
That's the sad part. It only takes one bad math teacher to give someone a lifelong distaste for math, and I'm sorry to say I met quite a few over my short tenure as a math professor before moving to private industry.The phrase worked beautifully until I had a wonderful student with a learning disability who honestly could not do the problems. I ended up personally marching her to the disability center to get accommodations. And in two weeks she'd gone from an F to an A.
I teach like I GM: You're there to cooperate with your students/PCs to accomplish something, NOT to prove you're "smarter" than them by ridiculing them or showing off your skills/power.
Maybe that's just me...

thejeff |
Peter Wason (1966), paraphrased, wrote:Only 10% of people get this question right. This number is more or less culture-independent.
You are shown a set of four cards placed on a table, each of which has a number on one side and a colored patch on the other side. The visible faces of the cards show 3, 8, red and brown. Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on one face, then its opposite face is red?
I'll chime in on this one to increase the numbers.
8 & brown.
Orfamay Quest |

Before I reveal the correct answer, I'd like to present another situation:
You are a police officer, responsible for enforcing the drinking laws at a local festival. Basically, anyone can come to the festival (and can drink soft drinks), but if you want an alcoholic beverage, you need to be of legal age.You encounter the following people in the food tent:
* Someone around the age cutoff drinking out of a beer bottle
* Someone around the age cutoff drinking a can of lemonade
* An obviously-underage person drinking a glass of juice (that may or may not be spiked)
* An obviously-legal person drinking a glass of juice (again, possibly spiked).Whom do you need to check?
It should be obvious to most people that the only potential lawbreakers are the person drinking the beer and the underage person. (Anyone can drink lemonade, and a 40 year old can drink anything she likes. Typical accuracies for this problem are around 90%.)
This problem is formally identical to the one I posed earlier.
The answer, of course, is 8 and brown. (Congratulations to the people who said that.) An odd number, like a 40 year old, can have anything it likes on the other side, and if the opposite face is red, like lemonade, any number can have that color.)

thejeff |
Before I reveal the correct answer, I'd like to present another situation:
Puzzle #2 wrote:It should be obvious to most people that the only potential lawbreakers are the person drinking the beer and the underage person. (Anyone can drink lemonade, and a 40 year old can drink anything she likes. Typical accuracies for this problem are around 90%.)
You are a police officer, responsible for enforcing the drinking laws at a local festival. Basically, anyone can come to the festival (and can drink soft drinks), but if you want an alcoholic beverage, you need to be of legal age.You encounter the following people in the food tent:
* Someone around the age cutoff drinking out of a beer bottle
* Someone around the age cutoff drinking a can of lemonade
* An obviously-underage person drinking a glass of juice (that may or may not be spiked)
* An obviously-legal person drinking a glass of juice (again, possibly spiked).Whom do you need to check?
Though, if you didn't watch closely, the can of lemonade could have been spiked too. Pouring a shot of booze into a can of soda was a fairly common way of walking around with alcohol back in my college days.

Electric Wizard |

Before I reveal the correct answer, I'd like to present another situation:
Puzzle #2 wrote:
You are a police officer, responsible for enforcing the drinking laws at a local festival. ...
This one is easier because cops have to follow rules; and the only
Probable Cause (in the description) is the dude with the open beer.Otherwise, you end up with these situations.
.

Orfamay Quest |
1 person marked this as a favorite. |

Orfamay Quest wrote:Before I reveal the correct answer, I'd like to present another situation:
Puzzle #2 wrote:
You are a police officer, responsible for enforcing the drinking laws at a local festival. ...This one is easier because cops have to follow rules; and the only
Probable Cause (in the description) is the dude with the open beer.
Otherwise, you end up with these situations.
Probable cause doesn't apply. It's just a Terry stop (q.v.)

BigDTBone |

BNW wrote:... (mathematicians round down. Carpet dealers want every penny!).
If the next digit is 5 or higher round up, else round down <-- how mathematicians do it.
e.g. rounding to the penny158.66666 ~ 158.67
158.55555 ~ 158.56
158.94411 ~ 158.94
158.94511 ~ 158.9599.9999 ~ 100.00
99.9949 ~ 99.99
.
I was taught that when rounding 1-4 rounds down, 6-9 rounds up, 5 followed by an even rounds down, and 5 followed by an odd rounds up. This is to ease the drift up effect from always rounding five up (ie 5/9 times rounding up inflates numbers). The problem still exists but by using this method you make it 2 orders of magnitude smaller than your lowest significant figure.

Freehold DM |

Grand Magus wrote:I was taught that when rounding 1-4 rounds down, 6-9 rounds up, 5 followed by an even rounds down, and 5 followed by an odd rounds up. This is to ease the drift up effect from always rounding five up (ie 5/9 times rounding up inflates numbers). The problem still exists but by using this method you make it 2 orders of magnitude smaller than your lowest significant figure.BNW wrote:... (mathematicians round down. Carpet dealers want every penny!).
If the next digit is 5 or higher round up, else round down <-- how mathematicians do it.
e.g. rounding to the penny158.66666 ~ 158.67
158.55555 ~ 158.56
158.94411 ~ 158.94
158.94511 ~ 158.9599.9999 ~ 100.00
99.9949 ~ 99.99
.
My grandmother was taught that way, and it resulted in a lot of arguments when she helped me with my homework.

Freehold DM |

Freehold DM wrote:are you sure you're a math teacher? You seem to be against trick questions and not too into ridiculing others who get answers wrong.It gets worse. I used to tell my students, "I've never met anyone who's bad at math. I've only met people who have had a bad math teacher."
That's the sad part. It only takes one bad math teacher to give someone a lifelong distaste for math, and I'm sorry to say I met quite a few over my short tenure as a math professor before moving to private industry.The phrase worked beautifully until I had a wonderful student with a learning disability who honestly could not do the problems. I ended up personally marching her to the disability center to get accommodations. And in two weeks she'd gone from an F to an A.
I teach like I GM: You're there to cooperate with your students/PCs to accomplish something, NOT to prove you're "smarter" than them by ridiculing them or showing off your skills/power.
Maybe that's just me...
ivory tower thinking causes problems in all fields, but it is the worst in mathematics, in my experience.

thejeff |
Grand Magus wrote:I was taught that when rounding 1-4 rounds down, 6-9 rounds up, 5 followed by an even rounds down, and 5 followed by an odd rounds up. This is to ease the drift up effect from always rounding five up (ie 5/9 times rounding up inflates numbers). The problem still exists but by using this method you make it 2 orders of magnitude smaller than your lowest significant figure.BNW wrote:... (mathematicians round down. Carpet dealers want every penny!).
If the next digit is 5 or higher round up, else round down <-- how mathematicians do it.
e.g. rounding to the penny158.66666 ~ 158.67
158.55555 ~ 158.56
158.94411 ~ 158.94
158.94511 ~ 158.9599.9999 ~ 100.00
99.9949 ~ 99.99
.
But it's just weird to think that 1.51 is bigger than 1.58 :)

Kirth Gersen |
1 person marked this as a favorite. |

You are shown a set of four cards placed on a table, each of which has a number on one side and a colored patch on the other side. The visible faces of the cards show 3, 8, red and brown. Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on one face, then its opposite face is red?
How many card's aren't on the table? If there's a stack of them, we may have an awful lot of cards to turn over.

Grand Magus |

Grand Magus wrote:I was taught that when rounding 1-4 rounds down, 6-9 rounds up, 5 followed by an even rounds down, and 5 followed by an odd rounds up. This is to ease the drift up effect from always rounding five up (ie 5/9 times rounding up inflates numbers). The problem still exists but by using this method you make it 2 orders of magnitude smaller than your lowest significant figure.BNW wrote:... (mathematicians round down. Carpet dealers want every penny!).
If the next digit is 5 or higher round up, else round down <-- how mathematicians do it.
e.g. rounding to the penny158.66666 ~ 158.67
158.55555 ~ 158.56
158.94411 ~ 158.94
158.94511 ~ 158.9599.9999 ~ 100.00
99.9949 ~ 99.99
.
.
Get back to work. You have bills to pay.
.

Electric Wizard |

I'm serious, though -- sample =/= population.
Now you've done it. You have crossed the line! The line between
Induction and Deduction. (Smooth move ex lax.)On a related note (note the word related) check out The Problem of Induction. [ url=Problem of Induction ]
.

thejeff |
Peter Wason (1966), paraphrased, wrote:How many card's aren't on the table? If there's a stack of them, we may have an awful lot of cards to turn over.
You are shown a set of four cards placed on a table, each of which has a number on one side and a colored patch on the other side. The visible faces of the cards show 3, 8, red and brown. Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on one face, then its opposite face is red?
It's just a logic problem. There are only the four cards.

![]() |
1 person marked this as a favorite. |
Kirth Gersen wrote:It's just a logic problem. There are only the four cards.Peter Wason (1966), paraphrased, wrote:How many card's aren't on the table? If there's a stack of them, we may have an awful lot of cards to turn over.
You are shown a set of four cards placed on a table, each of which has a number on one side and a colored patch on the other side. The visible faces of the cards show 3, 8, red and brown. Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on one face, then its opposite face is red?
Everyone knows there are five cards, citizen.

Hitdice |

Kirth Gersen wrote:It's just a logic problem. There are only the four cards.Peter Wason (1966), paraphrased, wrote:How many card's aren't on the table? If there's a stack of them, we may have an awful lot of cards to turn over.
You are shown a set of four cards placed on a table, each of which has a number on one side and a colored patch on the other side. The visible faces of the cards show 3, 8, red and brown. Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on one face, then its opposite face is red?
And that's great as far as it goes, but the OP was about how 82% (God, it'd be really embarrassing if I got the arithmetic wrong) of the adult US population can't translate a word problem from text into arithmetic. I think that a practical problem, not a hypothetical one.
Edit: Whereas the card problem was purely hypothetical. Then again the first post ended with, "I can do it, what's the big deal?"
We're probably just dealing with Grand Magus wanting to talk about how smart he is again.

Orfamay Quest |

And that's great as far as it goes, but the OP was about how 82% (God, it'd be really embarrassing if I got the arithmetic wrong) of the adult US population can't translate a word problem from text into arithmetic. I think that a practical problem, not a hypothetical one.
We don't know that. I don't think there's any information about why they got the problem wrong. I'm sure that some of the people simply did the math wrong; I'm sure others got the translation wrong, and I'm sure others just couldn't remember the conversion factors.
It's a complex problem. Not unreasonably so, in my view, in the same way that I expect any cook who to be able to figure out what time to put a 1.5 kg roast in the oven, when it cooks for an hour and fifteen minutes per kilogram and you want to serve at 1900.
The fact that a number of people on this thread didn't realize the carpet was L-shaped suggests to me that the formality of the problem was at least part of the issue. Anyone would have looked at the actual carpet and known that.....

Hitdice |

Hitdice wrote:
And that's great as far as it goes, but the OP was about how 82% (God, it'd be really embarrassing if I got the arithmetic wrong) of the adult US population can't translate a word problem from text into arithmetic. I think that a practical problem, not a hypothetical one.We don't know that. I don't think there's any information about why they got the problem wrong. I'm sure that some of the people simply did the math wrong; I'm sure others got the translation wrong, and I'm sure others just couldn't remember the conversion factors.
It's a complex problem. Not unreasonably so, in my view, in the same way that I expect any cook who to be able to figure out what time to put a 1.5 kg roast in the oven, when it cooks for an hour and fifteen minutes per kilogram and you want to serve at 1900.
The fact that a number of people on this thread didn't realize the carpet was L-shaped suggests to me that the formality of the problem was at least part of the issue. Anyone would have looked at the actual carpet and known that.....
I'm not saying you're wrong, Orf, I'm just remembering how my Algebra/Geometry teacher (same woman, small school) drilled into my head that reading the problem carefully is the most important step.
Of course, with my hippy Montessori education, I have to believe that reading comprehension is the key to everything. :P