Author Topic: A Quick Puzzle to Test Your Problem Solving  (Read 11941 times)


deborah

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #1 on: July 02, 2015, 05:56:35 PM »
Great information once you have completed the puzzle - whether rightly or wrongly

MDM

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #2 on: July 02, 2015, 06:35:01 PM »
Great puzzle, thanks!

Brings to mind an oldie but goodie (see http://www.esg.montana.edu/biol525/platt.pdf).  One should try the puzzle first before reading the quote below.  Full marks, however, if you decide to read the whole journal article before trying the puzzle. ;)
Spoiler: show
"a theory is not a theory unless it
can be disproved.   That is, unless it can
be falsified by some possible
experimental outcome."

Cathy

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #3 on: July 02, 2015, 07:18:09 PM »
The "puzzle" is impossible to solve because, regardless of how many trials are performed, there will always be an infinite set of possible rules that will still work.

I'll leave it to somebody else to write an essay on how the limitations of inductive reasoning are handled in the philosophy of science. I'm not up for that tonight.

dragoncar

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #4 on: July 02, 2015, 07:36:39 PM »
I tried 6 yes, 4 no before I guessed

jambongris

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #5 on: July 02, 2015, 07:42:51 PM »
There was also a video on the Veritasium YouTube channel about this puzzle.

Dee

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #6 on: July 02, 2015, 07:58:57 PM »
I tried 4 yes and 5 no before I tried.

dungoofed

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #7 on: July 02, 2015, 08:10:47 PM »
The "puzzle" is impossible to solve because, regardless of how many trials are performed, there will always be an infinite set of possible rules that will still work.

: ) so one could say it has similarities with the SWR Problem?

(or should I keep plugging numbers into cFIREsim?)

forummm

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #8 on: July 02, 2015, 08:22:20 PM »
The "puzzle" is impossible to solve because, regardless of how many trials are performed, there will always be an infinite set of possible rules that will still work.

I'll leave it to somebody else to write an essay on how the limitations of inductive reasoning are handled in the philosophy of science. I'm not up for that tonight.

It's impossible to know for certain that you've reached the exact intended solution. For example, the solution could be
Spoiler: show
a huge list of sets of 3 numbers, where any of those sets is correct and anything not in the set is not correct
. But given a sufficient number of intelligently selected test cases you can eliminate a huge amount of possible solutions. And combining with the fact that this is setup in the context of a quick game for the general public, the likelihood that the solution is readily obtainable would lead one to believe they could be reasonably confident in a solution. It could be a trick, but that's not too likely given the medium.

Cathy

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #9 on: July 02, 2015, 08:41:50 PM »
Okay, here's a far more challenging and also far more interesting version of the original puzzle. I'm sure MDM will enjoy it.

As before, suppose we are considering a function F with four arguments a, b, c, and d. For a given set of values, F returns either 0 or 1. Suppose that a, b, c, and d are integers within the domain [e, f], where e<f are both integers. We say that two possible functions f_1 and f_2 are considered to be the same if they have the same "image", i.e., f_1 and f_2 are the same iff f_1(a,b,c,d) = f_2(a,b,c,d) Ɐ a,b,c,d ∈ [e,f].

Given the above premises, there are (obviously) only a finite set of possible functions F that could characterise the unknown rule. Suppose that as part of the setup, one of the possible functions F is chosen at random, with each possibility having an equal chance of being chosen.

Now here is the problem:
  • What algorithm will on average minimise the number of trials required to determine F?
  • How many trials does that algorithm take on average?

MDM

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #10 on: July 02, 2015, 10:18:47 PM »
Ok, I'll bite - or maybe just a nibble.

It's not obvious (to me) why there is only a finite set of possible functions.  E.g., one could have a function F = if (a^n + b^n + c^n + d^n) > 0 then 1 else 0.  With an infinite number of "n"s to choose from, that gives an infinite number of possible functions.

If we're talking about a finite set of function inputs, then yes.  Let p = f - e.  Depending on whether a, b, c, and d can have the same value or not, and if the order matters or not, then there are (p+1)^4 or (p+1)^4/4! or (p+1)! / (p+1-4)! or (p+1)! / (4! * (p+1-4)!) possible inputs.

Anyway, I still need help with the rules of the game that limit the number of possible functions....

Cathy

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #11 on: July 02, 2015, 10:21:12 PM »
It's not obvious (to me) why there is only a finite set of possible functions.

This comes from my definition of what it means for two functions to be the same:

We say that two possible functions f_1 and f_2 are considered to be the same if they have the same "image", i.e., f_1 and f_2 are the same iff f_1(a,b,c,d) = f_2(a,b,c,d) Ɐ a,b,c,d ∈ [e,f].



Another reader pointed out to me that my problem is actually trivial and has a simple solution. I may reply later with a revised, more interesting version.

Coercivity

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #12 on: July 02, 2015, 11:46:03 PM »
Good puzzle
16 yes 10 no for the record
Spoiler: show
numbers get rounded to 16 decimal places
0.9999999999999999 1 2 is a yes
0.99999999999999991 0.99999999999999995 2 is a yes
0.99999999999999991 0.99999999999999992 2 is a no ;p

YoungInvestor

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #13 on: July 03, 2015, 01:02:45 AM »
The "puzzle" is impossible to solve because, regardless of how many trials are performed, there will always be an infinite set of possible rules that will still work.

I'll leave it to somebody else to write an essay on how the limitations of inductive reasoning are handled in the philosophy of science. I'm not up for that tonight.

I think that in the context this was given in, it's reasonable to assume that the answer was at least somewhat simple. Greatly restricting the possibilities.

dragoncar

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #14 on: July 03, 2015, 02:27:26 AM »
The "puzzle" is impossible to solve because, regardless of how many trials are performed, there will always be an infinite set of possible rules that will still work.

I'll leave it to somebody else to write an essay on how the limitations of inductive reasoning are handled in the philosophy of science. I'm not up for that tonight.

I think that in the context this was given in, it's reasonable to assume that the answer was at least somewhat simple. Greatly restricting the possibilities.

Yeah, it's a "puzzle" not a graduate level mathematics exam

Kalergie

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #15 on: July 03, 2015, 02:29:22 AM »
I believe this YouTube blogger, Veritasium, has a great bit on the puzzle.
http://youtu.be/vKA4w2O61Xo

forummm

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #16 on: July 03, 2015, 08:01:22 AM »
Here's another one:



Plug in the numbers 1 through 9, using each digit just once, filling each of the blank spaces. Using the identified operations and the inserted values, a correct answer would lead to a solution of 66 (those double dots stand for division).

tyir

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #17 on: July 03, 2015, 10:55:31 AM »
Okay, here's a far more challenging and also far more interesting version of the original puzzle. I'm sure MDM will enjoy it.

As before, suppose we are considering a function F with four arguments a, b, c, and d. For a given set of values, F returns either 0 or 1. Suppose that a, b, c, and d are integers within the domain [e, f], where e<f are both integers. We say that two possible functions f_1 and f_2 are considered to be the same if they have the same "image", i.e., f_1 and f_2 are the same iff f_1(a,b,c,d) = f_2(a,b,c,d) Ɐ a,b,c,d ∈ [e,f].

Given the above premises, there are (obviously) only a finite set of possible functions F that could characterise the unknown rule. Suppose that as part of the setup, one of the possible functions F is chosen at random, with each possibility having an equal chance of being chosen.

Now here is the problem:
  • What algorithm will on average minimise the number of trials required to determine F?
  • How many trials does that algorithm take on average?


Spoiler: show
One way to think about is to consider each function a "picture" in 4-dimensional space, with the 4D pixels being coloured depending on boolean output. In this language, we need to determine the picture which was generated randomly.
In that case there is no strategy other than iterating through the entire (f-e+1)^4 space, which takes (f-e+1)^4 steps.


bdbrooks

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #18 on: July 03, 2015, 11:10:58 AM »
Here's another one:



Plug in the numbers 1 through 9, using each digit just once, filling each of the blank spaces. Using the identified operations and the inserted values, a correct answer would lead to a solution of 66 (those double dots stand for division).

1+13*2/6+4+12*7-8-11+3*5/9-10=66

With 362,880 different options it was probably too difficult to do by hand. However, writing a macro in 5 minutes will solve the problem fairly quickly. I know that if I were guessing the order I wouldn't have picked to have fractions (13*2/6 and 3*5/9).

For MDM, it is obvious there is a finite set to Cathy's question due to the fact that ABCD as well as E and F are INTEGERS. At a surface level, I agree that there is no strategy other than iterating through all of the possible steps. So the average steps would be (f-e+1)^4/2. If there is a strategy, it is not apparent to me.

MDM

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #19 on: July 03, 2015, 12:05:34 PM »
For MDM, it is obvious there is a finite set to Cathy's question due to the fact that ABCD as well as E and F are INTEGERS. At a surface level, I agree that there is no strategy other than iterating through all of the possible steps. So the average steps would be (f-e+1)^4/2. If there is a strategy, it is not apparent to me.
Agree that there is a finite set of inputs to the to-be-determined function F.

So F is this black box that, given four integers, spits out either 0 or 1 as answer.  And any two of those black boxes are the same if, for every permutation of integer values between the integers e and f inclusive, the black boxes return the same value of 0 or 1.

But...what puts a limit on the number of black boxes?  E.g., let's say F = 0 for all inputs.  One black box could have =0*a*b*c*d, another could have =0*(a^2+b^2+c^2+d^2), etc., leading to an infinite set of possible functions F that could characterise the unknown rule.

At least, that's the path on which my brain is stuck.  Quite possible that I'm misinterpreting something.

Cathy

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #20 on: July 03, 2015, 12:16:53 PM »
E.g., let's say F = 0 for all inputs.  One black box could have =0*a*b*c*d, another could have =0*(a^2+b^2+c^2+d^2), etc., leading to an infinite set of possible functions F that could characterise the unknown rule

As restated above, those two black boxes are actually the same function F because of my definition of what it means for two functions to be the same function.

All you are saying is that for a given function F, there are infinitely many notations that can be used to write a definition of the function. However, that does not change the number of unique functions F.

StacheEngineer

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #21 on: July 03, 2015, 12:23:18 PM »

At a surface level, I agree that there is no strategy other than iterating through all of the possible steps. So the average steps would be (f-e+1)^4/2. If there is a strategy, it is not apparent to me.


IMO, the number of steps is deterministic. After (f-e+1)^4/2 steps, you have only tried half of the possible inputs. Therefore you don't know whether half the inputs map to a 1 or 0. You must iterate over the entire input space since the output of the function is random.

As far as I can tell Cathy, tyir is right that you effectively have a 4 dimensional bitmap and since each voxel is randomly colored, you must sample at each voxel. There isn't a randomized algorithm for this problem. Luckily, the number of steps is finite because the range is limited to a finite number of integers.

Cathy

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #22 on: July 03, 2015, 12:31:12 PM »
As far as I can tell Cathy, tyir is right that you effectively have a 4 dimensional bitmap and since each voxel is randomly colored, you must sample at each voxel. There isn't a randomized algorithm for this problem. Luckily, the number of steps is finite because the range is limited to a finite number of integers.

To be clear, this doesn't mean that the problem is flawed in any way. It just means, as I said above, that the problem "is actually trivial and has a simple solution".

MDM

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #23 on: July 03, 2015, 12:33:21 PM »
what it means for two functions to be the same function.

Thus tying this thread to the marriage thread(s): "...and the two shall become one."

Cathy

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #24 on: July 03, 2015, 12:40:31 PM »
what it means for two functions to be the same function.

Thus tying this thread to the marriage thread(s): "...and the two shall become one."

I included an explicit definition of sameness for greater clarity, but it was unnecessary because a function is a relation between a domain and codomain; the notation used to express that relation is not part of the "essence" of the function.

For example, if the domain and codomain of an arbitrary function Q(x) are both restricted to {0}, then there is only exactly one possible function Q(x). The notations x, x**2, x+x, or sin(x) are just different notations for the same function Q : {0} → {0}.
« Last Edit: July 03, 2015, 12:49:21 PM by Cathy »

Kaspian

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #25 on: July 03, 2015, 01:52:08 PM »
http://www.nytimes.com/interactive/2015/07/03/upshot/a-quick-puzzle-to-test-your-problem-solving.html?_r=0

I liked that!  Not only did I show confirmation bias right off the bat with my guesses, then follow through to the end with them, it showed me another bias:  I didn't want any of my guesses to be wrong.  There was actually no harm in running endless scenarios yet my mind was concerned that maybe I'd get one wrong and see a red message.  ....Or that maybe a box would appear telling me I'd run out of guesses. 

I like to think of myself as an "easily try anything and if it doesn't work? No harm done, just try something else", sort of guy.  Maybe when something presented as a challenge the need to succeed fully at it is more prevalent than I imagined.

Kaspian

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #26 on: July 03, 2015, 01:59:19 PM »
This also reminds me a little bit of the "Monty Hall Problem".  When Monty would present 3 closed doors (with a big prize hidden behind one) and tell an audience member to choose a door.  He'd then remove one of the other doors and tell the viewer they could either stick with the first door they'd chosen or switch to the remaining one.  People almost *always* stuck with the closed door they'd chosen even though it's been mathematically proven that they'd be better off switching.

...And also how it's been proven soccer players would have a better chance at blocking a penalty shot by just standing in the middle of the net doing almost nothing rather than leaping to a side when the ball comes at them. 

Humans often seem to make some choices and we stick with them regardless of what data is presented.  Is it the need to feel personally in charge?  Or a need to try and show we were right--even about a random guess?

https://en.wikipedia.org/wiki/Monty_Hall_problem


robotclown

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #27 on: July 03, 2015, 03:34:42 PM »
Interesting, but the puzzle jumps to weird conclusion.  People don't like hearing 'no' so they can't solve the puzzle?  What? 

MDM

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #28 on: July 03, 2015, 04:02:03 PM »
Interesting, but the puzzle jumps to weird conclusion.  People don't like hearing 'no' so they can't solve the puzzle?  What?

What about that conclusion seems unreasonable?

Conservatives like to talk with other conservatives, and liberals like to talk with other liberals, because they all agree with each other.

TheOldestYoungMan

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #29 on: July 03, 2015, 04:42:12 PM »
Well, talk about confirmation bias.

Create a puzzle, present it in a form that you know almost all of us saw during Geometry (at the latest).  Then communicate what you want us to do poorly.  And then be smug when we don't come up with the answer you expected.

Also have your puzzle be wrong.

If you put in 1.9999999999, 2, and 3, then it returns "no".  So their "solution" is wrong.  All possible iterations of the solution I picked, will still cause it to say Yes.

I'd say that instead of "But most people start off with the incorrect assumption that if we’re asking them to solve a problem, it must be a somewhat tricky problem."  It should say "Most of our readers were hoping we were asking them to do something difficult or challenging.  We used that to trick them into coming up with answers that may or may not be correct, but rest assured, we suck at programming."

I just like the circularity of the whole thing.  There was no way their puzzle wasn't going to produce the results they are reporting.  Confirmation bias is a thing ->  Create puzzle that will cause many many people to guess solutions that no amount of trials will ever present a No -> Conclude you were correct, and confirmation bias is a thing.

MDM

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #30 on: July 03, 2015, 05:25:45 PM »
what it means for two functions to be the same function.
Thus tying this thread to the marriage thread(s): "...and the two shall become one."
For example, if the domain and codomain of an arbitrary function Q(x) are both restricted to {0}, then there is only exactly one possible function Q(x). The notations x, x**2, x+x, or sin(x) are just different notations for the same function Q : {0} → {0}.
And the many shall become one - now it's polygamy! ;)

robotclown

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #31 on: July 03, 2015, 10:17:10 PM »

What about that conclusion seems unreasonable?

Conservatives like to talk with other conservatives, and liberals like to talk with other liberals, because they all agree with each other.

It doesn't follow from the data.  If people are guessing an answer without receiving a 'no' response, then they aren't reasoning out the solution correctly.  Maybe because it's an online puzzle with no consequences, so they don't care, or maybe they misinterpreted the question, or various other reasons.  Drawing the conclusion that people would be disappointed seeing the red 'no' boxes, and would rather fail to solve the puzzle (which is also receives a negative response) than have to see them, is quite a stretch in logic.

I got a little irritated when I kept seeing 'yes' pop up.  Entering the wrong sequence is required to confirm the answer.

MDM

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #32 on: July 03, 2015, 10:35:13 PM »
If people are guessing an answer without receiving a 'no' response, then they aren't reasoning out the solution correctly.
Agreed.

Quote
Maybe because it's an online puzzle with no consequences, so they don't care, or maybe they misinterpreted the question, or various other reasons.  Drawing the conclusion that people would be disappointed seeing the red 'no' boxes, and would rather fail to solve the puzzle (which is also receives a negative response) than have to see them, is quite a stretch in logic.
Maybe, maybe not.  There seems a difference between "would rather fail to solve" vs. "don't think to test a counterexample."  I've seen a preponderance in real-life of people looking to "prove their idea correct", with much fewer looking to "run a test that would prove their idea wrong."

Quote
I got a little irritated when I kept seeing 'yes' pop up.  Entering the wrong sequence is required to confirm the answer.
Good!  Just remember to apply the same thought process in real life.  Although, perhaps you do so already, in which case - keep up the good work!

MDM

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #33 on: July 03, 2015, 11:51:08 PM »
I included an explicit definition of sameness for greater clarity, but it was unnecessary because a function is a relation between a domain and codomain; the notation used to express that relation is not part of the "essence" of the function.

For example, if the domain and codomain of an arbitrary function Q(x) are both restricted to {0}, then there is only exactly one possible function Q(x). The notations x, x**2, x+x, or sin(x) are just different notations for the same function Q : {0} → {0}.

Cathy, kidding aside - thanks, I get what you were saying now.  From https://en.wikipedia.org/wiki/Function_%28mathematics%29, "Strictly speaking, a function is properly defined only when the domain and codomain are specified."

The wiki entry goes on to say "A function can be defined by any mathematical condition relating each argument (input value) to the corresponding output value. If the domain is finite, a function f may be defined by simply tabulating all the arguments x and their corresponding function values f(x)."  Seems that is the case here, and as tyir and StacheEngineer noted, one has to go through all (f - e + 1)^4 table entries.

The wiki entry also notes "More commonly, a function is defined by a formula, or (more generally) an algorithm — a recipe that tells how to compute the value of f(x) given any x in the domain."  That's related to what I thought you were asking: how does one determine if two algorithms give the same results in the specified domain.

Jack

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #34 on: July 04, 2015, 12:19:33 AM »
Good puzzle
16 yes 10 no for the record
Spoiler: show
numbers get rounded to 16 decimal places
0.9999999999999999 1 2 is a yes
0.99999999999999991 0.99999999999999995 2 is a yes
0.99999999999999991 0.99999999999999992 2 is a no ;p


I can almost guarantee (without checking the Javascript) that that isn't actually rounding to 16 decimal places. Floating-point data types are rounded to a fixed [total] number of base-2 digits (or a variable number to the right of the decimal point), which means that the base-10 precision can vary.

Also have your puzzle be wrong.

If you put in 1.9999999999, 2, and 3, then it returns "no".  So their "solution" is wrong.  All possible iterations of the solution I picked, will still cause it to say Yes.

Aside from the computer limitations I mentioned above, it's also true that 1.999999... (the repeating decimal) actually is mathematically equal to 2. See these proofs.

dungoofed

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #35 on: July 04, 2015, 08:44:43 AM »
Not on a pc now but I think 2 2.5 3 didn't work either so "whole number" might be a requirement.

TheOldestYoungMan

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #36 on: July 05, 2015, 08:57:34 PM »
Good puzzle
16 yes 10 no for the record
Spoiler: show
numbers get rounded to 16 decimal places
0.9999999999999999 1 2 is a yes
0.99999999999999991 0.99999999999999995 2 is a yes
0.99999999999999991 0.99999999999999992 2 is a no ;p


I can almost guarantee (without checking the Javascript) that that isn't actually rounding to 16 decimal places. Floating-point data types are rounded to a fixed [total] number of base-2 digits (or a variable number to the right of the decimal point), which means that the base-10 precision can vary.

Also have your puzzle be wrong.

If you put in 1.9999999999, 2, and 3, then it returns "no".  So their "solution" is wrong.  All possible iterations of the solution I picked, will still cause it to say Yes.

Aside from the computer limitations I mentioned above, it's also true that 1.999999... (the repeating decimal) actually is mathematically equal to 2. See these proofs.

I couldn't figure out how to put in an infinite number of 9's, I put in a finite number, so I put in a solution that met the "rule" but returned a "no."  It would have been possible to to program it such that what I put in wouldn't have returned a "no."  They didn't do that though, because they were afflicted with confirmation bias.

dragoncar

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #37 on: July 06, 2015, 08:14:14 AM »
Good puzzle
16 yes 10 no for the record
Spoiler: show
numbers get rounded to 16 decimal places
0.9999999999999999 1 2 is a yes
0.99999999999999991 0.99999999999999995 2 is a yes
0.99999999999999991 0.99999999999999992 2 is a no ;p


I can almost guarantee (without checking the Javascript) that that isn't actually rounding to 16 decimal places. Floating-point data types are rounded to a fixed [total] number of base-2 digits (or a variable number to the right of the decimal point), which means that the base-10 precision can vary.

Also have your puzzle be wrong.

If you put in 1.9999999999, 2, and 3, then it returns "no".  So their "solution" is wrong.  All possible iterations of the solution I picked, will still cause it to say Yes.

Aside from the computer limitations I mentioned above, it's also true that 1.999999... (the repeating decimal) actually is mathematically equal to 2. See these proofs.

I couldn't figure out how to put in an infinite number of 9's, I put in a finite number, so I put in a solution that met the "rule" but returned a "no."  It would have been possible to to program it such that what I put in wouldn't have returned a "no."  They didn't do that though, because they were afflicted with confirmation bias.

Confirmation bias?  Their conclusion is still supported by the data-- far too many people are guessing the rule before getting a "no" result.  This is true regardless of what the exact "rule" is (rising numbers vs rounded numbers rising). 

Will anyone here admit to guessing without getting "no" answer?  If so can you tell us about it?

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #38 on: July 06, 2015, 09:48:24 AM »
Will anyone here admit to guessing without getting "no" answer?  If so can you tell us about it?

Not me, I'm afraid. I had 4 "yes" and 3 "no" answers before guessing correctly.

TheOldestYoungMan

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #39 on: July 06, 2015, 09:52:52 AM »


Confirmation bias?  Their conclusion is still supported by the data-- far too many people are guessing the rule before getting a "no" result.  This is true regardless of what the exact "rule" is (rising numbers vs rounded numbers rising). 

Will anyone here admit to guessing without getting "no" answer?  If so can you tell us about it?

I'm not saying that they are wrong.  I'm saying that using their puzzle as a way of demonstrating that their conclusions have merit is in and of itself confirmation bias.  They believed it would work, they constructed it to work, and when it did work they didn't bother to see if it was a problem with the puzzle or if what they were seeing was good data.

So the experiment they concocted to demonstrate confirmation bias is an experiment that is itself afflicted with all kinds of confirmation bias.

I was amused is all.  Confirmation bias is a thing, no argument here.  It isn't limited to laypeople either.

But that really isn't what that article is about.

If I were peer reviewing the experiment, I'd have made the following recommendations:

1.  You need to randomize the starting prompt.  Your choice of numbers will subject you to a confirmation bias.  Preferably the rule for the puzzle will automatically generate 3 numbers that fit the rule.  As it is being administered on a computer there is no rational reason to not do this.
2.  You need to randomize the formula.  Choosing one and having every participant try to guess that one is not going to be relevant to your thesis.  You need at least four, preferably more, and they need mathematical variability.  Consult a mathematician for possible issues with mathematically literate folks taking your test.
3.  You need a control.  Barring that you need an antithesis test.  How do the results compare when almost all initial guesses would be a No?  How about when there is an actual reward for getting it correct?  Consider mturk with a $10 award to the first 100 people to guess the right answer.

So the puzzle in that article was not about science, it was about providing positive feedback that confirmation bias is a thing.  Which isn't science, it's confirmation bias!

I've given a few minutes thought to how to present it such that 50% of the initial guesses return a yes, and 50% a no.  That would provide a much better data set, as you could follow up with how many people think they nail it after their first strings of yes, vs. how many keep going until they get more no.  I think limiting it to numbers in itself is going to cause problems.  A lot of us have had quite a bit of math training, so simple rules may tend to have that first guess come out a yes most of the time regardless.

So, yea.  The data from the puzzle is total garbage, it illustrates nothing like what the article is claiming.  The claims in the article happen to be supported by other, independent, superior data.

There is an outside possibility, and just throwing it out there, that confirmation bias isn't necessarily a bad thing.  It's not good either.  It's just there.  It's a thing to be aware of.  I think that's what the point of the article was.  So it was deeply amusing to me how unaware it was of how crappy their puzzle was.  They missed an opportunity to be meta.

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #40 on: July 06, 2015, 10:33:26 AM »
SO I did 6 tests and then did the expected confirmation bias guess.  All of mine were yes.  So I "failed miserably" by doing the common expected thing.  I actually learned something from the article and kicked myself for not thinking critically (just assumed if this is for stupid people, it's a stupid solution).

My guess were (if I remember):
1/2/4 (double each number)
3/6/12 (to see if it was a factor of 2 issue)
5/10/20 (another random set to verify the double each number)
500/1000/2000 (to see if it was just doubling or a weird exponent)
.25/.5/1 (figured factions might mess them up, it didn't)
.66/1.33/2.66 (2/3, 4/3, 8/3... just more factions to try to confuse it)

Guessed the rule (wrong).

Things I should have done: negative numbers!  I just assumed positive, but now realize I have no reason to think that.  -1/-2/-4 would have failed and started to raise questions.  Obviously something like 1/2/3 that would purposely fail my thoughts, but hey, "no's are bad!"  But hey, I learned something today - so that means I get to go home now, right?

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Re: A Quick Puzzle to Test Your Problem Solving
« Reply #41 on: July 06, 2015, 10:50:21 AM »
7 yeses, 5 nos. At first I thought it was f(n)=C^(n-1), but then I tried simply doubling the constant and that worked as well. Then I tried 1, 2, 3, then 2, 3, 4. All rising sequences seem to work... tried a couple descending sequences, halving, rooting, and finally random. Tried negative numbers. Tried making the middle number lower or higher than both the first and second number. Finally decided that the rule was extremely simple. The little article after seemed kinda random.