

07142014, 11:28 PM

#41

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Before I forget, I must post this followup.
Marilyn publicly acknowledged her error in her column in Parade issue 7/13/2014, which is yesterday. She said that she was puzzled at first when people contacted her to say that she was wrong, but she finally saw the fault in her thinking.
Of course we have all seen people (including ourselves) being wrong in a logic or math problem that may be obvious to others (of course I am not talking about problems that are unprovable like EMH ). So, it's not schadenfreude but relief that I felt when I found that a smart person like Marilyn could be so wrong, and that so many people could spot her error like I did.
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07142014, 11:36 PM

#42

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I forgot to show answers to the variations I posed.
Quote:
Originally Posted by NWBound
Case 2: A and B working together can finish the job in 2 hrs. Individually, A takes 3 hrs less than B. How long does each take individually?
Case 3: A and B working together can finish the job in 3 hrs. Individually, A takes 8 hrs less than B. How long does each take individually?

Case 2: A takes 3 hrs and B takes 6. To verify: 1/3 + 1/6 = 1/2.
Case 3: A takes 4 hrs and B takes 12. To verify: 1/4 + 1/12 = 1/3.
Nice numbers in this case. No crazy square roots like in the case that stumbled Marilyn.
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07152014, 08:19 AM

#43

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I look back at the above post, and see that there's an easier way to see that the answers work.
Quote:
Case 2: A and B working together can finish the job in 2 hrs. Individually, A takes 3 hrs less than B. How long does each take individually?

Answer: 3 and 6 hours.
To verify, we can see that when working together for 2 hours, A will perform 2/3 of the job, and B 2/6 or 1/3 of the job. And 2/3 + 1/3 = 1 job.
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07232014, 09:41 AM

#44

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I get A = 1 hour and B = 5 hours.
Mike D.
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08032014, 12:29 PM

#45

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03  AUG  2014 column
While these may not be full on 'goofs', I really think someone of such high IQ should do a better job with educating people and using more precise language in her column. Three quibbles I have with today's column:
1) " Aluminum is the most abundant metal on the planet"  OK, that is true, but it's not the whole truth. Like hydrogen, almost all aluminum is bound up with other elements. We don't come across a 'vein' of aluminum like we do coal, gold, silver or lead, etc. It takes a large amount of energy to break that bond and make useful aluminum. That is why it is important to recycle it  melting it and reforming takes takes far less energy than refining it in the first place. So not a goof, but a lost opportunity, and maybe misleading to a lot of people.
2) " stainless steel conducts heat so unevenly"  makes me cringe. Stainless steel is a material, and conductivity is a property. For a given grade of stainless steel, its thermal conductivity is not uneven at all, its the same throughout.
It would be better to say "the temperature in a stainless steel pan can be uneven, due to the relatively poor conductivity of that metal" . Nitpicking maybe, but she is dumbing down her audience by using terms so loosely, IMO.
3) On using a magnet to distinguish an aluminum pot from a SS pot (some SS is somewhat magnetic, aluminum is not)  If a magnet doesn't stick "... you still don't know. The pot is either aluminum or it's not magnetized".
'Magnetized' is not the right term. 'Magnetic' is. 'Magnetic' means it can be 'magnetized', but it might not be. You can magnetize a steel screwdriver by swiping a magnet against it. You can demagnetize it in several ways. In either case, a magnet will stick to it.
I might let one nitpick slide, but three in one short column is sloppy, and I don't think it reflects well on someone who makes a living based on publicizing her high IQ.
Anyone disagree? Did I get something wrong (never got my IQ tested, but I'm certain I'd test lower than Marilyn)?
edit/add: I see clintonwylie agrees with the last item ('magnetized')  no, I'm not him.
http://parade.condenast.com/323423/m...pansharmful/
ERD50
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10  AUG  2014 column  More multiple manglings
08102014, 11:13 AM

#46

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10  AUG  2014 column  More multiple manglings
Sloppy, sloppy, sloppy! I really do expect more from such a bright person. She is giving 'High IQ' a bad name!
Today's questions was:
Quote:
The railway industry claims that it achieves incredible gas mileage—about 450 miles per gallon. How is this possible?

Marilyn starts out OK (but not fully correct):
Quote:
... railroad mileage is cited in tonmiles per gallon (instead of simply miles per gallon), it means something different. Rather, the term is more specific:

but then makes a slip that she should not have:
Quote:
A freight train can move one ton of weight about 450 miles on a single gallon of gas.

Bzzzzzt! A freight train uses diesel, not 'gas'. Heck, even Elton John knows that ( 'get about as oiled as diesel train...')! You could say a gallon of 'fuel' or 'gasoline equivalent', but not 'gas'. Sloppy.
But it gets worse...
Quote:
To match this mileage, a oneton car would have to get 450 mpg, and a twoton vehicle would have to get 225 mpg.

She is now off the tracks (keeping with the railroad theme). She got it mostly right above with 'tonmilespergallon', but the real measure used in the example question is 'freighttonmilespergallon'. No one cares what the train weighs (or the truck/car for comparison) when we look at efficiency of moving freight, we care about the amount of freight it can move.
So a comparison with a one or two ton vehicle is meaningless  the real comparison is how much freight they carry, and their mpg to calculate an equivalent 'freighttonmilespergallon'.
Her explanation of the train efficiency isn't so hot:
Quote:
Still, trains are extremely efficient. They have the benefit of an infrastructure that includes steel wheels rolling on steel rails (instead of soft rubber tires making frequent turns on pavement), highly sophisticated braking systems, and far, far fewer starts and stops.

Yes, rolling resistance is important. I'm not sure how 'highly sophisticatedbraking systems' play into efficiency, AFAIK most freight trains do not have a storage batteries for regenerative braking (though they are hybrids of a sort  the diesel generator drives electric motors), so the brake energy is going to waste.
Fewer starts/stops, maybe. But a truck on the highway isn't stopping either, and it's highway 'freighttonmilespergallon' is still far lower at that point.
I'm pretty sure she is ignoring a big factor  wind resistance. For a train, the long line of cars are effectively 'drafting' each other.
I didn't quickly find any good comparisons, but this link has lots of interesting data:
http://www.istc.illinois.edu/about/S...s/20091118.pdf
A truck's rolling resistance is 610x a trains, and they show that at 60 mph, the wind resistance of a train makes up about half of the total drag.
Lot's of other factors  just the size of a train brings 'economy of scale', you really don't need twice as much material to carry twice as much freight (the surface area of two onegallon containers is greater than the surface area of one twogallon container, etc).
And a BIG engine is more efficient (all things being equal) than a small engine  again, less surface area per HP to lose heat, etc. Probably less bearing surface area as well. Hmmm, but those big locomotives lose some eff% in converting mechanical to electrical to mechanical....
I expect better of her. I wonder if her mental acquity is diminishing with age, or some other factor. She really ought to do better than this.
ERD50
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08102014, 07:09 PM

#47

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I did a little more looking into whether freight trains were starting to use regenerative braking, found some interesting stuff.
Turns out that a freight train that might have one or two locomotives for 100 cars or so, could not depend on braking only from the locos. All the cars have brakes. So to utilize regen, they would need motor/generators on every car. Since they don't stop that often, it sounds like it just does not make economic sense.
They are putting batteries into the small switching units in the yards. These sit and idle for long periods, then need bursts of energy to switch trains around, and start and stop. So it can work for that application.
I have considered that maybe commuter trains could take advantage of having an electric line installed for maybe 1/4 mile before and after stations. If they picked up the electric power for their motors directly from the line for acceleration out of the station, they would not burn so much diesel fuel. And if they injected the power back into the grid when they brake going into the station, that would essentially give them regenerative braking w/o having to carry batteries on the train. Even if they needed batteries to absorb and release the power surges that maybe the grid could not handle, they would be stationary batteries, which can be made much cheaper since they don't need to be small and lightweight. And they only need to store enough for that same train to leave the station.
Could I raise $1M by putting that on some crowdsourced funding site and finding a catchy phrase to put on coffee mugs and teeshirts to give to 'donors'?
ERD50
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02NOV2014  Ooooopppps! She did it again!
11022014, 11:24 AM

#48

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02NOV2014  Ooooopppps! She did it again!
Do Gifted Children Need Special Support? Plus, a Dandy Logic Puzzle
Quote:
David Price in Decatur, Ill., writes:
If x and y are counting numbers (whole numbers above zero), what numbers do the terms (x÷2), (3÷y), and (yx) represent?

Marilyn gives the answer as "1". But she makes the mistake of explaining this as the 'terms' needing to be 'counting numbers', but the problem only states that ' x and y are counting numbers'.
So there are infinite solutions. She has several comments to that effect already.
ERD50
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11032014, 11:43 AM

#49

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The problem is poorly defined. It never states that (x÷2), (3÷y), and (yx) are equal.
If (x÷2) = (3÷y) = (yx), then solving for x and y we will get two solutions.
1) x = 2, and y = 3, then (x÷2) = (3÷y) = (yx) = 1
2) x =2, and y = 3, then (x÷2) = (3÷y) = (yx) =  1
If we now impose the condition that x and y are positive integers (counting numbers), then we retain only solution 1) above, and that's Marylyn's answer.
However, the problem statement never specifies that (x÷2), (3÷y) should be equal. Then, what does this mean? If x is a positive integer, then (x÷2) is either an integer or an odd multiple of (1/2). There is nothing else that we can conclude.
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11032014, 11:57 AM

#50

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I don't think she went down the path of thinking they would be equal  I think the mistake she made was thinking that the terms ((x÷2), (3÷y), and (yx)) must be counting numbers, but the problem only states that x and y are counting numbers.
ERD50
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06282015, 09:13 AM

#51

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I happened to pick up Parade, June 28 2015 issue, and saw that Marylin had another goof. Here's the puzzle.
"Say you have four specially marked dice. Two players each select one, and the player who rolls a higher number wins. The faces are 111555, 222266, 333333 and 444400. You’re given the first choice. Which die should you choose?" Marylin's answer follows.
"Here’s a brainbending challenge for you, puzzlers: I’ll tell you which die is the strongest (meaning that it will score more wins when played against all the other dice), and you figure out why it doesn’t matter which die you choose! The answer will appear June 29 here. OK, ready? The strongest die is 222266." I don't know what Marylin meant for the challenge. If we assume that the opponent can also figure out and pick the best dice, then it remains 50/50 chance that we can win. So, we can only expect to break even by choosing the best dice. But of course we do not want to pick a weak dice and have a higher chance of losing, do we?
Marylin's choice of 222266 is bad, if you consider a toss against the 333333 dice. You would have 66.66% chance of losing.
So, what's the best dice? I shall leave it as an exercise.
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06282015, 10:48 AM

#52

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Quote:
Originally Posted by NWBound
I happened to pick up Parade, June 28 2015 issue, and saw that Marylin had another goof. Here's the puzzle. "Say you have four specially marked dice. Two players each select one, and the player who rolls a higher number wins. The faces are 111555, 222266, 333333 and 444400. You’re given the first choice. Which die should you choose?" Marylin's answer follows. "Here’s a brainbending challenge for you, puzzlers: I’ll tell you which die is the strongest (meaning that it will score more wins when played against all the other dice), and you figure out why it doesn’t matter which die you choose! The answer will appear June 29 here. OK, ready? The strongest die is 222266." I don't know what Marylin meant for the challenge. If we assume that the opponent can also figure out and pick the best dice, then it remains 50/50 chance that we can win. So, we can only expect to break even by choosing the best dice. But of course we do not want to pick a weak dice and have a higher chance of losing, do we?
Marylin's choice of 222266 is bad, if you consider a toss against the 333333 dice. You would have 66.66% chance of losing.
So, what's the best dice? I shall leave it as an exercise.

I don't think Marylin has goofed, but rather I believe you have discovered the answer to the puzzle. The reason it doesn't matter that there is one "strongest" die is that you have to select your die first. No matter which die is selected, there is another die which is stronger ŕ la RockPaperScissors.
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06282015, 10:51 AM

#53

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A very interesting possibility. Can you provide proof?
And if so, then why did Marylin say that her choice "will score more wins when played against all the other dice"?
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06282015, 10:59 AM

#54

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***SPOILER ALERT**** (similar to MildlyEccentric's post, which I crossposted)
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I was racking my brain on this until I considered that the second player (B) gets to pick from the remaining three, and we assume they get to see what the first player (A) picked. Once you focus on that, relative strength really has no bearing, the only question is  can player B pick a die that will beat A? Or stated another way, as long as each die has at least one other die that is stronger than it, player A cannot win (assuming Player B chooses wisely). Player B will just choose the one that is stronger.
It's a bit (maybe a LOT?) like playing rockpaperscissors if you get to see what hand the other is playing. Each play has a winning defense, but no single defense is any stronger than any other.
the 662222 does 'win' in 2 of 3 cases, I have not checked if that makes it 'strongest', but it has the highest average (and the lowest median!) of the group. I wonder if that is her definition of 'strong'?
Is there a mathematical way to decide this (other than brute force comparisons)? I don't think averages or medians help, 6 > 2 has no more 'value' than 6>5 in this game. I guess just a count of the number of 'greater thans' is all you need?
I haven't commented on Marilyn in a while, most of the columns I've seen are more opinion or interesting fact rather than a 'puzzle'. One was striking to me  that human use of water is a considerable factor in sea level rise. I would have thought that to be near impossible, that we tend to use water that would have flowed to the oceans anyhow, but apparently we actually pump enough ground water that makes it to the oceans to have a (major, IIRC  40% ?) effect on the sea level rise. I guess most of the irrigation water evaporates/transpires, and makes its way to the ocean eventually?
ERD50
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06282015, 11:12 AM

#55

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OK, reflecting on this, I will qualify this as a 'goof' on Marilyn's part. Because her column is based on her supposed high intellect, she should be held to a high standard. So based on this quote:
"Say you have four specially marked dice. Two players each select one, and the player who rolls a higher number wins. The faces are 111555, 222266, 333333 and 444400. You’re given the first choice. Which die should you choose?"
I will say the 'goof' is she did not specify that player B observes Player A's pick. You can infer it, but I think that makes it weak.
'Two players each select one' and 'You’re given the first choice' does not indicate that player B sees your pick, and putting 'Two players each select one' first actually forms the image in your mind that they picked at the same time. It's awkward to contradict this later. As I have said about many of her puzzles, the wording could and should be better.
ERD50
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06282015, 11:20 AM

#56

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OK, there's some ambiguity here. When I wrote
Quote:
If we assume that the opponent can also figure out and pick the best dice, then it remains 50/50 chance that we can win. So, we can only expect to break even by choosing the best dice.

I was thinking that the opponent could use the same dice that you picked. In that case, if both of the players could figure out the best dice, then it would always be 50/50.
Now, thinking about the puzzle statement, I agree that it implies that if you get to pick first, your dice is no longer available to your opponent. That will make it more interesting.
But, but, but, that does not change how you solve the puzzle.
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06282015, 11:26 AM

#57

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Quote:
Originally Posted by ERD50
OK, reflecting on this, I will qualify this as a 'goof' on Marilyn's part. Because her column is based on her supposed high intellect, she should be held to a high standard. So based on this quote:
"Say you have four specially marked dice. Two players each select one, and the player who rolls a higher number wins. The faces are 111555, 222266, 333333 and 444400. You’re given the first choice. Which die should you choose?"
I will say the 'goof' is she did not specify that player B observes Player A's pick. You can infer it, but I think that makes it weak.
'Two players each select one' and 'You’re given the first choice' does not indicate that player B sees your pick, and putting 'Two players each select one' first actually forms the image in your mind that they picked at the same time. It's awkward to contradict this later. As I have said about many of her puzzles, the wording could and should be better.
ERD50

This would come into play if it was as MildlyEccentric observed, that there is not a definite order. "RockPaperScissors" becomes a trivial game if one has to make the 1st move. First player always loses. Of course, if there is only one dice of each type, the 2nd player would note the missing dice and know which the 1st picked.
But can you prove it?
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06282015, 11:42 AM

#58

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Quote:
Originally Posted by NWBound
...
But can you prove it?

I'm not sure what you are asking to be proved?
You can do the comparison  take any die and you will see that you can find another die among the remaining three with a higher chance of winning. I went through each, and found:
I pick A; B can beat me
I pick B; C can beat me
I pick C; D can beat me
I pick D; A can beat me
Conveniently in order! Of course, 'beat' only means on average.
ERD50
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06282015, 01:20 PM

#59

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That's what I meant by proof. One enumerates all the possibilities and counts the chances.
The order you describes is correct, and the chance of the 2nd picker winning is 2/3 or 66.67% for all 4 cases.
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06282015, 01:24 PM

#60

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Perhaps what Marylin meant when she said 222266 was the "strongest" was that if between any two randomly chosen pairs, 222266 has the best chance.
If so, then she was thinking ahead and about a different question. I still have to verify her answer though.
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