Goofs by a famous person

[NW-Bound]

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 1-1-1-5-5-5, 2-2-2-2-6-6, 3-3-3-3-3-3 and 4-4-4-4-0-0. You’re given the first choice. Which die should you choose?"​

Marylin's answer follows.

"Here’s a brain-bending 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 2-2-2-2-6-6."​

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.

 

[MildlyEccentric]

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 1-1-1-5-5-5, 2-2-2-2-6-6, 3-3-3-3-3-3 and 4-4-4-4-0-0. You’re given the first choice. Which die should you choose?"​

Marylin's answer follows.

"Here’s a brain-bending 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 2-2-2-2-6-6."​

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 à la Rock-Paper-Scissors.

 

[NW-Bound]

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"?

 

[ERD50]

***SPOILER ALERT**** (similar to MildlyEccentric's post, which I cross-posted)
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

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 rock-paper-scissors 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

 

[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 1-1-1-5-5-5, 2-2-2-2-6-6, 3-3-3-3-3-3 and 4-4-4-4-0-0. 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

 

[NW-Bound]

OK, there's some ambiguity here. When I wrote

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.

 

[NW-Bound]

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 1-1-1-5-5-5, 2-2-2-2-6-6, 3-3-3-3-3-3 and 4-4-4-4-0-0. 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. "Rock-Paper-Scissors" 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?

 

[ERD50]

...

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

 

[NW-Bound]

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.

 

[NW-Bound]

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.

 

[scrinch]

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. -ERD50

Assuming that what is meant by "on the planet" really means "in the planet", aluminum is nowhere near the earth's most abundant metal. It is the crust's most abundant metal, but the crust represents less than 1% of the volume (and mass) of the earth. In the bulk earth, magnesium is the most abundant metal by molecular abundance, and iron is the most abundant by mass. Magnesium is stored mostly in the mantle in silicate minerals, and iron is mostly stored in the core as an alloy with Ni and other heavy elements. They are both more abundant than aluminum by a factor of 10-20x (depending on whether we are talking about molar or mass abundances). Aside from iron in the core, the overwhelming majority of all elements in the earth are bonded ionically or covalently within minerals, and when geochemists talk about abundances, we always refer to bulk abundances, not abundances of free, unbonded elements, of which there are almost none.

 

[38Chevy454]

Assuming that what is meant by "on the planet" really means "in the planet", aluminum is nowhere near the earth's most abundant metal. It is the crust's most abundant metal, but the crust represents less than 1% of the volume (and mass) of the earth. In the bulk earth, magnesium is the most abundant metal by molecular abundance, and iron is the most abundant by mass. Magnesium is stored mostly in the mantle in silicate minerals, and iron is mostly stored in the core as an alloy with Ni and other heavy elements. They are both more abundant than aluminum by a factor of 10-20x (depending on whether we are talking about molar or mass abundances). Aside from iron in the core, the overwhelming majority of all elements in the earth are bonded ionically or covalently within minerals, and when geochemists talk about abundances, we always refer to bulk abundances, not abundances of free, unbonded elements, of which there are almost none.

Also, as a metallurgical engineer (or at least that is what my degree says.... not really what I do for work anymore) almost all metals in the earth are in the form of metal oxides. Alum oxide is very difficult to break the bond, recycling and remelting takes approx 10% of the energy of making new alum metal from the alum oxide. This is why most all alum production facilities are located by low-cost hydro-electric power sources. The reduction of the alum oxide is very electric power intensive. There is not any lower cost or easier chemical reduction process.

In comparison, new iron (steel) from iron oxide is approx the same cost as remelting scrap. Of course recycled scrap is better environmentally, but from cost standpoint it does not have as much incentive compared with aluminum.