Rule translation Help. URGENT Forum

[cdj588]

The two rules I have issue with are below.

1. PT 63 G4 Rule #2 - There is a box containing a green ball that is lower in the stack than any box that contains a red ball.
2. PT 38 Rules #1 - Frank demonstrates exactly one task before Glady's demonstrates any of the task.

1. I read this and thought, "Okay, I know there is at least one green ball and I know it is below all and any of the red balls." Similarly, "The Hulk, a very powerful actor, comes before any of the Avengers in a foot race." If that were a rule on a test that would tell me, The Hulk would have to be first in the race because he comes before all of the other Avengers.

However, this rule allows for greens to be above the reds as along as "at least one green" is below the "lowest red." I just do not see how that translation can possibly come from that rule. Why does a green have to be below the "lowest red?" Why can't atleast one green just be lower than the 2nd to lowest red?

Similarly,

2. I see this rule as F-G-G-F and no other way. Reason being, Frank demonstrates "exactly one" task before Gladys demonstrates "any" task. To me "any" = "all." I see this similar to a rule that may say, "A is faster than any of the other runners." If that were a rule on a test that would tell me A would have to be first in the race, because A is faster than any of the other runners.

However, this rule allows for F-G-F-G.


Thank you

[humbugger]

1. "any box that contains a red ball" so if your lowest balls are [low] rgrrr [high], well, now you don't have a green ball in a box that's lower than the lowest red. The lowest box contains a red ball, hence it counts as "any box that contains a red ball"

Good luck in your studying.

[marlo45]

1. PT 63 G4 Rule #2 - There is a box containing a green ball that is lower in the stack than any box that contains a red ball.

Why does a green have to be below the "lowest red?" Why can't atleast one green just be lower than the 2nd to lowest red?

Because of the highlighted word. The first lowest red is a red, after all. Therefore, if green has to be before 'any' red, then at least 1 green has to be before the first red as well. Furthermore, that at least 1 green must appear before any red does not mean green has to be first (assuming there are other colors).

Similarly, Frank must be before any of G. Frank does not have to be before anyone else, according to that rule, and once he's already before the first G, he can appear anywhere again. I don't remember the games you reference. I'm only responding as per the rules you stated.

On a different note, be careful about even partially reproducing LSAT questions here.

[flem]

The two rules I have issue with are below.

1. PT 63 G4 Rule #2 - There is a box containing a green ball that is lower in the stack than any box that contains a red ball.
2. PT 38 Rules #1 - Frank demonstrates exactly one task before Glady's demonstrates any of the task.

However, this rule allows for greens to be above the reds as along as "at least one green" is below the "lowest red." I just do not see how that translation can possibly come from that rule. Why does a green have to be below the "lowest red?" Why can't atleast one green just be lower than the 2nd to lowest red?

Similarly,

2. I see this rule as F-G-G-F and no other way. Reason being, Frank demonstrates "exactly one" task before Gladys demonstrates "any" task. To me "any" = "all." I see this similar to a rule that may say, "A is faster than any of the other runners." If that were a rule on a test that would tell me A would have to be first in the race, because A is faster than any of the other runners.

However, this rule allows for F-G-F-G.


Thank you

A green ball has to be before any red balls. There could be one green ball before the reds, or there could be five green balls. It doesn't matter, the rule just has to satisfy the bare minimum logical requirement.

The other rule is similar. The rule you're comparing it to is wrong. That would mean that A HAS to be first. But in the Frank/Gladys rule, Frank shows up EXACTLY ONCE before Gladys for the first time. (I don't know any of the rules other than the one you listed, so let's assume that's the only one)

So if the variables are A B C D E F G

This rule still allows for any of the other variables to show up before Frank, just as long as Frank shows up ONLY ONE TIME BEFORE Gladys shows up for the first time.

Helpful at all?

[cc.celina]

You might be trying to symbolize a little TOO much. I would probably diagram these as
1. 1 G - All the Rs
2. 1 F - 1st G

when a rule says "any of the" it means any ONE of them, not ALL of them. If it said "all of the" your interpretations would be correct.

Your race analogies don't work because they don't allow for repeats. In both these cases, there are repeats - multiple green balls, multiple Franks. If the Hulk had a clone, and either Hulk #1 or Hulk#2 could win the race by finishing first, the situation would be more analagous.

[cdj588]

Awesome. Thanks for the replies. I think I am really getting close to full understanding.

This little tid bit helped

when a rule says "any of the" it means any ONE of them, not ALL of them. If it said "all of the" your interpretations would be correct.

I can now see how greens can go above reds so long as atleast one green is below atleast any ONE of the reds. HOWEVERRRR, this gets me half way there. I still do not see the requirement in the rule for at least one green being below the "lowest" red.

[marlo45]

I can now see how greens can go above reds so long as atleast one green is below atleast any ONE all of the reds.

That's how it should be read.

When they say 'any' box that contains a red, it is meant to include even the very first red. So if there is red present anywhere, then at least 1 green must be before it. HTH

edit: elaborated.

[cdj588]

Alright well whos right cc.celina or marlo45?

[marlo45]

Alright well whos right cc.celina or marlo45?

Both.

In the case of the boxes, 'any' means anytime red appears. Think about it. If at least 1 green must be present before 'any' red appears, then it rules out red from being #1 because anywhere a red appears there must be at least 1 green before it. Therefore, at least 1 green must make an appearance before even the very first appearance of a red, so every red will have at least 1 green that appears before the very first red.

RGR is unacceptable because there isn't at least 1 G before that first R.
GGRGR works
GRRRG works

No matter the configuration, the rule makes it impossible for any 1 red to go before at least 1 green. Maybe i'm bad at explaining lol. Sorry about that.

edit: tried to explain better.

[cc.celina]

Haha ok I'll try to clarify a bit better:

There is a box containing a green ball that is lower in the stack than any box that contains a red ball.

We know that there's a box containing a green ball. We'll call these boxes G.
We know it's lower than "any box that contains a red ball." We'll call those R.

The rule reads: One G must be lower than any R.
"Any R" here means absolutely any R that is in the stack. If there's an R, there has to be a G lower than it. That means you can randomly select any one of the Rs in the stack, ANY of them, and there's guaranteed to be a G somewhere under it. You're never going to find an R without a G before it.

Let's try some hypotheticals (for the sake of simplicity, I'm going to call any box without a green or red ball X).

X G R X X R -- correct
Pick any of the Rs. Is there a G before it? Yeah, there is. There's no R that doesn't have a G before it. This hypo is OK.

X X R G R R R -- incorrect
Let's take a look at the Rs. Three of them have a G before them. Cool! But one of them, that first one, doesn't have a G before it! That's not ok. The "G" in this hypo is not lower than "any of the reds," just lower than "some of the reds."

Notice, however, that if there are multiple G's, it doesn't matter where you put the rest of 'em. It just matters that every R has a G before it.

G X R R G R X -- correct
So, we have an extra G thrown in there. Does it matter? Look at all the Rs - they all have a G lower than them somewhere. So this hypo is OK. One of the G's is before any of the Rs.

Frank demonstrates exactly one task before Glady's demonstrates any of the task.

The wording here changes the rule a slight bit, because Frank can only demonstrate "exactly once" before Gladys does. So G's first task has to have at exactly one -- no more, no less -- F before it. If it has 0, wrong. If it has 2, wrong.

F G G F - as you pointed out, correct
F F G G - incorrect, because Frank goes exactly TWICE before any of Gladys's, not exactly ONCE
G F F G - incorrect, because frank goes exactly ZERO times before any of gladys's
F G F G - correct, because Frank goes once and only once before Gladys even begins to demonstrate her tasks.

Does that clear anything up?

[abc12345675]

I'm curious as to why this was URGENT

[marlo45]

I'm curious as to why this was URGENT

There's a test around the corner. I guess OP wants to straighten out his/her confusion before sitting it.

[cdj588]

LSAT is monday and I cannot get this out of my head so do other stuff.

[Micdiddy]

I'm curious as to why this was URGENT

There's a test around the corner. I guess OP wants to straighten out his/her confusion before sitting it.

No, no, no. Clearly OP has a gun to his head with a crazed theoretical gamesman at the trigger, whose spent the last 40 years trying to adequately understand the nature of these governing rules.

[cc.celina]

a crazed theoretical gamesman at the trigger

where can i find one of those!? #motivationproblems

[marlo45]

I'm curious as to why this was URGENT

There's a test around the corner. I guess OP wants to straighten out his/her confusion before sitting it.

No, no, no. Clearly OP has a gun to his head with a crazed theoretical gamesman at the trigger, whose spent the last 40 years trying to adequately understand the nature of these governing rules.

[Micdiddy]

a crazed theoretical gamesman at the trigger

where can i find one of those!? #motivationproblems

Idk, but I imagine he looks like Doc from Back to the Future?? OP, does he look like Doc from Back to the Future??

[cdj588]

Thank you for the very detailed replies. I do appreciate the effort and time.

However, there is still one issue cc.celina.

One of the questions in the game goes against what your saying, I think.

The game stipulates there are 6 boxes, we have balls in 3 colors - R/G/W, there are more R than W, and atleast once we will see a W immeditately below a G.

One of the questions in this particular game says, If there is a G in boxes 5 and 6 what could be true?

So, we must have: W G R R G G but as you can see not all of the G are below an R, infact 2 G will never be below an R.

[Micdiddy]

Thank you for the very detailed replies. I do appreciate the effort and time.

However, there is still one issue cc.celina.

One of the questions in the game goes against what your saying, I think.

The game stipulates there are 6 boxes, we have balls in 3 colors - R/G/W, there are more R than W, and atleast once we will see a W immeditately below a G.

One of the questions in this particular game says, If there is a G in boxes 5 and 6 what could be true?

So, we must have: W G R R G G but as you can see not all of the G are below an R, infact 2 G will never be below an R.

That's not an issue with what she is saying, in fact she pointed out multiple scenarios where R's were below some G's.

ETA: I'm sorry, not "multiple," one scenario. But still everything she is saying is correct.

2ETA: and you never answered my question.

[cc.celina]

So, we must have: W G R R G G but as you can see not all of the G are below an R, infact 2 G will never be below an R.

Remember, I emphasized that the proper way to interpret the rule is "All R's have a G below them," not "All G's have an R above them."

[Micdiddy]

Oh wait, it WAS multiple.

[cc.celina]

If you're good at formal logic, the statement goes like this:

R --> G-R

(where G-R means a G is before an R)

If you have an R a G must be before it. The contrapositive is:

G-R --> R

What YOU'RE inferring is:
G -> G-R
Which is an incorrect representation of the statement and, logically, is not necessarily true.

[cdj588]

Got it now. Thanks alot cc.celina. I really appreciate it.

I also noticed, which I should have sooner, I can make these "any" rules into if/then statements correct?

Frank demonstrates exactly one task before Glady's demonstrates any of the task. Can be, If G demonstrates a task, then F demonstrates exactly one before it.

[cdj588]

Just saw your pst after I posted mine.

Once again, thank you for your help. Much appreciated.

[cc.celina]

No problem! Hope that satisfied your crazy gunman