Most who know how to play are not exceptionally good
Therefore, some at the NYP are not exceptionally good
This is giving us A -> B -m> C
So this is invalid as it's making C sufficient for A which is out of line with our diagram. For all we know there could be no violinists at NYP that aren't exceptionally good. This would lead to the argument being invalid as some is too vague and isn't supported
I'm getting stuck on where it says "Yet, it's crazy to suggest that some violinists at the New York Philharmonic are not exceptionally good. They are all exceptionally good."
I thought the valid conclusion would be that we can't say for certain that they're "not exceptionally good" nor "exeptionally good", we just don't know either way. If there's no premise for us to say with confidence that ALL violinists in the NY Philharmonic are exceptionally good then does this text assume that? Why can the text say it's crazy to think some (at least one) is not "exceptionaly good" (outside of the fact that one would have to assume they'd have to be "exceptionally good" to be in the NYPhilharmonic).
@hajmcl95 My read of this section is that the authors inclusion of that whole paragraph (“In fact, there are are only…”) is not meant to be read as a valid conclusion to the prior claims. Rather, it is being used to emphasize that the point the flaw is making is so obviously incorrect (claiming that NYP violinists are not exceptionally good).
@Tannercho06897 Being chronological in the text doesn't make a difference. The "before" in the situation refers to your Lawgic chaining as written out from left to right following the arrow direction.
Example: Most birds sing songs. All storks are birds. Thus, some storks sing songs.
Most birds sing songs: B -m-> Si
All storks are birds: St --> B
Birds (B) is the interlocking point. So when we chain together it looks like:
St --> B -m-> Si
Now you can identify that this is an "all before most" situation which does not yield a valid inference. Even though the "all" statement was the second premise given by the example text chronologically.
This made more sense when I remembered the question from the last skill builder:
"All surgeons enjoy the sight of blood. Most vampires enjoy the sight of blood."
I had incorrectly inferred that the "shared" concept of enjoying the sight of blood meant that the other groups (surgeons and vampires) also had overlap, and concluded that "Some surgeons are vampires."
That was wrong-- just because something enjoys the sight of blood, it does not mean they are either a surgeon, a vampire, or both. Enjoying the sight of blood is not necessary to be a surgeon or vampire.
@RISHABHKASHYAP1 I was basing this on the "Let's Review" section at the end of the lesson. I assume that the contrast between those two is the takeaway since that's what they gave us.
So how would the argument of "Most people who know how to play the violin are not exceptional. All violinists at the New York Philharmonic know how to play the violin. So, some violinists at the New York Philharmonic are not exceptional." work compared to the original argument?
@HenryLehmann my issue is with the use of the scoop analogy, as it suggests that "most" implies a random sample of a specific size that may or may not pick up an element from a very small subset. Most need not be a random sample. It can be a very specific subset that happens to be larger than 50%. A -> B -m-> C is invalid not because A might be tiny, but rather because A might be 49.999999% of B, and the C that do not contain any A are defined by the absence of A. Any particular subset containing most of the elements of a superset is not required to have the same general characteristics as that superset.
@dh2303 I think that makes sense to me. So given our real world knowledge we could say that A <–s–> C is invalid because we know A –> /C? Do I have that right? But isn't the point of understanding the lawgic to assume we don't know anything about the real world? Like couldn't you only say that C is defined by the Absence of A if you bring with you some sort of knowledge about C that exists outside of the premises?
@HenryLehmann close. The point is that you do not know the distribution or characteristics of C, so it is POSSIBLE that A -> /C. The scoop analogy acts as if any >50% subset of B is a random sample of >50% of B. That is the ecological fallacy. It need not be a random sample. It COULD be a random sample. It could require A. It could require A be absent. We do not know.
I think one of the primary reasons that this section uses A B and C is so they can do less hand holding. So instead of them actually writing out like you did, they just substitute it for letters.
That being said, you're Lawgic is right! So is the one listed in this section!
If we translate their Lawgic to your Lawgic, and let's use NEG, for "not exceptionally good" to substitute 'C' (since EG is exceptionally good in your Lawgic):
Their Lawgic becomes this:
NYP -> PV -m-> NEG
NYP <-s-> NEG
If we want to negate your /EG and their 'C', which is NEG, we get:
Yours:
/EG -> EG
which is exceptionally good
Theirs:
NEG -> /NEG
which is not not exceptionally good,
but the 'not's cancel out! which means;
which is exceptionally good
So it's the same! You got your Lawgic right! I think they're going simple with the letters to make us just understand the basic flaws.
@tiramistudy Thank you for bringing me back to this!! I understood why they used ABC which is really helpful for just understanding the basics. It was definitely a long study day for me cause I easily understood that right now.
@tiramistudy There is absolutely no way a simple argument like this should be this confusing in explanation. The formulas and letters make this a lot harder to navigate and understand. Will I have to process all of these easy to figure out flaws in formula on the lsat or can I just answer without all the alphabet soup and acronyms?
Here's some examples I made for myself, hope this provides some clarity!
Trap 4: Swapping "most" and "all" arrows
Remember: a valid conclusion about intersecting sets can only be made if the most arrow appears before the all arrow.
Ex.
All wrestlers in the WMMA know how to fight. Most people who know how to fight suck at it. Therefore, some fighters in the WMMA suck at it
WMMA -> F --m-> Suck@F
____
WMMA <-s-> Suck@F
The above is NOT a valid conclusion. The first set is not indicative of any other set, even if their qualities (fighting) overlap. So, that quality cannot be used to draw valid conclusions .
Ex.
Most WMMA women know how to fight. All women who know how to fight are strong. Therefore, most WMMA women are strong.
WMMA -m-> F -> S
____
WMMA -m-> S
The above conclusion IS valid
Remember, it's not "all most" chocolate is good, its "most all" chocolate is good
I don't want to "almost" ace the test, i want to get "most-all" questions correct
To break this down I like to think about it as numbers.
1st claim: Let's say there are 30 violinist at the NY Philharmonic that know how to play violin.
2nd claim: There are 1 million violinists in the entire world and more than 50% of them do not play the violin well.
Can we logically make the conclusion that the 30 violinists at the NY Philharmonic are part of the more than 50% of violinists who do not play violin well? Well no we cannot as doing so would require us to assume that the 30 violinists belong to the more than 50% group of violinist who do not play the violin well. The conclusion is based on an assumption that we cannot logically deduce or infer.
@robbietonie It truly does not. I'm sitting here trying to understand how sets play into this and it's not adding up.
If every single A is a B and over half of those Bs are Cs, then it's extremely possible that "scooping" the B will also give you an A. This flaw is incredibly difficult to put into "sets" and "buckets."
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123 comments
I'll add my input-
To simplify the violinist argument consider this:
All violinists at NYP know how to play
Most who know how to play are not exceptionally good
Therefore, some at the NYP are not exceptionally good
This is giving us A -> B -m> C
So this is invalid as it's making C sufficient for A which is out of line with our diagram. For all we know there could be no violinists at NYP that aren't exceptionally good. This would lead to the argument being invalid as some is too vague and isn't supported
I'm getting stuck on where it says "Yet, it's crazy to suggest that some violinists at the New York Philharmonic are not exceptionally good. They are all exceptionally good."
I thought the valid conclusion would be that we can't say for certain that they're "not exceptionally good" nor "exeptionally good", we just don't know either way. If there's no premise for us to say with confidence that ALL violinists in the NY Philharmonic are exceptionally good then does this text assume that? Why can the text say it's crazy to think some (at least one) is not "exceptionaly good" (outside of the fact that one would have to assume they'd have to be "exceptionally good" to be in the NYPhilharmonic).
@hajmcl95 My read of this section is that the authors inclusion of that whole paragraph (“In fact, there are are only…”) is not meant to be read as a valid conclusion to the prior claims. Rather, it is being used to emphasize that the point the flaw is making is so obviously incorrect (claiming that NYP violinists are not exceptionally good).
@BenWillGold Oh my gosh I think you're right. It's possible I'm too literal for the LSAT.. thank you! :)
Example:
All people who drive in Canada have passed driving test.
A --> B
P --> PDT
Most people who passed driving test are bad drivers.
Therefore, some people are bad drivers.
B -- most --> C
PDT -- m --> BD
A --> B -- m --> C
P --> PDT -- m --> BD
-----------------------------
P <--S--> BD
Invalid!!!!
Amazing we’re training for the LSAT and people are complaining about having to read
Adding this into the comment thread as this is what helped me understand this lesson since there are no videos:
Think:
❌ All → Most = “leakage”
You lose track of A
Nothing guaranteed
✅ Most → All = “capture”
Some A gets picked
Then ALL of them get carried forward
Thanks Chatgpt.
HELP: So is it just what comes up first in the passage that dictates if the situation is a "Most before All" or "All before All"???
@Tannercho06897 Being chronological in the text doesn't make a difference. The "before" in the situation refers to your Lawgic chaining as written out from left to right following the arrow direction.
Example: Most birds sing songs. All storks are birds. Thus, some storks sing songs.
Most birds sing songs: B -m-> Si
All storks are birds: St --> B
Birds (B) is the interlocking point. So when we chain together it looks like:
St --> B -m-> Si
Now you can identify that this is an "all before most" situation which does not yield a valid inference. Even though the "all" statement was the second premise given by the example text chronologically.
This made more sense when I remembered the question from the last skill builder:
"All surgeons enjoy the sight of blood. Most vampires enjoy the sight of blood."
I had incorrectly inferred that the "shared" concept of enjoying the sight of blood meant that the other groups (surgeons and vampires) also had overlap, and concluded that "Some surgeons are vampires."
That was wrong-- just because something enjoys the sight of blood, it does not mean they are either a surgeon, a vampire, or both. Enjoying the sight of blood is not necessary to be a surgeon or vampire.
Tricky one. I'll give putting this into words a shot.
A - NY Phil violinists
B - People who know how to play
C - People who are not exceptionally good
1.... (valid)
"A —m→ B → C yields valid conclusions via the chain.":
Premise 1 (A —m→ B) - Most violinists at the NY Phil (A) are people who know how to play the violin (B)
Premise (B→C) - All people who know how to play (B) are not exceptionally good (C)
Conclusion: Some (most) violinists at the NY Phil (A) are not exceptionally good at playing the violin (C)
2.... (invalid - this is originally given to us)
"A → B —m→ C yields no valid conclusions via the chain."
Premise 1 (A→B) - All violinists at the New York Philharmonic (A) know how to play the violin (B).
Premise 2 (B —m→ C) - Most people who know how to play the violin (B) are not exceptionally good at playing the violin (C)
Conclusion: Some violinists at the New York Philharmonic (A) are not exceptionally good at playing the violin (C).
(Incorrect because we don't know if the NY Philharmonic violinists are part of the "most people who know how to play violin" group.)
@kriskanya_ So, where should we applied this and should we then just cut the second rule as it does not provide us valid conclusion via chain??
@RISHABHKASHYAP1 I was basing this on the "Let's Review" section at the end of the lesson. I assume that the contrast between those two is the takeaway since that's what they gave us.
Why not videos this is frustrating
#FEEDBACK We need the videos back
So how would the argument of "Most people who know how to play the violin are not exceptional. All violinists at the New York Philharmonic know how to play the violin. So, some violinists at the New York Philharmonic are not exceptional." work compared to the original argument?
Can this general idea of A->B -m-> C; A<-s->C be valid if set sizes are within reasonable quantities?
where are the videos I dont like reading 😠
Does anyone else draw Venn Diagrams for these problems? It makes so much more sense when you can just visualize the buckets.
@Jurcis Yes I agree. When I think of the venn diagrams, it all comes into place!
This explanation uses a flawed argument itself, the ecological fallacy.
@dh2303 If I understand your comment you are referring to the conclusion that to be in the PY Phil you must be exceptionally good?
@HenryLehmann my issue is with the use of the scoop analogy, as it suggests that "most" implies a random sample of a specific size that may or may not pick up an element from a very small subset. Most need not be a random sample. It can be a very specific subset that happens to be larger than 50%. A -> B -m-> C is invalid not because A might be tiny, but rather because A might be 49.999999% of B, and the C that do not contain any A are defined by the absence of A. Any particular subset containing most of the elements of a superset is not required to have the same general characteristics as that superset.
@dh2303 I think that makes sense to me. So given our real world knowledge we could say that A <–s–> C is invalid because we know A –> /C? Do I have that right? But isn't the point of understanding the lawgic to assume we don't know anything about the real world? Like couldn't you only say that C is defined by the Absence of A if you bring with you some sort of knowledge about C that exists outside of the premises?
@HenryLehmann close. The point is that you do not know the distribution or characteristics of C, so it is POSSIBLE that A -> /C. The scoop analogy acts as if any >50% subset of B is a random sample of >50% of B. That is the ecological fallacy. It need not be a random sample. It COULD be a random sample. It could require A. It could require A be absent. We do not know.
A->B-m->C
-----
A <-s-> C
There could be a chance that all none of the A that are B were scooped into the C bucket when most B became C. So we cannot conclude that Some A are C
I am really confused by the "let's review" portion. The "Most before all" argument DOES say that:
A -m-> B --> C
Therefore, A -m-> C
But in the "let's review" part of this lesson it says that:
A -m-> B --> C yields no valid conclusions via the chain.
Can someone explain this please?
@KatherineElizabethMarkella Following because this is confusing me too bro
@KatherineElizabethMarkella There's a difference between
A -m-> B --> C
and
A --> B -m-> C
The first one allows us to conclude that most As are C.
But the second one doesn't allow us to make any valid conclusions.
Example
All non-caffeinated teas are relaxing. Most teas help with sleep. Therefore, most relaxing teas help with sleep.
@Gertabarentos Great example! Thanks for sharing, but I believe you meant to invalidly conclude that "therefore, some relaxing teas help with sleep".
I understand the logic behind this stim. My issue is with the lawgic. If I did it myself, I would've done:
Premise: NYP → PV ‑m→ /EG
Conclusion: NYP ←s→ /EG
Though, in the lesson, 'C' was not negated. I'm not sure if i'm not getting it or i've been reading too long today. #help
@QarimatOgunneye
I think one of the primary reasons that this section uses A B and C is so they can do less hand holding. So instead of them actually writing out like you did, they just substitute it for letters.
That being said, you're Lawgic is right! So is the one listed in this section!
If we translate their Lawgic to your Lawgic, and let's use NEG, for "not exceptionally good" to substitute 'C' (since EG is exceptionally good in your Lawgic):
Their Lawgic becomes this:
NYP -> PV -m-> NEG
NYP <-s-> NEG
If we want to negate your /EG and their 'C', which is NEG, we get:
Yours:
/EG -> EG
which is exceptionally good
Theirs:
NEG -> /NEG
which is not not exceptionally good,
but the 'not's cancel out! which means;
which is exceptionally good
So it's the same! You got your Lawgic right! I think they're going simple with the letters to make us just understand the basic flaws.
Hope this helps, and good luck studying!
@tiramistudy Thank you for bringing me back to this!! I understood why they used ABC which is really helpful for just understanding the basics. It was definitely a long study day for me cause I easily understood that right now.
Good luck with studying as well!
@tiramistudy There is absolutely no way a simple argument like this should be this confusing in explanation. The formulas and letters make this a lot harder to navigate and understand. Will I have to process all of these easy to figure out flaws in formula on the lsat or can I just answer without all the alphabet soup and acronyms?
Here is a rule of thumb that could help to understand this concept.
Most --> All chains can work.
All --> most chains don't preserve logic.
Hope this helps!
Has anybody found an actual LSAT question where this flaw is done, so we can see how the answers would look?
Here's some examples I made for myself, hope this provides some clarity!
Trap 4: Swapping "most" and "all" arrows
Remember: a valid conclusion about intersecting sets can only be made if the most arrow appears before the all arrow.
Ex.
All wrestlers in the WMMA know how to fight. Most people who know how to fight suck at it. Therefore, some fighters in the WMMA suck at it
WMMA -> F --m-> Suck@F
____
WMMA <-s-> Suck@F
The above is NOT a valid conclusion. The first set is not indicative of any other set, even if their qualities (fighting) overlap. So, that quality cannot be used to draw valid conclusions .
Ex.
Most WMMA women know how to fight. All women who know how to fight are strong. Therefore, most WMMA women are strong.
WMMA -m-> F -> S
____
WMMA -m-> S
The above conclusion IS valid
Remember, it's not "all most" chocolate is good, its "most all" chocolate is good
I don't want to "almost" ace the test, i want to get "most-all" questions correct
I am actually crying. I have no idea what this is saying
me. this is so confusing
I understood it with this analogy, maybe it'll help you:
Premise 1: All Olympic sprinters know how to run.
Premise 2: Most people who know how to run are not exceptionally fast.
Invalid conclusion: Therefore, some Olympic sprinters are not exceptionally fast.
To break this down I like to think about it as numbers.
1st claim: Let's say there are 30 violinist at the NY Philharmonic that know how to play violin.
2nd claim: There are 1 million violinists in the entire world and more than 50% of them do not play the violin well.
Can we logically make the conclusion that the 30 violinists at the NY Philharmonic are part of the more than 50% of violinists who do not play violin well? Well no we cannot as doing so would require us to assume that the 30 violinists belong to the more than 50% group of violinist who do not play the violin well. The conclusion is based on an assumption that we cannot logically deduce or infer.
Would you switch the "All →" to a "most ‑m→" and then you would have the "Two Mosts split" argument? Is this possible?
@isabella.pliska You cannot switch all for most as a rule of thumb!
this argument flaw does not lend itself well to his style of logic.
@robbietonie It truly does not. I'm sitting here trying to understand how sets play into this and it's not adding up.
If every single A is a B and over half of those Bs are Cs, then it's extremely possible that "scooping" the B will also give you an A. This flaw is incredibly difficult to put into "sets" and "buckets."