If I tell you that most law students study past midnight, and,
uh, and most people who study past midnight drink espresso...
what can you actually conclude with absolute logical certainty?
That, uh, that at least some law students are drinking espresso late at night?
Absolutely nothing.
Zero.
You can conclude precisely zero.
Wait, really?
But...
but it feels so intuitive!
Most do A to B, most do B to C, so at least a couple of them must go all the way
from A to C, right?
That right there is probably the single most common trap on the Logical Reasoning
section.
Test takers build a linear chain in their heads.
They see A most B, and then B most C, and their brain just fills
in the gap.
But mathematically?
The guaranteed overlap between A and C in that chain is exactly zero.
Okay, wait.
Walk me through the numbers on that.
How do you get zero overlap out of two mosts?
Alright, picture a group of 100 people who study past midnight.
That is group B, our middle term.
Now, say most of those 100 people are law students.
That could be, uh, 51 law students.
That leaves 49 people in that group who are not law students.
Right?
Right, 51 law students, 49 non law students.
Now take the second statement: most people who study past midnight drink espresso.
Again, out of our 100 late night studiers, that only requires 51 people to drink
espresso.
Where do those 51 espresso drinkers come from?
Oh!
They could all be the 49 non law students...
plus just 2 law students!
Wait, no, they could be the 49 non law students, and...
wait, why could it be zero?
Because the 51 espresso drinkers could easily be the 49 non law students plus 2 law
students, sure, but if group B was larger, or if...
let us take a group of 50 people.
Suppose group B is 50 people.
If 26 of them are law students, and 26 of them drink espresso...
wait, if B is the middle term, where does the overlap actually become guaranteed?
Oh, when B is the shared starting point!
The reference group!
Precisely.
Look at what happens when B radiates outward to both A and C at the same time.
If B is a group of 50 items, and most B are A, that means at least 26 items
in group B are A.
If most B are also C, that means at least 26 items in group B are C.
Ah!
26 plus 26 is 52!
But there are only 50 items in group B total!
So by the pigeonhole principle, at least 2 items in group B must be both
A and C at the exact same time.
When two most claims share the same reference group, their overlap guarantees a some
conclusion.
Let me rephrase that to make sure I have it clean.
If I say, most law students study past midnight, and most law students drink
espresso...
now law students is the shared hub.
It is the left side term for both claims.
Exactly.
Law students is the reference group of 100 percent in both statements.
So if most of them study past midnight, that is over 50 percent.
If most of them drink espresso, that is also over 50 percent of the exact same
group.
So those two majority groups have to crash into each other.
You are guaranteed that at least some law students do both.
Spot on.
That is the Shared Hub Rule.
You can only draw a valid some inference from two most statements if they radiate
outward from the exact same reference group.
If you try to chain them end to end like dominoes, A to B, B to C,
the chain snaps immediately.
Okay, but what if we do want to chain statements together?
We are not always stuck with two most claims, right?
What happens when we mix in an all claim?
Ah, the universal statement.
I call All the ultimate amplifier on the test.
Suppose you have A, most B.
And then you add, B, all, C.
Every single B is inside set C.
So if most A are B...
and every single B is C...
then all those B that were A are automatically swept into C as well!
Exactly.
The universal statement acts like an inescapable funnel.
So A, most, B, all, C validly yields A,
most, C.
The quantifier passes right through.
But wait, does the direction of the arrow matter for All?
It matters completely.
If you flip that arrow...
say you have A, most, B, but then C, all, B...
what do you know about A and C?
Um, C is inside B, but...
but the A items that are in B could be in the other part of B!
The part outside C!
Right.
C is just a small puddle inside the big lake of B.
The most A items might be swimming on the completely opposite side of the lake.
Reversing that arrow completely severs the inference.
It is so easy to fall for that when you are rushing through a Must Be True question
under time pressure.
What about Some?
How does Some play with All versus Most?
It is the same principle, but even tighter.
If you have A, some, B, and B, all, C...
then yes, those specific A items that hit B are guaranteed to be carried into C.
So A, some, C is valid.
Because All is 100 percent.
It captures everything in B.
Exactly.
But if you try A, some, B, and then B, most,
C...
think about where that some element lands.
A, some, B just gives you, like, maybe one or two items in B.
And B, most, C means more than half of B is C.
But those one or two items from A could easily be in the minority of B that is NOT
C!
You just dodged a trap that kills thousands of test takers every year.
That target element gets stranded in the unrepresented minority every single time.
I remember seeing those in my prep!
In the heat of the section, your brain sees Some and Most and thinks,
well, Most is pretty big, so they probably overlap.
But probably is an absolute poison pill on the LSAT.
In my litigator days, when we audited contracts, we looked for ambiguity designed to
trap the unmindful.
Test writers draft question choices the exact same way.
On Flaw and Must Be True questions, they love subtly swapping a Some for a Most,
or turning an All arrow backward, knowing rushing students read for broad context
rather than structural rigor.
So how do you stop yourself from making those silly errors when the clock is
ticking?
You run what I call the 3 second Arrow Head Audit.
Before you even look at the answer choices, you draw your terms on scratch paper and
audit the arrows at the shared junction.
Give me the checklist for that audit.
Step one: find the shared term.
Step two: check the arrow directions meeting at that term.
If you have two Most claims, are both arrows pointing AWAY from the shared term?
If yes, Shared Hub Rule applies, valid Some conclusion.
If one points in and one points out, stop right there, zero conclusion.
And if it is an All statement in the chain?
Step three: verify the All arrow points AWAY from the middle term toward the final
target.
It has to act as an outgoing funnel.
If the arrow points toward the middle term, the chain is broken.
It takes literally three seconds on scratch paper, but it prevents you from falling
for three different trap choices in a single stimulus.
Logical Reasoning does not reward speed or vibe reading.
It rewards a calm, repeatable audit.
Learn where the overlaps are guaranteed, and the traps just look like obvious noise.
Alright, shared hubs point out, funnels point away.
Good rule to live by.
Talk to you next time, Adrian.