Games
FORMAT OF THE GAME SECTION
The game section of the LSAT consists of four games, each of which
has about six questionssometimes 5 and sometimes 7. Thus, there
are usually twentyfour questions. The section is 35 minutes long.
FORMAT
Game 1 (about 6 questions)
Game 2 (about 6 questions)
Game 3 (about 6 questions)
Game 4 (about 6 questions)
This means that you have a little less than nine minutes for each
game. Or if you skip the most difficult game, as most people should,
then you have a little less than 12 minutes for each game. If this
sounds fastpaced, you're right. The LSAT is a highly "timed" test,
and the game section is the most highly "timed" part.
SKIPPING A GAME
Because games are difficult and time consuming, you should consider
skipping the hardest one.
ORDER OF DIFFICULTY
Unlike most standardized tests, the questions on the LSAT are not
listed from the easiest to the hardest. If they were, then deciding
which game to skip would be easyskip the last game. However, this
much can be said: the first game will not be the hardest and the
last game will not be the easiest; do not, therefore, skip the first
game. This is also true of the questionset to a game.
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THE THREE MAJOR TYPES OF GAMES
Ordering Games
These games require you to order elements, either in a line or around
a circle. The criteria used to determine order can include size,
time, rank, etc. Ordering games are the easiest games on the LSAT.
Luckily, they are also the most common.
Grouping Games
Grouping games, as the term implies, require you to separate elementstypically
peopleinto groups. Some conditions of the game can apply to entire
groups only, some to elements within a group only, and some to both.
This added complexity makes grouping games, in general, harder than
ordering games.
Assignment Games
These games involve assigning characteristics to the elements, typically
people. The most common task in these games is to assign a schedule.
You probably have had some experience with schedules; you may have
written the weekly workschedule for a business. If so, you know
how difficult the task can become, even when only a few conditions
are placed on the employees: Bob will work Monday, Tuesday, or Friday
only. Susan will work evenings only. Steve will not work with Bob.
Add to this that the company must have a full staff weekdays, but
only three people can work weekends. Scheduling games on the LSAT
are similar to this.
LOGICAL CONNECTIVES
While no training in formal logic is required for the LSAT, essentially
it is a logic test. So some knowledge of formal logic will give
you a definite advantage.
To begin, consider the seemingly innocuous connective "if..., then...."
Its meaning has perplexed both the philosopher and the layman through
the ages. The statement "if A, then B" means by definition "if
A is true, then B must be true as well," and nothing more. For
example, we know from experience that if it is raining, then it
is cloudy. So if we see rain falling past the window, we can validly
conclude that it is cloudy outside.
There are three statements that can be derived from the implication
"if A, then B"; two are invalid, and one is valid.
From "if A, then B" you cannot conclude "if B, then A." For example,
if it is cloudy, you cannot conclude that it is raining. From experience,
this example is obviously true; it seems silly that anyone could
commit such an error. However, when the implication is unfamiliar
to us, this fallacy can be tempting.
Another, and not as obvious, fallacy derived from "if A, then B"
is to conclude "if not A, then not B." Again, consider the weather
example. If it is not raining, you cannot conclude that it is not
cloudyit may still be overcast. This fallacy is popular with students.
Finally, there is one statement that is logically equivalent to
"if A, then B." Namely, "if not B, then not A." This is called
the contrapositive, and it is very important.
If there is a key to performing well on the LSAT, it is the contrapositive.
To show the contrapositive's validity, we once again appeal to
our weather example. If it is not cloudy, then from experience we
know that it cannot possibly be raining.
We now know two things about the implication "if A, then B":
1) If A is true, then B must be true.
2) If B is false, then A must be false.
If you assume no more that these two facts about an implication,
then you will not fall for the fallacies that trap many students.
We often need to rephrase a statement when it's worded in a way
that obscures the information it contains.
On the LSAT, as in everyday speech, two negatives make a positivethey
cancel each other out.
not(not A) = A
Example:
"It is not the case that John did not pass the LSAT"
means the same thing as "John did pass the LSAT."
The statement "if A, then B; and if B, then A" is logically equivalent
to "A if and only if B." Think of "if and only if" as an equal sign:
if one side is true, then the other side must be true, and if one
side is false, then the other side must be false.
(If A, then B; and if B, then A) = (A if and only if B)
A if and only if B 
A 
B 
True 
True 
False 
False 
Example:
"If it is sunny, then Biff is at the beach; and if Biff is at the
beach, then it is sunny" is logically equivalent to "It is sunny
if and only if Biff is at the beach."
"A only if B" means that when A occurs, B must also occur. That
is, "if A, then B."
A only if B = if A, then B
Example:
"John will do well on the LSAT only if he studies hard" is logically
equivalent to "If John did well on the LSAT, then he studied hard."
(Note: Students often wrongly interpret this statement to mean
"if John studies hard, then he will do well on the LSAT." There
is no such guarantee. The only guarantee is that if he does not
study hard, then he will not do well.)
The statement "A unless B" means that A is true in all cases, except
when B is true. In other words if B is false, then A must be true.
That is, if not B, then A.
A unless B = if not B, then A
Example:
"John did well on the LSAT unless he partied the night before"
is logically equivalent to
"If John did not party the night before, then he did well on the
LSAT."
The two statements "if A, then B" and "if B, then C" can be combined
to give "if A, then C." This is called the transitive property.
("If A, then B" and "if B, then C") = ("if A, then C")
Example:
From the two statements "if John did well on the LSAT, then he
studied hard" and "if John studied hard, then he did not party the
night before the test" you can conclude that "if John did well on
the LSAT, then he did not party the night before the test."
DIAGRAMMING
Virtually every game can be solved more easily and efficiently
by using a diagram. Unless you have a remarkable memory and can
process reams of information in your head, you must draw a diagram.
Because of the effectiveness of diagrams, games are the best candidates
for improvement. A wellconstructed diagram can change a convoluted,
unwieldy mass of information into an easily read list. In fact,
from a wellconstructed diagram, you can often readoff the answers
without any additional thought.
Symbols
The ability to symbolize sentences is one of the most important
skills you need to develop for the LSAT.
Five basic symbols are used throughout this discussion. They are
Symbol 
Meaning 
& 
and 
or 
or 
~ 
not 
> 
If..., then... 
( ) 
parentheses 
ORDERING GAMES
Ordering games are the easiest games, and fortunately they also
appear the most often.
Lineup Game
There are five peopleBugsy, Nelson, Dutch, Clyde, and Gottiin
a police lineup standing in spaces numbered 1 through 6, from left
to right. The following conditions apply:
There is always one empty space.
Clyde is not standing in space 1, 3, or 5.
Gotti is the third person from the left.
Bugsy is standing to the immediate left of Nelson.
"Clyde is not standing in space 1, 3, or 5" is symbolized as C = not(1,3,5).
"Gotti is the third person from the left" is naturally symbolized
as G = 3rd. Note: the fact that Gotti is third
does not force him into space 3he could stand in spaces 3 or 4.
"Bugsy is standing to the immediate left of Nelson" is symbolized
as BN. The diagram will consist of 6 dashes:
_1_ _2_ _3_ _4_ _5_ _6_
When placing the elements on the diagram, first look for a condition
that fixes the position of an element. There is none. Next, we look
for a condition that limits the position of an element. The second
condition, "Gotti is the third person from the left," limits Gotti
to spaces 3 and 4. This condition, as often happens with ordering
games, generates two diagrams: one with the empty space to Gotti's
left and one with the empty space to his right:
Diagram I ___ ___ ___ _G_ ___ ___
Diagram II ___ ___ _G_ ___ ___ ___
Next, we look for a condition that connects two or more people.
The last condition, BN, connects B with N. However, at this
stage we cannot place it on the diagram. Finally, we look for a
condition that states where a person cannot be standing. The first
condition states that Clyde cannot be standing in space 1, 3, or
5. Noting this on the diagram yields
Diagram I _~C_ ___ _~C_ _G_ _~C_ ___
Diagram II _~C_ ___ _G_ ___ _~C_ ___
(Note: D is "wild" because the conditions do not refer to him.
Thus D can stand in more positions than any other person.)
This diagram is selfcontained. There is no need to refer to the
original problem. If possible, avoid rereading the problem.
1. Nelson CANNOT stand in which one of the following spaces?
(A) 2 (B) 3 (C) 4 (D) 5 (E) 6
The method of solution to this problem is rather mechanical: We
merely place Nelson in one of the spaces offered. Then check whether
it is possible to place the other people in the lineup without
violating any initial condition. If so, then we eliminate that answerchoice.
Then place Nelson in another space offered, and repeat the process.
To that end, place Nelson in space 2 in Diagram II:
_~C_ _N_ _G_ ___ _~C_ ___
From the condition BN, we know that B must be in space 1:
_B_ _N_ _G_ ___ _~C_ ___
Now D could stand in space 4, and C could stand in space 6both
without violating any initial condition:
_B_ _N_ _G_ _D_ _X_ _C_ (Where X means "empty.")
This diagram is consistent with the initial conditions. So N could
stand in space 2. This eliminates choice (A).
Next, place Nelson in space 4. Then Diagram I is violated since
G is already in space 4, and Diagram II is also violated since there
is no room for the condition BN:
The answer is (C).
2. Which one of the following spaces CANNOT be empty?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
Assume that space 1 is empty. Then in Diagram I, the condition
BN can be placed in spaces 2 and 3, D can be placed in space
5, and C can be placed in space 6all without violating any initial
condition:
_X_ _B_ _N_ _G_ _D_ _C_
Thus space 1 could be empty. This eliminates (A).
Next, assume that space 2 is empty. In Diagram I, this forces BN
into spaces 5 and 6:
___ _X_ ___ _G_ _B_ _N_
However, this diagram does not leave room for C [recall C = not(1,
3, 5)]. Diagram I is thus impossible when space 2 is empty.
Turning to Diagram II, we see immediately that space 2 cannot be
empty, for this would make G second, violating the condition G = 3rd.
Hence Diagram II is also impossible when space 2 is empty. Thus
space 2 cannot be empty, and the answer is (B).
PATHS AND FLOW CHARTS
Because this type of game typically has many conditions, the chart
can easily get out of control. Charting is an art. However, there
are some guidelines that help:
1. Look for a condition that starts the "flow" or that contains
a lot of information.
2. Look for an element that occurs in many conditions.
3. Keep the chart flexible; it will probably have to evolve
with the changing conditions.
As you work through these games be alert to any opportunity to
apply the contrapositive rule of logic. Often negative conditions
can be expressed more clearly by rewording them in the contrapositive.
For example, the statement
"if it is not sunny, then Biff is not going to the beach"
can be reworded more directly as
"if Biff is going to the beach, then it is sunny."
We need to review two common fallacies associated with the contrapositive.
From the statement "if A, then B" we can conclude, using the contrapositive,
"if not B, then not A." It would be fallacious, however, to conclude
either "if not A, then not B" or "if B, then A." Also note that
some means "at least one and perhaps all."
Flow Chart
Six debutantesAlison, Bridgette, Courtney, Dominique, Emily,
Francinemeet at a party. During the time they have been at the
party some girls have come to like certain other girls.
Amiable Alison likes every girl at the party.
Aloof, yet popular Bridgette likes no one at the party, but everyone
likes her.
Courtney likes only two girls, one of whom is Dominique.
Dominique likes three girls, none of whom are Courtney or Francine.
Emily and Francine each like only one girl.
Alison likes every girl, so we start the flow chart with her:
Next, every girl likes Bridgette, but she does not like any of
them. So we end the "flow" with Bridgette. (Note how the diagram
evolves.)
Next, since Dominique likes three girls, two of whom are neither
Courtney nor Francine, she must like both Alison and Emily, in addition
to Bridgette. Adding this result plus the third condition, "Courtney
likes Dominique," to the diagram gives
Finally, since Emily and Francine each like only one girl and everyone
likes Bridgette, Emily and Francine each must like Bridgette only.
So there is nothing else to add to the diagram.
Note A,C,D forms a "loop", because from A the arrows can be followed
all the way around the "loop" back to A. But A,D,E does not form
a loop, because from A you cannot get back to A, whether you go
first to D, or first to E.
1. A "click" is a group of two or more girls who like one another.
How many clicks are formed amongst the six girls?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4
There is only one. In the chart, a twoway arrow connects A and
D, so they form a click. The loop A,C,D does not form a click because
it's not twoway: A likes C, but that feeling is not reciprocal.
The answer is (B).
2. How many girls at the party like at least one girl whose feelings
are not reciprocal?
(A) 2 (B) 3 (C) 4 (D) 5 (E) 6
In the chart, there are 5 arrows pointing to B, so 5 girls like
B. There are no arrows emanating from B, so none of those feelings
are reciprocal. The answer is (D).
GROUPING GAMES
Because grouping games partition elements into sets, the number
of elements is often an issue. Counting may have been one of man's
first thought processes; nevertheless, counting possibilities is
deceptively hard. This tends to make grouping games more difficult
than ordering games.
Pay close attention to the maximum or minimum number of elements
in a group; this is often the heart of the game.
Selection Game
The starting lineup for the Olympic basketball "Dream Team" is
chosen from the following two groups:
Group A: Johnson, Drexler, Bird, Ewing
Group B: Laettner, Robinson, Jordan, Malone, Pippen
The following requirements must be meet:
Two players are chosen from Group A, and three from Group B.
Jordan starts only if Bird starts.
Drexler and Bird do not both start.
If Jordan starts, then Malone does not.
Exactly 3 of the four fastbreak specialistsJohnson, Bird, Jordan,
Pippenmust be chosen.
It is best to solve this problem without a diagram; however, we
will still symbolize the conditions for clarity and easy reference.
The condition "Jordan starts only if Bird starts" implies only that
if Jordan is starting then Bird must be starting as well. So we
symbolize it as Jordan>Bird. The condition "Drexler and
Bird do not both start" means that if one starts then the other
does not. So we symbolize it as Drexler>~Bird. Students
often misinterpret this condition to mean that neither of them starts.
To state that neither starts, put both at the beginning of the sentence:
Both Drexler and Bird do not start.
The condition "If Jordan starts, then Malone does not" is naturally
symbolized as Jordan>~Malone. It tells us that if J starts
then M does not, but tells us nothing when M does not start. Such
a condition, where the two parts of an ifthen statement do not
similarly affect each other, is called a nonreciprocal condition.
On the other hand, a condition such as Jordan<>~Malone affects
J and M equally. In this case, we are told that if J starts then
M does not as before, but we are told additionally that if M does
not start then J does. It is important to keep the distinction between
reciprocal and nonreciprocal relations clear; a common mistake is
to interpret a nonreciprocal relation as reciprocal. The remaining
conditions cannot be easily written in symbol form, but we will
paraphrase them in the schematic:
Jordan>Bird
Drexler>~Bird
Jordan>~Malone
2 from Group A
3 from Group B
fastbreak specialists: Johnson, Bird, Jordan, Pippen
3 fastbreak specialists
Ewing, Laettner, Robinson are "wild"
Note: Ewing, Laettner, and Robinson are independent because there
are no conditions that refer directly to them. We now turn to the
questions.
1. If Jordan starts, which of the following must also start?
(A) Malone or Johnson
(B) Drexler or Laettner
(C) Drexler or Johnson
(D) Johnson or Pippen
(E) Malone or Robinson
From the condition Jordan>Bird, we know that if Jordan
starts, then Bird must start as well. Now both Jordan and Bird are
fastbreak specialists, and three of the four fastbreak specialists
must start. So at least one of the remaining fastbreak specialistsJohnson
or Pippenmust also start. The answer is (D).
2. All of the following pairs of players can start together EXCEPT:
(A) Ewing and Drexler
(B) Jordan and Johnson
(C) Robinson and Johnson
(D) Johnson and Bird
(E) Pippen and Malone
Begin with choice (A). Both Ewing and Drexler are from Group A,
so the remaining 3 starters must be chosen from Group B. Additionally,
they must all be fastbreak specialists since neither E nor D isthere
are exactly 3 fastbreak specialists. But Jordan and Pippen are
the only fastbreak specialists in Group B. So the third fastbreak
specialist cannot be chosen. The answer therefore is (A). This type
of question can be time consuming because you may have to check
all the answerchoicessave these questions for last.
3. If Malone starts, which one of the following is a complete and
accurate list of the players from Group A any one of whom could
also start?
(A) J (B) J, D (C) J, B (D)
J, D, B (E) J, E, B
Jordan cannot start with Malone according to the condition Jordan>~Malone.
To play three fastbreak specialists, therefore, Johnson, Bird,
and Pippen are all required to start. Since both Johnson and Bird
are from Group A and exactly two players from that group start,
these two players comprise the complete list of starters from Group
A when Malone also starts. The answer is (C).
4. Which one of the following players must start?
(A) Pippen (B) Johnson (C) Jordan (D) Malone (E) Bird
Suppose Bird does not start. Then the 3 fastbreak specialists
must be Johnson, Jordan, and Pippen. But if Jordan starts, then
from the initial conditions Bird must also start. Hence Bird must
always start. The answer is (E).
ASSIGNMENT GAMES
Assignment games match a characteristic with an element of the
game. For example, you may be asked to assign a schedule: Bob works
only Monday, Tuesday, or Friday. Or you may be told that a person
is either a Democrat or a Republican.
Because the characteristics are typically assigned to groups of
elements, assignment games can look very similar to grouping games.
Additionally, in grouping games the groups are often identified
by their characteristics. However, in assignment games you pair
each element with one or more characteristics, whereas in grouping
games you partition the elements into two or more groups.
Many assignment games can be solved very efficiently by using an
elimination grid. An example will illustrate this method of diagramming.
Elimination Grid
Dean Peterson, Head of the Math Department at Peabody Polytech,
is making the fall teaching schedule. Besides himself there are
four other professorsWarren, Novak, Dornan, and Emerson. Their
availability is subject to the following constraints.
Warren cannot teach on Monday or Thursday.
Dornan cannot teach on Wednesday.
Emerson cannot teach on Monday or Friday.
Associate Professor Novak can teach at any time.
Dean Peterson cannot teach evening classes.
Warren can teach only evening classes.
Dean Peterson cannot teach on Wednesday if Novak teaches on Thursday,
and Novak teaches on Thursday if Dean Peterson cannot teach on Wednesday.
At any given time there are always three classes being taught.
We indicate that a teacher does not work at a particular time by
placing an X on the elimination grid. Placing the conditions on
a grid yields
To answer the following questions, we will refer only to the table,
not the original problem.
1. At which one of the following times can Warren, Dornan, and
Emerson all be teaching?
(A) Monday morning
(B) Friday evening
(C) Tuesday evening
(D) Friday morning
(E) Wednesday morning
The table clearly shows that all three can work on Tuesday night.
The answer is (C).
2. For which day will the dean have to hire a parttime teacher?
(A) Monday
(B) Tuesday
(C) Wednesday
(D) Thursday
(E) Friday
Dornan and Novak are the only people who can work Monday evenings,
and three classes are always in session, so extra help will be needed
for Monday evenings. The answer is (A).
3. Which one of the following must be false?
(A) Dornan does not work on Tuesday.
(B) Emerson does not work on Tuesday morning.
(C) Peterson works every day of the week except Wednesday.
(D) Novak works every day of the week except Wednesday.
(E) Dornan works every day of the week except Wednesday.
The condition "Dean Peterson cannot teach on Wednesday if Novak
teaches on Thursday, and Novak teaches on Thursday if Dean Peterson
cannot teach on Wednesday" can be symbolized as (P not W)<>(N = TH).
Now, if Novak works every day of the week, except Wednesday, then
in particular he works Thursday. So from the condition (P not W)<>(N = TH),
we know that Dean Peterson cannot work on Wednesday. But from the
table this leaves only Novak and Emerson to teach the three Wednesday
morning classes. Hence the answer is (D).
4. If Novak does not work on Thursday, then which one of the following
must be true?
(A) Peterson works Tuesday morning.
(B) Dornan works Tuesday morning.
(C) Emerson works on Tuesday.
(D) Peterson works on Wednesday.
(E) Warren works on Tuesday morning.
If you remember to think of an ifandonlyif statement as an equality,
then this will be an easy problem. Negating both sides of the condition
(P not W)<>(N = TH) gives (P = W)<>(N not TH).
This tells us that Dean Peterson must work on Wednesday if Novak
does not work on Thursday. The answer, therefore, is (D).
Caution: Not all scheduling games lend themselves to an
elimination grid. It's sweet when this method can be applied because
the answers typically can be read directly from the table with little
thought. Only onethird of the assignment games, however, can be
solved this way. Most often the game will require a more functional
diagram, and you will need to spend more time tinkering with it.
When you first read an assignment game, you need to quickly decide
whether or not to use an elimination grid. You may decide to use
a table. Then spend three minutes trying to set it up, only to realize
you have taken the wrong path and have wasted three minutes. Unfortunately,
exact criteria cannot be given for when to use an elimination grid.
But this much can be said: if only two options (characteristics)
are available to the elementsyes/no, on/off, etc.then an elimination
grid is probably indicated.
