cldavid
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December 2015 Game Help

by cldavid Mon Jan 11, 2016 7:03 pm

I have never seen a game like this one, and it really hurt me on the previous LSAT. Would love a response on how to solve this game in a timely fashion!


Questions 13–17
Four employees—Jackson, Larabee, Paulson, and Torillo—are
to select from among four offices—W, X, Y, and Z. The order
in which they select, from first to fourth, is to be decided by a
random drawing. Each employee has ranked the offices from
first (highest) to fourth (lowest) as follows:
Jackson: Y, X, Z, W
Larabee: X, Z, W, Y
Paulson: Y, Z, X, W
Torillo: X, Y, Z, W
The following restrictions must apply:
Each employee selects an office that has not been selected
previously.
Each employee selects only one office.
Each employee selects the office that he or she ranks
highest among the unselected offices.

13- Which one of the following is a possible matching of
employees with the offices they select?
(A) Jackson: W; Larabee: Y; Paulson: X; Torillo: Z
(B) Jackson: Z; Larabee: X; Paulson: W; Torillo: Y
(C) Jackson: X; Larabee: W; Paulson: Z; Torillo: Y
(D) Jackson: Y; Larabee: W; Paulson: X; Torillo: Z
(E) Jackson: Y; Larabee: Z; Paulson: X; Torillo: W

14. Which one of the following must be true?
(A) At most one of the employees selects the office
he or she ranks first.
(B) At most one of the employees selects the office
he or she ranks second.
(C) At least one of the employees selects the office he
or she ranks first.
(D) At least one of the employees selects the office he
or she ranks second.
(E) At least one of the employees selects the office he
or she ranks third.

15. Which one of the following could be true?
(A) Exactly two of the employees each selects the
office he or she ranks third.
(B) Exactly two of the employees each selects the
office he or she ranks fourth.
(C) Exactly three of the employees each selects the
office he or she ranks second.
(D) Exactly three of the employees each selects the
office he or she ranks third.
(E) Exactly three of the employees each selects the
office he or she ranks fourth.

16. If Paulson selects office W, then which one of the
following could be true?
(A) Exactly two of the employees each selects the
office he or she ranks second.
(B) Exactly two of the employees each selects the
office he or she ranks third.
(C) Exactly three of the employees each selects the
office he or she ranks first.
(D) Jackson selects office X.
(E) Larabee selects office Z.

17. Which one of the following must be true?
(A) Jackson does not select office X.
(B) Larabee does not select office W.
(C) Larabee does not select office Z.
(D) Torillo does not select office X.
(E) Paulson does not select office X
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ohthatpatrick
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Re: December 2015 Game Help

by ohthatpatrick Thu Jan 14, 2016 4:24 pm

Just for future reference, you never need to (or should) post the text of the questions into your posts. We'll look at our copy. :)

This was definitely a weird, new game. It felt like a Draft to me. A four person draft.

Someone gets first pick and already knows who they want most. So they pick that person.

Someone gets second pick and knows who they want most. If that person is still available, they will pick that person. If their top pick already got snatched by the person who picked 1st, then they'll move onto selecting their second fave option.

For those of us who usually got picked near-last for any playground draft for soccer / kickball teams, this game is personal. :evil:

So here is everybody's order of preference.

Jackson: Y, X, Z, W
Larabee: X, Z, W, Y
Paulson: Y, Z, X, W
Torillo: X, Y, Z, W

Which kid is the total superstar that everyone wants first?

Well, it turns out it's a tie between X and Y.

But no matter what, pick 1 will be X or Y. And it looks like W is almost everyone's last pick. If W ever goes anywhere but last, we'll know that L picked 3rd.

It's terms of a diagram, we basically need 4 blanks to represent the four selections that will occur in a certain order.

__ __ __ __

For each blank, there are two pieces of info we need to record.
J,L,P,T (which person got to pick)
X,Y,Z,W (which option they picked)

So you might draw two rows
__ __ __ __ (J,L,P,T)
__ __ __ __ (X,Y,Z,W)

or just draw one row and write above/below it

Since this game is so unfamiliar and they don't really give us any rules we wouldn't have already assumed, there doesn't seem to be any front-end work to do.

But since we have no "feel" for this kind of game yet, we should either find a Conditional "IF" question or just write a couple hypotheticals of our own ...

We need to try to write a couple legal scenarios, just to ferret out the hidden constraints we might not yet see ... or conversely just to demonstrate for ourselves how wide open this game board really is

The Inference Chain we perform when we do so many questions in Games is some rhythmic series of questions we ask ourselves, "Where does the chunk go? Is H gonna be 1 or 7? Who's left? Let me put them in a cloud."

When we write hypotheticals, even 2 or 3, we start to get a feel for "what needs to be considered" in order to concoct a legal scenario.

So I would do Q16 first.

Q16
P picked W. Let's see ...

Jackson: Y, X, Z, W
Larabee: X, Z, W, Y
Paulson: Y, Z, X, W
Torillo: X, Y, Z, W

W is P's last choice. So P went 4th

__ __ __ P/w

Does that tell me anything else? Don't think I know much else. I know my one big inference that X/Y is always first.

x/y __ __ P/w

Here's where I would immediately switch in to trying stuff. I just wanna write ANY legal scenario that ends with P/w.

I'll arbitrarily pick an order
J L T P/w

Okay if J is picking first, he picks Y.

J/y L T P/w

L picks 2nd, and X is still available so L will pick X.

J/y L/x T/z P/w

That of course left T with Z, which seems fine. So let's see if we get lucky with this scenario.

Jackson: Y, X, Z, W
Larabee: X, Z, W, Y
Paulson: Y, Z, X, W
Torillo: X, Y, Z, W


J/y L/x T/z P/w

(A) No luck. No one had 2nd choice.
(B) T got 3rd choice. No one else.
(C) 3 people got their first pick? That'll never happen, right? Everyone wants X or Y for their first pick. So there's no way that THREE people could be happy. Eliminate this.
(D) J didn't have X, but could I just switch it over? No, I'd have to move J out of 1 in order to get him X.
(E) Could I just switch L and T? Sure.
J/y T/x L/z P/w
seems to work

(E) is the correct answer.


That's a tough start to the game. We got lucky with a hypothetical that was close enough to an answer choice that we could adapt our way to proving (E) was possible. But only (C) had an obvious Must Be False ring to me.

I would move on to Q17 and see if I can use the two hypotheticals we now have to eliminate any answers for which we have a counterexample.

Q17

Our two hypotheticals from Q16

J/y L/x T/z P/w
J/y T/x L/z P/w

(C) is gone. We've seen L/z
(D) is gone. We've seen T/x

(A) Could J have X? Seems like it. 2nd pick. Defer
(B) Could L have W? Seems like it. 3rd pick. Defer
(E) Could P have X? Wait. It would be 3rd pick? Is X gonna survive that long? Everyone else has X in the top 2, so X is definitely going to be gone by round 2. Here is our answer.

(E) is correct.


Q14


Our two hypotheticals from before:
J/y L/x T/z P/w
J/y T/x L/z P/w

(A) Could TWO people get their top pick? Sure, like in our hypothetical
J gets Y and T gets X. Eliminate.

(B) Could TWO people get their 2nd pick? Maybe. Defer.

(C) Could NO ONE get their 1st choice? Definitely not! The person who picks first, by definition, gets their 1st choice.

Since a counterexample is impossible, this answer MUST BE TRUE.

The correct answer is (C).


Q15

Four of these are impossible, so let's approach them looking for reasons they would break.

Given that (A) and (B) are about TWO people getting something in common while (C) / (D) / (E) are about THREE people getting something in common, I'll start with the latter. THREE matches is harder than TWO, so THREE is more likely to be impossible (if I'm planning to make eliminations ... if I'm planning to try to write legal hypotheticals, I would turn to A/B).

(C) If three people get their 2nd choice, J gets X, T gets Y, and L or P gets Z.

In order for J to get X or for T to get Y, they would each have to have 2nd pick. That's impossible. Eliminate this answer.

(D) If three people get their 3rd choice, L gets W, P gets X, and J or T gets Z.

P would only pick X with the 3rd or 4th pick. But X would never make it that far. The other three are going to snatch up X with the 1st or 2nd pick. Since P can't get X, eliminate this answer.

(E) Three people can't get their 4th choice. There are only two options available, W and Y. Eliminate

Returning to (A) and (B), (B) looks easier to think about because there are only two options for 4th pick.

(B) L gets Y and J/P/T gets W

Can Y survive to round 4? No. SOMEONE would have picked up Y in the first three rounds. Eliminate (B).

The correct answer is (A).
(for example: P/y T/x J/z L/w)

Q13


This isn't quite an Orientation question, but let's hope the dealbreakers are fairly easy and superficial.

(A) If J has W, it's 4th. If L has Y, it's 4th. Contradiction. Eliminate.

(B) If J has Z, it's 3rd. If P has W, it's 4th. The other two look doable.
L/x T/y J/z P/w

The correct answer is (B).


For the record
(C) If J has X, it's 2nd. If L has W, it's 3rd or 4th. If P has Z, it's 3rd. That forces T to be 1st. But T wouldn't pick Y 1st.

(D) If J has Y, it's 1st. If L has W, it's 3rd or 4th. If P has X, something impossible has happened. X will never make it to round 3.

(E) Also crushed by the impossible P/x.