June 2007 Game 2

June 2007 LSAT, Game 2 Recap

Having answered all the questions in Game 2 of the June 2007 LSAT, a few final notes: 1)  This game was a bit harder than Game 1. On average, the first game in any section of Logic Games is the easiest, and each subsequent game is harder (sometimes a little harder, sometimes a lot harder) than the last. (This is true on average--occasionally a section will throw you a curveball, but most of the time it's true. Trust me, I've done every section of games that's ever been released.) This implies two things:

a)  When you're doing a test, you should attempt the games in order. There's a lot of bad advice out there about selecting which games you want to do, and I really think you should avoid doing this. It wastes time, and you can't always tell which games are easiest just by looking at them. Furthermore, the easier games are usually first anyway. So why waste 5 minutes picking through the games at the beginning of the section? It's smarter just to dive into Game #1. After a couple minutes on Game 1, if it happens to be really confusing, then consider aborting that Game and proceeding directly into Game #2. Trust me on this one: If you do Game 3 or Game 4 first, you are definitely digging yourself a hole.

b)  When you're studying Logic Games, make sure you understand earlier, easier games before busting your own balls attempting to understand the later, more difficult games. You must walk before you can run. Make sure you are sure you understand how Game #1 works before bothering with Game #2. It'll get easier with practice, but you risk killing your own mojo if you try the harder games before you understand the easier ones. The hard games can be ten times harder than the easy games. Start with the basics and work your way up.

2)  Unlike Game 1, I didn't make Worlds on Game 2. This was because I didn't see a very limited set of options where I'd get to fill out a lot of stuff. Make Worlds only when 1) there are only a couple basic ways to do the game, and 2) you get to fill out a lot of stuff with certainty (in one or both Worlds) if you pencil out the basic scenarios.

3)  This game doesn't fit neatly into any sort of fixed, previously known template. The games are largely improvisational. My goal is to give you a set of tools for different types of common operations:  Putting things in order, putting things in groups, drawing out Worlds, making maps--but on any given game, you're going to have to cobble together your own solution. Here, we were both putting things in order and putting things in groups. I didn't use any predetermined template--I just figured out something that worked. Please don't try to force a square peg into a round hole. Use common sense on each game, making a drawing that 1) works for you and 2) fits the given facts. You'll get better at this after lots and lots of practice.

4)  Here's my finished test page. (Sorry about my kindergarten-style handwriting.)

Call me if you'd like to chat! 415-518-0630. I'm always here to help.



June 2007 LSAT, Game 2, #10

Final question in Game 2 of the June 2007 LSAT. Here's the basic setup for the game. Question 10 asks "If Limelight is shown exactly three times, Harvest is shown exactly twice, and Greed is shown exactly once, then which one of the following is a complete and accurate list of the films that could be shown on Thursday?" Since the question is giving me new information, I'm going to answer this one just like I did Question 8 and Question 9--I'm going to make a new diagram.

The starting position, you'll remember, is this:

Thursday:                                                            H   |

Friday:                                                             G/L    | (not both G&L)

Saturday:                                                         G/H   | (not both G&H)

Question 10 adds new information about each of L, H, and G. We'll start with L, since that's the one we know most about. The new rule says L has to be shown three times. Since a movie can't be shown twice on a single day, this means that L has to be shown every day of the festival. Therefore, the diagram looks like this:

Thursday:                                                  L       H   |

Friday:                                                                  L    | G

Saturday:                                                 L     G/H   | (not both G&H)

In the above diagram, L has to go before H on Thursday, since H must be last on Thursday. L must go last on Friday, since either G or L must go last on Friday and both G and L can't go on Friday. L must go before G or H on Saturday, since either G or H must go last on Saturday. (All I'm doing here is applying the initial rules.)

The question asks "which one could be a complete and accurate list of the films that could be first on Thursday." Immediately, it's apparent that H cannot go first on Thursday, since L must go before H on Thursday. So any answer with H is eliminated--this gets rid of A, C, and E.  Sweet! The only remaining question is whether or not G can go first on Thursday, since L is mentioned in both remaining answer choices.

So can G go first on Thursday? Yeah, I'm pretty sure it can. For Question 10, G must be shown exactly once. It hasn't been shown yet, so why can't it be shown first on Thursday? If you can't tell me why it can't go there, then it can go there. So our answer is D.

June 2007 LSAT, Game 2, #9

Onward through Game 2 of the June 2007 LSAT.  Last week, I created a setup for the game. Question 9 adds three new rules that apply only for this question: 1) Greed is shown exactly three times; 2) Harvest is shown exactly twice; 3) Limelight is shown exactly once. The question asks "Which one of the following must be true" but I'll start with the new rules before looking at the answer choices. Let's start with the big one first: Greed is shown three times. Since a movie can't be shown twice on a given day, and there are only three days, that means G has to be shown every day. On Thursday, that means G is shown sometime before H, since H has to be last on Thursday:

Thursday:   G  H |

On Friday, G will have to be shown last, because either G or L has to be shown last on Friday and both can't be shown on Friday. This means L can't be shown at all on Friday:

Friday:              G | L

On Saturday, G will have to be shown last, because either G or H has to be shown last on Saturday and both can't be shown on Saturday. This means H can't be shown at all on Saturday:

Saturday:         G | H

Wow, that's a lot. Before I even look at the new rules about H and L, my diagram looks like this:

Thursday:     G  H |

Friday:                G | L

Saturday:           G | H


Turning to the rule about Harvest, if Harvest has to go twice (and if it's already going on Thursday, and can't go on Friday, then H will have to go sometime before G on Friday.

Thursday:     G  H |

Friday:          H  G | L

Saturday:           G | H

So that's G and H both completely determined. The final issue is L, which must go exactly once. It can't go on Friday, but it can go either Thursday or Saturday. At this point, I'll turn to the answers. Remember, we're looking for a must be true.

A)  All three films could be shown on Thursday, but they don't have to be shown on Thursday because L could go on Saturday instead. So this isn't the answer.

B)  No, Saturday could have either one or two films.

C)  No, L doesn't have to be shown Thursday.

D)  No, L could also be shown Saturday.

E)  Yep--see the diagram and reasoning above--both H and G have to be shown on Friday. So this is the answer to our "must be true" question.

(Note that our diagram for this question applies only to this question, since we were given new rules that only apply to this question.)

June 2007 LSAT, Game 2, #8

Let's continue through Game 2 of the June 2007 LSAT.  Last week, I created a setup for the game and answered questions six and seven. Today we'll tackle number eight, which says "If Limelight is never shown again during the festival once Greed is shown, then which one of the following is the maximum number of film showings that could occur during the festival?" To me, that's a pretty nasty question, because it involves a new rule and a lot of potential different scenarios where we'll have to try to maximize the number of films shown. I'm going to use a bit of intuition here, because I don't want to waste too much time on any one question. The problem with a question like this is that there's no very clean and certain way to test it. (Contrast this with question number six, which could be answered with an idiotproof process of elimination.) I'm going to try to squeeze in as many movies as possible, but in doing so I will have to make some guesses. Every time I take a guess, there's a path that I didn't take that might have worked out better.

I might start with the answers here, and work backward. I know, because seven is the maximum number of movies listed in any answer choice (E), that that's the absolute limit for the number of movies. So if I can figure out a way to make seven work, then E is the answer. If I can't make seven work, but I can make six work, then I'll choose D. And so on.

Can I make seven work? Let's see:

If I'm trying to maximize the movies shown, why wouldn't I go ahead and show all three films on the first day?

Thursday:  L  G  H

The problem with this approach is that, once I show G, I now can't show L anywhere in the remainder of the schedule. So Friday I could show H, then G:

Friday:  H  G

And Saturday, I could show just ONE of G and H:

Saturday:  H/G (doesn't matter which one).

Hmm. That's a total of only six movies. I was hoping to get to seven. So I'm going to try one more scenario:

On Thursday, if I want to avoid the problem of showing G and being unable to show L ever again, then I've got to avoid showing G. So I'll start my Thursday schedule with just L and H:

Thursday:  L  H

On Friday, I'll again want to avoid showing G early so I'll show just H and L:

Friday:  H  L

On Saturday, I'll now show L with G:

Saturday:  L  G

And once again, I can only get to six.

As I said earlier, this is a supernasty question, because I still haven't explored all the scenarios. What if I showed just L and H on Thursday, for example, but then showed all three films on Friday? Would I be able to get to seven then? (The answer is no. If I showed all three films on Friday then I'd only be able to show one film on Saturday, so once again I'd be stuck at six.)

Rather than exhaust all possibilities here, once I sensed that I was getting stuck at six I'd be happy going ahead and picking D. I might miss a question like this once in a while, but I'll 1) usually get it right and 2) never waste five minutes trying to be sure. I have bigger fish to fry in the remaining of the section, so I'll make an educated guess here and move on. Our answer is D.

June 2007 LSAT, Game 2, #7

Let's continue through Game 2 of the June 2007 LSAT.  Yesterday, I created a setup for the game and answered the first question. I answered #6 pretty confidently, but didn't make all that many inferences in the setup. So I'm still a bit apprehensive--have I missed something? Maybe Question 7 will help me find out. The question says "Which one of the following CANNOT be true?" There's no new information here, so I can't build a new diagram for this question. Instead, I'll just have to evaluate the answer choices. In doing so, I'll be relying on my main setup, which looked like this:

Thursday:                                                            H   |

Friday:                                                             G/L    | (not both G&L)

Saturday:                                                         G/H   | (not both G&H)

Since this is a "must be" question (must be false), it should be easier to positively identify the CORRECT answer than to eliminate the incorrect answers. (Contrast this approach with the process of elimination I used on the "could be true" question #6.)

On #7, if there is one (correct) answer that must be false, then there are four (incorrect) answers that could be true. The key is to not spend too much time proving that the incorrect answers are actually possible. I"ll just say "I don't see why not," and move on. Ideally, the correct answer is going to jump out at me and say "I'm a problem!" I need to move fairly quickly through all five answers in the hopes that the right answer will jump out at me.

A)  Well, that was easy. H must be shown last on Thursday, and can be shown last on Saturday. But Rule 2 said that either G or L has to be shown last on Friday. Based on that one rule alone, Answer A must be false. So this will turn out to be our answer. Just to be sure, I'll scan through B-E and make sure none of them look problematic.

B)  Sure, why not? The only potential problem here is if L is shown on Friday then G can't be shown on Friday... but there's no reason why G has to be shown on Friday. So this is possible, and that means it's not our answer.

C)  Sure, why not? If Thursday was L, G, H, then G would be second that day. If Friday was H, G then G would be second that day. If Saturday was L, G then G would be second that day. No problem, so this is out.

D)  Many ways to do this. No big deal, so this is out.

E)  H on Thursday, G on Friday, L on Saturday is the only way to do this. But it works, so this is out. Our answer is certainly A.

At this point, you might be asking whether I really needed to go through B-E here, having positively identified A as the correct answer. I definitely understand your point. but I've learned, through long practice on the LSAT Logic Games, that cutting corners is really not the best strategy. Going through B-E took me about 20 seconds total, and allowed me to answer the question with 100% certainty. I have learned that attempting to save 20 seconds can sometimes cost 5 minutes. On later questions (more difficult questions) you might see me take a calculated risk here and there. But on this one, since it's still early in the game, I wanted to make sure that none of the other answers seemed correct. Having narrowed it down to one and only one answer, I'll proceed through the rest of the questions on this game with a bit more certainty. Thanks for reading!


June 2007 LSAT, Game 2, #6

In my last post, I created a setup for Game 2 of the June 2007 LSAT. I didn't make as much progress as I might have liked, but it just feels like one of those games where you can't make a lot of inferences. As much as I'd like to make a ton of inferences and predict the answers before I've even seen the questions, I don't think that's possible here. So I'm going to turn to the questions and see if I can sort them out. This is where the rubber meets the road--I should fairly quickly be able to learn whether I've missed something or I'm on the right track.

Question 6 is a simple list question. It says "Which one of the following could be a complete and accurate description of the order in which the films are shown at the festival?"

Each answer provides a complete film festival calendar, and we're asked to pick the one that works. Since there is only one correct answer to each question, this means four of the answer choices will not work. And I know from experience that it's much easier, on a question like this, to identify the four that will not work than it is to be sure that the right answer really will work. It's a "could be true," and to me that usually means I'm going to use a process of elimination.

And I'm going to be a stickler for how the process of elimination is applied. We are not going to go through the answer choices in order and try to see what's wrong with them. if we did it this way, we'd run the risk of getting stuck for too long on the correct answer, trying to figure out what's wrong with it. Instead, we're going to take the rules, one at a time, and use them to eliminate as many answer choices as possible. (This is the fastest way to do it, and the Logic Games require efficiency. This question type appears on almost every game, so shaving off 30 seconds every time we see it makes a big difference in the long run.)

When we're done, we will have tested all of the rules. If we understand the rules properly, we'll be left with only one answer choice. That will be our answer.

Rule 1: H is the last film on Thursday.

This gets rid of Answer D, which incorrectly places L as the last show on Thursday. All the other answer correctly place H last on Thursday, so they're still in the running. We won't consider Answer D again, since it's eliminated.

Rule 2: Either G or L is last on Friday, and G and L can't both be shown on Friday.

This gets rid of Answer B, since both G and L are listed for Friday. It also gets rid of Answer E, since H is incorrectly listed as the last show on Friday. We're already down to just A and C.

Rule 3: Either G or H is last on Saturday, and G and H can't both be shown on Saturday.

This gets rid of neither A nor C. What do we do now?

First thing:  Don't panic. Second thing: Check the body of the game description (the opening paragraph) to make sure we haven't forgotten any rules. Let's see: Did A and C each show every film at least once? Ahhh, that's it. Answer A fails to show G on any day. Since every film has to be shown at some point, that eliminates A.

Since we've tested all the rules and we're left with one and only one answer, the correct answer must be C. I'm very confident in this answer, and I'll use that confidence as we proceed through the rest of the questions.

June 2007 LSAT, Game 2 Setup

Game 2 of the June 2007 LSAT is a bit different, and a bit more complicated, than Game 1. In that game, all we had to do was put five digits in order. Here, we have to select which things are going to be chosen AND put them in order. It's not an impossible game, but it's definitely a step up in complexity. In Game 1, the digits 0-4 were each used exactly once. Here, the films G, H, and L are all going to be shown at least once during a three-day festival, and no more than once on any given day. In other words, every film must be shown 1, 2, or 3 times, and can never be shown twice on any one day. Furthermore, we have to show at least one film every day. All of these rules are hidden in the introductory paragraph of the game. Make sure you don't ignore them.

My basic setup will look like this:




I'll fill in the films as horizontal lists, from earliest to latest, for each day. I'm doing it this way simply because it seems the most natural to me. If I were creating a film festival calendar in real life, it might look something like this. So that's how I'll do it on the LSAT as well. Your calendar might look different, and that's perfectly fine. I'm improvising a solution that makes sense to me. That's always the first step.

The rules:

Rule 1)  This is a two-part rule:  First, Harvest has to be shown on Thursday.  Second, Harvest has to be the last film shown on Thursday. I can put that into my calendar like this:

Thursday:                                                             H |



Here, I'm using the | to denote that nothing else can come after the H on Thursday. (Because H has to be last on Thursday). I'm sure there are other ways of doing this, but this will work for me.

Rule 2)  This is a three-part rule:  First, G or L has to be shown on Friday. Second, G and L can't both be shown on Friday. Third, G or L has to be the last film shown on Friday. I'll add that to my diagram, like this:

Thursday:                                                             H   |

Friday:                                                             G/L     | (not both G&L)


Is it perfect? I don't know, maybe not. But it's rarely going to be perfect. The games are highly improvisational, and I think it's a huge mistake to think that you can simply memorize all the different "types" of games and then force every game you encounter on the test into one of your templates. If you do that, you're going to be trying to jam square pegs into round holes. Far better, in my view, to learn a set of skills and then practice improvising solutions using those skills. We're adding tools to your toolbelt, so that you can cobble together a solution to whatever you might face on test day.

Rule 3)  Another three-part rule, just like Rule 2. First, either G or H must be shown on Saturday. Second, G and H can't both be shown on Saturday. Third, G or H must be the last film shown on Saturday. I'll add that to my diagram, like this:

Thursday:                                                            H   |

Friday:                                                             G/L    | (not both G&L)

Saturday:                                                         G/H   | (not both G&H)

Okay. Let's take a step back and think about where we're at. My thoughts, in the order they come to me:

  • H has already been used, (on Thursday) so we don't have to use it again (although we certainly can). G and L still both need to be used at least once.
  • There will be no problem making sure we have at least one film every day, since the rules gave us a film for Thursday and a choice for the last film for both Friday and Saturday. We're never going to have to worry about accidentally putting zero films on any day.
  • The only day on which L can be the final film is Friday.
  • H can be last on both Thursday and Saturday.
  • G can be last on Friday and Saturday.
  • Any film can be first on any day, although if H is first on Thursday, it's also the only film shown on Thursday.  Similarly, if G or L is first on Friday, it's also the only film shown on Friday. Similarly, if G or H is first on Saturday, it's also the only film shown on Saturday.
  • Every film can be shown on every day.
  • Every film can be shown on all three days consecutively.
  • But it's not possible to show BOTH G and L on all three days simultaneously, or BOTH G and H on all three days simultaneously, because of Rules 2 and 3.

Okay. What I'm doing here is just considering the landscape of the game, and all the rules that apply to the game, in the context of one another. I was hoping that some big inferences were going to appear, but they're really not happening. That's fine. As it turns out, there's a lot of flexibility in this game. I've practiced enough on the games to know that you can't always make big inferences. This seems to be that type of game, soI'm just going to have to go on to the questions and see what happens.

A note about "worlds":  On Game 1, I ended up realizing that there were only two basic templates that would account for all possible outcomes for the game. The code had to start either 1-2 or 2-4, and by penciling out those two options I was able to come up with two highly constrained templates that allowed me to quickly answer the questions. I call this "Making Worlds."

It's an extremely powerful technique when correctly applied at the right time. But on this game, I see no such opportunity to make worlds. When I'm making worlds, I look first for variables or spots that have very limited possibilities. For example, in Game 2, because of Rule 2, we know that either G or L has to go last on Friday. So it's true that we could make World 1 where G goes last on Friday, and World 2 Where L goes last on Friday. In World 1, L wouldn't be able to go at all, and in World 2, G wouldn't be able to go at all. The problem is that, beyond what I've just stated, we would know nothing more about the game. We'd be left with two Worlds that were essentially empty, which would be of no use to us at all. (Alternatively, we could make our Worlds using Rule 3's requirement that either G or H goes last on Saturday... we'd end up with a similarly unsatisfying result.)

So if I'm going to make Worlds, I'm looking for:

1) Very limited options (preferably just 2 options). 2) The ability to fill out a lot of additional stuff in at least one World (preferably both Worlds).

In this Game, requirement 1 is met but requirement 2 is not. So I won't make Worlds here--I suspect it would be a waste of time.