Setup and Rule Diagrams

[Eric Ockert]

Hi!

The language of the rule says, "If Fleure left any message, Pasquale and Theodore did also, all of Pasquale's preceding any of Theodore's. This shows the rule is essentially giving us three necessary conditions that follow if F left a message:

1. P is in.
2. T is in.
3. P's are before T's.

So, if any one of those conditions is not met, then F cannot be in.

Hope that helps!

[yrivers]

Wonderful! Very helpful. Thank you!

[LSAT2018]

I struggled with the first rule 'At most one person left more than one message' in that I didn't think of the possibilities (numerical distribution) until Question 8 or 9.
Is there any way I can spot out numerical distribution games more easily? This game took me a very long time and it affected me in the later games which were easier but the time restrictions just ruined it

[Malila Robinson]

Hi LSAT2018,
Look for numerical words/phrases to help you to see numerical distribution games. Things like "at most" "at least" "no more than" "twice as many", etc. These all have numerical aspects to them and likely mean that there is a numerical distribution aspect to the game.
Hope that helps!
Malila

[agt185]

I had a question regarding the contrapositives for two of the rules here. "If F left any message P and T did also, all of P preceding any of T." and "If P left any message, H and L did also, all of H preceding any of L."

Here are my diagrams along with what I think are the contrapositives:

F ---> P + T (P's ----T's) Contra: /P or /T or (T's ----P's) -----> /F

P ----> H + L (H's ---- L's) Contra: /H or /L OR (L's ---- H's) -----> /P

For these two rules if the sequence specified changes does that mean that the necessary condition was failed?

Also if we start linking up some of these conditionals how do you suggest we incorporate the sequencing?

Sorry for the long question, but this game set really threw me for a loop. Thanks for your help!

Amy

[Charlie Melman]

Hi Amy,

On your first question: good intuition--you're exactly right that if the sequence is wrong (for example, if L is before H) the necessary condition fails and P can't have left a message.

Second question: If F leaves a message, we know for a fact that H-L (in other words, that H appears and L appears, and H appears before L). So I think the most concise and clear way to diagram the chain (which you are right to want to link!) is as follows:

F P-T H-L

Hope this helps!

[Kelly R]

Hi PS,

Quick question about conditional rules involving sequencing. Is it the case that if any component of the sequence serving as the necessary condition fails (either the sequence is reversed, or one of the variables involved in the sequence is out), the sufficient condition fails, as well? For example, when rule 3 is linked with rule 4 and rule 5, the chain: G--> F––>(P–T) AND P-->(H–L) results. If T is not present, if P is not present, or if a T-P sequence results, then F and G are not present either, correct? Similarly, if L is not present, if H is not present, or if an L-H sequence results, then neither P, F (which requires P) or G can be selected, right? It would not necessarily be the case, though, that if T is not selected, P isn't selected, since P is only involved in the sequencing relationship and isn't serving as a sufficient condition, right? Sorry for the onslaught of questions, but thanks so much for the help.

[j199393]

Hi Moshe,

I've always thought this is a deceptively hard game. It's linear with some sequencing, and that usually seems easy, but they throw everything for a loop when it turns out that someone can leave more than one message (and thus not everyone has to leave a message).

I'm not sure about the distributions you created (especially #4 and #5--are you saying that just 3 or 2 people left messages? That's not possible). I actually prefer to set this game up by looking at the distribution of number of messages to people. There are three Numerical Distributions under that framework:

• 1. 1-1-1-1-1-1
  
  Here, each person leaves exactly one message, and all variable-specific rules are in effect (P > T, H > L). The possibility of a distribution such as this one answers question #9.
  
  
  2. 2-1-1-1-1-0
  
  One person leaves exactly two messages (and thus this is the distribution in play for a question like #7), four people leave one message each, and one person leaves no message.
  
  In this distribution, everyone leaves a message except G (thus, G is the "0" above). If G is selected, every other variable must also be selected, thus, if someone isn't going to leave a message, it has to be G.
  
  
  3. 3-1-1-1-0-0
  
  One person leaves exactly three messages, three people leave one message each, and two people leaves no message.
  
  In this distribution, everyone leaves a message except F and G (thus, F and G are the "0s" above).

In every distribution, at least P, T, H, and L leave messages.

Please let me know if that helps. Thanks!

Hi,

I did not set this game up using a numerical distribution like this. I'm not sure how I would know to do it this way when initially setting a game up. Can someone walk me through this decision and how to know when to do it and why it is most beneficial? Thank you!

[Paul Marsh]

Hi j199393! Numerical Distributions can be a bit tricky to spot at first. If you have the PowerScore LSAT Course Book, I'd recommend you check out Lesson 9 where it talks about this. In addition, this is a great Forum post from Jon that touches on this:

https://forum.powerscore.com/lsat/viewt ... 257#p19257

To re-iterate some of what Jon's talking about in that post, on every game I start with considering all of the numerical options and then whittling down from there.

For Linear Games: do all of the variables match up with a "spot" (Is the game Balanced)? Does every variable get used once only? Do my rules limit that at all?

In a Grouping Game, do I know what the exact size of all the groups are? If not, what's the max/minimum of each group? how do the rules further limit the group sizes?

For this Game, as soon as I read my first rule I realize that this is not necessarily a game where every variable is used exactly once (it leaves open the possibility of one [and only one] of the people leaving multiple messages). From that point on, I want to constantly re-evaluate all the numeric possibilities of the game. Can that one person leave all six messages? Our second rule quickly clarifies things: that one person can leave only up to 3 messages. I re-evaluate the possibilities and realize there are only 3: everybody leaves one message, one person leaves 2 messages and four people leave one message, or one person leaves 3 messages and three people leave one message. Now I've got my 3 possibilities!

Hope that helps!

[JocelynL]

Hi Moshe,

I've always thought this is a deceptively hard game. It's linear with some sequencing, and that usually seems easy, but they throw everything for a loop when it turns out that someone can leave more than one message (and thus not everyone has to leave a message).

I'm not sure about the distributions you created (especially #4 and #5--are you saying that just 3 or 2 people left messages? That's not possible). I actually prefer to set this game up by looking at the distribution of number of messages to people. There are three Numerical Distributions under that framework:

• 1. 1-1-1-1-1-1
  
  Here, each person leaves exactly one message, and all variable-specific rules are in effect (P > T, H > L). The possibility of a distribution such as this one answers question #9.
  
  
  2. 2-1-1-1-1-0
  
  One person leaves exactly two messages (and thus this is the distribution in play for a question like #7), four people leave one message each, and one person leaves no message.
  
  In this distribution, everyone leaves a message except G (thus, G is the "0" above). If G is selected, every other variable must also be selected, thus, if someone isn't going to leave a message, it has to be G.
  
  
  3. 3-1-1-1-0-0
  
  One person leaves exactly three messages, three people leave one message each, and two people leaves no message.
  
  In this distribution, everyone leaves a message except F and G (thus, F and G are the "0s" above).

In every distribution, at least P, T, H, and L leave messages.

Please let me know if that helps. Thanks!

I was able to make the inference that in every distribution P, T, H, and L leave messages (please confirm the reasoning is because they are all necessary conditions and by removing one of these, it removes two variables in addition to G). But I fail to see why F (being that its a necessary condition) isn't always in every distribution.

is it because removing F only removes 2 variables (G and F) and removing any of the other NC variables removes 3 variables and at most you can only remove 2 variables to make the distribution work?