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Setup and Rule Diagram Explanation

This is a Grouping Game: Defined-Fixed, Unbalanced: Overloaded.

This game features a fixed group of four selections, with an overloaded group of seven variables available to fill those four spaces. Thus, four variables are always selected and three variables are not selected.

The game contains only four rules. The first two rules are quite powerful and “reserve” two of the four available spaces. These rules are represented directly on the game diagram with dual-options. The third and fourth rules are both simple conditional rules, and are represented with arrow diagrams in our setup:
Oct 03_M12_game#2_L5_explanations_game#3_setup_diagram_1.png
Oct 03_M12_game#2_L5_explanations_game#3_setup_diagram_1.png (5.59 KiB) Viewed 3745 times
Of course, there are also inferences that can be made in the game:
  • Because J, K, N, and P will collectively occupy exactly two of the spaces, the remaining two spaces are occupied by the group of L, M, and Q. Thus, we can infer that two people from the group of L, M, and Q must always be selected. This will be represented directly on the diagram using a parenthetical notation. Note that any time one of the members of the group of L, M, and Q is not selected, the other two must be selected.

    From the last rule, when Q is selected then K must be selected, and from the first rule when K is selected J cannot be selected. Thus, Q and J cannot be selected together. This rule is shown with a double-not arrow.

    Note that some students attempt to draw an inference between P and L by combining the second and third rules. There is no usable inference that can be drawn from these two rules (an inference is present, “Some Ls are not Ps” but that inference has no value in this game).
The final entire setup then appears as:
Oct 03_M12_game#2_L5_explanations_game#3_setup_diagram_2.png
Oct 03_M12_game#2_L5_explanations_game#3_setup_diagram_2.png (7.15 KiB) Viewed 3745 times
With the information above, we are ready to attack the questions.
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I got this game correct, but got really nervous when initially solving it because the rules seemed to conflict with me.

October 1993 (8-13)

The first rule says Either J or K must be selected, but J and K cannot both be selected. I wrote it as

Not J---->K

Same goes with the next rules

Either N or P must be selected, but N and P cannot both be selected

Not N---->P

The reason why I ask is because it has the either/or and the Not Both in there. If I reversed it, and did

J---> Not K or N--------->Not P

Would my answers be the same? More specifically, If I see this again, what is my thought process in order to diagram it efficiently?
Last edited by JKing on Fri Apr 19, 2013 2:39 pm, edited 1 time in total.
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 Dave Killoran
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Hi J,

Good question. A rule such as "Either J or K must be selected, but J and K cannot both be selected" is really two-rules-in-one, so let's break it down:
  • The "first" rule: "Either J or K must be selected, "

    In this instance, at least one of J and K is selected, which is diagrammed as:

    J :arrow: K
    K :arrow: J

    Combined, that yields: J :dblline: K

    Thus, one of the two (J or K) is always selected. By itself, this portion of the rule would allow both to be selected. But, the test makers didn't want that to occur, and that's where the second part of the rule comes into play.

    The "second" rule: "but J and K cannot both be selected"

    In this instance, both of J and K cannot be selected, which is diagrammed as:

    J :arrow: K
    K :arrow: J

    Combined, that yields: J :dblline: K

    Thus, both cannot be selected.
So, what's the bottom line when those two rules are combined? That one and exactly one of J and K is selected, and that the other one is not selected. So, every game solution contains J or K, and either K or J is always among the variables not selected. A tricky rule overall!

Please let me know if that helps. Thanks!
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I will appreciates your kind reply to the above game from the Lesson book (5) the #3 game..regarding the proper set-up….i got all the answers correct but it took me about 14minutes…which is an indication to me that my setup is not the best……
Please help me with the setup….. All i did with the rules is :
1) J or K
2) N or P
3) If N then L not P
4) If Q then K not J

with my interpretation of the rules and inferences ….it still took me about 14mins…It makes me feel…. there are still other inferences i didnt make….pls help…..will greatly appreciates your usual assistance
 Adam Tyson
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First, your diagram looks great to me. Near perfect, in fact. What needs to happen next is that you need to draw a few inferences. For example, once we have our first two slots set aside for J/K and N/P, we have two slots left and only three variables that can go there. So, what if L is out? M and Q are in; if M is out, then L and Q must be in; and finally, if Q is out, L and M must be in. In other words, you need to be aware that every solution will have exactly two of those variables (and that understanding helps you quickly answer question 9!)

What other inferences can we draw? For example, ask yourself, what happens when L is out, or what happens when J is in? You've got one partway already - if Q is in, so is K, so J is out. What solutions does that leave that include Q? (test them out - I think you will only find three solutions that include Q).

Drawing inferences is crucial to success on the games. It's all about asking yourself a series of "what if" questions, like those above, focusing on the variables that have a lot of influence on the game due to severe restrictions or multiple rules. The time spent doing this is usually well spent, because you will more than make up for it in answering the questions quickly, confidently and accurately.

Try that out and see what happens. Good luck!
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I have a question in regards to the set-up of the game.

I was wondering why the conditions: N<-/->P and N->L cannot be combined to form: P<-/->L?

I combined the first condition with the last condition and it worked. When I was reviewing the in-depth answers for this logic game, I noticed that it didn't show this conditional phrase. It makes it hard to answer q.11 because I'm left with either B or E (which can fit if my conditionals are correct).

Seeing as how I know I'm wrong (because the LSAT doesn't allow for two correct answers :( ), I was wondering if you would be so kind as to explain the double-not arrow for both of the conditionals.

Thank you!
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After reviewing the question for what seems like a hundred times(!!), I noticed that I made an erroneous assumption when tackling q.11. I presumed that because L was there, that N would be there. This is erroneous because the condition is: N->L not L->N.

I can't believed I made that mistake!

However, can my previous question in regards to how to properly set up the double-not arrow for linking together P<-/->L still work? Is it correct?

Thank you!
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 Dave Killoran
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Hi Sotor,

This is one of those situations where when you look at in the abstract, it is difficult to see why it doesn't work, but when you look at it with a concrete example, it becomes easier to understand.

Imagine for a moment that N stands for motorcycles, and P stands for planes. Well, motorcycles and planes are different and never overlap, so N :dblline: P. Next, let's say that L stands for wheels. Every motorcycle has wheels, so we can say that N :arrow: L. Does that mean that P :dblline: L (in other words, that planes don't have wheels?)? No, it doesn't.

In the simplest terms, the arrow between N and L is going the "wrong" way to make the inference in question. The group of motorcycles is actually relatively small compared to the group of things with wheels, and so even though no motorcycles are planes, and all motorcycles have wheels, that doesn't mean that planes don't have wheels.

If it were reversed to L :arrow: N, then yes, the inference P :dblline: L would stand.

Let me know if that helps, and if not we can look at it some more. Thanks!
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Can you not draw the inference that Not L :arrow: P? If you have the contrapositive not L :arrow: Not N, and if not N :arrow: P then we get Not L :arrow: P. Is that right?
 Jon Denning
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Hey jrc,

Thanks for the question and welcome to the Forum!

That inference about N, L, and P is spot on! If L is not selected then N cannot be either, meaning P must be:

..... Not L :arrow: Not N :arrow: P

Nice work! Keep in mind of course that L and P can still go together if N is gone, but L's absence knocks N out and forces P in.

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