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In/out games

samirakhafasamirakhafa Alum Member
edited September 2018 in Law School Admissions 64 karma

Hey guys,
I’ve been practicing in/out games for a while and although I understand most of it, there are a few things that still confuse me enough to throw me off. Hopefully someone can help clear that up :)
I normally get confused with conditionals. For example:

I understand /A—>B is an either or rule so at least 1 has to be in. I would put a place holder in the in group to visualize that.

B—>\C is a not both rule so at least 1 has to be out, so I would put a placeholder in the out group.

I seem to get confused when there is either a chain or a biconditional.

Ie: /A—>B—>C ( I would usually put 2 placeholders for /A—>B and /A—>C ) I get confused when A is in because then b and c become floaters and the 2nd placeholder doesn’t necessarily need to be there.

Or

If I’m given a couple of rules and they happen to link up as a biconditional. I.e -pt 83 game 3

We have
/N—>R
N—>L
R—>M
L—>/R

before I link them up I would place my placeholders. 1 in the in group (/N—>R) and one in the out group (L—>/R)

After you link them however, L, N and R become a biconditional so we know LN Always together and R is alway apart.

When I compare this to my original placeholders, I’m not sure what happened lol
Any advice would be appreciated. Please let me know if you guys would like me to clarify anything :)

Thanks!

Comments

  • SamiSami Live Member Sage 7Sage Tutor
    edited September 2018 10774 karma

    Everything above this looked really good.

    I seem to get confused when there is either a chain or a biconditional.

    Ie: /A—>B—>C ( I would usually put 2 placeholders for /A—>B and /A—>C ) I get confused when A is in because then b and c become floaters and the 2nd placeholder doesn’t necessarily need to be there.

    I think the problem with splitting it into two placeholders, "/A—>B and /A—>C " and when A is in B and C don't become floaters. A being in, negating the sufficient, does nothing to the rule about B--->C. Only the part about /A--->B becomes irrelevant. B can still be in, and if it's in, C has to be in.

    Usually my strategy for "/A—>B—>C" would be to represent /A-->B on the master game board and write B--->C as a rule on the side. Trying to represent each aspect of rule is complicating the matter. Sure its an inference that when /A, C is in but we don't have to represent it on the game board, especially since we already have a rule about /A-->B represented.

    Or

    If I’m given a couple of rules and they happen to link up as a biconditional. I.e -pt 83 game 3

    We have
    /N—>R
    N—>L
    R—>M
    L—>/R

    before I link them up I would place my placeholders. 1 in the in group (/N—>R) and one in the out group (L—>/R)

    The more efficient representation of a not both rule on master game board would be to put N and R with a switch between the groups. Because the other rules link directly to N and R, seeing that, its more efficient to then split the game board and that split on not both would take care of all the other rules.

    For example in one game board N is in, R is out. Then the other, N is out and R is in.

    For the game board, when N is in, R is out: our second rules gets used up and L is in and the third and fourth rules become irrelevant as R is already out.

    When N is out and R is in: Our second rule is irrelevant, M is in and L is out.

    If those are the only rules of the game, we basically have no more rules to represent.

    A general rule about representing rules that I like to adhere to is to do what makes going through the questions easy. There are better ways to represent inferences without having the same variable repeat in in and out games in two slots.

    I hope this helped.

  • samirakhafasamirakhafa Alum Member
    64 karma

    Yes that was really helpful! Thank you!

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