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We've written a couple of games - a hard and simple version to explore how the LSAT messes with standard forms. Have fun! http://www.manhattanlsat.com/logic-games-practice.cfm

Big Disclaimer since TLS is a tough crowd: NOT that you should do fake logic games in the place of real ones (and it's always great to do games over and over again), but if you've done them all...

On first read I expected that 'walk into a bar' was going to complete the thread title!

/just saying

Alright, who's got the joke figured out? All I can think of are dirty ones about ordering, and that wouldn't befit the LSAT...

Fixed already. Fun game, I can see these being worthwhile if someone were out of actual games to play.

Phew - I was working hard to figure out if that was the "a pair of games walks into a bar" joke. Yeah, for a few hours, the site was showing questions for both the easy and hard version - that's not a twist you'll see on the LSAT!

Thanks for posting another game, Noah! These are always fun.

If I am not mistaken, the seventh question of the hard version has three correct answer choices: B, C, and D.

7. Which of the following, if substituted for the rule that P presents after L if and only if P does not present after M, would have the same impact on determining the order of presentations?

The original rule:

L-P <--> P-M
Contrapositive: M-P <--> P-L

From the original rule, there are two possible arrangements of L, P, and M: L-P-M or M-P-L.

Answer choices B, C, and D all appear to create this same dichotomy.

(B) If L is before P, P is before M.
L-P -> P-M (L-P-M)
M-P -> P-L (M-P-L)

(C) If P is after M, L is after P.
M-P -> P-L (M-P-L)
L-P -> P-M (L-P-M)

(D) Either L or M, but not both, is before P.
M-P-L
L-P-M

Kurst wrote:From the original rule, there are two possible arrangements of L, P, and M: L-P-M or M-P-L.

Answer choices B, C, and D all appear to create this same dichotomy.

(B) If L is before P, P is before M.
L-P -> P-M (L-P-M)
M-P -> P-L (M-P-L)

(C) If P is after M, L is after P.
M-P -> P-L (M-P-L)
L-P -> P-M (L-P-M)

Choices B and C are incorrect for the same reason. You've neglected the third possibility for each choice: that in which neither the replacement condition nor its contrapositive is triggered. Note that choices B and C are actually contrapositives of each other, and are logically equivalent. Each one allows for the possibility that P is before both L and M. Thus, neither choice is a perfect replacement for the original condition.