Logic Games Tips: Partying with Brutal Deductions
- Mar 03, 2010
- General LSAT Advice, LSAT
- Reviewed by: Matt Riley
Since some crazy folks have started circulating rumors about the Logic Games on the LSAT becoming more difficult over the last year or so, I thought it might be appropriate to outline some brutal deductions.
Say hello to the final game from the September 2009 LSAT. In order to get through this game in less than an hour (and not slice your wrists in the process), some crucial deductions were needed.
Let’s take a look (the subject of the game has been slightly altered):
A young princess (not literally) is selecting friends to invite to her big Sweet 16 bash. She must invite at least three friends from among the following seven: Harriett, Liza, Margaret, Penelope, Sigourney, Tabitha, and Wilma. The birthday girl’s parents have placed the following restrictions on the invitations that can go out:
Doesn’t sound too bad, right? This is one of the basic and very common type of games on the LSAT. It involves selecting one group from a larger group. Feeling good, feeling strong, here comes the rules:
If Harriett, is invited, then neither Sigourney nor Margaret can be invited.
If Margaret is invited, then neither Penelope nor Tabitha can be invited.
If Wilma is invited, then neither Penelope nor Sigourney can be invited.
Again, nothing is screaming, “I am a terrible, bloody game that is going to bend you over and ruin your LSAT score.” However, the worst games don’t always have to scream; a whisper can be even more dangerous.
The rules here essentially rule out certain combinations of girls from being invited. This should sound familiar, we all have to follow similar rules in our everyday life. If I invite my cop friend, then I can’t invite my drug dealer friend. If the new girl that I am dating is going to show up, then the ex should be left off the invite list. Seems straightforward enough.
Trust me, if you dove into the questions on this game without doing some work, the road was similar to a quick trip through Assault on American Gladiators. Not the lame new version, the old school ass-kicking version against Sky and Nitro.
There are actually two elements that are working together on this game to create a scary combination: there are a bunch of combinations that do not work, and yet you still must meet a minimum requirement of invitees. This leads to a situation where certain combinations will not work because you will not be able to meet the minimum number of girls that must be invited. Here is the breakdown:
Harriett: Boots out 2 girls (Sigourney and Margaret)
Liza: Totally random
Margaret: Boots out 3 girls (Harriett, Penelope, and Tabitha)
Penelope: Boots out 2 girls (Margaret and Wilma)
Sigourney: Boots out 2 girls (Harriett and Wilma)
Tabatha: Boots out 1 girl (Margaret)
Wilma: Boots out 2 girls (Penelope and Sigourney)
The first thing that you should notice I already highlighted. Since there must be at least three girls that are invited, at most four girls can be left off the lucky list. If Margaret is invited, at least three girls are definitely gone.
Party time: M
Not so much: H P T
But wait, there’s more. Since both Wilma and Sigourney cannot be invited, and there is only one spot left in the un-cool group, one or the other would have to be invited and Liza would have to take the final spot.
Party time: M S/W L
Not so much: H P T W/S
This leads to the huge deduction that Margaret can only be invited if Liza is invited. And if Liza is not invited, then Margaret cannot be invited.
Armed with this deduction and the list of how many girls hate each other (who ever said that Logic Games were not like real life?), the questions were smooth sailing.
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