In-Out (Selection) Logic Games are definitely in this year.

The September 2009 LSAT had two of them, and then there was another on the December exam. There goes the popular theory that they wouldn't have an In-Out Game again so soon.

Particularly, LSAC loves In-Out Games where all the rules can be connected (see #1). It's not hard to understand why. They require a solid understanding of conditional reasoning and indicator words.

Despite the fact that In-Out games with similar rules have been administered several times in the past, people still have trouble with them.

I've never diagrammed an In-Out game of this type on the blog - it's time I did.

Birds in the Forest - PrepTest 33, Game 2

(p177 in Next 10)

We know this is an In-Out game because we want "to see which of the following...it contains." In any given scenario, the birds that it contains are "In," and the birds that it does not contain are "Out."

We can connect the rules to form two long chains (which are the contrapositives of each other). We do this by looking for cases where the necessary condition of one rule and the sufficient condition of another are identical. If they are the same, we can link them.

For example, suppose we're given the rules:

If X -> Y

If Y -> Z

In this case, Y is necessary in the first rule and sufficient in the second.

This means we can form the chain X -> Y -> Z.

Now that we have that out of the way, let's go look at the Birds in the Forest game:

First rule:

If H is in, then G is not.

H --> NOT G

I diagram this as:

Contrapositive:

If G is in, then H is not.

G --> NOT H.

Second rule:

If J, M, or both are in, then so is H. The both part is redundant, so we can just say that either one is sufficient to bring about H.

This means that if H is out, then both J and M must have been out.

I immediately connect this rule (and its contrapositive) with what I've written down for the first rule (and its contrapositive).

We now have:

Third rule:

If W is in, then G is in.

Contrapositive:

If G is out, then W must be out.

Connecting this to what we already have, it becomes:

Fourth rule:

If J is out, then S must be in.

Contrapositive:

If S is out, then J must be in.

Connecting these to what we already have, it becomes:

Now we have every rule connected in one big chain - in both the original form of each rule, and its contrapositive. I simply connected each new rule (and its contrapositive) with a previous rule (or its contrapositive) as I read it.

With this big conditional chain, we know that when W is in, both G and S are in, and H, J, and M will be out.

When S is out, both J and H are in, and G and W are out. However, we don't know where M is.

On the chain, proximity/closeness doesn't matter.

Anytime we have a positive variable followed by a negative variable, at least one of those two variables is out. (Maybe both are out.) We don't care whether the negative is immediately after the positive or several variables later.

Anytime we have a negative variable followed by a positive variable, at least one of those two variables is in. (Maybe both are in.) We don't care whether the positive is immediately after the negative or several variables later.

The final question of this game requires a new rule (and its contrapositive) to be added to the diagram. On the diagram, it would look like the following:

Since this is the last question of the game anyway, and assuming you’ve answered all the others, it's okay to modify the main diagram since you won’t be using it for anything else. Whenever a Logic Game requires a new rule to be added, taken away, or substituted for an old rule, it'll be in the last question of the game.

Why the technique in this blog post is better than other techniques:

Some companies recommend listing all the rules in their original form, then writing the contrapositive, then looking for connections. Other methods just fail to adequately represent all the inferences. Either way, they take up a lot of space and time.

Both methods require writing out all the rules and then looking for connections. This leads to a moment of panic when you're not confident whether you've made all your inferences or not. You're never confident that you've done everything.

The method I've laid out above does not require any extra writing, which saves space and time. As long as you can do the contrapositive, you can do this. Just remember that you can only follow an arrow in the direction it points.

The chain perfectly represents the relationships between the variables. All the information from the rules already appears on the chain.

A sufficient condition appears to the left of a given arrow, and a necessary condition appears to the right of that arrow.

Just follow the arrows in the direction they're pointing in, and you'll be fine.

The difficulty that many of you have with these conditional chains is you're trying to read them backward, which is not "permitted" in logic. Be careful not to confuse necessary and sufficient conditions.

***

To see this technique demonstrated on another game, check out the diagram I posted for my own In-Out Logic Game.

To see more games like this, check out the first category in 7 Logic Games Repeated on Future PrepTests.

Photo by deniscollette / CC BY-NC-ND 2.0

(Those birds in the photo are grosbeaks, which is what "G" stands for in the game.)

The September 2009 LSAT had two of them, and then there was another on the December exam. There goes the popular theory that they wouldn't have an In-Out Game again so soon.

Particularly, LSAC loves In-Out Games where all the rules can be connected (see #1). It's not hard to understand why. They require a solid understanding of conditional reasoning and indicator words.

Despite the fact that In-Out games with similar rules have been administered several times in the past, people still have trouble with them.

I've never diagrammed an In-Out game of this type on the blog - it's time I did.

Birds in the Forest - PrepTest 33, Game 2

(p177 in Next 10)

We know this is an In-Out game because we want "to see which of the following...it contains." In any given scenario, the birds that it contains are "In," and the birds that it does not contain are "Out."

We can connect the rules to form two long chains (which are the contrapositives of each other). We do this by looking for cases where the necessary condition of one rule and the sufficient condition of another are identical. If they are the same, we can link them.

For example, suppose we're given the rules:

If X -> Y

If Y -> Z

In this case, Y is necessary in the first rule and sufficient in the second.

This means we can form the chain X -> Y -> Z.

Now that we have that out of the way, let's go look at the Birds in the Forest game:

First rule:

If H is in, then G is not.

H --> NOT G

I diagram this as:

Contrapositive:

If G is in, then H is not.

G --> NOT H.

Second rule:

If J, M, or both are in, then so is H. The both part is redundant, so we can just say that either one is sufficient to bring about H.

This means that if H is out, then both J and M must have been out.

I immediately connect this rule (and its contrapositive) with what I've written down for the first rule (and its contrapositive).

We now have:

Third rule:

If W is in, then G is in.

Contrapositive:

If G is out, then W must be out.

Connecting this to what we already have, it becomes:

Fourth rule:

If J is out, then S must be in.

Contrapositive:

If S is out, then J must be in.

Connecting these to what we already have, it becomes:

Now we have every rule connected in one big chain - in both the original form of each rule, and its contrapositive. I simply connected each new rule (and its contrapositive) with a previous rule (or its contrapositive) as I read it.

With this big conditional chain, we know that when W is in, both G and S are in, and H, J, and M will be out.

When S is out, both J and H are in, and G and W are out. However, we don't know where M is.

On the chain, proximity/closeness doesn't matter.

Anytime we have a positive variable followed by a negative variable, at least one of those two variables is out. (Maybe both are out.) We don't care whether the negative is immediately after the positive or several variables later.

Anytime we have a negative variable followed by a positive variable, at least one of those two variables is in. (Maybe both are in.) We don't care whether the positive is immediately after the negative or several variables later.

The final question of this game requires a new rule (and its contrapositive) to be added to the diagram. On the diagram, it would look like the following:

Since this is the last question of the game anyway, and assuming you’ve answered all the others, it's okay to modify the main diagram since you won’t be using it for anything else. Whenever a Logic Game requires a new rule to be added, taken away, or substituted for an old rule, it'll be in the last question of the game.

Why the technique in this blog post is better than other techniques:

Some companies recommend listing all the rules in their original form, then writing the contrapositive, then looking for connections. Other methods just fail to adequately represent all the inferences. Either way, they take up a lot of space and time.

Both methods require writing out all the rules and then looking for connections. This leads to a moment of panic when you're not confident whether you've made all your inferences or not. You're never confident that you've done everything.

The method I've laid out above does not require any extra writing, which saves space and time. As long as you can do the contrapositive, you can do this. Just remember that you can only follow an arrow in the direction it points.

**A hint for those having trouble:**

The chain perfectly represents the relationships between the variables. All the information from the rules already appears on the chain.

A sufficient condition appears to the left of a given arrow, and a necessary condition appears to the right of that arrow.

Just follow the arrows in the direction they're pointing in, and you'll be fine.

The difficulty that many of you have with these conditional chains is you're trying to read them backward, which is not "permitted" in logic. Be careful not to confuse necessary and sufficient conditions.

***

To see this technique demonstrated on another game, check out the diagram I posted for my own In-Out Logic Game.

To see more games like this, check out the first category in 7 Logic Games Repeated on Future PrepTests.

Photo by deniscollette / CC BY-NC-ND 2.0

(Those birds in the photo are grosbeaks, which is what "G" stands for in the game.)

Thank you for taking the time to help us tackle the lsat! Your explanations and strategies are very clear and have been extremely helpful. Thanks!

ReplyDeleteFor grouping and even complex linear advanced games, do you have any advice on thinking and diagraming quicker. In a recent example for myself, their was an ordering game, which required the test taker to match two musical instruments with players and it was not a difficult game but under time pressure I could not adequately do the game. But when I did it not timed, I was able to do it and got everyone one of the questions right. I know practice is all I can do but any other advice would be much appreciated.

ReplyDeleteI have a really dumb question....can you define what the symbols mean -> and --> .... I get the greater than part....I just don't get what the dashes define to.

ReplyDelete@Anonymous - Thank YOU!

ReplyDelete@Losingtouchwithmymind - Being aware of Limited Options/Templates/Possibilities and Numerical Distributions can help you get through many games more quickly. See my logic games-related posts for more on that.

@Dana - Those are not "greater-thans." They're just arrows. (If X --> Y = If X then Y)

Thank you!!!!! Wow I am so happy I found this site :)

ReplyDeleteAs someone who previously learned the PowerScore double not arrow approach, I will testify that Steve's conditional chain approach is MUCH better. First, you know when you've completed chains (versus the uncertainty of whether or not you have made enough inferences to proceed). Second, creating these chains takes much less time. Third, your information is organized in a much more compact, concise manner, allowing you to access it and therefore answer questions much quicker than you would if you have double not arrows written all over your test booklet. I have applied both approaches to a couple of the classic In and Out games now, and Steve's method is clearly better.

ReplyDeleteAGREED. I also found the double not arrow approach messed me up too often.

DeleteI like this method. Powerscore's version is kind of redundant.

ReplyDeleteThis is a much clearer method. However, isn't the chain slightly too simplistic for this question? Your chain does not illustrate, primarily, that when S is in the forest, J, M & H can also be in the forest. To the contrary: the chain makes it look like the three could not go together. This ends up being an integral recognition for the second question in this game. Perhaps a solution is to note to yourself that while sufficient conditions (J is out) can indicate necessary ones (S is in), the fact that S is in does not say anything about whether or not J is in. Do you have a suggestion for including this inference in your chain? I really love your blog and have just begun to follow the 5-Month study plan. Thanks!

ReplyDeleteHi Emma,

ReplyDeleteThe chain is not too simplistic at all. It perfectly represents the relationships between the variables. All the information from the rules already appears on the chain.

A sufficient condition appears to the left of a given arrow, and a necessary condition appears to the right of that arrow.

Just follow the arrows in the direction they're pointing in, and you'll be fine.

The difficulty you're having with these conditional chains is you're trying to read them backward, which is not "permitted" in logic. You're confusing necessary and sufficient conditions.

Steve, the way I read it is that if S or any other variable doesn't point to anything to the right, then the presence (or lack thereof) of any other variable is fair game. Do I have that right?

DeleteHi Steve,

ReplyDeleteThanks for the site and this tip--it works really well. One question: how would you diagram a rule like, "If A, then B or C or both"?

It's easy to diagram the contrapositive of that, since it's just NotB and NotC separated and both pointing to NotA. But when it goes in the other direction I'm not sure how to clearly indicate that, in this particular case, B and C don't BOTH have to happen just because they're downstream from A.

I've tried diagramming it as "A --> 'B or C or BC'", but that's not as elegant as the rest of this method. It could also get confusing if that statement comes in the middle of the diagram, making it necessary to have separate branches extending from each of the possibilities.

It's very possible that I'm just thinking about this in the wrong way, or that there's a simple diagramming fix that I'm not seeing. In any case I'd appreciate any advice.

Thanks again!

Glad to help!

ReplyDeleteA rule like that is annoying, but it does come up on occasion.

Your description of the rule is pretty accurate. HERE's how I diagram the "W -> F or T or both" rule in PrepTest 36 (December 2001), Game 1 (fruit stand). It's not super-elegant, but it's the best I've been able to do, given the annoyingness of the rule.

W requires at least one of F or T.

If both F and T are out, then W is out.

Hope this helps!

Steve

Hey Steve or anyone,

ReplyDeleteSay the Second rule said "If J, M, but NOT both are in, then so is H." How would we represent that?

Thanks in advance.

Steve, I am confused about question #9 on pg 177 for this game. Wouldn't the max # of birds be 3, not 4? Any way I diagram, I am left with 3; am I missing something?

ReplyDeleteI spent hours yesterday being baffled by the birds. I got 6 of 7 wrong! After reading the explanation in LGB, I still got 5 of 7 wrong. It wasn't a good day. Today, I discovered how you deal with birds, and it's so much better than PS's ponderous grocery list of arrows and inferences.

ReplyDeleteThank you!!

Could someone explain what new implications are formed with the new condition from question #12? Thank you!

ReplyDeleteI spent sometime pondering this extra rule on my own and I determined that the extra condition merely implies that S and H cannot both be in the forest. This will affect J and M. Thus, JS nor MS can be in the forest at the same time.. If anyone has anything else to add please do so. Happy New Years!

ReplyDeleteSimple question: does anyone else find it much easier and better to write out the regular logic chain AND then write the contrapositive logic chain simply going backwards from the original?

ReplyDeleteI find this much more effective than writing out the contrapositive while I write out the regular version. I can go a lot faster, and it's actually easier to not make a mistake with the contrapositive.

I am having problems with the game. Rule #4 specifically says that J and S cannot be in the forest together but questions #8 and #9 both require you to place J & S in the forest together to arrive at the answers the testmakers want..a clear violation of Rule #4 (If J is not in the forest the shirkes are). Help!!!

ReplyDeleteTo Ananymous from 3/15/2011,

ReplyDeleteThis is the key error many test takers make. PowerScore spent a paragraph explaining it. This rule merely says AT LEAST ONE of either J or S has to be in, but it does not forbid both of them being in.

Steve,

After struggling with the PowerScore diagram method, I can truly appreciate how much clearer and efficient yours indeed is. Thank you!

I have a question regarding question 6: do you recommend coverting all the choices into "a list of birds IN the forest"? If not, how should I approach this question? Thanks.

Steve,

ReplyDeleteYour method is so much clearer than others. Thank you for all you do.

I am stuck on question 7. I don't understand if M and H are in then how are there at most two other kinds of birds in the forest. Based on the logic chain if M and H are in, G and W are out. Could you please explain how to tackle this question?

Vicky,

ReplyDeleteJ and S can be in the forest together. Remember, the rules only give information for whether J or S is NOT present. There's no reason you cannot have both as opposed to just one of them. So, at most two other types can be in the forest (in addition to the given MH). Just for clarification, D is not correct because it doesn't allow for just one of the two (J,S) being in the forest.

Hope that helps!

you really find that helpful? i mean, i appreciate linking all the rules in a chain and showing the concomitant contrapositive chain, but if you tweek the powerscore approach, or use the testmasters approach, I'd argue that you save a ton of time. by constructing one double-not arrow

ReplyDeleteg,w <-|-> j,m,h

you can clearly see that at any given time, all variables from one of the two sides need to be out. this makes the hardest questions on this game a breeze. all in-out games hinge on the out group and getting it set up in this way codifies it. i like where yours was headed but i think this is superior.

sincerely, a colleague in the always interesting field of lsat

Steve,

ReplyDeleteI purchased the 3 month plan, and I appreciate everything that you have put together to help us get ready for the LSAT. I understand how this method worked out for the birds selection game, but can you diagram the next game with the university library budget committee reducing the areas of expenditures?? Seems with the birds game everything kind of fit together, but with this other game, it seems much tougher...for me at least. Any help would ne much appreciated. Thank you again.

Jeff

Steve,

ReplyDeleteIs there anyway that you can use your methods and examples on more problems?? In your day to day study plan, you are literally telling me to diagram in your style, but have not even gone over your style yet. Then when you do go over your style it is hard to understand, and you only do it for one problem. I have followed along to your study plan perfectly until now, and am having the worst time with the Grouping: In/Out diagram chain. Any help would be appreciated. I am getting real frustrated. Thanks.

Jeff

thanks for this explanation...I took about 10 minutes solving the entire set with too many sets of rules and your way is so much more efficient.

ReplyDeleteAlso, thanks to the commentor that explained the J/S rule. I was tripped up on problem 9 as well.

How would you diagram for harriers and grosbeaks both not being in the forest?

ReplyDeleteSteve,

ReplyDeleteI have the same question as Sarah on #9. Looking at your diagram, it is clearly that there could only be maximum three birth at a given time in forest. I couldn't see why the answer is C four birds. Please help!!!! Thank you.

Hi, I was wondering if you could explain #12 in a little more detail. I still don't understand how the new rule affects the relationship between J and M.

ReplyDeleteThank you! Jenn

I read through the explanation for it in LGB too. I understand all the rest of them. I just can't make this click. Maybe there's some logic rule I'm not understanding?

ReplyDeleteJenn

"Whenever a Logic Game requires a new rule to be added, taken away, or substituted for an old rule, it'll be in the last question of the game."

ReplyDeleteThank you.

ALL HAIL STEVE SCHWARTZ

ReplyDeleteSteve,

ReplyDeleteCould you explain which two variables the following refers to if somebody else doesn't first? Would those be any two or just the first two? Or maybe somebody else could rephrase it? Thanks.

"Anytime we have a positive variable followed by a negative variable, at least one of those two variables is out. (Maybe both are out.) We don't care whether the negative is immediately after the positive or several variables later.

Anytime we have a negative variable followed by a positive variable, at least one of those two variables is in. (Maybe both are in.) We don't care whether the positive is immediately after the negative or several variables later."

Steve,

ReplyDeleteI have a question about the second rule. Why is it interpreted that J & M HAVE to be in the forest in order for H to be in the forest? The rule states that jays, martins, or both need to be there in order for H to be present, so doesn't that mean that J and H can be present together even without M?

Any advice is appreciated.

Thanks!

Jen,

DeleteYes. You are right. The diagram is not omnipotent. You have to use your head too, to fill in some of the holes it seems. This method is not a sliver bullet. You should read my post I made to get a better understanding why and how.

Steve,

ReplyDeleteI have come to realize that a number of the rules have loose interpretations. I only got two questions out of the seven correct. I got question 1 and 7 correct following your diagram, but all the rest I got incorrect. I feel like what is more important is, while performing your diagramming technique, that people should keep in mind the rules as they are stated.

Question 2 asks if both M and H are in the forest then which of the following must be true. I answered (B), which reads J is the only other bird in the forest. In the diagram clearly if both M and H are in than the only other possibility is J. I realize the mistake I made was eliminating S; in reality S and J could both be included because of how Rule 4 is written: if J not in forest than S is, which could actually read to say that both J and S are in, therefore, there are at most two other kinds of birds in the forest or answer (E).

Question 3, which I got wrong, asks: If J not in forest, then which one of the following must be false. The first two answers state that either M or H are in the forest, however, both are actually out of the forest if J is out so technically both are false according to the diagram. The reality is H and M can be in the forest because Rule 2 only states that if J, M or both are in the forest then so is H. Only G and W have to be excluded. This diverges a lot from your original and contrapositive diagrams. Your diagram made it harder to see all of this going on since it stated that if J is out so is H and M but that is incorrect as demonstrated.

I got question 4 wrong for the same reason I got both question 2 and 3 wrong it asks me how many different kinds of birds was the maximum that could be inside the forest. I realized quickly how I got this one wrong because I just looked at the diagram to see which were in; this suggests that only three different kinds of birds could be in the forest, which was answer (C). This was, of course, wrong because Rule 4, yet again, stated only that if J is not in the forest, then S has to be in the forest it does not say both cannot be in the forest at the same time. Therefore, the correct answer actually is answer (D).

However, Question 5 I struggled with the most and still do not know why I got it wrong. Question 5 asks: Which one of the following pairs of birds CANNOT be among those birds contained in the forest? Well if you look at the diagram it has no answer to this question since the original and the contrapositive forms seem to suggest that no pair needs to be excluded; however, I inferred the contrapositive was false (I do not know if that was a mistake or not and whether or not I am thinking about this the wrong way) and answered with answer (C) stating that S and W had to be excluded. However, this was wrong and the correct answer was answer (A), therefore, J and W were the pair that had to be excluded. The diagram does not suggest that J or W cannot be in the forest. Just looking at this on its face it’s possible for (A) to be correct since J and W can be excluded but that does not HAVE TO BE true.

As for question 6 I got this one wrong for the same reason I got question 2 and 4 wrong. It asked that if G was in then which of the following must be true. I was distracted by the contrapositive of Rule 3 if W is in, then G is in and put my answer as answer (B), which states that W is in the forest. The contrapositive suggests this to be true, if G is not in the forest than neither can be W, however, this is too rigid of an assumption and is ultimately wrong. Answer (A), which suggests that S must be in the forest, is correct because Rule 3 does not state that G HAS to be in the forest with W.

Anyways, I still am very confused about Question 5 but getting this opportunity to write out why got the others wrong helped me understand better why I may have gotten them wrong just because I wrote this out. So, for that, thank you!

I just did the same exact thing! Glad to know I'm not the only one with this issue. Will keep working on it..

DeleteSteve, I watched your video explanation, but I don't understand why B cannot be the answer to #10: if J is out, we know that S is in and vice versa. This is no different from the knowledge that if W is in then J is out. Thanks so much for clarifying this.

ReplyDeleteFrom the chain, we know that when J is out, S is in. But, we do not know that when S is in, then J is out. J and S can both in the forest together.

DeleteFrom the stated rule "~J --> S" alone, we can only infer "~S-->J" but not "S-->~J." "S-->~J" is a mistaken reversal.

This method is good but it will confuse people IE: #10.

ReplyDeleteYou need to write the chain and still write the conditionals and contrapositives or hold them in you head.

Question 10: which birds cant be in the forest together.

the correct answer is A.

B is wrong because the original rule is: notJ-S contra notS-J

so if J is out S is in

If S is out J is in

this rule breaks down to j/s/or both can be in; only 1 can be out.

Now go look back at the chain and you will notice its hard to get that inference within the chain.

Also just making the chain will make you confidently get #9 wrong.

DeleteHey Steve,

ReplyDeleteI just moved on to this section in the study guide, and was (initially) thoroughly confused when I got 5 of the 7 questions wrong (which seems to be a semi-common occurrence, reading through the thread).

It took me a bit to notice your post about not being able to apply the rules "against the flow" of the arrows. I didn't realize that you could ONLY apply the rules in one direction, and that if a relationship wasn't addressed in the "forward moving" direction in either of the two diagrams, then no assumptions could be made about that particular relationship. Once I understood this, I retook the game, and easily got 7 out of 7.

It seems that many other readers/posters made the same mistake I did, so I thought I might humbly suggest modifying the walk-through up top to address this.

Thanks for pointing this out! I've added the text of my earlier comment to the end of the blog post to make it more prominent.

DeleteWould you be able to provide any examples of in/out games in the newer LSAT exam books? In test 52 or more recent?

ReplyDelete