Setup and Rule Diagrams

[Dave Killoran]

Setup and Rule Diagram Explanation

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

The game scenario indicates that six new cars are each equipped with at least one of three options. Prior to reading the rules, either the cars or the three options could be chosen as the base of the game. However, after reviewing the rules—each of which starts by listing a car and its related options—the cars are the better choice:

PT35-Oct2001_LGE-G2_srd1.png

The first two rules establish options for V and W:

PT35-Oct2001_LGE-G2_srd2.png

The third rule indicates that W and Y do not have any options in common. Because W already has two options from the second rule, Y must then have the remaining option, S. The combination of the second and third rules also numerically sets W’s options at 2, and Y’s options at 1:

PT35-Oct2001_LGE-G2_srd3.png

The fourth rule states that X has more options than W:

PT35-Oct2001_LGE-G2_srd4.png

Because W already has two options, X must have all three options:

PT35-Oct2001_LGE-G2_srd5.png

The fifth rule indicates that V and Z have exactly one option in common:

PT35-Oct2001_LGE-G2_srd6.png

The operational effect of this rule is that Z cannot have all three options (otherwise Z would have both P and S, a violation of the rule).

The sixth rule is another numerical rule, this time indicating that T has fewer options than Z:

PT35-Oct2001_LGE-G2_srd7.png

Because each car has at least one option, T must have at least one or two options under this rule. But, if T has two options, Z would have to have three options, which is not possible as discussed in the analysis of the fifth rule. So, Z must have two options, and T must have exactly one option. Because Z has two options, and can only have one of P and S, we can also infer that Z must have L as an option. Finally, because Z has two options, by applying the fifth rule we can infer that V cannot have all three options, and thus has only two options. Combining all of the prior information results in the final setup:

PT35-Oct2001_LGE-G2_srd8.png

Because the Numerical Distribution is so critical, let’s review the steps that create the distribution:

1. Since W has at least two options, and W and Y have no options in common, it follows that
W must have exactly two options and Y must have exactly one option.

2. Since X has more options than W, X must have exactly three options.

3. Since T has fewer options than Z, Z must have at least two options. And since V has at least
two options and V and Z have only one option in common, it follows that Z cannot have three
options. Therefore, Z must have exactly two options. Because of the VZ rule it follows that V
cannot have three options and therefore V must have exactly two options.

4. Since Z must have two options, it follows that T must have exactly one option.

5. The above steps create the 1-2-2-3-1-2 Fixed Distribution.

With the distribution, determining most of the options is simple. The only uncertainty in the game is what option T will have, and whether Z will have P or S.

[stsai]

Hi,
This grouping game appeared to be easy, yet I found myself stuck with #9 with all answer choices could be chosen. I also got #10 and 12 wrong. So there must have been something essential that I missed when I did my set up. Could you tell me where went wrong? Thanks!

Here's how I set this up:

W<-\->Y
X>W
V<--exact 1-->Z
T<Z

T: __ (1)
V: PS
W: PL
X: PLS
Y: S
Z: P/__ or __/S (2)

[Dave Killoran]

Hi Stsai,

The one thing I see missing in your setup is that Z must have a leather interior. Thus, Z is either LS or LP. With that in place, the only uncertainty in this game is which option T has, and whether Z has P or S. That should put you on the path to more easily solving #9 and #10.

The one other thing I will note about questions #9, #10, and #12 is that they are each Cannot Be True questions. Perhaps the language in the question stems threw you off a bit too?

Please take a look at those questions again and let me know if the above information helps. Thanks!

[srcline@noctrl.edu]

Hello

So I do not really know where I am going wrong with this game, is it an advanced linear game? I do not know if my set up is correct.


PW: 1 2 3 4 5 6

LI: 1 2 3 4 5 6

S: 1 2 3 4 5 6

I was also completly lost on the rules as well.

Thankyou
Sarah

[David Boyle]

Hello

So I do not really know where I am going wrong with this game, is it an advanced linear game? I do not know if my set up is correct.


PW: 1 2 3 4 5 6

LI: 1 2 3 4 5 6

S: 1 2 3 4 5 6

I was also completly lost on the rules as well.

Thankyou
Sarah

Hello Sarah,

It's more of a grouping and numerical distribution game, actually.
Let me give you some quick pointers:

Try using the six letters of the cars themselves as the base: t v w x y z. With that, and using the idea of numerical distribution, that may help you work out a usable template for the game.

Hope this helps,
David

[avengingangel]

This game was really difficult for me, in fact, I couldn't get through the distribution.

I had it as: 2-2-2; 4-1-1; 3-2-1, because, in the course book it says: the numbers add up to an amount equal to the total number of variables in the set being allocated; the number of separate numbers is equal to the number of elements "receiving" the allocated set."

Isn't it the 6 new cars that are being allocated TO the 3 options ??

Could someone please walk me through the distribution process here, because I am clearly not getting it. Thanks!!

[Adam Tyson]

Two issues here to consider, angel. First, your numerical distributions appear to have an assumption built in that is not supported by the scenario and rules, and that is that each car gets exactly 1 option. In fact, that is directly contradicted by rules like X gets more than W (so they cannot both get just one). I may be misunderstanding your distributions, but that's what it looks like.

Second, when you see numerical relationships between variables, such as there being more of one than another or that they cannot have anything in common, that suggests that those variables should be your base. X having more than W should be thought of as "group X is bigger than group W". Take a look at our diagram at the top of this thread and see if it doesn't work better for you than using the options as the base.

Give it a try!

[avengingangel]

Adam,

No, yeah, I definitely see that I am using the wrong set of variables as my base, but it't just not clicking for me (in this particular game setup) why. Beyond the 'indicator' language from the rules ("X has more options that W" etc.) I'm not really understanding what my first step would be in figuring out the distribution. And the set-up explanation provided here is not helping me... it seems to speed past the explanation of why you want to switch it up. Thanks!

[avengingangel]

For example, I just ran into the sameish problem (I think) in the June 2005 Subcommittees Game . I read thru all the rules and had my distribution as 3-3-3. But I'm not getting anywhere with that, so clearly I am wrong. I feel like these are trickier distribution set-ups... blegh.

[Adam Tyson]

The distribution in this case doesn't really lend itself to being discovered in advance, but rather becomes apparent as you complete the setup. It's not like where you have 6 things going into 3 groups and you know up front that the options are 4-1-1, 3-2-1 and 2-2-2. Instead, the distributions are created as a byproduct of the rules - X has more than W, W and Y have none in common, T has fewer than Z, etc. As you work through those, step by step, you discover that X has to have all three options, W has to have exactly 2, Y has just 1, and so on.

Rather than focusing on the distribution, focus on the interplay of the rules, and see what comes up. The distribution - the size (whether the exact size or the range of sizes) of each group - will unfold along the way.

We've seen other games just like this one, and you might want to take a look at them. For example, December 2005, Game 3, is virtually identical to this game. Focus on the numerical relationships between the variables to determine the best base and then go from there.

Good luck!