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 Dave Killoran
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#46076
Setup and Rule Diagram Explanation

This is a Grouping, Numerical Distribution, Underfunded game.

This game was widely considered the most difficult of the June 2005 exam. After three linear-based games, the test makers saved a Grouping game for last.

At first, this game appears to be a straightforward Grouping game: six committee members serving on three subcommittees. However, the game scenario and rules indicate that there are nine subcommittee spaces, and that some committee members serve on multiple subcommittees. If each member served on only one subcommittee, the game would be considerably easier because the assignment of member to a subcommittee would eliminate that member from further consideration. Thus, the Underfunded aspect of this game gave test takers more to consider as they looked at the composition of each subcommittee.

Note that since the committees do not have formal names, we have labeled them as “SC.”

The game scenario provides us with the following basic setup:
J05_Game_#4_setup_diagram 1.png
Now, let us examine each rule.

Rule #1. With six committee members filling in nine spaces, and with the game scenario stating that each member serves on at least one subcommittee, you should be on the lookout for rules that reveal a specific Numerical Distribution. The first rule is just such a rule.

Because one of the committee members serves on all three subcommittees, the remaining five members must fill the remaining six subcommittee spaces. Since every member must serve on at least one subcommittee, the five members automatically fill five of the six spaces. The remaining subcommittee space can be assigned to any one of the committee members (aside from the one member serving on all three committees, of course). Thus, the game is controlled by a 3-2-1-1-1-1 distribution of spaces to committee members.

Rules #2 and #3. These two rules establish the following not-blocks:
J05_Game_#4_setup_diagram 2.png
While these not-blocks may appear standard, when combined with the 3-2-1-1-1-1 numerical distribution there is an important implication. First, because F, G, H and I are involved in the not-blocks, none of those four members can serve on all three subcommittees (otherwise there would be an overlap that would conflict with the not-block). Thus, only M or P can be the member who serves on all three subcommittees.

With the information on the previous page, the complete game diagram is:
J05_Game_#4_setup_diagram 3.png
 oktos92
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#25179
Hello,

I want to thank you for the detailed explanation in the book which has helped me realize my weaknesses while solving this question. The only thing I am yet to properly understand is why we have that numerical distribution: 3-2-1-1-1-1. I thought we've already been given the numerical distribution that "THERE ARE THREE SUBCOMMITTEES, EACH HAVING THREE MEMBERS". Instead of just sticking with the 3-3-3, please could you explain why you are suggesting we use 3-2-1-1-1-1?

Thanks

A.
 Robert Carroll
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#25194
Hi A,

The Numerical Distribution being referenced in the text is not the size of the subcommittees (which you correctly identified as 3-3-3), but rather the number of subcommittees each person is on. There are 6 people, so there are going to be 6 different numbers:

..... ..... ..... X-X-X-X-X-X-X

Next, there are 9 spaces available in total on the three subcommittees, so those 9 spaces will be distributed among the 6 people:

..... ..... ..... X-X-X-X-X-X-X = 9

According to the first rule, there is one person who serves on all three subcommittees. So exactly one "3" exists, and we can start building the distribution:

..... ..... ..... 3-X-X-X-X-X-X = 9

The other numbers follow from the other implications about how many people serve on 2 subcommittees, and on 1 subcommittee, which results in the final distribution:

..... ..... ..... 3-2-1-1-1-1 = 9

Please let me know if that helps explain that one. Thanks!

Robert Carroll
 avengingangel
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#32501
OK - After reading the explanation in the book, I came here to ask the same question as oktos92, and so thanks, Robert, bc the "but rather the number of subcommittees each person is on" is a huge help & gave me that "ooohh" moment. HOWEVER, at "There are 6 people, so there are going to be 6 different numbers" I am stumped again! My understanding/the way it's explained is: "the number of separate numbers is equal to the number of elements 'receiving' the allocated set." There are are 3 subcommittees—those 3 subcommittees are "receiving" the 6 members—so, still not quite understanding the WHY of why you would write out the distribution w/ each separate number representing the # of subcommittee each person is on, to total 6.

I can clearly see why that works better than 3-3-3, but like, if I was faced with a similar game again, I still don't think I would be able to recognize that I should craft my distribution in such a way. I hope that makes sense. In other words, once I looked up the correct distribution, 3-2-1-1-1-1, that's all I needed to then go back and solve the whole game myself. But I am still not able to, on my own, come to that distribution after reading the game rules.

Any help is appreciated. This is really stumping me and I feel like I just need to see "the light."

Thanks !!
 Adam Tyson
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#32522
Yeah, angel, that language about receiving the allocated set is a little confusing in this particular context, since the base is clearly the three subcommittees with 3 spaces each, and it's that base that "receives" the committee members. Think of it instead as being "the ways in which the variables can be spread around". In some games, that's about the size of the base (the groups). For example, in December 1997, Game 1, about appetizer and main dish, we can see that the base is either a 2-5 distribution (two things in the app, five in the main dish) or a 3-4 distribution (3 in the app, 4 in the main dish). There, that language about receiving the allocated set makes more sense - you need 7 spaces to receive the 7 ingredients.

In a game like this one, where the base is fixed in place, the distribution is about the repetition of the variables over that base. 6 variables need to fill 9 spaces, and one of those 6 goes 3 times. That leaves the other 5 variables to fill the remaining 6 spaces, so one of them goes twice and the rest go once each. For another game like that, check out June 1995, Game 3, with 4 drivers in a carpool driving to work 6 days a week. It seems odd to think of the committee members "receiving" the committees or the drivers "receiving" the days of the week on which they each drive, but if you look at it just as "how do I spread things around to make it all fit" you might have an easier time of it.

Keep pounding!
 avengingangel
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#32696
Adam, yes, thanks so much! That explanation ("the ways in which variables can be spread around) is MUCH clearer. I just completed the June 1995 Carpool game and found it much easier. I'm going to try the Dec 1997 game now though.... I hope I don't get confused with the distribution and do it the way I did this game... :0
 avengingangel
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#32697
OKAY-- yay! I got it totally right.

I'm trying to come up with a distribution "rule" for myself that makes sense to me, since that first one I got from the book can become confusing, as you said, when encountering certain types of problems. This is now how I'm thinking about it, can you tell me if it might not be true in all cases ???:

"Whichever variable set can only be used once (per the rules), the total number of those variables is what your #s in your distribution should add up to. And the remaining variable set dictates how many separate numbers you should have."

I think that makes sense for both the June 1995 carpool game [each of the 6 days is only 'used' once, thus the total = 6], as well as the Dec 1997 recipe game [each of the 7 recipes is only used once, thus the total = 7]...

Thanks!
 avengingangel
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#32698
Hm, and it kind of works for this June 2005 Subcommittee game, if you take each individual subcommittee member slot as the variable that is only 'used' once, thus making the total # add up to 9, and making the remainder variable set (the members) the variable that dictates how many individual numbers you should have (6)! [Per my rule] Hmm.. I'm going to test it on additional distribution games in the course book and see...
 Adam Brinker
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#63162
We have received the following question from a student. An instructor will respond below. Thank you!
Why: 3-2-1-1-1-1 distribution? Why if out of 6 candidates in subcommittee o3 which will be 9 obviously therefore some of variables will be used more than once which ones and, why it was not the distribution of: 3-3-3 I thought?
 Robert Carroll
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#63175
Adam Brinker wrote:We have received the following question from a student. An instructor will respond below. Thank you!
Why: 3-2-1-1-1-1 distribution? Why if out of 6 candidates in subcommittee o3 which will be 9 obviously therefore some of variables will be used more than once which ones and, why it was not the distribution of: 3-3-3 I thought?
It's not entirely clear which variables will be used more than once, although one must definitely be used three times and another twice, as discussed in the original post. Thus, the distribution is not "fixed" because it cannot be determined which person represents the "3", which the "2", and which all the rest.

There's an imbalance in this game - there are 6 people but 9 slots. The "3-3-3" distribution would reflect the distribution of slots to the subcommittees, but that's already determined by the scenario, so it's not as useful. What is useful to identify as a separate inference is how many "repeats" you'll have - if 6 people are filling 9 slots, and all those slots must indeed be filled, then some number of people must fill multiple slots. So it makes sense to consider what to do with the "excess" slots - if each person filled only a single slot, there would be 3 left over. How are these leftovers going to be allocated? The 3-2-1-1-1-1 distribution is the result of that inquiry.

Robert Carroll

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