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- Sat Oct 21, 2017 10:36 am
#40808
Setup and Rule Diagram Explanation
This is a Basic Linear: Balanced, Numerical Distribution game.
After the June 2011 LSAT, this was the most talked about game, and many students felt this was by far the most difficult game of the exam. While this is indeed a fascinating game, with the right approach it is not overly difficult.
The game scenario establishes a vertical Basic Linear setup: six boxes are stacked on top of one another and each box contains a single colored ball that is green, red, or white. This creates the following initial setup:
Heading into the rules, one of the areas of concern is the selection of colors for the balls. We know there are three colors, but we do not know the numbers of balls in each color. Focusing on this aspect is one of the keys to successfully completing this game.
The first rule immediately gives us some information regarding the numbers of each ball color. In this case, the number of red balls is greater than the number of white balls:
#R 
#W
Without considering the other rules, this leaves a number of possible combinations. More on these possibilities will be discussed after all three rules have been examined.
The second rule first establishes that there is a green ball (this is important because nothing in the game scenario states that there is actually a ball of each color), and that the box containing this green ball is lower in the stack than all of the red balls. This rule can be diagrammed as:
This rule does not state that there is only one green ball; there can be other green balls. The rule just indicates that one of the green balls always will be lower than any red ball. Thus, we can also infer that no red ball can be placed in box 1, which must then be green or white:
A green ball could be placed in box 6 as long as another green ball was lower than all of the red balls.
The third rule establishes that there is at least one white ball, and that a white ball is immediately below a green ball:
Because there can be multiple balls of each color, no Not Laws can be drawn from this rule; for example, a white ball could still be in box 6 as long as the conditions for this rule are met somewhere among the other five boxes (hence the “1” next to the block). Note that in the second rule we could draw a Not Law because the rule referenced “any box that contains a red ball.”
With the basic rules diagrammed, let us turn towards the possibilities for the numbers of each ball color. The relationship of the number of balls to colors is a classic Numerical Distribution scenario, and thus we will first establish the minimums of each color.
The second rule establishes that there is at least one green ball, and the third rule establishes that there is at least one white ball. From the first rule, then, we can deduce that there are at least two red balls. This information leads to the following minimum distribution:
Thus, 4 of the 6 balls’ colors are already established, leaving only 2 ball colors as uncertain. These can be added to the minimums, as long as the first rule is taken into account.
First, because red has the greatest number of balls, go to the extreme by adding the two extra balls to the red group:
Next, remove one red ball from the prior distribution, and first add it to the green group, and then next add it to the white group:
That exhausts the possibilities for adding extra balls to the red group. Because more balls cannot be added to the white group when there are only two red balls, the only remaining option is to add two balls to the green group:
This exhausts all possibilities, meaning that there are a total of four distributions in this game:
With these four distributions in hand, we have the final setup for the game:
This is a Basic Linear: Balanced, Numerical Distribution game.
After the June 2011 LSAT, this was the most talked about game, and many students felt this was by far the most difficult game of the exam. While this is indeed a fascinating game, with the right approach it is not overly difficult.
The game scenario establishes a vertical Basic Linear setup: six boxes are stacked on top of one another and each box contains a single colored ball that is green, red, or white. This creates the following initial setup:
Heading into the rules, one of the areas of concern is the selection of colors for the balls. We know there are three colors, but we do not know the numbers of balls in each color. Focusing on this aspect is one of the keys to successfully completing this game.
The first rule immediately gives us some information regarding the numbers of each ball color. In this case, the number of red balls is greater than the number of white balls:
#R #WWithout considering the other rules, this leaves a number of possible combinations. More on these possibilities will be discussed after all three rules have been examined.
The second rule first establishes that there is a green ball (this is important because nothing in the game scenario states that there is actually a ball of each color), and that the box containing this green ball is lower in the stack than all of the red balls. This rule can be diagrammed as:
This rule does not state that there is only one green ball; there can be other green balls. The rule just indicates that one of the green balls always will be lower than any red ball. Thus, we can also infer that no red ball can be placed in box 1, which must then be green or white:
A green ball could be placed in box 6 as long as another green ball was lower than all of the red balls.
The third rule establishes that there is at least one white ball, and that a white ball is immediately below a green ball:
Because there can be multiple balls of each color, no Not Laws can be drawn from this rule; for example, a white ball could still be in box 6 as long as the conditions for this rule are met somewhere among the other five boxes (hence the “1” next to the block). Note that in the second rule we could draw a Not Law because the rule referenced “any box that contains a red ball.”
With the basic rules diagrammed, let us turn towards the possibilities for the numbers of each ball color. The relationship of the number of balls to colors is a classic Numerical Distribution scenario, and thus we will first establish the minimums of each color.
The second rule establishes that there is at least one green ball, and the third rule establishes that there is at least one white ball. From the first rule, then, we can deduce that there are at least two red balls. This information leads to the following minimum distribution:
Thus, 4 of the 6 balls’ colors are already established, leaving only 2 ball colors as uncertain. These can be added to the minimums, as long as the first rule is taken into account.
First, because red has the greatest number of balls, go to the extreme by adding the two extra balls to the red group:
Next, remove one red ball from the prior distribution, and first add it to the green group, and then next add it to the white group:
That exhausts the possibilities for adding extra balls to the red group. Because more balls cannot be added to the white group when there are only two red balls, the only remaining option is to add two balls to the green group:
This exhausts all possibilities, meaning that there are a total of four distributions in this game:
With these four distributions in hand, we have the final setup for the game:
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