The setup for this game is fairly straightforward even though there are few initial deductions:
We have six boxes with one ball each, either g, r, or w
___ ___ ___ ___ ___ ___ 1 2 3 4 5 6
(1) There are more red balls than white
R >W
(2) There is a box containing a green ball that is lower than any box that contains a red ball
G(1) -> R
What this rule is telling us is that there is at least 1 green ball that is below all the red ones. There could be more green balls that are above red ones, hence we cannot rule out box #6 as not green, but we can conclude that box #1 is not red. ___ ___ ___ ___ ___ ___ 1 2 3 4 5 6 ~R
(3) There is a white ball in a box that is immediately below a box with a green ball.
WG
Now, the question asks us if we have two white balls, then which one could contain a green ball?
Let's think about this setup:
Per rule (1), if we have 2 W, 3R, and 1G ball. Per rules (2) and (3) we can conclude that this G ball must be below all the R balls, and directly above the W ball.
/R /G /G /W /W /W W /W /W R R R ___ ___ ___ ___ ___ ___ 1 2 3 4 5 6
These restrictions demonstrate that the lowest possible box for the G ball could be #3 if all 3 R balls are consecutive and are in boxes #3, #4 and #5.
W W G R R R ___ ___ ___ ___ ___ ___ 1 2 3 4 5 6
The only other alternative for the G ball is the box #2 in a scenario where boxes 3,4,5, and 6 are filled with some combination of 3 red and 1 white ball, the exact order of which is not restricted by any of the rules Box #1 must be the W ball per rule (3).
/R /W /W /W W G /W /R /R / R ___ ___ ___ ___ ___ ___ 1 2 3 4 5 6
Does this make sense?
Let me know if you have any further questions.
Crookon May 20, 2020
The maximum amount of R's could be four, can't the minimum be two? For example WGRRGG or WGRGRG or GRRGWG, etc.
BenMingovon May 31, 2020
Hi Crook, thanks for reaching out.
You are right. The maximum R's is 4 and the minimum is 2. There are multiple ways we can have 2 Rs. The ones you have shown are acceptable.
