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Daishain's 5e D&D Rules, Q&A, And Support - Page 921

As I said, the first scenario, you find out the coins are heavier or lighter when you weigh the first pair of the known bad four or the single of the last pair. You don't need to know this fact to resolve that part. The leftover two coins, if bad, get checked by the last weighing against a good coin. Again, that part doesn't need to know if it's heavier or lighter. The only time the actual variance needs to be known is if the two stacks of three mixed coins contain the bad one. Then the one lighter or two heavier rule comes into play, and that last weighing will determine if the solution is a lighter or heavier - because you know the pair you weigh has to be heavy if it has the bad coin in it.
What I need to know is if there is a way that the bad coin escapes being found, the way I describe it.
EDIT - just to clarify: If the lighter side stays light, then the bad coin is now either the moved light one, or the heavy pair. If the other side becomes light, then it's the moved single there (A, in this case) or the heavy pair on the other side. I'm ready to post my solution, unless you can prove it missed the coin somehow.

Edited: Gknightbc on 26th Sep, 2018 - 3:02pm

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Just go for it, I'd rather not sit here and deal with this riddle forever.

Daishain's 5e D&D Rules, Q&A, And Support Archive Pathfinder / D&D

Okay, then, if no one has any other objections, I'll post what I have and let's proceed.
For riddle number two, I see how that question works better, as it eliminates two variables for each answer. Whew! But I admit it, I had fun working these out *smile*.

Edited: Gknightbc on 26th Sep, 2018 - 3:47pm

Page 921 Support QampA Rules Dragons and Dungeons e Daishains

As stated, the solution is incorrect, but it looks like the error was a slip of the metaphorical tongue rather than one of logic.

I'll give you a chance to correct it, hint, the statements presented on this page and their counterparts in the main thread don't quite match.

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So did the party get a long rest while the wizard worked all of this out and explained all if/then statements to the sphynx? I picture everyone passed out on the ground. (Laugh)

Kudos to Gnight and Seath for working these out. (... Regardless of who was right or made errors in the recent posts about riddle #3. I'm sparing myself the pain of going through it all again.)

It can, potentially, be easy to get 90% of the way there with your logic and all of your 'if/then' statements, conditional questions for #2, etc. Etc. But still fall short of taking all possibilities into account, so thanks again guys for working through it all.

(Without looking back, I do think both guys were correct about riddle #3, 'scenario A' where the first use of the scale reveals that the rogue coin is in stack (9,10,11,12). I think it is valid to solve that one either way (Stack of 2 vs. Neutral stack) or (Stack of 3 vs neutral stack). It is the rest of the puzzle that gets even more complex.).

Daishain's 5e D&D Rules, Q&A, And Support

Dai, please confirm the more elaborate edit. I see where I put the wrong statement/words in, and stretched it out further to be clear.

Daishain's 5e D&D Rules Q&A Support - Page 921

Okay I don't want to cheat anyone, so let's compare logic.

Skip up until the point where you are weighing A, 2, 3 versus 1, B, C. With A, B, C having been previously on the heavy side and the mixed piles not matching.

If a23 is now heavy, then the bad coin was transferred and is either a heavy A or a light 1. Test one of those two against a known good coin. This works.

But if 1BC is heavy, then the bad coin was not transferred, and is a light 2/3, or a heavy B/C. With only one weigh left, there's no chance to be sure which

What I think you needed to do, if light side was 1 2 3 4 and heavy side was a b c d, would be to discard 3 and 4. Then weigh 1ab vs 3bc. Assuming 1ab is heavier, the bad coin is either a light 2 or a heavy a/b. Weighing a vs b would determine which of the three.

If I misunderstood, by all means point it out.

Edited: daishain on 26th Sep, 2018 - 5:28pm

Daishain's 5e D&D Rules Q&A Support D&D / Pathfinder Archive - Page 921

Guys, please. You can't put two off to the side from the pile of eight and still have this work. You need to put THREE off to the side and throw in one from the third "Known good" pile. So using your labeling, if 1234 is heavier than ABCD then the initial piles of 3 are, A,B,1 and C,2,5. You need one coin in here from the known good pile to make this whole problem work. Please re-read my post on the last page.

EDIT Just have the sphinx tell us we are wrong on this one already.

EDIT2: I'm going to suffer an aneurysm if this continues.

Edited: seath on 26th Sep, 2018 - 6:07pm