Congratulations to @GarethMcCaughan for his victory on the long unsolved Part II of this mystery! Kudos!

Part VIII will hold a not-so-fun twist! Solve this one quickly, and we can get to it!

New Hints Below!

The hints should have made this obvious, by now!

You go to


(Solution to Part VI)

as per the hiker’s note. Now, this is getting annoying! You’ve gone to two different continents just because of his obscure clues... but, at the same time, you’re proud of yourself for managing to solve these tough riddles!

Unfortunately, as soon as you arrive at your hotel, you notice a display of cards on your bed! card display

There is also a note:

enter image description here

Get into the files as soon as possible to discover more clues. You’re starting to regret hiking in those woods... Who knows what could happen next?

Card Images Credit: Random.org

Our hiker will return in Part VIII with a not-so-fun twist!


Hint 1

If the hiker only wants whether it is red or black (and not the suit), shouldn't it make sense that the pattern in the cards is also dependent on the color and not the suit? Hmm...

Hint 2

This hiker has been leading you on so many wild goose chases! He must be a very mean person, right?

Hint 3

You know what they say, right? Red cards are twice as good as black cards!

Hint 4

Man, after eating all this much, your weight must be twice as much as average!

(^ That is a HUGE hint!)

  • $\begingroup$ @Ak19 You shouldn’t delete your answer just because it was wrong. Someone might be able to build off of it and get to the right answer. $\endgroup$ May 14 '19 at 16:49
  • $\begingroup$ I'm editing it. $\endgroup$
    – Ak.
    May 14 '19 at 16:58
  • $\begingroup$ Ok till then I'll let the old one remain as it was. $\endgroup$
    – Ak.
    May 14 '19 at 17:00

This puzzle doesn't seem to be reasonably solvable without the hints (unless I'm missing some clues hidden in the letter, perhaps). Hint 3 contains vital information which I didn't see anywhere else. Anyway, the solution is



in each row, the value of the 3rd card is a weighted average of the 1st and 2nd, where (for some reason) red is worth twice as much as black.

  • Row 1: $\frac{9\times2+4\times1}{3}=7.333...\approx7$.

  • Row 2: $\frac{13\times2+2\times1}{3}=9.333...\approx9$.

  • Row 3: $\frac{5\times2+3\times2}{4}=4$.

  • Row 4: $\frac{11\times1+8\times2}{3}=9$.

  • $\begingroup$ This is correct... Regarding your comment about it not being reasonably solvable without the hints, I disagree. Users need to be able to figure out a pattern within the setup of cards. The hints just gave away the pattern. $\endgroup$ May 23 '19 at 17:01
  • $\begingroup$ @INTERESTING You may be right ... IMO, using averages in a row is a natural thing to figure out, but the weighting just seems to come out of nowhere. (Also, is there anything to say the last card in row 3 must be black 4 rather than red 2?) $\endgroup$ May 23 '19 at 17:06
  • $\begingroup$ The weighting is to make it a little more difficult to find the pattern. The cards in the last columns being all black is part of the pattern that can be found by looking at the other rows. $\endgroup$ May 23 '19 at 17:07
  • $\begingroup$ That's disappointing, I discarded this solution because the averages didn't actually work for the first two rows. Why not a black 3 instead of a 4 to make the first weighted average actually 7, or make the third row consistent with the other two (as in also off by 1/3)? $\endgroup$
    – JProblems
    May 23 '19 at 21:01
  • 1
    $\begingroup$ @Rand al'Thor, I agree the hints were pretty explicit. I think it was obvious even before the 4th hint (PuzzlingPlatypus also mentions it in their partial answer), but the OP also seemed surprised/frustrated that the easy question was unsolved for so long, and I think that's because most of us were expecting a uniform rule. It's especially confusing because uniformity would have been really easy to implement in this particular case, and I don't think having two approximate weighted averages and two exact weighted averages added anything. $\endgroup$
    – JProblems
    May 24 '19 at 12:50

Probably a far-fetched answer but the card could be a

Black King

Logic being .....

1st (if Number of R > Number of B , "+", "-") (3rd - 2nd) = 6
9 - (7 - 4) = 6
13(K) - (9 - 2) = 6
5 + (4 - 3) = 6
Hence ...
11 - (13 - 8 ) = 6

  • $\begingroup$ secretly hoping that rot13 (vg vf abg gur pbybhe, ohg gur fhvgf gung znggref) $\endgroup$
    – Kryesec
    May 17 '19 at 3:43
  • $\begingroup$ You sort of have the right idea, (+1), but you’re not correct. $\endgroup$ May 17 '19 at 11:17

Partial answer

The card is an Ace (A).


A = 1, J = 11, Q = 12, K = 13

Pattern followed:

first card - second card $\pm$2 = third card


Row 1: 9 - 4 + 2 = 7 \ Row 2: K - 2 - 2 = 13 -4 = 9 \ Row 3: 5 - 3 +2 = 4 \ Row 4: J - 8 -2 = 11 -8 -2 = 1 = A

  • $\begingroup$ Sorry, that’s not it. I don’t think you are close enough for an upvote, either. $\endgroup$ May 14 '19 at 16:42

Partial The final card is



The first column has

3 reds 1 black

The second column has

2 reds 2 blacks

So the third column should have

3 Blacks, which it does, and 1 red, which is missing

Also, I believe the emphasis in the hint refers to

mean in the math sense, as in averages.

However, I tried and none of the ways i could think to do this resulted in the same number.

  • $\begingroup$ Hmm... Interesting thought. $\endgroup$ May 19 '19 at 19:04
  • $\begingroup$ Was my approach to the color correct? $\endgroup$ May 20 '19 at 16:31
  • $\begingroup$ No, it was not. $\endgroup$ May 20 '19 at 21:14

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