Stolen Rubies Puzzle, How many rubies can you guarantee to win?

Question

I just saw this youtube video https://www.youtube.com/watch?v=2QJ2L2ip32w

Introducing a puzzle with the following rules.

There are 30 Rubys, which have to be distributed between three boxes. Every box has to contain atleast 2 rubys. One box has to contain 6 more rubys then one of the remaining two.

Are the rubys hidden, I have to guess the number of rubys inside each box. If I guess to high, I get none. If my guess is $\leq$ the actually number, I get my guess.

I thought about it for a while, and I got stuck. I also came up with guessing 8 for every box pretty quickly, as I know that at least one box has to contain at least 8 rubys.

But I disagree with the presented solution. They claim you can always guarantee that you get 16 rubys, which is not true. You can guarantee to get 8 rubys, but not more.

I could always play differently and for example distribute like this: 2 2 26. So when the guess is 8 8 8 I would just lose 8 rubys.

So as the puzzle is presented, as a game where I know that you know that I know ... I could not come up with a sensable solution, because I could always play differently, and distribute arbitrarly, and try my luck, instead of losing 16 rubys for sure.

So in real life this game should play out very differently. Would you really dare to guess 26 26 26?

Why would you distribute like 8 8 14 when you know this is going to lose you 16 rubys for sure? For me it seems like a more sensable way to play, is to really just play random numbers, where each number is different (obeying the rules of the game of course), so guessing three times the same number doesnt work as well.

So let me ask a different question:

When you really play arbitrarly, what is the expected value of rubys you would lose? Would this expected value still be 16?

So we would go through all possible combinations of playing, and we would also guess every possible combination from 2 to 26 (guessing the same number for every box).

Also I am interested in what you think of the presented solution.

Thanks in advance.

Answer 1

Let the boxes have $a,b,c$ rubies respectively, with $a\le b\le c$.

Claim:$\;b\ge 8$.

Proof:

It's given that one of $a,b,c$ is exactly $6$ more than one of the others, hence one of the relations $$ b=a+6,\;\;\;c=a+6,\;\;c=b+6 $$ must hold.

• If $b=a+6$ then $a\ge 2$ implies $b\ge 8$.$\\[4pt]$
• If $c=a+6$ or $c=b+6$, then $c\le b+6$, hence $$ b=30-a-c\ge 30-b-(b+6) $$ so$\;b\ge 24-2b$, hence $b\ge 8$

Thus in all cases, we have $b\ge 8$, which proves the claim.

It follows that guessing $8,8,8$ guarantees a win of at least $16$, as claimed.

Moreover, the guess sequence $8,8,8$ is the unique optimal strategy, in the following sense . . .

Claim:$\;$If a guess sequence $x,y,z$ guarantees a win of at least $16$, then $(x,y,z)=(8,8,8)$.

Proof:

Suppose the guess sequence $x,y,z$ guarantees a win of at least $16$.

Without loss of generality, assume $x\le y\le z$.

Let $A,B,C$ be the box count sequence corresponding the guess sequence $x,y,z$.

If $x < 8$, then for $(A.B,C)=(20,8,2)$, the strategy $x,y,z$ wins at most $x+8$, which is less than $16$, contradiction.

Hence we must have $x\ge 8$.

If $z\ge 16$, then for $(A.B,C)=(6,9,15)$, the strategy $x,y,z$ wins at most $9$, contradiction.

If $8 < z < 16$, then for $(A.B,C)=(20,2,8)$, the strategy $x,y,z$ wins exactly $x$, which is less than $16$, contradiction.

Hence we must have $z\le 8$.

Thus $8\le x\le y\le z\le 8$, which yields $(x,y,z)=(8,8,8)$, as claimed.