8
$\begingroup$

Playing an online game I came across a small puzzle (a mini-game in game). The game is about getting jewels in a row - a player needs to get 5 in a row / column or a 5-square-diagonal. Let me show you a screenshot:

enter image description here

So, there's a square 5x5 on which you place 6 types of jewels (4 of each). Then a random jewel is drawn (14 times) and a player marks a chosen one on a board. The goal is to get the most draughts. The center square is neutral and can be used as any jewel.

And here come my questions:

  1. Is there any difference in placing jewels?
  2. If answer for 1 is yes, then how should one place them strategically?

EDIT I have found a video on YouTube showing a player playing this mini-game. I think it'll clarify. (from 10s to 1min15sec) https://www.youtube.com/watch?v=4g7RK3B8uT4#t=10s

$\endgroup$
  • 1
    $\begingroup$ So.. how does the game work again? Do you get to choose the initial positions of the gems? And then you get a random gem which you can use to replace an existing gem in the board, and you have to do this 14 times? $\endgroup$ – votbear Feb 19 '18 at 2:00
  • 4
    $\begingroup$ I think strategy tag would fit well here. Please feel free to remove it if you disagree. $\endgroup$ – North Feb 19 '18 at 2:40
  • 1
    $\begingroup$ @Votbear I have just pasted a Youtube link showing how the game looks like. There is some language barrier I find hard to defeat. Hope it makes the question clear. $\endgroup$ – Aea Reth Feb 19 '18 at 6:42
  • $\begingroup$ @North Thank you, fits very well! I'm a newcomer here :) $\endgroup$ – Aea Reth Feb 19 '18 at 6:43
  • $\begingroup$ Does the random jewel drawn depend on how many are left on the board? E.g. if there is one blue jewel left it still has the same chance as the others? If all four jewels of a type have been marked can it still be drawn? $\endgroup$ – Jeff Feb 19 '18 at 7:14
1
$\begingroup$

The way I see it, placing doesn't matter too much but I wouldn't place all jewels of the same type on the same line as drawing all 4 of them is statistically less likely than drawing just 3. So I'd go at most 3 per type per line. That's of course if the chance of drawing a stone depends on the number of remaining stones. If not then placing is completely irrelevant.

Then we have the marking phase. The center square is a big bonus so I'd always go for that first - you can make 2 lines with 8 stones instead of 10. I'd go for a single row/column and a single diagonal rather than 2 diagonals or a row and column. Why? Because then you can cross them with a 3rd line using only 3 stones rather than 4. And then go with the last 3 stones for a 4th line crossing either two of the previous three lines. I think that's the only way to achieve 4 lines with just 14 stones. Of course you can do the lines in a different order but the end goal should always be the same.

$\endgroup$
0
$\begingroup$

So this seems to basically be a BINGO game. Since the jewels are randomly chosen, it doesn't matter where exactly they are placed.

The actual problem is choosing which one of the positions to highlight when the jewel is chosen. Clearly the most one can get is 4 lines, so you want to work towards that. Otherwise I don't think there's much of a specific 'best order'.

Edit: Here is an example of 4 completed lines.

4 Row example

$\endgroup$
0
$\begingroup$

Answer to Question 1

It doesn't matter if and only if for every pair of possible placements every random sequence of jewels results in the same maximum number of bingos. I can prove that it does matter by providing a counter example: two possible placements that have different maximum number of bingos for the same sequence of random jewels.

Random sequence of 14

AAAABBBBDDDDCC

Placement #1

 AAAAF
 BFBBB
 CCXCF
 DDDDF
 EEEEC


After filling in all of the As, Bs, and Ds:

 XXXXF
 XFXXX
 CCXCF
 XXXXF
 EEEEC

We still get to choose two Cs but we can see that we can't make any bingos with them because the Es and Fs block them all.

Placement #2

 AAAAB
 BBBCC
 CCXDD
 DDEEE
 EFFFF

We can quickly see that we can fill in the first row so there is at least one bingo.

So the placement DOES matter.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.