Dice Puzzle

Here's a solo dice puzzle for you. It's an oddly hypnotic way to spend a few minutes before giving up in frustration.

Gather a block of 36 dice.
Roll a die.
Place it on the table.

On Your Turn...
Roll a die. Place it adjacent to a die already on the table. Gradually, you'll create a branching dice formation. As soon as you create a contiguous chain of 3 or more with matching results, remove all the matching dice in that chain. Chains do not count diagonally adjacent dice. Only vertically and horizontally adjacent dice count.

1. Your formation can't extend past a 6x6 grid. Note: There is no board. The overall formation of dice simply can't extend taller or wider than 6 dice. Thus, the first die you place is technically the center. As you add or remove dice, the outerbounds of your formation can shift dramatically. Indeed, over time, the formation may seem to crawl like an amoeba over the table.

2. You may not remove any dice that would create "islands" of disconnected dice. Thus, you might have a chain that contains more than 3 matching results, but you may not remove those dice if it would leave behind even a single die disconnected from the rest.

3. If you already have a contiguous chain of 3 or more matching dice on the board, it stays on the board until you can add a freshly rolled matching result to that chain. Then, you may remove that chain.

If you manage to remove all the dice from the table, you have solved the puzzle. If you find a reliable strategy or solution, leave it in the comments!


  1. Must you remove dice if you have 3+ in a row that won't otherwise cause an island, or are you allowed to leave them as desired?
    If you have, say, 4 in a row, must you remove all 4, or can you remove only 3? (If not, then having a layout of 123 will guarantee a loss if you proceed to only roll 2s - you can't remove them since that will make 1 and 3 islands, and eventually you'll fill up the board.)
    If you have three 1s in a row and removing them won't create an island, and you roll and place a 2, can you remove the 1s, or must you wait until you roll a 1 again?

    Assuming various answers to these questions, the following strategy guarantees that you'll never lose, but still might not have a great shot at victory:
    Place your first die in the top left corner. Every time you roll a new die that you haven't rolled before, place it in the top-left-most spot that's free. Since it's only a d6, at worst you'll fill the top row.
    If you roll a value you've gotten before, place it beneath a matching die, building down a column.
    If you MUST remove dice if possible: a column never grows beyond 4. Columns on the ends will get removed after the third die. Eventually, you might get lucky. Or you might roll a single 1 in millions of throws.
    If you MAY remove dice if possible, wait until a column gets up to six before removing. If you may remove dice at any time, remove the bottom three. Otherwise, you might as well remove the bottom 4, since you'd need another die anyway to trigger a removal.
    If you may remove dice at any time, you're just waiting to roll each die at least three times.
    If you can only remove after adding a die of that value, you're waiting to roll everything twice, then rolling everything once and only once.

    This strategy can probably be improved using a space-filling pattern such that you can remove dice at more places than just the ends. E.g.

  2. I should clarify: there is no board. The outerbounds are simply that the formation cannot extend longer than 6x6. So, the first die you place is technically the center, but the overall shape of the formation can shift and grow in different directions.

    You must remove all matching dice if you create a contiguous group of 3 or more matching dice.

    If you already have a contiguous group of matching dice on the table, you cannot remove it until you place a newly rolled matching die on that group.

  3. This layout
    might work better. All numbers are adjacent to the 1s, so as long as you don't remove them, you can remove any other number safely. I think this gives you a bit more flexibility in cleaning up.

    Without removal at will, winning at any given point is always going to be a bit unlikely, since you need to roll six dice with each number showing up only once: 6! / 6^6 = 1.5%
    In light of that, your best strategy is probably to keep playing as long as possible. In light of that, it might be best to not have all the 1s contiguous, and once you have two in the center, start making non-contiguous blocks of them that you can trim as necessary. Once you've trimmed all other numbers, add a third 1 to the center 1s to trim them.

  4. What about somethig like that:


    Every number touches at least 2 others so removing one won't be a problem.

    Almost any chain can be expanded if needed.

    And the lower right 3x3-corner it's still empty allowing you to solve dificult situations!

  5. So, you distract me with yet another relatively easy to develop computer game.
    Except that I find myself instantly modifying it in the following ways:
    First, I ditch the die rolls. I replace them with randomly generated sprites. This allows me to make varying levels of "difficulty" with the number of sprites available. Also allows me to add weird things like "wild" sprites or whatever.
    Second, I ditch the win condition. If your first die is a 1, your second die is a 1, and your third die is a 1, you win after two moves. No good. Based too much on luck.
    Third, I replace the win condition with a scoring metric(allows for comparisons in multiple games and between players). Placing a die gives 1 point. If a chain is made with that die, another point (assuming a 3 die chain). Larger chains are worth more points (adds the "risk/reward" decision). Game ends when a 6x6 grid has been completely filled. Theoretically infinite in duration.
    Fourth, I add in a "random seed" that allows me to replicate the order of sprites in a game again exactly.
    This game is basically Triple Town with an elimination mechanic rather than a promotion mechanic.


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