I’m designing a board game which uses hexagonal tiles (hex[es]) to form the game area the players move about in. There is a single hex which is the players starting point, and one “layer” of hexes around that centre one. This is the starting layout:

7 hexagons in formation

After this, players can add new hexes to the structure in any direction. The only rules are they fit in to the hex ‘grid’ and they touch an existing hex. Therefore they do not need to fully complete concentric layers, they can go one direction in a line of hexes for example.

I would like to introduce a mechanic that targets a random hex, and this is where I’m struggling to come up with a good way of determining that random hex.

Ideally, the method should give an equal chance for any hex to be targeted; and it should only use components which are generally found in board games. It would be very easy to do this with the addition of a mobile phone, for example, but I would like to try and keep this classic. Common components are items such as dice, spinners, tokens, cardboard components, etc. Feel free to suggest anything that might help. The other requirement would be that the process is not overly complicated - but as that’s subjective I’m open to anything.

TL;DR Is there a method of choosing a random hex from any number of hexes, in any formation, using common board game components?

  • Are you aware of a mechanic that works for squares? If so, it can be adopted to hexes.
    – John
    May 23, 2019 at 18:16
  • 1
    In a non-regularized arrangement, you'll have to provide an artificial/external way to order them somehow. Something like Catan's chits numbered 1 to N (well, Catan's are lettered, but you get the idea), where N is the most number of hexes that can ever be in play, would work. May 23, 2019 at 18:32
  • @L.ScottJohnson No need for that you can still break it down to a square grid and the example shown it would be a 3x5 grid and just need to make a rule to deal with the 4 corners that don't have tiles.
    – Joe W
    May 23, 2019 at 18:35
  • @JoeW I think you're overlooking the irregular growth of the board allowed by the game's mechanics (or eliding the issues that come up from that. It'll be a great deal more than "four corners" in the general case). May 23, 2019 at 18:50

3 Answers 3


The best way to deal with variable count randomization is to use a drawing bag

Create two sets of tiles. Whenever you add a tile to the board, add its copy to a drawing bag.

Now whenever you need a random tile, draw it from the bag.

This technique requires all your tiles to be unique. If they are not unique, and can't be given a unique identifier, then create a double set of numbered tokens, and assign a number to each tile as it is placed.

Deck vs Draw Bag

StartPlayer mentioned using a deck, which is also an excellent suggestion. The two options have slightly different use cases, so which one you choose should depend on how you need them.

Draw bags are better if you are frequently adding items between draws, because it is easier to shuffle a bag, especially if there are only a few items.

On the other hand, a deck is better if you need to remove specific items in a manner besides randomly drawing them, because it is easier to search a deck than a bag.

  • I did question whether this would be a little unwieldy for players - finding a token that matches the tile they've just placed - although it certainly works. Inspired by the concept though, I think you could print a unique two digit reference on each tile, such as [A-Z]-[1-n] and have two bags (one for letters, one for numbers) and a pool of A tokens, B tokens, 1 tokens, 2 tokens etc. Although the process is slightly longer on average it would rule out the annoyance of not finding one specific token or it being lost.
    – djfw
    May 24, 2019 at 14:18
  • @djfw a two-token method runs a danger of screwing up the probabilities if the tiles aren't evenly distributed. Adding the tokens is a bit fiddly, but if you can sort the tokens ahead of time it isn't so bad. May 24, 2019 at 14:36
  • To clarify, there would be pools of each type of token and two bags. When you draw a tile, you put a matching letter token in the letter bag, and a number token in the number bag. Each bag would have the same number of tokens with the same distribution as on the board. To choose your random tile you would then draw a letter and a number and find that tile. I'll prototype and see what's better, two bags and the extra time drawing two tokens, or one bag and the extra time finding the exact matching token.
    – djfw
    May 24, 2019 at 14:50
  • @djfw If I understand you correctly, if you had the tiles A1, A2, and B2, you'd have one bag with A A B and one bag with 1 2 2. Is this right? This gives you a 44% chance of drawing A2, a 22% chance each of drawing A1 or B2, and a 11% chance of drawing B1 and needing to redraw. May 24, 2019 at 17:07
  • Depending on how many of your tiles are usually in play, it might also be more efficient to keep all the tokens in the bag/all cards in the deck and just keep drawing until you hit a valid one each time you need to pick a random hex. May 28, 2019 at 20:25

There is a print and play on BGG called Beehive (full disclaimer I'm the designer).

Throughout the game I needed to select random hexes. This was achieved by each hex having a unique number and a deck of cards with each number as well. Throughout the game cards would be removed and added back to the deck as required. This meant is was always possible to randomly choose any hex. If each of your hexes is numbered and placed you can add its matching card to a deck when you need to choose one randomly.


Eclipse has a similar hextile placement method. It also has a unique number printed on every hextile - the galactic center is tile 001, the next ring is 100-115 or so, then 200-230, and 300+.

If you implemented a similar numbering scheme that was printed on the hextiles, you could then have 3 dice that are rolled to determine which random hex is selected (like how d% is two d10s, one for the tens place and one for the ones place.) So a roll of 2, 6, and 1 would be hex number 261. If the hex does not exist on the board, reroll.

  • 1
    The problem with this method is that you're going to be rerolling constantly. You'll have only a dozen or so tiles out, with the potential to roll any of 400 numbers. May 23, 2019 at 21:25
  • @ArcanistLupus sure, that is a problem with it. It could be mitigated by matching the dice to the actual number of hexes available, i.e if there are 40 tiles, a d4 and a d10. Still more rerolling than ideal (I like your draw bag personally) but a good option for people who get scared away by fiddly bits. I love fiddly bits myself :)
    – James Wade
    May 24, 2019 at 15:49

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