For LEGO's board game line, they have moulded a LEGO die, with a 2x2 stud pattern on each side.

A picture of a LEGO die; a standard 5-pip on one side, multicolored studless plates on another; and a third empty side are visible.

Obviously, weight distribution is going to be thrown off if you use a full studless plate on one side and four 1x1 studless plates on another because of the kerf.

Experimentally (by constructing different dice and rolling hundreds of times) or analytically (by analyzing the weight distribution of different plate configurations), how unfair can this die be assuming all studs are covered by a studless plate? What can be done to minimize the unfairness?

  • 2
    If you want to be torough, your assumption about all studs being covered has de be rescinded. Indeed, some of the games make heavy use of the "build your dice" feature, and add or remove tiles during the game itself. This adds a whole new dimension to the dice mechanics as event probabilities change over time, even if you ignore the unfairness introduced by the weight difference. Minimizing the unfairness can be of course by using only 2x2 tiles, but where's the fun in that?
    – Joubarc
    Commented Oct 29, 2011 at 19:27
  • I don't think I've played a Lego game yet that allows empty die sides? Or perhaps I forgot - some of the games are fairly unmemorable. I'm also interested in any unfairness that may manifest for a plain die covered in 2x2 as well.
    – user23
    Commented Oct 29, 2011 at 19:28
  • I don't expect much unfairness if all sides are covered by 2x2 tiles. From what? The paint? The LEGO dice is probably much more balanced than a regular Monopoly engraved dice. As for games where sides are built during the game, a lot of them do, and I think Lava Dragon is a good example: when you roll the dice, if there's room on the upper side, you may add a 1x1 tile of your color, then you can make as many moves as there are tiles of your color on that face (or something like that).
    – Joubarc
    Commented Oct 29, 2011 at 19:51
  • Indeed, Lego's manufacturing genius has probably resulted in an extremely fair die - but like I said in the question, a good answer will be something empirical or analytic, and not assumed. (Lava Dragon is one I haven't played yet.)
    – user23
    Commented Oct 29, 2011 at 19:54

1 Answer 1


Conditions for unfairness:

  • Different shaped tiles applied to each side resulting in different weight wich maked the dice "biased."


The maximum difference in weight (when all studs get covered) is one side using a 2x2 tile, and another side is using 4 1x1 tiles. the weight difference in this case is calculated based on using these measurements. To show the difference:

X       . .       X
X       . .       X
X       . .       X
X . . . . . . . . X
X . . . . . . . . X
X       . .       X
X       . .       X
X       . .       X

X = borders
. = border from the 1x1 tiles that are missing using a 2x2 tile

So there are ~ 28 legounits³ = 44.8 mm³ difference in material used. ABS, the material used for LEGO bricks has a density of between 1.04 and 1.12 g/cm³ so 44.8 mm³ have a maximum height of:

1.12g/cm³ / 1000mm³ * 44.8mm³ = 0.050176g

0.05 g would be way too much for a precision-dice used in a casino, but for a "toy" it's almost nothing (it's less than 1% of the die's total weight - I'd like to calculate that, but I can't find information about its weight), so I won't expect anyone to notice the LEGO-dice is "unfair".

Found the dice on bricklink. weight: 8.02 g, so the 0.05 g are just ~ 0.6 % of the total weight.

  • 4
    Actually, BrickLink will give you the weights for tiles too. A 2x2 is 0.48g and a 1x1 is 0.16g so the difference between one 2x2 and 4 1x1 is 0.16g. Which is not the highest unfairness you could have; as I said you can have empty sides, so the highest unfairness is empty side against 4 1x1 tiles, or a difference of 0.64g, or 8% of the total weight, which is bound to make a difference. Although usually when an empty side comes up you may add stuff on it so the situation is unlikely to happen.
    – Joubarc
    Commented Oct 30, 2011 at 19:46
  • 3
    i've chosen to compare 2x2 against 4 1x1 because that was one of the things asked ("a full studless plate on one side and four 1x1 studless plates on another"). the difference in this cas eas also much more interesting as the dice gets filled quickly and as long as there are empty sides its obvious the weight-difference is much higher.
    – oezi
    Commented Oct 31, 2011 at 6:35

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