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I bought a Grab & Go edition of monopoly and realized they changed the Chance and Community cards.

Instead of physical cards, they have you roll 3 dice, and pick out of a list (3-18) of effects.

Right off the bat, I know this will bell curve the effects toward the middle.

How would I use 3 dice to best replicate the 16 card distribution of the original game?

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  • 12
    What if the game is designed around that statistical distribution? Have to consider they designed it so that certain cards are more likely than others.
    – Joe W
    Aug 23 '20 at 16:34
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    Are chance and community cards distributed evenly in the regular deck? Aug 23 '20 at 16:37
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    @AzorAhai--hehim They are as there are 16 cards and 1 of each type, however that does not mean that the game developers didn't take that into account when designing this new version of the game.
    – Joe W
    Aug 23 '20 at 17:24
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    @JoeW Unlikely. The "Get Out Of Jail" card has a 0.5% chance in this new format (a roll of a 3 with 3 dice), equivalent to 1 out of 216. The four highest occurring values (9 to 12, almost 50% of total possibilities), results in $10 collect tax from others, $20 reward, $200 fine, or go to jail. This will basically make the game even less fun.
    – Nelson
    Aug 24 '20 at 4:52
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    The get out of jail free card being a 3 doesn't mean it wasn't designed without that thought in mind. It could very well be the fact that they designed it so that it would go quicker due to the fact that you are more likely to lose money or not get very much back when landing on chance. Remember it is the house rules that people add that make the game take longer and they could be working on making sure that games finish in a timely manner.
    – Joe W
    Aug 24 '20 at 12:09
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If I understand it correctly, you're looking to generate a uniformly distributed random integer in the range 1-16 using only regular d6s.

You'll need two distinguishable dice; if you don't have distinguishable dice, then roll one die, note the result and then one die again. Call the dice "die A" and "die B".

  1. Map the result of die A as follows: 1-2 => 0, 3-4 => 6, 5-6 => 12
  2. Add die B to the result of step 1 - you now have an integer in the range 1-18.
  3. If your total is 17 or 18, go all the way back to step 1. Do not just re-roll die B until you get a value of 1-4.
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If you aren't opposed to buying different dice, you can obtain the result directly from two consecutive d4 rolls.

The first roll tells you whether you are looking at the first, second, third or fourth quarter of the list. The second roll tells you whether you want the first ... or fourth item in that quarter. No dice need to be refilled (no empty results) and there are no difficult rules involved (beyond knowing that 16 ÷ 4 = 4).

If you don't want to seek out special equipment, but like the idea of splitting the list gradually, you need just one coin.

At each flip, you take the first or second half of what remains, and "cross out" the rest. After four flips, you have eliminated 15/16 items and are left with one result.

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  • It's also pretty trivial to turn a d6 into a d4 simply by rerolling if you get a 5 or 6.
    – Brady Gilg
    Aug 24 '20 at 16:47
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    Trivial, yes. But rerolling 1/3 of the time gets old fast, especially when you need to make two rolls. Aug 24 '20 at 19:53
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You could get yourself a d8 (eight-sided die, available from any well-equipped game shop). Use the result from that die (multiplied by 2), and the result of a second die to decide whether to add one (low roll, e.g. 1-3 on d6) or two (high roll, e.g. 4-6 om d6) to the result.

That will give you an even distribution from 3 (1 * 2 + 1) to 18 (8 * 2 + 2).

Of course, really well-equipped game stores will have actual d16. They're not common, but they do exist. That would relieve you of any math stunts (other than possibly adding 2 so you get the 3-18 range).

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    Hmm, if I'm getting an extra die, why not just go to a d20 and use that?
    – Nelson
    Aug 23 '20 at 15:11
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    @Nelson Depends on what's getting on your nerves more, a bit of maths or re-rolling 20% of the time. I don't like re-rolling.
    – DevSolar
    Aug 23 '20 at 15:17
  • Call me crazy, but a d16 seems like a better solution than two d8s...
    – TTT
    Aug 28 '20 at 3:39
  • @TTT I've yet to come across one of those, whereas a d8 is part of any D&D dice set. Of course, if you can get a d16 somewhere, that's ideal.
    – DevSolar
    Aug 28 '20 at 6:18
  • @DevSolar - I found a few in quick search. But yeah, it does seem like they are rarer than others, but still easy to get nowadays.
    – TTT
    Aug 28 '20 at 14:24
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The following grid assigns 13 results each to 8 of the 16 cards (ie: 1, 3, 5, 7, 10, 12, 14, 16) and 14 results to the other 8 (ie: 2, 4, 6, 8, 9, 11, 13, 15).

It requires one distinguished die, here indicated in green, cross-referenced with the sum of the other two die here marked in red.

It features a symmetrical distribution, to please the eye, with traditional die symmetry on the distinguished die: the sum of the values for the distinguished die as rolled and its inverse summing to 17, just as for a traditional 1d6 each pair of opposite faces sums to 7.

![enter image description here

This table gives a nicely (though not perfectly - see above) balanced distribution of the cards. It would require many games to differentiate this distribution from that of the original Monopoly game with physical cards.

Note: (14 * 8) + (13 * 8) = 112 + 104 = 216 = 63.

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  • While this is clever, I'm honestly struggling to see how it's better than my two die solution. Aug 24 '20 at 22:07
  • @PhilipKendall: It's better because: (1) it requires no re-rolls to obtain a balanced distribution; (2) It's an instant lookup,so faster; (3) it provides a more balanced distribution (a variation top-to-bottom of 7%; (4) No complicated re-roll rules so no arguments over interpretation: just roll, read result, read Chance/Community Chest Card; (5) I presented the lookup card ready for printing prior to game play. Aug 24 '20 at 23:28

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