<p>If you aren't opposed to buying different dice, you can obtain the result directly from two consecutive d4 rolls.</p>
<p>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).</p>
<p>If you don't want to seek out special equipment, but like the idea of splitting the list gradually, you need just one coin.</p>
<p>At each flip, you take the first or second half of what remains, and &quot;cross out&quot; the rest. After four flips, you have eliminated 15/16 items and are left with one result.</p>

CC BY-SA 4.0
2020-08-24T16:47:56.497Z
It's also pretty trivial to turn a d6 into a d4 simply by rerolling if you get a 5 or 6.
Brady Gilg
CC BY-SA 4.0
2020-08-24T19:53:04.770Z
Trivial, yes.  But rerolling 1/3 of the time gets old fast, especially when you need to make two rolls.
Arcanist Lupus
CC BY-SA 4.0
2020-08-24T05:43:02.057Z
2020-08-24T05:43:02.057Z
Nij
<p>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.</p>
<p>That will give you an even distribution from 3 (1 * 2 + 1) to 18 (8 * 2 + 2).</p>
<p>Of course, <em>really</em> 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).</p>

CC BY-SA 4.0
2020-08-23T15:11:57.740Z
Hmm, if I'm getting an extra die, why not just go to a d20 and use that?
Nelson
CC BY-SA 4.0
2020-08-23T15:17:13.553Z
@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
CC BY-SA 4.0
2020-08-28T03:39:19.350Z
Call me crazy, but a d16 seems like a better solution than two d8s...
TTT
CC BY-SA 4.0
2020-08-28T06:18:29.920Z
@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
CC BY-SA 4.0
2020-08-28T14:24:49.927Z
@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
CC BY-SA 4.0
2020-08-23T14:37:38.993Z
2020-08-28T07:41:01.527Z
2020-08-28T07:41:01.527Z
DevSolar
<p>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).</p>
<p>It requires one <em>distinguished die</em>, here indicated in green, cross-referenced with the sum of the other two die here marked in red.</p>
<p>It features a symmetrical distribution, to please the eye, with traditional die symmetry on the <em>distinguished die</em>: the sum of the values for the <em>distinguished die</em> as rolled and its inverse summing to 17, just as for a traditional 1d6 each pair of opposite faces sums to 7.</p>
<p><a href="https://i.sstatic.net/BKLeM.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/BKLeM.png" alt="![enter image description here" /></a></p>
<p>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.</p>
<p>Note: (14 * 8) + (13 * 8) = 112 + 104 = 216 = 6<sup>3</sup>.</p>

CC BY-SA 4.0
2020-08-24T22:07:16.700Z
While this is clever, I'm honestly struggling to see how it's better than my two die solution.
Philip Kendall
CC BY-SA 4.0
2020-08-24T23:28:34.350Z
@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.
Forget I was ever here
CC BY-SA 4.0
2020-08-24T15:12:33.337Z
2020-08-24T19:53:58.123Z
2020-08-24T19:53:58.123Z
Forget I was ever here
<p>If I understand it correctly, you're looking to generate a uniformly distributed random integer in the range 1-16 using only regular d6s.</p>
<p>You'll need two <em>distinguishable</em> dice; if you don't have distinguishable dice, then roll one die, note the result and then one die again. Call the dice &quot;die A&quot; and &quot;die B&quot;.</p>
<ol>
<li>Map the result of die A as follows: 1-2 =&gt; 0, 3-4 =&gt; 6, 5-6 =&gt; 12</li>
<li>Add die B to the result of step 1 - you now have an integer in the range 1-18.</li>
<li>If your total is 17 or 18, go all the way back to step 1. Do <em>not</em> just re-roll die B until you get a value of 1-4.</li>
</ol>

CC BY-SA 4.0
2020-08-23T17:08:39.893Z
This, for sure.  If you want to see a proof of it working, here is a quick example in python: https://github.com/snoopy369/Sports.SE/blob/master/rolls_16.py
Joe
CC BY-SA 4.0
2020-08-23T19:29:11.020Z
@Joe Possibly easier just to take all 36 possibilities of 6x6 and show the distribution is correct there :-)
Philip Kendall
CC BY-SA 4.0
2020-08-23T07:17:35.120Z
2020-08-23T07:17:35.120Z
Philip Kendall
<p>I bought a Grab &amp; Go edition of monopoly and realized they changed the Chance and Community cards.</p>
<p>Instead of physical cards, they have you roll 3 dice, and pick out of a list (3-18) of effects.</p>
<p>Right off the bat, I know this will bell curve the effects toward the middle.</p>
<p>How would I use 3 dice to best replicate the 16 card distribution of the original game?</p>

CC BY-SA 4.0
2020-08-23T16:34:44.603Z
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
CC BY-SA 4.0
2020-08-23T16:37:52.807Z
Are chance and community cards distributed evenly in the regular deck?
Azor Ahai -him-
CC BY-SA 4.0
2020-08-23T17:24:38.247Z
@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
CC BY-SA 4.0
2020-08-24T04:52:01.890Z
@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
CC BY-SA 4.0
2020-08-24T12:09:01.710Z
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
CC BY-SA 4.0
2020-08-24T12:48:38.870Z
This is simply a portable version of the classic monopoly game. They went so far as to replicate the 16 card effects, but obviously someone didn't understand high school statistics and completely skewed the effects. The "new" rules will drain money out of the players more frequently than normal. I laugh at their understanding of the statistic distribution because they put in 4 "Get out of jail" cards, which will happen 1 out of 216 ^ 4, or about 1 out of 2 billion.
Nelson
CC BY-SA 4.0
2020-08-24T12:51:28.243Z
Just in case someone doesn't know how the original monopoly worked. You have a deck of 16 Chance or Community chest cards. You draw from the deck until they're all gone, and you re-shuffle. This results in a 1/16 chance but WITHOUT replacement (e.g.: deck is not shuffled per draw). It'll be pretty darn hard to replicate it completely but at least the probability isn't skewed towards 4 cards out of the 16.
Nelson
CC BY-SA 4.0
2020-08-24T13:39:35.737Z
Draining money = shorter game. This is conceivably intended and doesn't necessarily mean "less fun".
L. Scott Johnson
CC BY-SA 4.0
2020-08-24T20:43:14.707Z
Why do you seem to keep ignoring the fact that they likely designed the outcomes of the game based on how often they want each outcome to happen? You seem to be making a very big assumption that they wouldn't know that the outcome would be impacted by rolling 3 dice instead of a single one. No reason they couldn't have included a 16 sided die or even a 20 sided one and figured out how to deal with the extra sides.
Joe W
CC BY-SA 4.0
2020-08-25T05:17:33.217Z
The evidence for a redesign isn't my question. I don't want a travel edition of a classic game to play differently. That's not what I'm asking nor what I'm interested in.
Nelson
CC BY-SA 4.0
2020-08-25T15:59:52.530Z
You not wanting it to play different doesn't mean you should insult the devlopers. Just because they made design decisions to change how the chance/community chest cards worked and you don't like them doesn't mean that they don't know what they are doing.
Joe W
CC BY-SA 4.0
2022-02-11T02:32:30.330Z
Yeah.. But where did u find the chance and community chest lists? Mine didn't come with them? Lol Thanks!!
Jessica
<house-rules><monopoly>
How do I fix the statistical distribution of using 3 dice to handle Chance and Community cards in Monopoly?
Nelson
# How do I fix the statistical distribution of using 3 dice to handle Chance and Community cards in Monopoly?

(Asked by: username_0 on 2020-08-23)

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?

**Tags:** house-rules, monopoly

> **Comments:**
> **username_1**: 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.  
> **username_2**: Are chance and community cards distributed evenly in the regular deck?  
> **username_1**: @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.  
> **username_0**: @username_12W 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.  
> **username_1**: 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.  
> **username_0**: This is simply a portable version of the classic monopoly game. They went so far as to replicate the 16 card effects, but obviously someone didn't understand high school statistics and completely skewed the effects. The "new" rules will drain money out of the players more frequently than normal. I laugh at their understanding of the statistic distribution because they put in 4 "Get out of jail" cards, which will happen 1 out of 216 ^ 4, or about 1 out of 2 billion.  
> **username_0**: Just in case someone doesn't know how the original monopoly worked. You have a deck of 16 Chance or Community chest cards. You draw from the deck until they're all gone, and you re-shuffle. This results in a 1/16 chance but WITHOUT replacement (e.g.: deck is not shuffled per draw). It'll be pretty darn hard to replicate it completely but at least the probability isn't skewed towards 4 cards out of the 16.  
> **username_3**: Draining money = shorter game. This is conceivably intended and doesn't necessarily mean "less fun".  
> **username_1**: Why do you seem to keep ignoring the fact that they likely designed the outcomes of the game based on how often they want each outcome to happen? You seem to be making a very big assumption that they wouldn't know that the outcome would be impacted by rolling 3 dice instead of a single one. No reason they couldn't have included a 16 sided die or even a 20 sided one and figured out how to deal with the extra sides.  
> **username_0**: The evidence for a redesign isn't my question. I don't want a travel edition of a classic game to play differently. That's not what I'm asking nor what I'm interested in.  
> **username_1**: You not wanting it to play different doesn't mean you should insult the devlopers. Just because they made design decisions to change how the chance/community chest cards worked and you don't like them doesn't mean that they don't know what they are doing.  
> **username_4**: Yeah.. But where did u find the chance and community chest lists? Mine didn't come with them? Lol Thanks!!  

---

## 4 answers

### Answer by username_11 on 2020-08-23. Score: 12 | Accepted answer

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.

> **Comments:**
> **username_12**: This, for sure.  If you want to see a proof of it working, here is a quick example in python: https://github.com/snoopy369/Sports.SE/blob/master/rolls_16.py  
> **username_11**: @username_12 Possibly easier just to take all 36 possibilities of 6x6 and show the distribution is correct there :-)  

---

### Answer by username_5 on 2020-08-24. Score: 7

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.

> **Comments:**
> **username_6**: It's also pretty trivial to turn a d6 into a d4 simply by rerolling if you get a 5 or 6.  
> **username_7**: Trivial, yes.  But rerolling 1/3 of the time gets old fast, especially when you need to make two rolls.  

---

### Answer by username_8 on 2020-08-23. Score: 3

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).

> **Comments:**
> **username_0**: Hmm, if I'm getting an extra die, why not just go to a d20 and use that?  
> **username_8**: @username_0 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.  
> **username_9**: Call me crazy, but a d16 seems like a better solution than two d8s...  
> **username_8**: @username_9 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.  
> **username_9**: @username_8 - I found a few in quick search. But yeah, it does seem like they are rarer than others, but still easy to get nowadays.  

---

### Answer by username_10 on 2020-08-24. Score: 0

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](https://i.sstatic.net/BKLeM.png)](https://i.sstatic.net/BKLeM.png)

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.

> **Comments:**
> **username_11**: While this is clever, I'm honestly struggling to see how it's better than my two die solution.  
> **username_10**: @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.  

---

