Dice Odds for Settlers of Catan

pair of dice This is a quick discussion of the distribution of rolls you get over time with a pair of 6 sided dice -- something you happen to care about quite a lot in the game Settlers of Catan. (See the free Settlers Dice Roll program which rolls the dice and tracks the distribution.) The probability of rolling a particular number with two dice is the number of ways the dice can fall that add up to that number divided by the total number of ways the dice can fall. So for example, the odds of a 4 roll are 3/36 since out of the total of 36 ways the two dice can fall, 3 of those ways add to 4: 1+3, 2+2, 3+1. Note that 1+3 and 3+1 count as 2 different ways along with 2+2, yielding 3 ways total.

odds for 7 6/36 =0.167
odds for 6 and 8
5/36 =0.139
odds for 5 and 9
4/36 =0.111
odds for 4 and 10 3/36 =0.083
odds for 3 and 11
2/36 =0.056
odds for 2 and 12
1/36 =0.028

In some versions of the Settlers, there are little dots on each of production circles that indicate how often that number comes up, and the dots correspond exactly to the odds above -- 1 dot on the 2 and the 12, 2 dots on the 3 and 11, and so on. We made fun of the dots when we first got the game -- "oh look, dots, ha ha, how useful could that be", but now we stare at them in quiet awe as we try to figure out our initial placements.

Eventually you will play a Settlers of Catan game where a certain number does not come at all and that can seem pretty strange, especially if it's a number like 4 or 5 which you expect to come up pretty often. This is especially noticeable if you have a computer program such as Settlers Dice Roll which tracks the distribution of the rolls through the game.

What are the odds of playing a Settlers game of, say, 50 turns without rolling any 3's?

The odds of rolling anything but 3 are just one minus the odds of rolling a 3...

The odds of rolling no 3's 50 times in a row is just that probability raised to the 50th power. It works out to around 57 times per 1000 games (here, the syntax 2^20 means 2 raised to the 20th power)...

Conceptually, the 3 and the 11 go together, since they occur with the same odds. To think about the odds of numbers not being rolled, I think it's easiest to take then numbers like 3 and 11 together. So for example, what are the odds of seeing a 50 turn game where either no 3's or alternately no 11's are rolled? To a close approximation, the odds of no 3's or no 11's is just double the odds of no 3's...
2*(1-2/36)^50 =0.115

Or put another way, that's about once every 8 or 9 games: 1.0/0.115=8.696

The only problem is, just adding the odds for the no 3's and no 11's is not quite right. Here's a quick introduction to the correct way to add probabilities. Suppose on any given day, there is a 20 percent chance of rain, a 30 percent chance of my oversleeping, and a 10 percent chance of both rain and oversleeping. What are the odds on a given day that either it will rain or I will oversleep? The simple answer looks  like 20 + 30=50, but that doesn't quite work right. The problem is that we're double counting a little. The 20 rain cases already include some cases where I oversleep, and the 50 oversleep cases already include some rain cases, so by adding 20+30=50, I get a number that's too big. The over-counted cases are exactly the cases where both rain and oversleep happen, so we just subtract the "both" cases to adjust for the double counting: 20+30-10=40.

For the dice, this means the odds of either no 3's or no 11's, is the no 3's odds, plus the no 11's odds, minus the odds of no 3's and no 11's. The odds of no 3's are (1-2/36)^50=0.057, the odds of no 11's are the same (1-2/36)^50=0.057, and the odds of a 50 rolls with no 3's and no 11's are (1-4/36)^50=0.003. Putting it all together, the odds of a game with either no 3's or  no 11's are..

2 * (1-2/36)^50 - (1-4/36)^50=0.112

The correct odds, 0.112, are just a little less than the 0.115 answer from before, since we're now subtracting out the double counted cases.

Now that we know how to compute the odds, below are the odds of a game with 50 rolls where one or the other of each pair of numbers is never rolled. As you would expect, a game with no 12's or no 2's is not uncommon (43%), but a game with no 5's or no 9's is extremely rare (0.6%)...

either no 2 or no 12: 2*(1-1/36)^50 - (1-2/36)^50 =0.432
either no 3 or no 11: 2*(1-2/36)^50 - (1-4/36)^50  =0.112
either no 4 or no 10: 2*(1-3/36)^50 - (1-6/36)^50  =0.026
either no 5 or no 9:   2*(1-4/36)^50 - (1-8/36)^50  =0.006
either no 6 or no 8:   2*(1-5/36)^50 - (1-10/36)^50  =0.0011
no 7: (1-6/36)^50 =0.00011

If you ever play a game of settlers where a number never comes up, you can think of it as a rough estimate of how many Settlers games you've played. One time, I played in a game that had no 9's, which seemed pretty strange at the time. However, it's actually about right, since I think I've played between 100 and 200 games, and no 5's or no 9's should happen about once in every 1/0.006=166.667 games. A game with no 6's or no 8's should come up about once every 1000 games. A game with no 7's should happen about once in 10000 games -- if that ever happens, it's a sign that you're playing too much settlers!

Nick Parlante, nick.parlante@cs.stanford.edu, March 2003