# dice probability issue

In the game Settlers of Catan, territories/tiles are each (effectively) randomly assigned a number from 2-12. When that number is rolled as the sum of two dice, the tile generates resources for the player or players who have settled it (technically: along it). A key element of the game is considering the probability of each of these numbers arising when choosing where on the board to settle. Players generally vie for tiles with the greatest probability of being rolled (depending also on what resources the tiles will generate and the anticipated scarcity of that resource).

Sometimes players are faced with the choice of settling two tiles of either the same number, or different numbers of equal probability -- eg: two tiles bearing the number 8, or two tiles bearing the number 6, or one of each. Mathematically, one would think the outcomes are likely to be equally good, as the two numbers should come up equally often, and so (again, leaving aside which resources they produce) it shouldn't matter. In reality, choosing same-number options can produce sub-optimal results, as the numbers rolled do not perfectly reflect theory -- ie: a given number may be rolled far more or far less frequently than would be expected across a game.

Is there a mathematical principle that accounts for this? Is this simply a case of small sample error, where a couple of hundred rolls may produce odd results but a million would more closely hew to expected probabilities?

• Do you actually have records showing this effect, or is this a subjective impression? – Brian M. Scott May 14 '12 at 19:08
• This is called diversification – Henry May 14 '12 at 22:13

Let $X_6=1$ if you toss a total of $6$, and $0$ otherwise. Let $X_8=1$ if you toss a total of $8$, and let $X_8=0$ otherwise.
We seem to be comparing the random variables $2X_6$ (or $2X_8$) and $X_6+X_8$. Since $P(X_6)=P(X_8)$, the random variables $2X_6$ and $X_6+X_8$ have the same mean. So in the sense of expected return, the strategies may be equivalent.
However, the variance of $2X_6$ is more than twice the variance of $X_6+X_8$. More precisely, the variance of $2X_6$ is $(4)(5/36)(1-5/36)$, while the variance of $X_6+X_8$ is $(10/36)(1-10/36)$. So for small samples, one would expect much more fluctuation if one plays $2X_6$ rather than $X_6+X_8$. That can be a good thing, of course, if you get "lucky," but it is a less conservative strategy.