What are the odds, in a standard break of billiard balls in a standard game of 15 ball rotation, that the aforementioned balls will end up in the same position in two games in a “row”. Before you answer (and don’t get me started on why the word “answer” is not pronounced “an-swer”,or for that mention, why “row” is in quotation marks, ) what are the odds?

I am going to guess that the odds of all the balls ending up in the same positions on two consecutive breaks (assuming a ‘good faith’ break and not just some little tap on the rack) is going to be about as close to (without being) zero as possible. The variables are enormous, variation in cue ball speed, angular momentum, the condition of the felt, the humidity, the tempurature, all of those (and I am sure a million more) variables will most likely be *slightly* different in every case. I am sure someone will be along shortly to describe it all in terms of chaos theory.

**Rhum Runner**, all of those variables mentioned make it realy impossible to **predict** a given layout before a break. Just coming up with the number of possibilities can be done.

First I would segment the table into squares, say a 1/4" (you can be more precise but a 1/4" works well). Then you need to run through all the combinations that the 16 (or less) balles could be situated in our grid.

**Frank** Good point. If we follow your aproach, don’t forget that some/all of the balls may be pocketed on the break, also don’t forget to account for the cue ball filling one position (i.e. 16 balls on the table) I also think a 1/4" grid is way too big. Think more like 1/16, or 1/64". One could do all the math, and I am willing to predict the odds of two consecutive breaks yeilding the same layout will be about 0.

It depends entirely on how precise you want to be about “same” layout.

In terms of the game, I suppose many experts would agree that even if the balls are visibly in slightly different locations the layout could be essentially the same – same shots to run the table, and the shots are not any more or less difficult. In other words, some balls could be “in the same place” within a 1" margin of error, and other balls might have a more stringent margin of error.

And obviously if by layout you mean absolutely identical positions relative to the table, then the probability is 0 because as you decrease the margin of error, so decreases to probability.

Finally, assuming a margin of error, the probability of duplicate positions is significantly more likely than a completely random distribution over the table, because the break settup does result in a sort of pattern, and there are many break results that actually have names, and experts break in such a way to try to achieve certain break conditions.

**Frank.** Upon reflection, I think one problem with just counting the number of possible positions for the balls is that it assumes the balls are randomly distributed, which they are not. For example the balls in the middle of the rack are not likely to end up at the far end of the table, so some locations are more likely to be used than others. Just another thought.