In my last post I discussed the game of hangman, aided by an algorithmic analysis of the winningest words. A few months ago I did something similar with the peg game they have on the tables at Cracker Barrel.  The one where you have a triangle with 15 pegs, with one missing. You remove a peg by jumping over it. The goal is to leave only one peg remaining. I think I won the game the very first time I ever played it, and I don’t think I’ve won since then.

It occurred to me that the game would be easily solved with brute force, and after an hour or two of coding I had done so. However, I never went much farther than that. I had hoped to look for patterns or simple rules that lead to a victory, but never really got very far. But I decided to post what I have here just for the sake of doing so.

         0
       1   2
     3   4   5
   6   7   8   9
10  11  12  13  14

Given the above peg positions, there are four unique starting configurations: you can start with peg 0, 1, 3, or 4 removed. Any other position is a mirror and/or rotation of those four. So I looked at which starting positions were the most likely to win.

Peg 0: 29,760 ways to win of   568,630 games (5.23%)
Peg 1: 14,880 ways to win of   294,543 games (5.05%)
Peg 3: 85,258 ways to win of 1,149,568 games (7.42%)
Peg 4:  1,550 ways to win of   137,846 games (1.12%)

So the moral of the story is: start with a middle edge peg removed, not the traditional configuration of top peg removed. Beyond that, I got nuthin.