My understanding was that the competitor's program collected a strategic number of stones N in such a way which forced the other program to be able to collect at most N-1 stones on its next turn. I'm not familiar with the rules of Mancala but I presume they allow for strategies such as this.

There's a difference between collecting the maximum amount of stones and collecting more stones than the opponent. I suspect the winning program also tries to minimize the stones the opponent will be able to collect, thus preserving its marginal advantage.

Edit: Apparently colanderman beat me to it.

Maybe the "one more stone" strategy leaves more options to the decider algorith and is therefore less likely to get stuck in a local maximum that ultimately leads to a loss.