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This question has puzzled me for quite a while. It was a statement on the box that I own that triggered it, translated it says:

The strategic player can outwit his opponent and win all his pebbles".

My 3 year old son loves the game since he wins every time (his victory condition is to be the first with no pebbles left to play). It gives me the opportunity to try to get as many as possible. The most I finished the game with, was 46 six pebbles, though I'm not sure if something was amiss there (like misplaced pebbles). And I am as of yet unsuccessful at recreating this end-game condition.

I'm talking about the normal 6 holes * two players * 4 pebbles a hole game = 48 pebbles.

So, is it possible to get all of the pebbles? Or do strategic players embody the mantra

If you are not trying to cheat, you are not trying to win.

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2 Answers

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First, the issues:

1: you've not defined which version. Yes, I know, yo've specified some starting conditions. But not the capture conditions.
2: There are literally hundreds of variants of Mancala, and at least 3 different capture mechanics I've played.
3: the normal setup is 12 pits... 6 per player...

Still, despite these, I think I can answer the question.

Given that I'm most familiar with Oh-wah-ree, the 3M bookcase game, which has a capture condition of "last stone played landed in an opponent's pit with 1, 2 or 3 stones already present, regress backward for additional pits captured"... And Oh-Wah-Ree's end is only one player can move, and can't provide a piece to another player (which is required if possible).

It is possible, in theory, to capture all the stones on the board in this variant.

In most variants, it's possible to capture all the stones, but not easy.

In the variants where you sow to the capture pit at each end, it's virtually impossible as your opponent must sow at least the starting number of pebbles worth into their own capture (if your capture is at the end of your side) or you must sow theirs (if your capture is at the beginning of your side).

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You are correct, I missed to write down capture rules. But they are, last pebble in empty pit = collection opponents pebbles in opposing pit + your last pebble; last pebble in kalaha (collecting pit?) = another turn. End game condition = no pebbles in your own pits => collect all pebbles in play. On point number 3: I wrote that down a bit dumb, but it is 6 pits per player = 12 pits times 4 pebbles = 48 pebbles. – Major Byte Sep 6 '11 at 20:30
Is your board O::::::O shaped? if so, do you sow to the large ones on the end as you go around? If so, and the pit you collect to and score is the one you sow to first, then it's not possible to capture all stones yourself, as your opponent must sow to it for you to hit an empty on their side that captures from pits 1-4 of your side. I've never played the capture opposite on last stone in empty pit variants, but the issue is that you can't capture them all before he has to drop one... If they play pit 8, that sows 9-12, leaves 8 open. Then 7, sows 8-11. then 8 again, sows 9, next hits capture. – aramis Sep 7 '11 at 7:19
It is shaped like that, and I realized something yesterday night when I wrote the endgame condition. I think it's not down to empty pits for collecting, but emptying your pits as soon as possible so you achieve end game condition. The second part of the requirement is a unskilled opponent of course. Like my son for instance, he always starts from his left side pit, and as long as I am able to put a pebble there he will handle that one first. I will give it a go this evening and see what I can achieve. – Major Byte Sep 7 '11 at 7:53
I accepted yours as the answer, as my question was incomplete from the beginning and your answer covers more rule-sets – Major Byte Sep 9 '11 at 12:55
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I tried it out just now, with the following rules:

  1. Sow your last pebble to the capture pit: get another turn
  2. Sow your last pebble in an empty pit that's your, collect opposite, opponents pit plus your own last pebble sown.
  3. If you have no pebbles left, the game ends and you gain all of the opponents pebbles in he has left (of course not the ones in his capture pit).

Requirements

  • I get to start.
  • A stupid player, so this will be a proof that it is possible, but it will probably never happen in real life.

boardlayout (I play 1 to 6)

 12         7
O : : : : : : O <-- my capture pit 
  1         6
  1. I start by playing pit 4 --> gives me an extra turn, in which I play pit 6. Opponent plays pit 8
  2. I play pit 5, which gives me an extra turn, in which I play pit 4. Opponent plays pit 7
  3. I play pit 1. Opponent plays pit 8.
  4. I play pit 2. Opponent plays pit 7.
  5. I play pit 3. Opponent plays pit 8.
  6. I play pit 2. Opponent plays pit 7.
  7. I play pit 1 which leaves me with no pebbles left on my side, fulfilling the end game conditions, the opponent has not capture any pebble, so it get all of his pebbles, which gives me 48 pebbles, Hurray!

But I wouldn't call that outwitting my opponent.

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