Updated 24 Feb 2005
I called this game is 'pakau' because it don't has a exact name. So, I used the word 'Pakau' from Hokkien named to this game.
The game started with 52 shuffled cards. Player deals card one by one. Each card contains the value itself. For card 'J', 'Q' and 'K' are value 10 each. Condition of the game is sum of three-pairs card can divided by 10.
[1 2 7], [2 J 8], [J 10 Q] can divided by 10.
For three-pairs pakau, there has 3 situation, which is
situation #1: ___XXX
situation #2: X___XX
situation #3: XX___X
XXX means three-pair cards can divided by 10.
___ means some cards you dealed.
If one of that situations occurs, that three cards MAY excerpt out(depend on u).
Winning Condition: Remaining 1 card.
How about four-pairs pakau?
For four-pairs pakau, condition is same. But it need sum of four cards can divided by 10.
situation #1: ___XXXX
situation #2: X___XXX
situation #3: XX___XX
situation #4: XXX___X
Winning condition for four-pairs pakau is no card remaining.
Here,i publish version for three-pairs pakau(pakau_gui) and four-pairs pakau(pakau_gui2).
Note: After you played this game, you may found out that four-pairs pakau is quite easier than three-pairs pakau to achieve winning. Why? Because of the combinations of cards so that theirs sum can be divided by 10. So for two-pairs pakau ,i think it is impossible to achieve winning.
(Thanks to my father for teaching me this game when i still a child.)
Chee Lam Ng (2023). Pakau (https://www.mathworks.com/matlabcentral/fileexchange/6971-pakau), MATLAB Central File Exchange. Retrieved .
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