Pai Gow Tiles tie scenario.
March 31st, 2013 at 8:45:17 PM
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I was playing Pai Gow tiles game on the wizard's site.. while waiting for someone. We ended up having a tie on the high hand. We both had 9.
Don't a tie in number go to the person with the higher tile being used? The computers 9 had a gee tile and mine had a Bon tile. My understanding is that I would win on tile rank because my Bon 9 would beat his 9 because any tile paired with a gee tile is automatically the lower ranked tile (I recall being told that gee joon together as a pair unbeatable- but separate as a wild 6/9 makes the pairing automatically lower in rank then any other of same amount (unless my 9 also had the other half of gee joon in which case it would have been a complete tie and the dealer would win all ties).
What did I miss or am I confusing?
BTW Ive come a long way in learning the game- buying my own tile set helped a lot, as well as learning the story behind the tiles- that helped me remember the tile rankings a lot.
Don't a tie in number go to the person with the higher tile being used? The computers 9 had a gee tile and mine had a Bon tile. My understanding is that I would win on tile rank because my Bon 9 would beat his 9 because any tile paired with a gee tile is automatically the lower ranked tile (I recall being told that gee joon together as a pair unbeatable- but separate as a wild 6/9 makes the pairing automatically lower in rank then any other of same amount (unless my 9 also had the other half of gee joon in which case it would have been a complete tie and the dealer would win all ties).
What did I miss or am I confusing?
BTW Ive come a long way in learning the game- buying my own tile set helped a lot, as well as learning the story behind the tiles- that helped me remember the tile rankings a lot.
"Although men flatter themselves with their great actions, they are not so often the result of a great design as of chance." - Francois De La Rochefoucauld
March 31st, 2013 at 9:13:04 PM
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The dealer must have had the Gee with a high 6 (Chong) tile, which beats the Bon 9 (9 with low 4) because Chong outranks Bon.
With a 2-tile hand where exactly one of the tiles is a Gee, that hand's high tile is always the non-Gee tile, because every non-Gee tile outranks an individual (unpaired) Gee tile.
With a 2-tile hand where exactly one of the tiles is a Gee, that hand's high tile is always the non-Gee tile, because every non-Gee tile outranks an individual (unpaired) Gee tile.
March 31st, 2013 at 10:44:41 PM
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simplest way to remember this is that gee joon is the highest pair but the tiles are never used to break a tie.
March 31st, 2013 at 11:20:20 PM
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Ok, I see my mistake then.. I cant remember if it had the chong or not- must have. I was thinking that by making a tile paired with the gee tiles it automatically made the hand lower then an equal point hand- but your saying it would be the higher of the individual tiles involved- so he would use the other tile paired with it (most likely, it would almost certainly have to be chong to win against my nine. Got it.
"Although men flatter themselves with their great actions, they are not so often the result of a great design as of chance." - Francois De La Rochefoucauld
April 1st, 2013 at 12:13:26 AM
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another easy way to think of it is that in the event of a tie hand, the single highest tile among the four tiles that make up the 2 tied hands is used to break the tie. if there are 2 of the same highest tiles, then it goes to the banker.
gee joon is never used to break a tie, but it doesnt affect the other tile in the hand. thus an 8 with a teen and a gee joon would beat an 8 with a day and a chong.
gee joon is never used to break a tie, but it doesnt affect the other tile in the hand. thus an 8 with a teen and a gee joon would beat an 8 with a day and a chong.
