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Say your playing ,oh, 5 card draw, and two players both get Royal Flushes.........Does it play out as Spades being the best suit, Hearts second, etc. or what? Or is it a tie?

2007-01-26 01:32:38 · 9 answers · asked by Anonymous in Games & Recreation Card Games

9 answers

The odds are better that hell freezes over

2007-01-26 05:02:28 · answer #1 · answered by Craig R 3 · 0 4

The great thing about card games is that there are many different versions. So, though it's true that according to the "official" rules (ie the rules used in most casinos) there are no ranks according to suit, if you and your buddies agree to a ranking of suits in order to determine who wins in cases of ties, then you are certainly able to do just that. Whats matters is that everyone involved in the game agrees to the set of rules.

In any event, here is a listing of hand rankings (though you may already know them by heart).

2007-01-26 09:52:24 · answer #2 · answered by Anonymous · 0 0

Split pot, but the odds of getting a Royal Flush are slim (not impossible), but even more slim (nearly impossible) for TWO players to get it.

2007-01-26 17:28:02 · answer #3 · answered by Poker_Chick 2 · 0 0

It would be a tie and a split pot. The suits are ranked in Bridge, not in Poker.

2007-01-26 12:40:47 · answer #4 · answered by Chris S 5 · 1 0

It would be a tie, also called a split pot. Suits have no rank in poker.

2007-01-26 09:38:06 · answer #5 · answered by LoneStar 6 · 1 0

If that ever happened, it's a rigged game bc the odds are in the quintillions.

But yeah, spades then hearts then diamonds then clubs... for your home-game.

Casino's and what you see on TV would split that.... GAY

2007-01-26 09:41:34 · answer #6 · answered by Anonymous · 0 3

i use to hear spades was higher, but that first guy is probably right

2007-01-26 09:40:38 · answer #7 · answered by ? 4 · 1 0

tie

2007-01-26 09:40:06 · answer #8 · answered by IGH3Rat 5 · 0 0

Split pot, and then i would yell at them for cheating

2007-01-26 09:40:45 · answer #9 · answered by Dave 2 · 0 0

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