thx
2007-02-06 19:17:44 · 10 answers · asked by Punkin 1 in Games & Recreation ➔ Card Games
I think it is because the decks we use today originate from the Tarot deck, which has 70-plus cards. 56 of these were the Minor Arcana or soemthing like that...and they also had 14 ranks in four suits, from Deuce to Ten, Knave, Knight, Queen, King, Ace...in Cups, Swords, Wands and Sticks I think...for some reason the Knight was removed but I have no idea why...and of course the Fool was also added and replaed by the Joker...that makes 52 normal cards plus a few jokers...
Hope this helps...though there may be mathematical explainations for this which I'm not sure about...
2007-02-06 23:20:07 · answer #1 · answered by Anonymous · 0⤊ 0⤋
How many cards are there in an abnormal pack? Actually there are 54 cards in a pack, if you include the 2 Jokers.
2007-02-08 10:40:01 · answer #2 · answered by Poker_Chick 2 · 0⤊ 0⤋
Because there are 14 cards per suit, all cards are used to play various games. There isn't any reason for 52 cards, there just is.
2007-02-06 19:23:51 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Becuz there are nine numbers from two to ten, aces, jacks, kings, and queens. That's 13 different cards for each suit, and there are four suits. 4 x 13 = 52.
2007-02-06 19:23:09 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Simple! If there weren't 52 cards it wouldn't be a normal pack!
Think about it, Einstein
2007-02-06 19:24:23 · answer #5 · answered by mad_mick001 5 · 1⤊ 1⤋
There arent ... theres usually 2 to 4 extra cards - 2 Jokers and / or 2 extras
2007-02-06 21:22:16 · answer #6 · answered by Truth D 4 · 0⤊ 0⤋
54 should be the normal pack as some games are played with the jokers
2007-02-06 20:13:36 · answer #7 · answered by ufo18 4 · 0⤊ 0⤋
Ace through King x 4, do the math.
2007-02-06 19:21:04 · answer #8 · answered by Anonymous · 0⤊ 0⤋
have just counted mine and theres 56 in my pack 2 extra aces and 2 extra kings, wait til my son gets home, the cheater! no wonder he always wins.
2007-02-06 19:59:04 · answer #9 · answered by Anonymous · 0⤊ 0⤋
so
2007-02-06 22:02:09 · answer #10 · answered by Anonymous · 0⤊ 0⤋