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Rummy Luck (Posted on 2005-01-07) Difficulty: 4 of 5
In the card game of Rummy, all players start with the same number of cards and the aim is to fill your hand such that all cards are in exactly one meld. Each individual meld is composed of 3 or 4 cards and can each can be completed two ways: cards of the same number/court or consecutive cards of the same suit. (This would mean you have a meld of 3 and a meld of 4 in 7 card rummy and 2 melds of 3 and a meld of 4 in 10 card rummy.) Each individual ace can count as higher than a king or lower than a 2, but not both. (This means K, A, 2 is not allowed.)

What are the probabilities of being dealt a winning hand when: (Note that all decks are without jokers)

- Playing seven card rummy with one deck?
- Playing seven-card rummy with two decks?
- Playing ten-card rummy with one deck?
- Playing ten-card rummy with two decks?
- One of the cards was inadvertantly dropped on the floor before dealing for seven-card rummy?

No Solution Yet Submitted by Rob    
Rating: 3.0000 (2 votes)

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Solution For the first case | Comment 3 of 9 |

I'll use some poker terminology, but this is what I see for the first case--seven-card rummy with one deck: 

Two "straight flushes" same suit: (2*(C(9,2)-1))*4 = 280 combinations of cards.
  (The *4 is for the 4 suits, the C(9,2) is for laying out A,2,3,...,Q,K,A, and taking a block of 3 as a straight and a block of 4 as a straight, so calling individual cards I, the 3-straight as 3 and the 4-straight as 4, one example is III3II4II, and we're choosing 2 out of the 9 to be the 3 and the 4. It can't be 3IIIIIII4 as both straights cannot include the A, thus the subtraction of 1. That whole thing is doubled to account for the possibility of the 4-straight being lower down than the 3-straight.)
  Two "straight flushes" of different suits: 11*10*4*3 combinations of cards. (The 11 is the number of face values the 3-straight can begin on; the 10 is the number of face values the 4-straight can begin on; the 4*3 is the number of possible pairs of suits.
  Three-of-a-kind and 4-of-a-kind: 13*4*12 combinations of cards. (13 choices for the 3-of-a-kind denomination times four possible missing suits times 12 choices of denomination for the 4-of-a-kind for each of the choices of the 3-of-a-kind.)
  3-straight-flush and 4-of-a-kind: 11*4*10 combinations (11 choices for denomination beginning the 3-straight times four possible suits times ten choices of denomination for the 4-of-a-kind -- as three denominations are unavailable in a given suit).
  4-straight-flush and 3-of-a-kind exclusive suit: 10*4*13 (10 choices for denomination beginning the 4-straight times four possible suits times 13 choices for 3-of-a-kind).
  4-straight-flush and 3-of-a-kind inclusive suit: 10*4*9*3 (10 choices for denomination beginning the 4-straight times four possible suits times 9 choices for 3-of-a-kind denomination time 3 choices for 3-of-a-kind suit).
  This totals to 280+11*10*4*3+13*4*12+11*4*10+10*4*13+10*4*9*3 = 4,264 combinations, out of the C(52,7) = 133,784,560 possible hands or a probability of about 1 in 31375 or about 0.000031872.

As to the last part, it shouldn't make a difference if a card had been dropped on the floor before dealing.  It's like any other random way of that card not being in the dealt hand.

  Posted by Charlie on 2005-01-07 18:35:43
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