# 65-card deck consisting of 13 ranks and 5 suits

** I FIGURED OUT 15 out of 16 cases. I don't understand the last case of RUNT. Anyone helps? I recently went to a math event and one person presented a weird card deck, consisting of 13 ranks and 5 suits, yielding a total of 65 cards. I later found out this card deck is here.

What I couldn't figure out is the counts of 5-card poker with super hands (these are hands consisting of cards from all five suits). The probability is easy to figure out, since it is the count divided by the number of ways to get 5 cards ($8,259,888$ in this situation). I worked out the first two, 5-of-a-kind and straight flushes but couldn't go further. Any suggestion is really appreciated.

A few to get you started:

• The number of hands in a Super 4 of a kind would be calculated as:

$${5 \choose 4} \cdot {12 \choose 1} \cdot {13 \choose 1} = 780$$ There are ${5 \choose 4}$ ways to get a four of a kind, 12 choices for the remaining card (it must be in the opposite suit, and cannot be the same type as the previous), and then 13 ways to choose the type of the original card.

• The number of hands in a super straight: Ignoring suits, there are 10 types of straights: Ace to $5$, $2$ to $6$, ..., $10$ to A. Then there are 5 choices for the suit of the first card, 4 choices for the suit of the second card, 3 choices for the suit of the third card, 2 choices for the fourth card, and 1 choice for the suit of the last card. This gives the number of super straights as $$10 \cdot 5! = 1200$$

• For Super full houses, you should count the number of ways to get a super full house consisting of three aces and two kings. This would be ${5 \choose 3} \cdot {2 \choose 2}$. These numbers come about since there are five aces to choose from, and then once the suits have been selected, you must choose kings from the remaining two suits. This means there are ${5 \choose 3} = 10$ ways to get a super full house consisting of 3 aces and two hearts. Now, there are $13$ choices for the first card (the type of card of which you have three) and tweleve choices for the remaining type. This gives:

$${5 \choose 3} \cdot {2 \choose 2} \cdot 13 \cdot 12 = 1560$$

• i got how they get super straight. we have 10 choices for the lowest card. then we just have 4 choices for the 2nd card,.... so in total we have 10x4x3x2x1 = 240 but we can rotate the choice of suite so in total we get 240x5 = 1200 counts – michael May 10 '13 at 2:22
• Could you look at the last case, RUNT? i don't understand this case. could you help me please? – michael May 10 '13 at 4:47
• RUNT seems to be five cards of different suits. $1\cdot \frac {52}{64} \cdot \frac {39}{63} \cdot \frac {26}{62} \cdot \frac{13}{61}$ – Ross Millikan May 10 '13 at 5:00
• @RossMillikan: please look at the count the website gives. i don't know how they can derive that. – michael May 10 '13 at 5:23
• The easiest way to calculate a runt hand is to take all of the possible hands (${65 \choose 5}$, and subtract all the hands you've counted before. The difficulty is that to have a runt hand is quite more than just having 5 cards of different types. You need to ensure they are not straights, flushes, etc. – JavaMan May 10 '13 at 5:29