MATH 105Q FIVE HAND POKER

1. Q: How many ways are there to draw a royal flush (10JQKA in the same suit.) ?

   A: The only possibilities are 10JQKA of harts, clubs, diamonds and spades. So there are only 4 ``Royal Flushes".
   

2. Q: Find the number of different poker hands with ``4 of a kind"?

   A: There are 13 values, so there are 13 different ways to get a ``four of a kind". In other words there are 4 twos, 4 three's, and so on to 4 kings and 4 aces. The fifth card can then be chosen out of the 48 cards left after taking out the 4 cards needed for the four of a kind. This means that in total, there are $13 \cdot 48 = 624$ different poker hands with a ``four of a kind".
   

3. Q: How many hands are there that contain exactly one 4, one 6, and one Jack (and whatever else)?

   A: The total number of possibilities is
   

   $$C(4,1) \cdot C(4,1) \cdot C(4,1) \cdot C(40,2) = 49920$$
   
   
   where the first $C(4,1)$ stands for your number of choices for a 4, the second $C(4,1)$ stands for your number of possible choices for a 6, and the third $C(4,1)$ stands for your number of possible choices for a Jack. Finally $C(4,2)$ stands for choosing 2 cards out of the remaining 40 (i.e. all cards but fours, sixes and jacks).
   

4. Q: How many different poker hands are there that contain exactly 2 Jacks?

   A: The total number of possibilities is
   

   $$C(4,2) \cdot C(48,3) = 103776$$
   
   
   where $C(4,2)$ stands for choosing 2 jacks out of the 4 available ones and $C(48,3)$ stands for choosing the other 3 cards out of the remaining (anything but a Jack) 48 available ones.
   

5. Q: How many different poker hands are there that contain exactly 3 Jacks?

   A: The total number of possibilities is
   

   $$C(4,3) \cdot C(48,2) = 4512$$
   
   
   where $C(4,3)$ stands for choosing 3 jacks out of the 4 available ones and $C(48,2)$ stands for choosing the other 2 cards out of the remaining (anything but a Jack) 48 available ones.
   

6. Q: How many different poker hands are there that contain at least 3 Jacks?

   A: At least 3 Jacks means either 3 or 4 Jacks, so we just have to add the number of poker hands with 3 Jacks and the number of poker hands with 4 Jacks:
   

   $$C(4,3) \cdot C(48,2) + C(4,4) \cdot C(48,1) = 4560$$
   
   
   
   

7. Q: Find the number of different poker hands with a straight flush (five cards in a sequence in the same suit, but not a royal flush).

   A: Within each suit you have the following options: $A2345$, $23456$, $34567$, $45678$, $56789$, $678910$, $78910J$, $8910JQ$ and $910JQK$. Note that $10JQKA$ is not in the list because that would be a royal flush. So within each suit, there are 9 options. As there are four suits, the total is $4 \cdot 9 = 36$.
   

8. Q: Find the number of different poker hands with 2 Jacks or 3 Jacks.

   A: Just add questions 4 and 5 and get
   

   $$C(4,2) \cdot C(48,3) + C(4,3) \cdot C(48,2) = 108288$$
   
   
   
   

9. Q: Find the number of different poker hands with 2 pairs (and not a ``4 of a kind" or a ``three of a kind").

   A: The total number of possibilities is
   

   $$13 \cdot C(4,2) \cdot 12 \cdot C(4,2) \cdot {1 \over 2} \cdot C(44,1) = 123552$$
   
   
   where $13 \cdot C(4,2)$ stands for the number of choices possible for the first pair, i.e. 2 cards out of 4 possible and this times 13 possible choices for the face value. Then, $12 \cdot C(4,2)$ stands for the number of choices possible for the second pair, i.e. 2 cards out of 4 possible and this times 12 possible choices for the face value (as one has already been chosen and we want to avoid a ``four of a kind"). This has to be divided by 2, because choosing, say a Jack and then a seven gives the same hand as a seven and then a Jack. Finally $C(44,1)$ stands for choosing one card out of the remaining 44 ($52-4-4$ to avoid ``three of a kind").
   

10. Q: Find the number of different poker hands that contain exactly 3 Jacks while the remaining 2 cards do not form a pair.

    A: The total number of ways to do this is
    

    $$C(4,3) \cdot {48 \cdot 44 \over 2!} = 4224$$
    
    
    $C(4,3)$ of course indicates the choice of three Jacks. Then the fourth card can be chosen in 48 ways (everything but a Jack). For the last card, there are 40 options left (not a Jack and not the same face value as the fourth card). $48 \cdot 44$ then has to be divided by the number of ways in which these cards can be permuted, namely $2!$. Note that the order has already been divided out of the pair because we use combination and not permutation.
    

11. Q: In how many ways can you get a pair (and nothing else, i.e. not a hand with 2 pairs, no three of a kind, and no four of a kind)?

    A: The total number of ways in doing this is
    

    $$13 \cdot C(4,2) \cdot {48 \cdot 44 \cdot 40 \over 3!}=1098240\ldots$$
    
    
    Where $13 \cdot C(4,2)$ stands for the total number of ways to choose your two cards that form a pair. Then for the third card, you have 48 choices left, because you want to avoid choosing the same face value as for the pair. The fourth card can then be chosen out of 44 cards, because you want to avoid both the face value of the pair and of the third card. Finally, the final card can only be chosen from 40 cards to avoid the face values of all the previous ones. $48 \cdot 44 \cdot 40$ then has to be divided by the number of ways in which these cards can be permuted, namely $3\!$. Note that the order has already been divided out of the pair because we use combination and not permutation.
    

12. Q: How many ways are there to draw a straight ? A straight consists of 5 following cards of different suits. The easiest way to do this is to calculate all possibilities and then deduct the number of straight flushes and royal flushes ?

    A: Any straight (including straight flushes and royal flushes) can be chosen in
    

    $$10 \cdot 4^5$$
    
    
    ways. 10 indicates the number of ways in which the first face value can be chosen, namely $2,3,4,\ldots ,10$. ``Jack" would already be to high, because this would result in $JQKA2$ which is excluded. Once the face value has been chosen, for the first card there are only 4 options, namely $\heartsuit$, $\clubsuit$, $\diamondsuit$, $\spadesuit$. For the second card there are the same for options and so on for the third, fourth and fifth card. The we still have to deduct the straight flushes and the royal flushes. This gives
    
    $$10 \cdot 4^5 - 36 - 4 = 10200$$
    
    
    
    

Remark: The probability for all of the above can be calculated by dividing the numbers obtained by the total number of ways in which a 5 hand poker can be chosen, i.e. $C(52,5) = 2598960$. The probability for e.g. ``four of a kind" is then

$$p(\hbox{\lq\lq four of a kind''}) = {624 \over 2598960} = 0.000240096\ldots$$