How many ways to draw consecutive fibonacci numbers from deck of cards In a deck of cards there are 4 suits of 13 cards each.  If the face value of the aces is defined as 1 and the jack, queen, and king are 11, 12, and 13 respectively, then:
1)
What is the probability of drawing 2 cards from the deck whose face values add up to 13?
What is the probability of drawing 3 cards whose face values add up to 13?
Is there a way to generalize this to $k$ cards adding up to $n$?
2)
What is the probability of drawing 3 cards that are 3 consecutive Fibonacci numbers?
What is the probability of drawing 4 cards that are 4 consecutive Fibonacci numbers?
(For eg.: ace, ace, 2, 3 or 3, 5, 8, 13.  But the order in which they are drawn is not important!)
Is there a way to generalize this to $k$ cards that are $k$ consecutive Fibonacci numbers?
 A: For the first part, you are just picking two cards and there is not much to it.  You have in this case, 6 different ways to produce 13;
$$(1,Q),(2,J),(3,10),(4,9),(5,8),(6,7)$$
Since $$(1,Q)=(Q,1)$$we have 2 ways of drawing each of them.  There are 4 cards of each number/face, so the probability of drawing two cards with replacement that add up to 13 is
$$6\cdot{2}\cdot{\frac{4\cdot{4}}{52\cdot{52}}}=.071$$
Without replacement it is just
$$6\cdot{2}\cdot{\frac{4\cdot{4}}{52\cdot{51}}}=.0724$$
For 3 cards it's a little more work, but it is the same technique.  THe ways of having 3 cards add up to 13 are as follows;
$$(1,1,J),(1,2,10),(1,3,9),(1,4,8),(1,5,7),(1,6,6)$$
$$(2,2,9),(2,3,8),(2,4,7),(2,5,6)$$
$$(3,3,7),(3,4,6),(3,5,5)$$
$$(4,4,5)$$
Now instead of two different ways for each 3-tuple, we have a couple of choices:  For the 3-tuples with 3 distinct digits, there are 3! different ways of selecting them, and for the 3 tuples with 2 distinct digits (i.e., $(1,1,J)$), there are $\frac{3!}{2!\cdot{1!}}=3$ ways of selecting.
So there are 8 different 3-tuples with 3 distinct digits and 6 different 3-tuples with 2 distinct digits so the probability of drawing 3 cards with replacement that sum to 13 is
$$8\cdot{6}\cdot{\frac{4^3}{52^3}}+6\cdot{3}\cdot{\frac{4^2\cdot{3}}{52^3}}=.028$$
Without replacement it is just
$$8\cdot{6}\cdot{\frac{4^3}{52\cdot{51}\cdot{50}}}+6\cdot{3}\cdot{\frac{4^2\cdot{3}}{52\cdot{51}\cdot{50}}}=.0297$$
Use the same general principles to work out your consecutive Fibonacci sequences.
A: A generalization for the second part of your question; consecutive fibonacci numbers:
As above we have k-tuples that are unordered; so for example, 4 and 5 consecutive fibonacci numbers can be
$$(2,3,5,8)$$and$$(1,1,2,3,5)$$
There will always be the k-tuple that has $(1,1,...)$ in it as well.  We also know that the maximum number k can be is 7, since
$$(1,1,2,3,5,8,K)$$is the longest sequence of fibonacci numbers we can draw from a deck of cards
In general, for each k-tuple, there are $(7-k)$ k-tuples with $k$ distinct digits and one k-tuple with $(k-1)$ distinct digits (the k-tuple with two 1's).  There are also $k!$ different ways of ordering the k-tuples and $(k-1)!$ ways of ordering the single $(1,1,...)$ k-tuple.  Thus, we can generalize the probability as such;
$$P(\text{sequence of k fibonacci numbers})=(7-k)\cdot{k!}\cdot\frac{4^k}{P(52,k)}+(k-1)!\cdot{\frac{4^{(k-1)}\cdot3}{P(52,k)}}$$
$$=\frac{(k-1)!\cdot{4^{k-1}}}{P(52,k)}\cdot[(3+28k-4k^2)]$$
Where $P(n,k)$ represents the permutation $\frac{n!}{(n-k)!}$
