poker probability a pack of poker contains 52 cards and we are going to flip through one by one so the probability of the following events are


*

*a king right after the first ace ?

*an ace right after the first ace ?

*the first ace is the 10-th card?

*the probability the next card is the ace of spades, if the  first ace is the 30-th card?

*The probability the next card is the jack of diamonds If the 
first ace is the 30-th card?


I have tried the (1)(2)(3) and not sure is right or not,but no idea with the (4) (5)
(1)$\frac{4}{52}*\frac{4}{51}$ and which is same prob as (2)
(3)$\frac{48}{52}*\frac{47}{51}*\frac{46}{50}*\frac{45}{49}*\frac{44}{48}*\frac{43}{47}*\frac{42}{46}*\frac{41}{45}*\frac{40}{44}*\frac{4}{43}$but any simpler way to write it out?
 A: Let me answer $1$ in detail, as my comment was inaccurate.  
Note:  I am unhappy with this argument, but I don't see a flaw in it. I'd be grateful if someone could either spot the blunder or supply a better argument.
Key Remark:  By symmetry, the probability that the card following the first $A$ has value $i$ is independent of $i$ for all values different from $A$. Call this common value $p$ and let $p_A$ denote the probability that the card following the first $A$ is also an $A$.  Of course we have $$12p+p_A=1\implies p=\frac 1{12}\times\left(1-p_A\right)$$
Let's compute $p_A$.  
If the first $A$ appears in slot $k$ then we get $p_A(k)=\frac 3{52-k}$, where $p_A(k)$ denotes the conditional probability that the card after the first $A$ is also an $A$ conditioned on the event that the first $A$ appears in slot $k$. Now, the probability that the first $A$ appears in slot $k$ is $$\frac {48}{52}\times\frac{47}{51}\times \cdots \times \frac {48-(k-2)}{52-(k-2)}\times \frac 4{52-(k-1)}$$  It follows that $$p_A=\sum_{k=1}^{49}\left(\frac {48}{52}\times\frac{47}{51}\times \cdots \times \frac {48-(k-2)}{52-(k-2)}\times \frac 4{52-(k-1)}\times \frac 3{52-k}\right)$$
Computing that sum numerically we get $p_A=\frac 4{52}$  Which implies that $p=\frac 4{52}$.
Sanity Check:  Suppose we only had a $4$ card deck, $\{A,A,K,K\}$.  Then there are six possible shuffles, namely $AAKK,AKAK,AKKA,KAAK,KAKA,KKAA$.  The card following the first $A$ is $\{A,K,K,A,K,A\}$ so the probability that it is $A$ equals the probability that it is $K$.
A: Here is a bijection on the set of all deck orders:
$$(c_1,c_2,\dots, c_{m-1}, A, c_{m+1},\dots, c_{52} )
\longleftrightarrow (c_{m+1},\dots, c_{52}, A, c_1,c_2,\dots, c_{m-1}),$$
where $A$ is an ace, and none of $c_1,\dots, c_{m-1}$ is an ace. 
Therefore the card following the first ace,
 that is $c_{m+1}$,  is equally likely to be any of the 52 cards, since the top card of a thoroughly shuffled deck has this property.
In other words, 
the statistical distribution of "card following first ace" and "top card" are the same.
A: King right after the first ace?  What you show is the chance of the first two cards being A,K.  It is inevitable that an A comes off sooner or later.
$\frac 4{52}$
2) While it would seem that 2 is not the same as 1, the more I calculate, the more that I think it is. 
$\frac 4{52}$
3)First Ace is the 10th card is correct, but this is cleaner.
$(\frac {48!}{39!}\frac {42!}{52!})4$
4) $(\frac 34)$ the first ace is not the ace of spades.
$(\frac 34)(\frac 1{22})$ the first ace is not the ace of spades, the 31 card is the ace of spades, given that the first 30 are not the ace of spades.
5) The rough approach.  The J of daimonds could be in one of 51 locations, including is space number 31.
However, that there are no aces in the first 29 cards flipped means that there is a greater than fair probability that the 31st card is an A, so it should be slighly less.
$P(\text{No A no J of D in first 29 cards}) = \frac {47!}{(47-29)!} \frac {52!}{(52-29)!}\\
P(\text{No A in first 29 cards}) = \frac {48!}{(48-29)!} \frac {52!}{(52-29)!}$
$P(\text{No A no J of D in first 29 cards|No A in first 29 cards}) = \frac {47!}{(47-29)!} \frac {(48-29)!}{48!} = \frac {19}{48}$
$P(\text{No A no J of D in first 29 cards|No A in first 29 cards})P(\text{J on card 31|A on card 30}) = \frac {19}{48}\frac {1}{22}$
A: 1) King follows First Ace.
Let's not worry about suits or values of the other cards.  Take a pack of four king of hearts, four ace of spades, and forty four jokers.
There are $52!/(4!4!44!)$ equally probable ways to arrange this deck.
Set aside one king, shuffle the remaining cards, stick that king after the first ace.  There are $51!/(3!4!44!)$ ways to do this.
So the probability of the event is $\quad\dfrac 1 {13}$

2) Ace follows First Ace.
As above, so below.  
Set aside one ace.   There are $51!/(4!3!44!)$ ways to sort 4 kings, 3 aces, 44 jokers (and stick the set aside ace behind the first ace).   Coincidentally yielding the same probability of $1/13$.

3) First Ace is tenth card.
Take 4 ace of spades and 48 jokers.  $52!/(4!48!)$ ways to arrange them in total.  Now count ways to arrange 9 jokers, an ace in tenth place, and then all the rest (3 aces, 39 jokers).

4) Ace of spades follows first ace when that is in 30th place.
The deck consists of three ace of hearts, one ace of spades, and 48 jokers.  The deck is shuffled so we have 29 jokers, an ace, in that order, and then the rest (3 aces, 19 jokers) in any order.
If the first ace is spades, it cannot follow itself.   Multiply the probability that it is not, by the conditional probability of the event given that.

5) Jack of Diamonds follows First Ace when that is in thirtieth place.
The deck consists of four ace of hearts, jack of diamond, and 47 jokers, shuffled so we have 29 cards, an aces, and then the rest.  
If the jack is among the 29 cards before first ace, then it cannot follow it.   Multiply the probability that it is not, by the conditional probability of the event given that.
