Probability: Is the next card the ace of spades or heart two? A card deck with 52 playing cards is well mixed and the cards are indicated successively, until the first ace appears. Is it more likely that the next card is the ace of spades or heart two?
My idea was that the probability of drawing the ace of spades or heart two is the same with $\frac{4}{52}$, but that doesn't seem right. Can anyone help me here?
Thanks in advance.
 A: The probability that the next card following the first ace is the two of hearts is 1/52.  Surprisingly, the probability that it is the ace of spades is also 1/52.
To find the probability that the next card is the ace of spades, imagine that we deal out all 52 cards.  This can be done in $52!$ ways, all of which we assume are equally likely.  We would like to count the arrangements in which the ace of spades immediately follows the first ace dealt. Each ordering of the 52 cards can be produced by first dealing out all the cards except the ace of spades, then inserting the ace of spades into that ordering.  There are $51!$ ways to arrange the first group of cards, and there is only one place to insert the ace of spades so it immediately follows the first ace.  So there are $51!$ arrangements in which the aces of spades follows the first ace.  Therefore the probability that the card following the first ace is the ace of spades is 
$$\frac{51!}{52!} = \frac{1}{52}$$
By the same argument, simply replacing the ace of spades with the two of hearts above, the probability that the card following the first ace is the two of hearts is also $1/52$.
A: This might shed some light.
You can get a well mixed deck of cards like this. 
Place a card. Then there are two places for the second card to be placed: at the left of the first card or at the right. Let a coin (or a virtual fair two-sided die) decide about that. Then there $3$ places for the third card: at the left, in the middle or at the right of the two cards that were allready placed. Going on like that we come to a point that $51$ cards are placed and $52$ places equiprobable places are there for the last card. There is only one spot for it such that - if it is placed there - we can say that this last placed card is the next of the first ace. This no matter what card it is: heart two, ace of spades or some other card.
Conclusion: in a well mixed deck of cards every card has probability $\frac1{52}$ to be the next card after the first ace.
A: It is more likely that the next card is heart two, and the probability of heart two versus ace of spades is 4:3.
Let $A$ denotes ace of spades, $B$ denotes heart two, and $a_1, a_2, a_3$ denote ace of other 3 colors. There are $3 \times 3!$ ways to let $A$ be after the first $a_i$:
$\{a_1, A, ...\}$$\{a_2, A, ...\}$$\{a_3, A, ...\}$
and there are $4 \times 3!$ ways to let B after the first $A$ or $a_i$:
$\{a_1, B, ...\}$$\{a_2, B, ...\}$$\{a_3, B, ...\}$$\{A, B, ...\}$
Till now, we can get the versus of probability between the two cards. But for the exact probability, we need more computation. For all the 52 cards, there are $52!$ ways to permute them. For the other 47 cards (exclude $A,B,a_1,a_2,a_3$), there are $47!$ ways to permute them. So the probability of $A$ after the first $a_i$ is 
$\frac{47! \times 3 \times 3! \times C}{52!}$
and the probability of $B$ after the first $A$ or $a_i$ is
$\frac{47! \times 4 \times 3! \times C}{52!}$ 
where $C$ is number of ways to insert the 5 cards (have been permuted) into 48 slots of 47 cards, requiring the first 2 cards to be in the same slot. One solution of $C$ is to consider 1,2,3,4 slots separately.
1 slot: $C_{48}^1 \times C_3^0$, like #####
2 slots: $C_{48}^2 \times C_3^1$, like ##|###, ###|##, ####|#
3 slots: $C_{48}^3 \times C_3^2$, like ##|#|##, ##|##|#, ###|#|#
4 slots: $C_{48}^4 \times C_3^3$, like ##|#|#|#
Sum them all, we get 249900 ways. So the probability of ace of spades next to the first ace is $\frac{3}{208}$, and heart two next to the first ace is $\frac{1}{52}$.
