So Many Cards!

Despite all of the complications of the problem, it is clear that there could not logically be any tend towards one specific card instead of another, and so by the end of the process the probability for each card would still be uniform. Thus the probability should be $\frac{4}{52} = \frac{1}{13}$ , and the answer is $1 + 13 = \boxed{14}$ .

However, if this seems too easy, the following could be a more satisfying proof. Let's keep track of just one specific card rather than four.

The probability that Card $\textbf{A}$ will be taken from Deck $1$ and moved to Deck $2$ is $\frac{1}{52}$ , as there is one desired event from 52 possible events. Now the probability that Card $\textbf{A}$ will be taken from Deck $2$ and moved to Deck $3$ is $\frac{1}{52} \cdot \frac{2}{53} + \frac{51}{52} \cdot \frac{1}{53}$ . This takes the sum of the probabilities that $\textbf{A}$ is chosen, whether it is in Deck $2$ or another card is. This simplifies to $\frac{1}{52}$ !

Iterated 1374 times, clearly the probability will remain $\frac{1}{52}$ for an individual card, and for four cards the probability is $\frac{4}{52} = \frac{1}{13}$ , and so the answer is $1 + 13 = \boxed{14}$

For any given deck of cards, we find the probability that an ace is picked.

For the first deck, it is easy. It is simply $\dfrac{4}{52}$

For the second deck, it is slightly more involved. The probability that an ace is picked depends on if an ace was picked from the 1st deck and placed into the second deck.

Thus, $P(\text{ace picked from second deck}) = P(\text{ace was picked in the first deck}) \times P(\text{random card from second deck is ace}) + P(\text{ace was not picked in the first deck}) \times P(\text{random card from second deck is ace})$

This is equal to $\dfrac{4}{52} \times \dfrac{5}{53} + \left( 1- \dfrac{4}{52} \right) \times \dfrac{4}{53}$

Similarly, for the third deck, the probability that an ace is picked depends on if an ace was picked from the 2nd deck.

Thus, $P(\text{ace picked from third deck}) = P(\text{ace was picked in the second deck}) \times P(\text{random card from third deck is an ace}) + P(\text{ace was not picked in the second deck}) \times P(\text{random card from third deck is an ace})$

By enumerating the probabilities for the first two decks we notice a pattern. So the probability that a random card from the $1374^{\text{th}}$ deck is an ace depends on the previous deck.

Thus, in general, the probability that an ace is picked from a deck $x$ is picked is

$p(x) = \left( p(x-1) \times \dfrac{5}{53} \right) + \left( \left(1 - p(x-1) \right) \times \dfrac{4}{53} \right)$ for $x > 2$ and

$p(2) = \dfrac{4}{52} \times \dfrac{5}{53} + \left( 1- \dfrac{4}{52} \right) \times \dfrac{4}{53}$

By solving this linear recurrence, we get $p(x) = \dfrac{1}{13}$ . (I did it in wolfram alpha due to laziness)

RSolve[{a[n] == (5/53) *a[n - 1] + (4/53)*(1 - a[n-1]), a[2] == 1/13}, a[n], n]

Thus, $p(1374) = \dfrac{1}{13}$

and we get our answer $1 + 13 = \boxed{14}$

Let $p_n$ denote the probability that the card selected from the $n$ th deck is an ace. Then we have $p_1=\frac4{52}=\frac{1}{13}\hspace{10pt} p_n=p_{n-1}\cdot\frac5{53}+(1-p_{n-1})\cdot\frac4{53}=\frac1{53}p_{n-1}+\frac4{53}$ Solving the difference equation we have $p_n=\alpha\left(\frac1{53}\right)^n+\beta$ . Substituting back and using $p_1=\frac{1}{13}$ we have $\beta=\frac{1}{13}$ , so $\alpha=0$ and $p_n=\frac{1}{13}$ for all $n$ .
So the required sum is $1+13=14$ .

The probability of choosing an ace out of the first deck is 4/52 and (assuming you draw an ace from the first deck and each consecutive deck) the probability of choosing an ace from the other 1373 decks is (5/53)^1373. Add 4/52 and (5/53)^1373 to get a probability of 1/13. 1 plus 13 = 14.

The probability of an ace being taken from the first deck is 4/52 . Now , there are 53 cards in the second deck and you have to take into consideration that this extra card might be an ace ( this would make 5 aces in the second deck ) , and you might think that its probability is 1/53 , so , the total probability of an ace taken to the third deck is 5/53 . This is wrong because this extra ace doesn't stand the same chances of existing in the second deck in the first place as the other 4 original second-deck aces. So , you have to firstly find its separated probability by multiplying its old probability by 1/53 , then add the result to 4/53 ( probability of the 4 original aces ) . If you kept using this same principle , you'll always get the same probability for infinite number of decks . Here's the general formula : P( 1/53 ) + (4/53) ( where P is the old probability of an ace in the previous deck )

A simpler way to explain this (although people have already said this in mathematical terms) is that when the extra card is added to the second deck (and all subsequent decks) it does not change the probability of drawing an ace. In plain English, you have a 4/52 chance (on your first draw) of increasing your chances of drawing an ace from the second deck substantially, but a 48/52 chance of slightly reducing the odds of drawing an ace. The calculations show that these probabilities cancel out - i.e. considering all possible outcomes, the probability of drawing an ace from the second deck is still 4/52. This property carries through every draw, and thus from the perspective of the complete sample space of every possible way you can draw, the odds for the final deck do not change.

Therefore, the probability of drawing an ace from the final deck is the same as from the first deck, 4/52 or 1/13.

4/52

Its pretty simple. There are 4 aces in a deck of 52 cards. and probablity of selecting an ace is 4/52 which on simplifying given 1/13, where both 1 and 13 are co-primes. So a+b = 1+13 = 14. Thats the answer

In the last pack there are 52 cards out of which 4 are aces. So the probability of picking an ace is 1/13. so a+b =14

Since cards are always randomly selected, the probability of any face/number is equal. So the probability is 4/52, or 1/13, so the answer is 14.

Seems a bit trivial!

There are 13 different ranks, and the ace is one of them. The probability of choosing an ace is no different from any other rank and hence is it 1/13. 1+13 = 14.

Let $P(k)$ the probability that the last card taken is $k$ . By bijection we have $P(2) = P(3) = P(4) = ... = P(Jack) = P(Queen) = P(King) = P(Ace)$ . Therefore we have $P(Ace) = \frac{1}{13} = \frac{a}{b} <=> a + b = 14$ .