If I draw a card from a pack of cards, then place it back in, shuffle and draw again, what is the chance it is the same card? Let's say I draw the $2$ of hearts from a standard pack of $52$ playing cards, place it back in and shuffle it, what's the chance I would draw it again?
I'm thinking the other cards are now twice as likely to be drawn, so it would be $\dfrac 1{104}$, is my logic correct?
And what if we introduce more cards, say I draw the $2$ of hearts and the $3$ of diamonds, then try to redraw them. How would I calculate this?
 A: Why should the others be more likely to be drawn.  This is the gambler's fallacy.  Assuming you shuffled well, each card has $1/52$ chance to be drawn the second time.  The deck does not remember the first time.  You might as well just name a card, say the $2$ of hearts and try to draw it.
A: Your logic is incorrect.
Let's take a simple example, heads or tails.
You might think the probability for getting tails is very high if you get $50$ heads in a row, but the probability for flipping the coin to get heads is always $50\%$, it will never change.
Summing up, "The coin does not remember what you flipped last time".
(For a fair coin)
Apply this same concept to cards, a probability to draw one specific cards out of a standard ($52$ card deck) is always $\dfrac 1{52}$.
A: Assuming you shuffle well, the fact that you just drew a $2$ of hearts is independent of what you'll draw next. The probability is still $\frac{1}{52}$.
In general, again assuming you shuffle very well:
The probability that you draw the same card twice in a row is $\frac{1}{52}$
The probability that you draw the same card three times in a row is $\left(\frac{1}{52}\right)^2$
$\vdots$
The probability that you draw the same card $n$ times in a row is $\left(\frac{1}{52}\right)^{n-1}$
since the first card can be anything with probability $\frac{52}{52}$ but then the next cards must be the same as the initial drawing.
A: The cards in the deck have no memory of what happened before the current draw.
So when you get a two of hearts the first time and put it back the chance that you draw it again is $1/52$.
If you draw the two of hearts and the three of diamonds (replacing in between or not) and then draw two cards with replacement the chance that you get them back in that order is $(1/52)^2 = 1/2704$. If you're happy with either order, double that.
A: Note that if that card now has a chance of $\frac{1}{104}$ of being drawn, while the other $51$ cards still have a chance of $\frac{1}{52}$ of being drawn, then the chance of the card being drawn from the deck of $52$ cards is one of those $52$ cards is $\frac{103}{104}$.   Hmmm, something is wrong ...
A: Thanks to idk making me think simply with a coin example I've calculated myself what I needed.
What I mean is let's say I ask what the probability of drawing a $2$ of hearts, shuffling the deck, then drawing it again is. I can use a simple example of card $A$, $B$, $C$. Let's ask the chance of drawing card $A$ twice. Every time we reshuffle we just expand the tree, so $A$ branches to make $AA$, $AB$, $AC$ and so on for the others, giving $9$ possibilities. So the chance of drawing $AA$ from the start is $\dfrac 19$.
And with two draws, say $A+B$, we will just change the tree to $A+A$, $A+B$, $A+C$, $B+A$ etc, or if we don't care about the order of cards then it will just be halved.
