How adding a joker makes events dependent? I have a question about probability. Imagine we have a deck of 40 cards with 4 different suits of ten cards. Define two events:

$A \equiv $ We take a card and it's an ace $\Rightarrow P(A)=\frac{1}{10}$

$B \equiv $ We take a card and it's a spade $\Rightarrow P(B)=\frac{1}{4}$

$P(A\cap B) = \frac{1}{40}$

We can see that, since we return the card each time, these events are independent:

$P(A\cap B) = P(A) \cdot P(B)$

If we now add a joker to the deck, which can take any card value, then:

$P(A) = \frac{5}{41}$

$P(B) = \frac{11}{41}$

$P(A\cap B) = \frac{2}{41}$

And we can see that now the events aren't independent!

$P(A\cap B) \ne P(A)\cdot P(B)$

My question is, why does adding the joker make the events independent? How is it different from just having one more ace of spades? I mean, before adding the joker we already had a card that belonged to both events (the ace of spades)...

Thanks!
Note:  the rules aren't entirely clear.  Is the Joker always interpreted as $A\spadesuit$?  If not, what rules determine how it is interpreted? 

To be clear:   If you added a second $A\spadesuit$ then the events aren't independent either.  You'd get $$P(A)=\frac 5{41}\quad P(B)=\frac {11}{41}\quad P(A\cap B)=\frac 2{41}$$ just as before. 

Indeed, having doubled the $A\spadesuit$, either directly or by the Joker, you make it so that drawing an Ace is evidence that the card is a spade, and drawing a spade is evidence that it is an Ace.

Specifically,  before you draw anything, the probability that a random draw will be a spade is $\frac {11}{41}=0.268$.  If I tell you that you have drawn an ace, however, the probability that it is a spade is now $\frac 25=.4$  so drawing an ace is strong evidence that you have drawn a spade.
We take a card and its an ace


This is not a description of an event to which you can assign probability.
What is the probability I will draw a card today and it will be an ace?
Maybe I will not draw any cards today. How do you compute the probability of that?

Now introduce a second description, "We take a card and its a spades,"
and the situation is even more murky.
Did we put the first card back and then draw another? Or are the two sentences talking about things that would occur simultaneously, that is we draw just one card from the deck one time, and look at that single card, which might be an ace and might be a spade.

A better way to describe probabilities is to set up exactly the experiment/trial (or experiments/trials) that we will perform, and then specify the results as events.
So either there is a single trial ("shuffle these $40$ cards, draw one, and look at it") or two trials ("shuffle these $40$ cards, draw one, look at it, then shuffle the cards again, draw a card, and look at it").

For a double trial, with one joker in the deck, there are not just $41$ possible equally-likely, distinguishable outcomes; there are $41\times41$ of those outcomes.
For each card we could draw the first time there are $41$ possible cards we could draw the second time. The number of ways to draw an ace the first time and a spade the second time is $5\times 11$ ways. So if we actually make two draws then the events are independent.