What is the probability that the last card is ace if we are dealt 10 cards from a well-shuffled deck? (Doubt in the solution given ) This is a similar question been asked before long back in which the reasoning behind the solution was given as following :
"As I am dealt the 10 cards, I could just set aside the first 9 face down, and then only look at the 10th card and not look at the first 9, and that would be equivalent. And the 10th card is random from the deck just like the 1st card is random from the deck. "
Progress:
I am not getting the right intuition behind the above reasoning , what if the 9 cards we set aside has all the aces , or say 1 ace or 2 aces and so on- why isn't this fact impacting the probability of the 10th card to be an ace? Even our sample space is affected when we are talking about 10 cards , contrary to the problem of choosing an ace from 52 cards
 A: I am going to show you a slightly easier example of this phenomenon. I know this issue was puzzling to my students when I taught probability a few years ago.
The probability of drawing the $3\diamondsuit$ from a complete deck is, of course, $1/52$. If I first draw a card and show it to you (say it's the ace of spades) and then ask you to draw, the probability of drawing the $3\diamondsuit$ will of course be $1/51$.
Now here is the kicker. What is the probability of drawing the $3\diamondsuit$ if someone has removed a card at random from the deck? It is still $1/52$. We'll use Bayes' Formula to verify this: Let $E$ be the event that we draw the $3\diamondsuit$ and let $F$ be the event that the card removed at random was the $3\diamondsuit$. We have
$$P(E) = P(E|F)P(F) + P(E|F^c)P(F^c) = 0\cdot\frac1{52} + \frac1{51}\cdot\frac{51}{52} = \frac1{52}.$$
As your discussion suggests, the intuition is that if we pull out two cards at random, we may as well do either one first and the results are the same.
A: There are several good answers here. I offer another, based on this part of your question

What if the 9 cards we set aside has all the aces , or say 1 ace or 2
aces and so on- why isn't this fact impacting the probability of the
10th card to be an ace?

Indeed, the fact that the first $9$ cards contained two aces would impact the probability - if you knew that fact. But you don't. Probability calculations are always based on the knowledge you have. If you want to take into account how another condition would affect those calculations you need the probability of that other condition. That's what conditional probability and Bayes' rule are for.
A: Remark
Let $E$ be the event "the 10th card is an ace".  Let $A_k$ be the event "there are $k$ aces in the first 9 cards".  Then
$$
P(E) = P(A_0)P(E|A_0)+ P(A_1)P(E|A_1)+ 
P(A_2)P(E|A_2)+ P(A_3)P(E|A_3)+ P(A_4)P(E|A_4)
\tag1$$
You can compute these if you want to.  You are correct that $P(E|A_k)$ is not equal to $1/13$ in most cases.  But this combination is $1/13$.  I can tell that from the reasoning given in the OP, so I do not need to do this computation.  But perhaps you should do the complutation to convince yourself.
$$
P(E|A_0) = 4/43 > 1/13\\
P(E|A_1) = 3/43 < 1/13\\
P(E|A_2) = 2/43 < 1/13\\
P(E|A_3) = 1/43 < 1/13\\
P(E|A_4) = 0/43 < 1/13.
$$
The weighted average shown in $(1)$ is exactly $1/13$.
A: Another way of looking at it:
Yes, if there are aces in the first 9, then the probability of the 10th being an ace is reduced.
But if there are no aces in the first 9, then the probability of the 10th being an ace is increased.
Overall, those effects balance, so that the chance of the 10th being an ace is exactly what you'd expect if you knew nothing about those first 9. (Calculating this exactly is hard, but we know it from the simple analysis you quote.)
