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
Best Answer
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.