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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

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    $\begingroup$ What applies to $10$th card applies to the $2$nd card too. What is the probability that the second card is an Ace? $\frac{4}{52} \cdot \frac{3}{51} + \frac{48}{52} \cdot \frac{4}{51} = \frac{4}{52}$ which is the probability of the first card being an Ace. $\endgroup$
    – Math Lover
    Commented Oct 2, 2021 at 18:12
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    $\begingroup$ If you compute all those "what if" cases and add their probabilities together, you get the answer given. But that is a lot of work. $\endgroup$
    – GEdgar
    Commented Oct 2, 2021 at 18:12
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    $\begingroup$ If you were to put the first 9 cards at the bottom of the deck instead, would that help your intuition? How about if you keep them in the deck, but you just pick out the card 10th from the top? Or what if you place all 52 cards face down, and then look at the 10th one? These are functionally the same as far as determining the card drawn is concerned. The only difference is where you put the other cards, which obviously shouldn't impact probability in any way. $\endgroup$ Commented Oct 3, 2021 at 3:55

4 Answers 4

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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.

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  • $\begingroup$ Thanks for the wonderful explanation, how shall I extend the fact which you stated at the end to more than 2 cards too , for example in my question we have 10 cards to be randomly pulled one by one so do you mean to convey that the order wont matter of whichever card is being picked first, second and so on like its fair to assume that 10th card is the first card to be pulled up therefore giving the probability of 4/52 , but this reasoning looks a bit off to me , am I interpreting it correctly ? $\endgroup$
    – Fin27
    Commented Oct 2, 2021 at 19:49
  • $\begingroup$ Yes, that’s right. GEdgar has written out the analogous computation with Bayes’ Formula with the ten cards and different numbers of aces drawn. Write it out carefully for yourself. $\endgroup$ Commented Oct 2, 2021 at 20:03
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    $\begingroup$ For completeness, here's the computation for your case with 10 cards: $$\frac{\binom{48}9}{\binom{52}9}\cdot\frac4{43} + \binom41\cdot\frac{\binom{48}8}{\binom{52}9}\cdot\frac3{43}+\binom42\cdot\frac{\binom{48}7}{\binom{52}9}\cdot\frac2{43}+\binom43\cdot\frac{\binom{48}6}{\binom{52}9}\cdot\frac1{43} = \frac1{13}.$$ $\endgroup$ Commented Oct 3, 2021 at 18:50
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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.

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  • $\begingroup$ So basically till the time we don't know whether the prior events have taken place or not , our probability of an event won't be affected even if the events are dependent on each other like here the outcome of 10th card is dependent on the outcome of the first 9 cards. Till the time we aren't given information explicitly about the occurrence of prior events our probability would be similar to if only our desired event would have occurred independent of all other events. $\endgroup$
    – Fin27
    Commented Oct 2, 2021 at 19:30
  • $\begingroup$ Exactly. That is the essential idea. You should probably stay away from the phrasing "independent of all other events". It helps you but doesn't quite use the term "independent" in its official sense. $\endgroup$ Commented Oct 2, 2021 at 19:48
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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 $1/13$.

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  • $\begingroup$ Thanks , I computed it and yes it did matched... but I am not quite understanding what you are showing through the weights and those inequalities, also shouldn't the weights be 4/43 , 3/43, 2/43, 1/43, and 0/43 $\endgroup$
    – Fin27
    Commented Oct 2, 2021 at 19:14
  • $\begingroup$ @Fin27 What is $P(A_0)$, for example? You should have ratios of binomial coefficients here. $\endgroup$ Commented Oct 2, 2021 at 20:11
  • $\begingroup$ Denominator corrected to $43$. $\endgroup$
    – GEdgar
    Commented Oct 2, 2021 at 21:07
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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.)

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