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A deck of $52$ cards is shuffled, and the cards are turned up one at a time until the first $A$ appears. Show that: $$P(\text{next hand is Ace of spade}) = P(\text{next hand is 2 of club}) = \frac{1}{52}$$

My try: I am not able to understand what they really want. Are they talking about the probabilities when not even a single card is turned(the probability for both cards will be $\frac{1}{52}$) but if we are turning cards one by one, the probability increases with every card which is turned. If one card is turned and it is not one of them then $51$ cards are remaining and probability for each will be $\frac{1}{51}$.

What I am doing wrong here?

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    $\begingroup$ I guess you mean P(next card is Ace of spade) = P(next card is 2 of club) $\endgroup$ – true blue anil Feb 11 '16 at 5:20
  • $\begingroup$ Next card and next hand are the same thing. $\endgroup$ – max Feb 11 '16 at 5:22
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    $\begingroup$ A hand, as in bridge, say, consists of $13$ cards. $\endgroup$ – true blue anil Feb 11 '16 at 5:29
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Shuffling the cards “to randomize them thoroughly” implies that every ordering of the 52 cards is as likely as any other ordering; there are 52! different orderings.
In how many of these orderings does the ace of spades follow immediately after the first ace?
We can count them by removing the ace of spades, ordering the 51 remaining cards, and then putting the ace of spades back immediately after the first ace; there are 51! such orderings.
Therefore the probability that the ace of spades will be found next is (51!)/(52!) = 1/52 .
The same argument yields the same probability 1/52 for finding the two of clubs (or any other card) next.
Surprisingly, each of the 52 cards in the deck is as likely as any other to go into the envelope.

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