I encountered the following problem:
From a shuffled deck the cards are flipped one by one until a queen shows up. Then again a card is flipped. Is it more likely that this card is the Jack of Spades or the Queen of Hearts?
I managed to solve this by first finding the probability that the card will be the Jack of Spades under the extra condition that the rank of the Jack of Spades is $n$. If $J_s$ denotes the event that the Jack of Spades is the card and $N$ denotes the rank of the Jack of Spades then:$$P(J_s\mid N=n)=\binom{52-n}3\times\binom{51}4^{-1}$$This because under condition $N=n$ there are $\binom{51}4$ possibilities for the $4$ queens of which $\binom{52-n}3$ are favorable.
Applying the hockey stick identity and the law of total probability we then find: $$P(J_s)=\frac1{52}$$From this it can be deduced easily that also $P(Q_h)=\frac1{52}$ so apparantly the chances are equal. This amazed me because I could not find an explaining symmetry for that.
My question arises from the fact that I am simply not satisfied with this solution and is:
Could someone give me a "nicer" solution that avoids calculations and rests on something like a smart perspective of the case?
I just feel that a nicer and more direct solution can be found.
Thank you for taking notice of my question, which is purely an effort to enrich my intuitions on probability.
Edit
Thank you for comments already, but if you have a real answer then please do not hesitate to provide one. And preferably a complete one. Also: more perspectives will imply a larger enrichment of my intuition.
Thank you in advance.
Best Answer
Take the card you are interested in, be it $Q\heartsuit$, $J\spadesuit$, or something else entirely, call it $X$. Now remove $X$ from the deck and sort the rest of the cards. There are, of course, $51!$ ways to sort the $X-$less deck, and then there is a unique spot in which to insert the $X$ so that it immediately follows the first queen.
In this way, we see that there are exactly $51!$ arrangements of the deck such that $X$ is immediately following the first Queen. As there are $52!$ ways to sort the deck without worrying about $X$, the probability that $X$ is in the desired slot is $$\frac {51!}{52!}=\frac 1{52}$$