In a deck of cards, do a queen and a non-queen have equal chances of being the card that follows the first queen?

Question

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.

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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.

Answer 1

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 $$\frac {51!}{52!}=\frac 1{52}$$

Answer 2

Here’s another perspective, by far not as elegant as the one in lulu’s answer, but perhaps it throws some more light on what misleads our intuition on this nice problem.

A quantitative form of the argument for the wrong intuition that the jack should be more likely would be: The probability that $Q\heartsuit$ has already been drawn after the first queen is drawn is $\frac14$, whereas the probability that $J\spadesuit$ has already been drawn is $\frac15$ (since that card and the four queens are all equally likely to be drawn first). Thus it’s more likely that $J\spadesuit$ is still in the deck than that $Q\heartsuit$ is still in the deck.

The mistake here is that these are just the marginal probabilities, but these events correlate with how many cards are still in the deck, and while $J\spadesuit$ is more likely to still be in the deck, in that case the deck is more likely to still contain more cards, which makes it less likely that $J\spadesuit$ will be drawn next.

The probability that, after the first queen is drawn, $Q\heartsuit$ has been drawn and there are $k$ cards left is

$$ \mathsf{Pr}(Q\heartsuit\text{ drawn}\land K=k)=\frac14\frac{\binom k3}{\binom{52}4}\;, $$

since there are $\binom{52}4$ ways to choose where the $4$ queens are, and for the event to occur $Q\heartsuit$ has to be in a particular place and the remaining $3$ queens must be among the remaining $k$ cards.

The probability that, after the first queen is drawn, $J\spadesuit$ has been drawn and there are $k$ cards left is

$$ \mathsf{Pr}(J\spadesuit\text{ drawn}\land K=k)=\frac{51-k}{48}\frac{\binom k3}{\binom{52}4}\;, $$

since one queen must be in a particular place, the other three must be among the remaining $k$ cards and $J\spadesuit$ must be among the $51-k$ cards drawn before the first queen, and there are $48$ places left where it could be.

Summing these over $k$ yields $\frac14$ and $\frac15$, respectively, as it must. But to get the probability that $Q\heartsuit$ or $J\spadesuit$ is drawn next, we need to sum the probability that that card hasn’t been drawn times $\frac1k$:

\begin{eqnarray} \mathsf{Pr}(Q\heartsuit\text{ is next}) &=& \sum_{k=3}^{51}\mathsf{Pr}(Q\heartsuit\text{ not drawn}\land K=k)\cdot\frac1k \\ &=& \sum_{k=3}^{51}\left(1-\frac14\right)\frac{\binom k3}{\binom{52}4}\cdot\frac1k \\ &=& \frac1{52} \end{eqnarray}

and

\begin{eqnarray} \mathsf{Pr}(J\spadesuit\text{ is next}) &=& \sum_{k=3}^{51}\mathsf{Pr}(J\spadesuit\text{ not drawn}\land K=k)\cdot\frac1k \\ &=& \sum_{k=3}^{51}\left(1-\frac{51-k}{48}\right)\frac{\binom k3}{\binom{52}4}\cdot\frac1k \\ &=& \frac1{52}\;. \end{eqnarray}

Here’s a Wolfram|Alpha plot of $\mathsf{Pr}(Q\heartsuit\text{ not drawn}\land K=k)$ and $\mathsf{Pr}(J\spadesuit\text{ not drawn}\land K=k)$ over $k$. You can see that the area under the curve for $J\spadesuit$ is a bit larger, but more of that area is at high $k$, whereas the probability for $Q\heartsuit$ is higher than the one for $J\spadesuit$ for $k\lt39$. The expected value of $k$ given that $Q\heartsuit$ or $J\spadesuit$ is still in the deck is

\begin{eqnarray} \mathsf E[K\mid Q\heartsuit\text{ not drawn}] &=& \frac43\sum_{k=3}^{51}\mathsf{Pr}(Q\heartsuit\text{ not drawn}\land K=k)\cdot k \\ &=& \frac43\sum_{k=3}^{51}\left(1-\frac14\right)\frac{\binom k3}{\binom{52}4}\cdot k \\ &=& \frac{207}5 \\ &=& 41.4 \end{eqnarray}

and

\begin{eqnarray} \mathsf E[K\mid J\spadesuit\text{ not drawn}] &=& \frac54\sum_{k=3}^{51}\mathsf{Pr}(J\spadesuit\text{ not drawn}\land K=k)\cdot k \\ &=& \frac54\sum_{k=3}^{51}\left(1-\frac{51-k}{48}\right)\frac{\binom k3}{\binom{52}4}\cdot k \\ &=&\frac{259}6 \\ &=&43.1\overline6\;, \end{eqnarray}

respectively, not that much of a difference, but what matters is the probabilities for small values of $k$, where it’s most likely to actually draw the card next if it’s still there, and since the ratio of the two probabilities is $\frac{k-3}{36}$, the queen has a considerable advantage at small $k$. This, of course, goes back to what’s already been discussed in comments, that $J\spadesuit$ can be gone due to having been drawn as any of the cards before the first queen, whereas $Q\heartsuit$ can only be gone due to being the first queen, which always has the same probability of $\frac14$ no matter how many cards have already been drawn.

Answer 3

Edited: We know absolutely nothing about the arrangement. Therefore each card is as likely to be any card, including the card after the first queen.