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According to many sources online, probability of no kings, queens and jacks before first ace = $\frac{1}{4}$, which makes sense because it's the same as answering which of the four honor cards is first. What confuses me is this post: Expected number of cards you should turn before finding an ace

They claim, as I've seen in some other textbook, that each of the $48$ non-ace cards has probability $\frac{1}{5}$ of being before the first ace, which makes sense because one can think of the four aces as splitting up the deck into $5$ buckets, $* A_1 * A_2 * A_3 * A_4 *$, where each $*$ represents a bucket and $A_1, A_2, A_3$ and $A_4$ are the $4$ aces.

Let $K_1$ be the event that King $1$ is after $A_1$ (the first ace), $K_2, K_3, K_4, Q_1, \cdots Q_4$, $J_1 \cdots J_4$ be defined the same for the other kings, queens and jacks. Then, by the logic in the earlier stackexchange post, $P(K_1) = \frac{4}{5}$ each.

Because they are independent, $P(K_1 \cap K_2 \cap K_3 \cdots \cap J_3 \cap J_4) = \left(\frac{4}{5}\right)^{12}?$ But that evidentally doesn't agree with the answer of $\frac{1}{4}$ - where could my logic have gone?

Edited: mistakenly put it as $\left(\frac{4}{5}\right)^{16}$ instead of $\left(\frac{4}{5}\right)^{12}$

19021605
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Nimaye Garodia
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  • See this question for a discussion of various approaches to this sort of question. Linearity of Expectation is an especially easy method, and avoids questions of independence (which you have inappropriately ignored in your method). – lulu Aug 20 '24 at 19:24
  • Does $K_1$ represent the first king? – David K Aug 20 '24 at 19:46

3 Answers3

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$P(K_1)=\frac{4}{5}$ is correct, but $P(K1\cap K2)$ is actually $\frac{4}{5}.\frac{5}{6}$ as an extra 'bucket' is created. $$\therefore P(K_1 \cap K_2 \cap K_3 \cdots \cap J_3 \cap J_4)= \frac{4}{5}.\frac{5}{6} \cdots \frac{15}{16} = \frac{4}{16} = \frac{1}{4} $$

Rishit Garg
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  • Quick question though, why should the extra bucket matter? All that matters is that the second King lies after the first ace right? – Nimaye Garodia Aug 20 '24 at 19:30
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    Correct. But now, the probability that the second king comes $\textit{before}$ the first ace changes. It is equivalent to the probability that one of six cards comes first ($=\frac{1}{6}$) as opposed to earlier, when it was $\frac{1}{5}$. – Rishit Garg Aug 20 '24 at 19:50
  • I think this ought to make it clear. $P(K_2 \cap K_1)=P(K_2 |K_1).P(K_1)$, where $P(K_2 |K_1)$ is the conditional probability of event $K_2$, i.e, the probability that $K_2$ will occur given the knowledge that event $K_1$ has already occurred. Thus, here $P(K_1)=\frac{4}{5}$, $P(K_2 |K_1)=\frac{5}{6}$, $P(K_3 | {(K_2 \cap K_1)})=\frac{6}{7}$ and so on. – Rishit Garg Aug 20 '24 at 20:32
  • But why would the position of $K_1$ matter, given that we only care about the position of $K_2$ relative to each of the aces? – Nimaye Garodia Aug 21 '24 at 05:20
  • It wouldn't matter if we simply wanted to find $P(K_2)$. It matters here since we want to find $P(K_1 \cap K_2)$. They are not independent as you mentioned in your post. – Rishit Garg Aug 23 '24 at 20:18
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I don't think events $K_1$ and $K_2$ are independent. As an extreme case, consider the probability of $K_1$ conditioned on the occurrence of all other events $K_2, K_3, K_4, Q_1,\dots,Q_4,J_1,\dots,J_4$. The sequence will have to start with some ace and followed by some permutation of the remaining $14$ cards (all $4$ Jacks, all $4$ Queens, $3$ aces, $3$ kings) in some order. Now, the remaining card (King $1$) can be placed in one of the $16$ places (buckets in your language). For event $K_1$ to occur, it can be placed in $15$ of the $16$ places. In a random shuffle, the probability of this occurring is $\frac{15}{16}$ which is larger than the original $\frac{4}{5}$ without conditioning. This shows that the events are not independent and your calculation of $(\frac{4}{5})^{16}$ (or actually raised to the power $12$ as you don't want to count aces here) doesn't hold.

Also, the notations are a bit unclear. For $A_1,A_2,\dots$, it is described as the first, second, $\dots$ Ace to appear. However, for $K_1,K_2,\dots$, it looks like they represent the King card of different suits with no ordering among them.

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Ramakrishnan K
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The usual assumption about shuffling a deck of cards is that every card is treated exactly the same as every other in the shuffle. As a result, every permutation of the $52$ cards is equally likely to occur. Namely, the probability of each permutation is $\dfrac1{52!}$.

We can imagine that we shuffle a deck as follows. Shuffle the four aces in a random permutation ($4!$ equally likely sequences) and place them in a row with spaces between them and on each end of the array so that there are five spaces. Assign each of the five spaces to one of the numbers on a five-sided fair die. For each other card from the two of clubs to the king of spades, roll the die and place the card in the indicated space. Finally, shuffle each of the piles in the spaces in the usual way (all permutations of that pile equally likely) and stack the cards in one deck by alternating cards-in-space, ace, cards-in-space, etc.

We will get a random arrangement of the $52$ cards in this way, but not all permutations will be equally likely. For example, the probability of the permutation $$\clubsuit A\diamondsuit A\heartsuit A\spadesuit A \clubsuit 2\diamondsuit 2\heartsuit 2\spadesuit 2 \ldots \clubsuit Q\diamondsuit Q\heartsuit Q\spadesuit Q \clubsuit K\diamondsuit K\heartsuit K\spadesuit K$$ (all cards appearing in increasing sequence of rank, ace low, with the cards of each rank always occurring in the same sequence of suits) is $$ \left(\frac15\right)^{48} \cdot \frac1{48!}, $$ because every card other than an ace has to be put in the last space initially, and then those $48$ cards need to be shuffled into the exact permutation shown above.

It is easy to see that $\left(\dfrac15\right)^{48} \cdot \dfrac1{48!} \neq \dfrac1{52!}$, from which we can conclude that this method of shuffling the cards does not respect the usual assumptions and is incorrect to use in this problem (since there is no indication that we are meant to shuffle the cards this way).

It also follows that any intuition this nonstandard method might give you about the probability of finding a $\diamondsuit K$ between the first two aces, given that $\clubsuit K$ is there, does not apply to a deck of cards shuffled under the usual assumptions.

There's no need to get all tied into knots trying to reconcile step by step how and why the model of putting cards in the five spaces between and around the aces has to be modified in order to make it a standard shuffle. Just don't start there.

On the other hand, if you identify any subset of the $52$ cards by their ranks and suits, each permutation of that subset is equally likely to occur when the entire deck is shuffled under the usual assumptions. Hence when you're thinking about whether a particular card, for example $\clubsuit K$, comes before the first ace you only have to look at $\clubsuit K$ and the four aces, a set of five cards, for which there are $5!$ equally likely permutations in which $\clubsuit K$ is the first card in $4!$ permutations. But when you ask about the order of the four aces, $\clubsuit K$, and $\diamondsuit K$, then there are $6!$ equally likely permutations of the six cards.

It is possible to show that one way to shuffle the entire deck is to put one card on the table, then randomly place the next card in the space to the left or right of the first, then randomly place the next card in one of the three spaces made by the two cards already placed, and so forth. (There are $n$ places to put the $n$th card.) Each permutation of all $52$ cards then occurs with probability $\dfrac1{52!}$. So if you want to shuffle the cards that way, starting with the aces and continuing with the other $12$ honor cards before inserting the others, you will get an answer consistent with the usual assumptions.

David K
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