# Probability of no jack, queen, king before the first ace

I am reviewing some probability puzzles, and trying to solve them under a standard timed duration. But, I think could be completely wrong in formulating the below exercise problem. So, I'd like someone to verify if my solution attempt is correct and if I've counted correctly.

[Blitzstein 1.37] A deck of cards is shuffled well. The cards are dealt one-by-one, until the first time an ace appears.

(a) Find the probability that no kings, queens or jacks appear before the first ace.

Solution.

Firstly, we are sampling without replacement. Moreover, I think that, since we are not dealing hands, the order in which the cards appear is important.

We are interested to form an $$r$$-tuple, $$(x_1,x_2,\ldots,x_r)$$ from the population $$\{\heartsuit,\clubsuit,\spadesuit,\diamondsuit \} \times \{1,2,\ldots,13\}$$, where all $$x_j$$'s are distinct.

Let $$A_r$$ be the event that the $$r$$'th draw is an ace. This last ace can be chosen in $$4$$ ways. Should there be no kings, queens and jacks before the first ace, we have $$36$$ cards to choose from. Thus, there are $$(36)_{r-1}=\frac{36!}{(36-r+1)!}$$ distinguishable choices for the first $$(r-1)$$ cards. Obviously, $$r=1,2,\ldots,36$$.

The desired probability is: $$P(\cup_{r=1}^{36} A_r) = \sum_{r=1}^{36} \frac{4 \cdot (36)_{r-1}}{(52)_r}$$

(b) Find the probability that exactly one king, exactly one queen and exactly one jack (appear in any order) before the first ace.

The desired probability is:

$$4^3 \sum_{r=1}^{36} \frac{4 \cdot (36)_{r-4}}{(52)_r}$$

You do not need to use $$52$$ or $$36$$ or your $$r$$ in the calculations for (a) or (b) as the positions of the other $$36$$ cards are irrelevant and you would get the same result if they were not there.

So for (a): the probability that at the first Ace appears before the first Jack, Queen or King is (by exchangeability) $$\frac{4}{16}=\frac14$$; one of the $$16$$ interesting cards must appear earlier than the others, each equally likely to do so, and $$4$$ of them are Aces;

and similarly for (b): the probability that exactly one king, exactly one queen and exactly one jack appear in any order before the first ace is $$\frac{12}{16}\times\frac{8}{15}\times \frac{4}{14}\times \frac{4}{13} = \frac{16}{455}$$.

You might want to check whether these correspond to your expressions.

• I'm still puzzled, how one can stack the $16$ interesting cards together. I am aware that the chronology of events does not matter in probability. For example, if I have $g$ green balls and $r$ red balls, and draw $2$ balls without replacement, the second ball is equally likely to be any of the $g+r$ balls, so the probability is $g/(g+r)$. Mar 29, 2023 at 9:23
• @Quasar It is less that you "stack the $16$ interesting cards together" and more that you ignore the positions of the other $36$ cards (you also ignore the positions of the second, third and fourth Aces apart from the point that they come after the first Ace). Mar 29, 2023 at 9:27

Alternative solutions which also could explain why we can ignore irrelevant cards.

a) There are $$52!$$ total orderings.

To build the event let's divide the cards into 3 groups:

• $$36$$ extra cards (everything except aces, kings, queens and jacks)
• $$1$$ ace that will be first (we have a choice from $$4$$)
• $$15$$ remaining cards

And we need the following:

• $$36$$ slots to put our $$36$$ cards: $$\binom{52}{36}36!$$
• after this we have $$1$$ slot for ace (it should be before remaining $$15$$ cards) and we need to choose one from $$4$$
• $$15$$ remaining slots for other aces, kings, queens and jacks: $$\binom{15}{15}15!$$

$$\frac{\binom{52}{36}36! \times 4 \times 15!}{52!} = \frac{4 \times 15!}{16!} = \frac{1}{4}.$$

b) Again, there are $$52!$$ total orderings.

For the event we need:

• $$36$$ slots for $$36$$ extra cards: $$\binom{52}{36}36!$$
• $$3$$ slots to put our jack, king and queen. We need to put these cards before the first ace and before any other remaining cards that should appear after the first ace. Therefore we reserve some $$3$$ slots for them that will allow us to do that. We select $$3$$ slots, put the cards to them (any out of $$4$$) and shuffle them, therefore: $$\binom{3}{3}3! \times 4^3$$
• one slot for the first ace and choose the ace: $$\binom{1}{1}4$$
• $$12$$ remaining slots for remaining cards that should be after the first ace: $$\binom{12}{12}12!$$

$$\frac{\binom{52}{36}36! \times 3! \times 4^3 \times 4 \times 12!}{52!} = \frac{4^4 \times 3! \times 12!}{16!}.$$

Hope this helps.

You draw cards until you get a Jack, Queen, king or ace. Each has the same probability 1/4, so the answer is 1/4.

There is absolutely no reason to use any two digit or larger numbers.