# A deck of cards is well shuffled. The cards are dealt one by one, until the first time an ace appears

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

1. Find the probability that no kings, queens or jacks appear before the first ace.
2. Find the probability that exactly one king, exactly one queen and exactly one jack appear (in any order) before the ace first.

## One: Find the probability that no kings, queens or jacks appear before the first ace.

Notation:

• Let b cards precede the first ace
• Let A be the first Ace
• Let a cards follow the first ace

Conditions:

• b will range from 0 to 36 ($52 - (12 + 4)$)
• A will be one of four Ace cards
• a will range from 51 to 15 ($12 + 3$)

In order to maintain equal likeliness of all configurations (an assumption), we will find the number of valid permutations and divide by the total number of permutations which will result in the probability of a valid configuration, denoted P(A).

$$P(A) = \frac{\textrm{number of valid permutations}}{\textrm{number of all permutations}}$$

$$P(A) = \frac{1}{52!}\sum_{i=0}^{36}\left(\frac{36!}{(36-i)!}\frac{4!}{(4-1)!}(51-i)!\right)$$ where within the summation, the first expression selects and permutes the number of cards preceding the first ace, the second expression selects the first ace, the third expression selects and permutes the cards which follow the first ace.

The expression reduces to: $$P(A)=\frac{4\times36!}{52!}\sum_{i=0}^{36} \frac{(51-i)!}{(36-i)!}$$ $$P(A) = 0.25$$

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

There are $\frac{4!}{(4-1)!}$ ways to select one of four elements, be it an ace, jack, queen or king.

Proceed in a similar fashion to part 1. by finding the quotient of the total number of permissible permutations and the total number of permutations, labelling it $P(B)$.
$$P(B) = \frac{1}{52!} \sum_{i=0}^{36}\left(\frac{4!}{(4-1)!}\frac{4!}{(4-1)!}\frac{4!}{(4-1)!}\frac{1}{3!}\times\frac{36!}{(36-i)!}\frac{4!}{(4-1)!}(48-i)! \right)$$ Where the first three expressions in the summation are the selection of a king, queen and jack, the fourth expression allows these three cards to be chosen in any order. The expressions following the multiplication symbol are:

• selection of a number of cards ( two, three, ... , ten) to follow the face cards and precede the first ace.
• selection of the first ace
• selection of the remaining cards

The expression collapses to: $$P(B) = \frac{4^4 \times 36!}{52!\times 3!} \sum_{i=0}^{36}\frac{(48-i)!}{(36-i)!}$$ $$P(B) = 2.4752\times{10}^{-5}$$

• This seems all well and good, but perhaps you might find it easier to work with a smaller deck. The only cards that matter at all are the aces and face cards, so how about throwing out all other cards from the deck. After all, their location in the deck doesn't affect the events we are interested in. This leaves a deck of just $16$ cards with four aces, four jacks, four queens, and four kings. What is the probability the first card is an ace? Do you see why this probability should be the same as the probability you tried to calculate in the first problem above? – JMoravitz Feb 22 '17 at 17:41
• Your first answer is overkill, however, because you can rephrase the question as "What is the probability that the first honor card (ace,king,queen, or jack) is an ace." Then it is clearly 1/4 - each is equally likely – Thomas Andrews Feb 22 '17 at 17:52
• Others have already said most of what I was going to say, but on your final calculation, first you should be multiplying by $3!$ not dividing, and you also need to take into account that the previous face cards, and the other cards before the ace, can be chosen in any order. This adds a binomial coefficient which complicates things further, @JMoravitz recommendation is one that I would second. – Pepe Silvia Feb 22 '17 at 18:01
• Thank you all. Your comments have helped tremendously. – DWD Feb 22 '17 at 20:14
• @DipeshGupta I think I did it computationally in Python3 or WXMaxima – DWD Sep 20 '17 at 19:31

You can rephrase the question as "What is the probability that the first honor card (ace,king,queen, or jack) is an ace." Then it is clearly 1/4 - each is equally likely.

Note that all the probabilities are the same if you remove the cards not in question - so that you can think of it as a deck of 16 cards. There are $16!$ orders for the deck. There are $3!4^4$ different ways to have the first four honors be all different and the fourth be an ace.

So the probability for the second case is:

$$\frac{3!4^412!}{16!}\approx 0.035$$

Which is significantly higher than your probability.

Did you assume that there were no cards other than the king, queen, and jack before the ace? Then you need all 52 cards, and you'd get:

$$\frac{3!4^448!}{52!}=\frac{3!4^4}{52\cdot 51\cdot 50\cdot 49}\approx 0.000236$$

Which is closer to your result, but still different.