# probability that you get one ace and the rest are face cards if you are dealt with 5 cards from a Standard Deck

You are dealt 5 cards from a standard deck. You keep careful track of the order of the cards you are dealt. What is the probability that you get one ace and the rest are face cards?

Am thinking I can solve the problem as P(AFFFF)+P(FAFFF)+P(FFAFF)+P(FFFAF)+P(FFFFA) Without replacement. But am also wondering if this is a case of a Binomial where n=5 x=1 P=4/52 1-P=48/52

• The probability doesn't depend on whether you keep track of the order. – joriki Apr 23 '16 at 12:15
• Since order is important here what does it mean? – Ben Apr 23 '16 at 12:17
• It seems odd to code Ace as $F$ and Facecard as $A$! – almagest Apr 23 '16 at 12:18
• @Ben Order is not important in the event: "the hand is one ace and four face cards." – Graham Kemp Apr 23 '16 at 12:27
• Independent of whether the order is important for the event (it isn't), certainly whether you keep track of it can't be important -- unless these are quantum cards. – joriki Apr 23 '16 at 12:28

This is not related to a Binomial distribution - which is that of the count of successes in a series of iid Bernoulli trials.

This senario is more similar to a Hypergeometric Distribution - that of the count of successes in a sample drawn without replacement from a population.   In this case we have two disjoint kinds of favored items in the population (deck).

$$\Pr(\textsf{5 Card Hand = 1 Ace & 4 Face}) ~=~\dfrac{\dbinom{4}{1}\dbinom{12}{4}\color{gainsboro}{\dbinom{52-16}{0}}}{\dbinom{52}{5}\qquad\qquad\quad}$$

Wondering if this is a case of a Binomial where $n=5,x=1,P=4/52,1-P=48/52$.

No it's not, because $A$ and $F$ are not complementary events.

Keep in mind that some of the cards are neither $A$ nor $F$.

In other words: $P(A)=4/52$, but $P(F)\neq48/52$.

• Another reason it's not binomial is that the chance of $A$ on the first card is not independent from the chance of $A$ on the second card: given that the first card is an ace, the second card has less than $1/13$ chance to be an ace; and given that the first card is not an ace, the second card has more than $1/13$ chance to be an ace. – David K Apr 24 '16 at 14:24
• @DavidK: Yes, but I've assumed that you choose all $5$ cards at once. – barak manos Apr 24 '16 at 14:26
• What I said has nothing to do with the order in which the cards are chosen; all it requires is that at some point you consider the joint distribution of two of those cards, labeled "first" and "second". But if we refuse to assign any such sequential arrangement to the cards at any time, that's yet one more reason not to consider a binomial distribution, which is usually defined in terms of a sequence of Bernoulli trials. – David K Apr 24 '16 at 14:33
• @DavidK: Neither what I said. What I meant was, that if you chose all $5$ cards at once and without replacement, then you'd still need to have events $A$ and $F$ complementary in order to consider using Binomial distribution as a solution to the problem at hand. – barak manos Apr 24 '16 at 14:46
• That is true too, of course. These are two things that make the binomial distribution unsuitable for this problem. That is why my comment is only "another" reason; also why it was only a comment and not an answer, because your answer was already the strongest reason. – David K Apr 25 '16 at 2:27

The probability of something happening is its $\frac {\text{successful outcomes}}{\text{possible outcomes}}$

So our successful outcomes for this problem would be $4\cdot 12\cdot 11\cdot 10\cdot 9$ because there are $4$ Aces to choose from, and $12$ face cards to choose from. But once you remove one of the face cards, you're left with $11$, then $10$, and so on...

And we have a total of $52\cdot 51\cdot 50\cdot 49\cdot 48$. So we have: $$P(\text{Choosing Ace and Choosing 4 face cards next})=\frac {4\times 12\times 11\times 10\times 9}{52\times 51\times 50\times 49\times 48}=\boxed{\frac {99}{649,740}}$$

There are 4 ace cards, you need 1. There are 12 face cards, you need 4. The number of ways to get 5 cards from 52 is $\binom{52}{5}$.