Question

A deck of cards is dealt out. What is the probability that the $14$th card dealt is an ace? What is the probability that the first ace occurs on the $14$th card?

Short answer

$P\left(14\text{th card is an ace}\right)=\frac{4·51!}{52!}=\frac{4}{52}=\frac{1}{13}$

$P(14thcardisthefirstace$$)\approx 0.03$

Step-by-step solution

Given Information.

A deck of cards is dealt out.

Explanation.

A deck$\left(52\right)$of cards is dealt.

Outcome space permutations of cards$\Rightarrow$each element is equally likely.

A deck has $52$cards so the outcome space has $52!$elements.

$P\left(14\text{th card is an ace}\right)=\frac{\text{\# events in which 14th card is an ace}}{\text{\# of events in the outcome space}}$

Fix one of the $4$aces in the $14$th place and permute the remaining $51$cards $-4·51$! events from the outcome space are such that$14$th card is an ace

$P(14$th card is th card is the first ace$)=\frac{4·51!}{52!}=\frac{4}{52}=\frac{1}{13}$.

$P(14$th card is the first ace$)=\frac{\text{\# events in which 14th card is the first ace}}{\#\text{of events in the outcome space}}$

Fix one of the four aces in the $14$th place and then choose without repetition$13$cards from the $48$non-ace cards for the first $13$delts. After choosing the first $14$cards, permute the remaining $38$of them.

$4·48·47·\dots 36·38$! possible card dealings where the card $14$is the first ace.

$P\left(14\text{th card is the first ace}\right)=\frac{4·48·47·\dots 36·38!}{52!}=\frac{4·38·37·36}{52·51·50·49}\approx 0.03$