# Basic Probability Question (Deck of Cards)

• Feb 18th 2013, 10:32 PM
Pourdo
Basic Probability Question (Deck of Cards)
Hi everyone, hoping you can all help me with this question.

A card is drawn from a shuffled deck of 52 cards, and not replaced. Then a second card is drawn. What is the probability that the second card is a king?

If possible could you please explain the process of arriving at the answer? The answer key gives me "1/13", hopefully that is correct.
• Feb 19th 2013, 04:04 AM
Plato
Re: Basic Probability Question (Deck of Cards)
Quote:

Originally Posted by Pourdo
A card is drawn from a shuffled deck of 52 cards, and not replaced. Then a second card is drawn. What is the probability that the second card is a king?

Notation: $K_1$ is first card is a king. $\overline{K_1}$ is first card is not a king.

We want $\mathcal{P}(K_2)=\mathcal{P}(K_1\cap K_2)+\mathcal{P}(\overline{K_1}\cap K_2)$.

Well what is $\frac{4}{52}\frac{3}{51}+\frac{48}{52}\frac{4}{51} =~?$
• Feb 19th 2013, 04:56 AM
Kmath
Re: Basic Probability Question (Deck of Cards)
you may also think like that:

P( 2nd card is a K)=P( 2nd card is a Q )=P( 2nd card is a J)=P( 2nd card is a 10)=...=P( 2nd card is an A)

and surely

P( 2nd card is a K)+P( 2nd card is a Q )+P( 2nd card is a J)+P( 2nd card is a 10)+...+P( 2nd card is an A)=1

that is $13 P(\text{2nd card is a K})=1$
• Feb 19th 2013, 03:41 PM
Soroban
Re: Basic Probability Question (Deck of Cards)
Hello, Pourdo!

This is a classic "trick question"

Quote:

A card is drawn from a shuffled deck of 52 cards, and not replaced.
Then a second card is drawn.
What is the probability that the second card is a King?

Let us alter the problem.

Cards are drawn one at a time from a shuffled deck without replacement.
What is the probability that the 4th card is a King?

. . $\begin{array}{cccccccccc}\square & \square & \square & \boxed{K} & \square & \square & \cdots \end{array}$

You are already thinking of the dozens of possibilities, aren't you?

What if the first card is a King?. What if it isn't?
What if the second card is a King?. What if it isn't?
. . . and so on.

We can disregard all of that!

Our only concern is: "Is the 4th card a King?"

Answer: . $P(\text{4th is King}) \:=\:\frac{4}{52} \:=\:\frac{1}{13}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Further proof:

There are $52!$ possible arrangements of the 52 cards.

How many of them have a King in the 4th position?
. . There are 4 choices of Kings for the 4th position.
. . The other 51 cards can be arranged in $51!$ ways.
Hence, there are $4\cdot51!$ ways

Therefore: . $P(\text{4th is King}) \:=\:\frac{4\cdot51!}{52!} \:=\:\frac{4}{52} \:=\:\frac{1}{13}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

In general, the probability that the $n^{th}$ card is a King is $\frac{1}{13}.$
• Feb 19th 2013, 04:38 PM
HallsofIvy
Re: Basic Probability Question (Deck of Cards)
This is how I would think about it: I presume you know that "if an event can happen in N "equally likely ways" and M of them give event "X" then the probability of event "X" is $\frac{M}{N}$.

For example, there are four kings in a standard deck of 52 cards, so the probability of getting a king on the first draw is 4/52= 1/13. But you want a king on the second draw. Since this is "without replacement", there will be 51 cards in the deck on the second draw. But how many kings? That depends on what happened on the first draw. If the first draw was a king, there will be, now, 3 kings left so the probability of a king on the second draw is 3/51= 1/17. If the first draw was not a king, there will be, now, 4 kings left so the probability of a king on the second draw is 4/51.

We have to "weight" those two probabilities by the probability they will happen: the probability of a king on the first draw is, as before, 4/52= 1/13. The probability of any thing other than a king on the first draw is 1- 1/13= 12/13. The "weighted average" will be (1/13)(1/17)+ (12/13)(4/51)= 1/221+ 48/663= 3/661+ 48/663= 51/663= 1/13.