Question

A poker hand consisting of 5 cards, is dealt from pack of 52 cards. Find the probability of obtaining: a) 1 pair b) 2 pairs c) 3 of a kind d) 4 of a kind e) a flush f) a royal flush?

Answer 1

See below:

Explanation:

When working with the probability of poker hands, we need to know the number of ways a hand can be dealt (this is the denominator) and the numerator is the number of ways to have a certain hand.

These will be combination problems (we don't care about the order of the cards dealt):

`C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!)` with `n="population", k="picks"`

First let's calculate the denominator (picking 5 random cards from a pack of 52 cards):

`C_(52,5)=((52),(5))=(52!)/((5)!(52-5)!)=(52!)/((5!)(47!))`

Let's evaluate it!

`(52xx51xxcancelcolor(orange)(50)^10xx49xxcancelcolor(red)48^2xxcancelcolor(brown)(47!))/(cancelcolor(orange)5xxcancelcolor(red)(4xx3xx2)xxcancelcolor(brown)(47!))=52xx51xx10xx49xx2=2,598,960`

Now to our numerators:

a) 1 Pair

There are 13 ordinals (Ace through King) in a standard deck of cards. We want 2 of the same ordinal (out of the 4 suits) and then the other 3 cards can't be that first ordinal, nor can we have an ordinal repeat (that would create a different hand - 2-pair or a full house).

So let's work this through:

• we want to pick 1 ordinal from the population of 13
• from that ordinal, we want to pick 2 cards that are available from a population of 4
• from the remaining 12 ordinals, we pick 3
• each one of these 3 cards can be chosen from a population of 4 suits

`((13),(1))((4),(2))((12),(3))((4),(1))^3`

`=(13!)/((1!)(12!))(4!)/((2!)(2!))(12!)/((3!)(9!))((4!)/((1!)(3!)))^3`

`=13xx6xx220xx(4)^3=1,089,240` different 1-pair hands

Giving a probability of

`1089240/2598960~~0.4226`

b) 2 pair

• we want to pick 2 ordinals from the population of 13
• from those ordinals, we want to pick 2 cards that are available from a population of 4 each
• from the remaining 11 ordinals, we pick 1
• that 1 card can be chosen from a population of 4 suits

`((13),(2))((4),(2))^2((11),(1))((4),(1))`

`78xx6^2xx11xx4=123,552` different 2-pair hands

Giving a probability of:

`123552/2598960~~0.0475`

c) 3 of a kind

• we want to pick 1 ordinal from 13
• we want 3 of the 4 suits in that ordinal
• from the remaining 12 ordinals, we pick 2
• we want 1 suit each from those 2 ordinals

`((13),(1))((4),(3))((12),(2))((4),(1))^2`

`=13xx4xx66xx4^2=54,912` different 3 of a kind hands

giving:

`54912/2598960~~0.0211`

d) 4 of a kind

• we want to pick 1 ordinal from 13
• we want 4 of the 4 suits in that ordinal
• from the remaining 12 ordinals, we pick 1
• we want 1 suit from the population of 4

`((13),(1))((4),(4))((12),(1))((4),(1))`

`=13xx1xx12xx4=624` different 4 of a kind hands

giving:

`624/2598960~~0.0002`

e) flush

• From the 4 suits, we pick 1
• we pick 5 ordinals of the 13 available

`((4),(1))((13),(5))`

`4xx1287=5148`

giving `5148/2598960~~0.0020`

NOTE - this is all hands that have a flush element to them, and so includes Royal and Straight flushes. We can eliminate those by first seeing that those hands are:

• From the 4 suits, we pick 1
• From the 13 ordinals available, we pick 5 consecutive ones. There are 10 ways to do that (AKQJ10 down to 5432A):

`((4),(1))xx10=40`

and so

`5148-40=5108`

gives

`5108/2598960~~0.0020`

f) Royal Flush

There are only 4 Royal Flushes available - AKQJ10 in each of the 4 suits.

`4/2598960~~0.000002`

For more information about poker hand probability, check out this wikipedia link:

https://en.wikipedia.org/wiki/Poker_probability