The probability of a flush A poker player holds a flush when all $5$cards in the hand belong to the same suit. We will find the probability of a flush when $5$ cards are dealt. Remember that a deck contains $52$ cards, $13$ of each suit, and that when the deck is well shuffled, each card dealt is equally likely to be any of those that remain in the deck.

(a) We will concentrate on spades. What is the probability that the first card dealt is a spade? What is the conditional probability that the second card is

a spade given that the first is a spade?

(b) Continue to count the remaining cards to find the conditional probabilities of a spade on the third, the fourth, and the fifth card given in each case that

all previous cards are spades.

(c) The probability of being dealt 5 spades is the product of the five probabilities you have found. Why? What is this probability?

(d) The probability of being dealt $5$ hearts or $5$diamonds or $5$ clubs is the same as the probability of being dealt $5$ spades. What is the probability of being dealt a flush?

There are $13$ spade cards in a deck. The total number of cards in the deck is $52$

$Probability\left(P\right)=\frac{Numberoffavourable}{Totalnumberofexhaustive}$

The likelihood of the first card being a spade is calculated as follows:

$P\left(Spade\right)=\frac{Numberofspadecards}{Totalnumberofcards}=\frac{13}{52}=0.25$

If the first card is also a spade, the conditional probability of the second card being a spade is:

$$\begin{gathered} P(Secondspade|Firstspade)=\frac{P(SecondspadeAndFirstspade)}{P(Firstspade)} \\ P(Secondspade|Firstspade)=\frac{\frac{13}{52}\times \frac{13}{51}}{\frac{13}{52}}=0.2353 \end{gathered}$$

As a result, the needed probabilities are $0.25$and $0.2353$, respectively.

The probabilities are:

$$\begin{gathered} P(Secondspade|Firstspade)=\frac{P(SecondspadeAndFirstspade)}{P(Firstspade)} \\ P(Secondspade|Firstspade)=\frac{{\displaystyle \frac{13}{52}}\times \frac{13}{51}}{\frac{13}{52}}=0.2353 \end{gathered}$$

$$\begin{gathered} P(Thirdspade|Secondspade)=\frac{P(ThirdspadeAndSecondspade)}{P(Secondspade)} \\ P(Thirdspade|Secondspade)=\frac{{\displaystyle \frac{13}{52}}\times \frac{12}{51}\times {\displaystyle \frac{11}{50}}}{\frac{13}{52}\times \frac{12}{51}}=0.22 \end{gathered}$$

$$\begin{gathered} P(Forthspade|Thirdspade)=\frac{P(ForthspadeAndThirdspade)}{P(Thirdspade} \\ P(Forthspade|Thirdspade)=\frac{{\displaystyle \frac{13}{52}}\times {\displaystyle \frac{12}{51}}\times {\displaystyle \frac{11}{50}}\times {\displaystyle \frac{10}{49}}}{{\displaystyle \frac{13}{52}}\times {\displaystyle \frac{12}{51}}\times {\displaystyle \frac{11}{50}}}=0.2041 \end{gathered}$$

$$\begin{gathered} P(Fifthspade|Forthspade)=\frac{P(ForthspadeAndFifthspade)}{P(Forthspade)} \\ P(Fifthspade|Forthspade)=\frac{{\displaystyle \frac{13}{52}}\times \frac{12}{51}\times {\displaystyle \frac{11}{50}}\times {\displaystyle \frac{10}{49}}}{{\displaystyle \frac{13}{52}}\times {\displaystyle \frac{12}{51}}\times {\displaystyle \frac{11}{50}}\times {\displaystyle \frac{10}{49}}}=0.1875 \end{gathered}$$

$0.22,0.20411$, and $0.1875$ are the probability, respectively.

The following formula can be used to compute the value:

$$\begin{gathered} P\left(Spade\right)=\frac{1}{4}\times \frac{12}{51}\times \frac{11}{50}\times \frac{10}{49}\times \frac{9}{48} \\ P\left(Spade\right)=0.000495 \end{gathered}$$

$0.000495$ is the value.

The likelihood of having a flush can be estimated as follows: $$\begin{gathered} P\left(Flush\right)=P(5spades)+P(5Hearts)+P(5Clubs)+P(5Diamonds) \\ P\left(Flush\right)=\frac{33}{66640}+\frac{33}{66640}+\frac{33}{66640}+\frac{33}{66640}=0.00198 \end{gathered}$$

$0.00198$ is the probability.