91Ó°ÊÓ

The term "probability" simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

First: In order to find the probability that out of \({\rm{5}}\) dealt car it will be a straight with high card \({\rm{10}}\), we will use

\({\rm{P(A) = }}\frac{{{\rm{\# of favorable outcomes in A }}}}{{{\rm{\# of outcomes in the sample space }}}}\)where # of outcomes in the sample space is number of ways to take \({\rm{5}}\) card out of \({\rm{52}}\) card deck, and for that we use combinations.

A combination is an unordered subset. For \({\rm{n}}\) individuals in a group, the number of combinations size \({\rm{k}}\) is denoted as \({{\rm{C}}_{{\rm{k,n}}}}\) or

\(\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{k}}\end{array}} \right)\)

which we read as from \({\rm{n}}\) elements we choose \({\rm{k}}\). The proposition:

\(\begin{array}{c}{{\rm{C}}_{{\rm{k,n}}}} \rm &=& \left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{k}}\end{array}} \right)\\ \rm &=& \frac{{{{\rm{P}}_{{\rm{k,n}}}}}}{{{\rm{k!}}}}\\ \rm &=& \frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}\end{array}\)

Therefore, we have

\(\begin{array}{c}{{\rm{C}}_{{\rm{5,52}}}} \rm &=& \left( {\begin{array}{*{20}{c}}{{\rm{52}}}\\{\rm{5}}\end{array}} \right)\\ \rm &=& \frac{{{\rm{52!}}}}{{{\rm{5!(52 - 5)!}}}}\\ \rm &=& 2,598,960 \end{array}\)

ways to select five cards out of \({\rm{52}}\) (five card hand), which is our number of outcomes in sample space.

If a \({\rm{10}}\) is high card, a straight consist of six, seven, eight, nine, ten. There are four different types of cards for each six/seven/eight/nine/ten, which means that we have

\({\rm{4*4*4*4*4 = }}{{\rm{4}}^{\rm{5}}}{\rm{ = 1024}}\)

straights, or our number of favorable outcomes.

The probability that out of \({\rm{5}}\) dealt car it will be a straight with high card \({\rm{10}}\) is

\(\begin{array}{c} \rm P(A) &=& \frac{{{\rm{\# of favorable outcomes in A}}}}{{{\rm{\# of outcomes in the sample space }}}}\\ \rm &=& \frac{{{\rm{1024}}}}{{{\rm{2,598,960}}}}\\ \rm &=& 0{\rm{.000394}}{\rm{.}}\end{array}\)

Second: The same way as above, we want to find the number of favorable outcomes (the number of outcomes in sample space stays the same).

There are straights with high card 5 , high card $6, \ldots$, high card $\mathrm{K}$ and high card $\mathrm{A}$, which makes it 10 different high cards. For each of the high cards, we have calculated the number of favorable outcomes above, 1024 , so

$$

P(\{\text { straight }\})=\frac{10 \cdot 1024}{2,598,960}=0.00394 .

$$

Third: The only difference in the straight and the straight flush is type of cards (they all have to be in same suit), so for every of the \({\rm{10}}\) different straights (with high card\({\rm{\;5}}\), high card\({\rm{6, \ldots }}\), high card \({\rm{K}}\) and high card A) we have \({\rm{4}}\) different straights, which makes it

\({\rm{10*4 = 40}}\) favorable outcomes. The Royal Flush (the four straight flushes with high card A) is included!

The probability that it will be a straight flush is

\(\begin{array}{c} \rm P(\{ straight flush \} ) &=& \frac{{{\rm{40}}}}{{{\rm{2,598,960}}}}\\ \rm &=& 0{\rm{.00001539}}{\rm{.}}\end{array}\))

Therefore, the probabilities are \({\rm{0}}{\rm{.000394,0}}{\rm{.00394 and 0}}{\rm{.00001539}}{\rm{. }}\)