What is the probability that you can get a straight?
Emily Wong 
Disclaimer: I am sure this question has been asked but I want to arrive to the answer on my own as it is important to first understand scenarios like this when first learning about Probability at the college level and really diving into the material. I encountered this question in an introductory reading and it has been rather irksome as I don't think the answer is so obvious.
Assuming the following: 1) The game is 5 card stud (not sure if game mechanics affect probability, ie holdem?) 2) Aces don't count as ones, they are only counted after the king 3) Both Jokers are left out of the deck in this game
There are 10 possible different ways (sets) you can get a straight:
A,2,3,4,5
2,3,4,5,6
3,4,5,6,7
4,5,6,7,8
5,6,7,8,9
6,7,8,9,10
7,8,9,10,J
8,9,10,J,Q
9,10,J,Q,K
10,J,Q,K,A
There are 10 different sets and each set can be ordered $4^5$ different ways so to take into account all the possible straights: $\frac{4^5\times 10}{52!}$? But wait this just accounts for the suits correct?
The 4^5 is from the basic property that you multiply all the elements from one set by all the elements from another set when you want to find all the different combinations of the two sets. So this tells me all the different combinations of suits I believe. Now I need the orders of the cards. Which would be 5! in each set so more or less im thinking: $\frac{(4^5\times5!) \times (10)}{52!}$ is much closer? Mark said I'm close but I'm not sure where my error is. The denominator I think would be closer to $4^14 \times 14!$ (the number of orders of suits times the number of orders of rank)so up to this point I have:
$\frac{(4^5 \times 5!) \times (10)}{4^14 \times 14!}. Is there possibly a different relationship between the two (rank and suit) that might call for a different operation other than multiplication
$\endgroup$ 112 Answers
$\begingroup$A-low is a straight.
$10$ possible values for the lowest card. $4^5$ possible arrangements of suits. $(4^5-4)$ if you want to keep the straight-flush separate.
$10*(4^5-4) = 10200$ ways to make a straight.
$52 \choose 5$$=\dfrac{52!}{(52-5)!(5!)} = 2.6$ million possible hands.
$0.4\%$ chance of pulling a straight on 5 cards.
$\endgroup$ 1 $\begingroup$Order doesn't matter, because A,2,3,4,5 is the same hand has 3,4,2,A,5. So you want to stick with $4^5*10$ in your numerator. To consider straights independently from straight flushes, remove the 4 possible straight flushes from each of the 10 initial positions, giving you $(4^5-4)*10$. This is done because a straight flush is much more valuable than a typical straight.
For the denominator, you should consider a random draw of your hand A,B,C,D,E. You have 52 possibilities for card A, 51 possibilities for card B, etc. for $(52*51*50*49*48)$ possible combinations. Another way of writing this is $\frac{52!}{47!}$. But remember, we said order doesn't matter. Just like you multiplied by $5!$ to account for the order of five cards, we will divide by $5!$ to discount the order, making your denominator $\frac{52!}{(5!)47!}$.
$\endgroup$
