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For counting total no of Straights =>

1 2 3 4 5 6 7 8 9 10 J Q K Ace(1)

Here we got to select 5 cards in row. So Total no of ways are 10 !

Now each card can take one of 4 suits

So total possibilites of straight = 10 * 45

Total ways of selecting cards i.e. Sample space = 52C5

Total Probability = 10 * 4/ 52C5

Correct me If I'm wrong not 100% sure here !

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1 votes
Step 1$\Rightarrow$Straight set=$\left \{ A,1,2,3,4,5,6,7,8,9,10 \right \}$ and 1st card can start from any of the element from the set and then continue contiguously,so number of ways we can do this is $\binom{10}{1}= 10$

Step 2$\Rightarrow$There are 4 suits ,so number of ways selecting 1 suit from this 4 is $\binom{4}{1}= 4$

Step 3$\Rightarrow$5 choices(5 cards) are there for this 4 possiblity making it $4^{5}$

Using Product rule $N\left ( E \right )=10*4^{5}$

N(S)=$\binom{52}{5}$

Therefore required probablity=$N\left ( E \right )/N\left ( S \right )$=$10*4^{5}/\binom{52}{5}$
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what i understood by your question a straight of 5 is a set of 5 cards where the cards are consecutive (like 2,3,4,5,6) and it belongs to a particular suit(diamonds,clubs,hearts,spade)

lets talk about a particular suit first say spades

the number of straights in spades is as follows

1-5

2-6

3-7

4-8

5-9

6-10                              consider 10 as J

7-11                              consider 11 as Q

8-12                              consider 12 as K

9-13                              consider 13 as A

so there are 9 straights 

now for 4 suits there will be 9*4=36 suits

so the probability must be 36/52C5

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number of stright is 10C1* (4^5)..

probability of straight (10C1* (4^5))/  52C5

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