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Five cards are dealt from a standard 52-card deck.

(a) What is the probability that we draw 1 ace, 1 two, 1 three, 1 four, and 1 five (this is one way to get a "straight")?

(b) What is the probability that we draw any straight (including "straight flush" and "royal straight flush" hands)?

 Feb 13, 2020
 #1
avatar+109740 
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(a) What is the probability that we draw 1 ace, 1 two, 1 three, 1 four, and 1 five (this is one way to get a "straight")?

 

In each case  we want to  select  1 of 4 cards from each of the specified ranks  = [ C(4,1)]^5  = 4^5

 

And the total  hands = C(52,5)

 

So.....the probability is just

 

4^5 / C(52,5)   ≈  .000394  = .0394%

 

Note that these hands include the possibility of a straight flush ....

 

cool cool cool

 Feb 13, 2020
edited by CPhill  Feb 13, 2020
edited by CPhill  Feb 13, 2020
 #2
avatar+109740 
+1

(b) What is the probability that we draw any straight (including "straight flush" and "royal straight flush" hands)?

 

This is similar  to the last one

 

If  we let  the ace be the lowest or highest card in the straight.....we have 10 possible straights within the ranks :

A-2-3-4-5

2-3-4-5-6

......

10-J-Q-K-A

 

So the probability  is just  

 

C(10,1) * C(4,1)/ C(52,5)  ≈  : .00394  = .394%

 

 

cool cool cool

 Feb 13, 2020

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