Response
"With 3 spades on your first three cards [in seven-card stud] and no other cards seen, you will make a spade flush in seven cards 18 percent of the time. Could someone show the calculation for this?"
Long answer :
There are 10 more spades remaining.
C(10,4)=210 combinations that give you 4 more spades, for a 7-card hand of all spades.
C(10,3)=120 , so (120*39 non-spade cards)=4680 combinations that have 3 spades and 1 non-spade card, which you give you a 6-card Flush.
C(10,2)=45 , so (45*741)= 33345 combinations that contain 2 spades along with 2 non-spade cards. That's because there are C(39,2)=741 combinations of 2 cards that the 39 non-spade cards can produce.
All in all, 210+4680+33345= 38235 total combinations that give you a Flush of at least 5 spades.
Dividing that number with all the possible combinations of 4 cards that can be made from the remaining 49 cards that are unseen, gives 38235 / C(49,4) = 38235/211816 = 0.180510443026022585640367111077539 or 18%
Short Answer : Sklansky said so.
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