Question 226299: Bob and Alice are playing Texas Hold 'Em. Bob is dealt cards 1 and 3 while Alice is dealt cards 2 and 4. Three of the five community cards are dealt (5of clubs, 6 of clubs, K of hearts). Bob has the 2 of clubs and 4of clubs.
a. What is probability that bob has a flush?
b. What is probability that bob has a straight?
c. What is probability of a straight flush?
Work:
a. Is it 25% because the next cards could be one of four choices?
b. Is the answer .5% because odds of an Ace is 1 out of 13 multiplied by odds of a 7 (1 out of 13)?
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! 52-5=47 cards left to choose from since you don't know Alice's cards.
a)
13-4=9 remaining clubs
(9C1*38C1+9C2)/(47C2)
=(9*38+36)/1081 The number of possible combinations of 47 cards drawn 2 at a time is 1081.
=(342+36)/1018
=.350... probability of a flush in the next 2 cards.
.
b)
there are 4 threes that will make a straight.
(4C1*43C1 +4C2)/47C2 [combination of 1 three out of 4, times the combination of 1 of the other 43 cards, plus the combination of 2 of the threes, divided by 1018.
=(4*43 +6)/1081
=178/1081
=.165... probability of a straight in the next 2 cards.
.
c)
Only 1 card will make a straight flush, 3 of clubs.
(1C1*46C1)/47C2
=46/1081
=.0426... Probability of a straight flush in the next 2 cards.
.
Ed
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