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Card Probability
- Thread starter dhs316
- Start date
Hello, dhs316!
\(\displaystyle \text{There are: }\;_{52}C_5 \:=\:\frac{52!}{51\,47!} \:=\:2,598,960\text{ possible poker hands.}\)
There are 13 choices for the value of the 4-of-a-kind.
There is 1 way to get the 4 cards of that value.
The fifth card is one of the remaining 48 cards.
Hence, there are: .\(\displaystyle 13\cdot 48 \:=\:624\) ways to get a 4-of-a-kind.
\(\displaystyle P(\text{4-of-a-kind}) \;=\;\frac{624}{2,598,960} \;\approx\;0.00024\)
\(\displaystyle \text{There are 13 choices for the value of the Triple.}\)
. . \(\displaystyle \text{There are: }\:_4C_3 \:=\:4\text{ ways to get the 3 cards of that Triple.}\)
\(\displaystyle \text{There are 12 choices for the value of the Pair.}\)
. . \(\displaystyle \text{There are: }\:_4C_2 \,=\,6\text{ ways to get the 2 cards of that Pair.}\)
Hence, there are: .\(\displaystyle 13\cdot4\cdot12\cdot6 \:=\:3744\) ways to get a Full House.
\(\displaystyle P(\text{Full House}) \;=\;\frac{3744}{2,598,960} \;\approx\;0.00144\)
There are 13 choices for the value of the Triple.
. . \(\displaystyle \text{There are: }\:_4C_3 \,=\,4\) ways to get the 3 cards of that Triple.
Suppose we have three Kings.
Then the fourth card must be one of the other 48 cards (not Kings).
. . Suppose it is the \(\displaystyle 7\spadesuit\).
Then the fifth card must be one of the other 44 cards (not Kings and not 7s).
. . Suppose it is the \(\displaystyle 5\heartsuit\)
It seems that there are \(\displaystyle 48\cdot44\) ways to get the two unmatched cards.
But the order of those two cards in not important.
. . \(\displaystyle (7\spadesuit\text{, then }5\heartsuit) \;=\;(5\heartsuit\text{, then }7\spadesuit)\)
So we must divide by 2.
\(\displaystyle \text{Hence, there are: }\:13\cdot4\cdot\frac{48\cdot44}{2} \:=\:54,912\text{ ways to get 3-of-a-kind.}\)
\(\displaystyle P(\text{3-of-a-kind}) \;=\;\frac{54,912}{2,598,960} \;\approx\;0.02113\)
\(\displaystyle \text{There are: }\:_{13}C_2 \,=\,78\text{ choices for the values of the Two Pairs.}\)
\(\displaystyle \text{There are: }\:_4C_2\,=\,6\text{ ways to get the 2 cards for one Pair.}\)
\(\displaystyle \text{There are: }\:_4C_2\,=\,6\text{ ways to get the 2 cards for the other Pair.}\)
The fifth cards is chosen from the other 44 cards.
\(\displaystyle \text{Hence, there are: }\:78\cdot6\cdot6\cdot44 \:=\:123,552\text{ ways to get Two Pairs.}\)
\(\displaystyle P(\text{Two Pairs}) \;=\;\frac{123,552}{2,598,960} \;\approx\;0.04754\)
There are 13 choices for the value of the Pair.
\(\displaystyle \text{There are: }\,_4C_2 \,=\,6\) ways to get the 2 cards of that Pair.
The last 3 cards are chosen from the other 48, 44, 40 cards.
. . Again, the order is not important, so we divide by \(\displaystyle 3! = 6.\)
\(\displaystyle \text{Hence, there are: }\:13\cdot6\cdot\frac{48\cdot44\cdot40}{6} \:=\:1,098,240\text{ ways to One Pair.}\)
\(\displaystyle P(\text{One Pair}) \;=\;\frac{1,098,240}{2,598,960} \;\approx\;0.42257\)
A poker hand is defined as drawing five cards at random without replacement from a deck of 52 playing cards
\(\displaystyle \text{There are: }\;_{52}C_5 \:=\:\frac{52!}{51\,47!} \:=\:2,598,960\text{ possible poker hands.}\)
Find the probability of the following power hands:
a) Four of a kind (four cards of equal face value and one of a different value)
Ans: 0.000024 . . . . too many zeros
There are 13 choices for the value of the 4-of-a-kind.
There is 1 way to get the 4 cards of that value.
The fifth card is one of the remaining 48 cards.
Hence, there are: .\(\displaystyle 13\cdot 48 \:=\:624\) ways to get a 4-of-a-kind.
\(\displaystyle P(\text{4-of-a-kind}) \;=\;\frac{624}{2,598,960} \;\approx\;0.00024\)
b) Full house (one pair and one triple cards with equal face value)
Ans: 0.00144
\(\displaystyle \text{There are 13 choices for the value of the Triple.}\)
. . \(\displaystyle \text{There are: }\:_4C_3 \:=\:4\text{ ways to get the 3 cards of that Triple.}\)
\(\displaystyle \text{There are 12 choices for the value of the Pair.}\)
. . \(\displaystyle \text{There are: }\:_4C_2 \,=\,6\text{ ways to get the 2 cards of that Pair.}\)
Hence, there are: .\(\displaystyle 13\cdot4\cdot12\cdot6 \:=\:3744\) ways to get a Full House.
\(\displaystyle P(\text{Full House}) \;=\;\frac{3744}{2,598,960} \;\approx\;0.00144\)
c) Three of a kind (three equal face values plus two cards of different values)
Ans: 0.02113
There are 13 choices for the value of the Triple.
. . \(\displaystyle \text{There are: }\:_4C_3 \,=\,4\) ways to get the 3 cards of that Triple.
Suppose we have three Kings.
Then the fourth card must be one of the other 48 cards (not Kings).
. . Suppose it is the \(\displaystyle 7\spadesuit\).
Then the fifth card must be one of the other 44 cards (not Kings and not 7s).
. . Suppose it is the \(\displaystyle 5\heartsuit\)
It seems that there are \(\displaystyle 48\cdot44\) ways to get the two unmatched cards.
But the order of those two cards in not important.
. . \(\displaystyle (7\spadesuit\text{, then }5\heartsuit) \;=\;(5\heartsuit\text{, then }7\spadesuit)\)
So we must divide by 2.
\(\displaystyle \text{Hence, there are: }\:13\cdot4\cdot\frac{48\cdot44}{2} \:=\:54,912\text{ ways to get 3-of-a-kind.}\)
\(\displaystyle P(\text{3-of-a-kind}) \;=\;\frac{54,912}{2,598,960} \;\approx\;0.02113\)
d) Two pairs (two pairs of equal face value plus one card of different value)
Ans: 0.04754
\(\displaystyle \text{There are: }\:_{13}C_2 \,=\,78\text{ choices for the values of the Two Pairs.}\)
\(\displaystyle \text{There are: }\:_4C_2\,=\,6\text{ ways to get the 2 cards for one Pair.}\)
\(\displaystyle \text{There are: }\:_4C_2\,=\,6\text{ ways to get the 2 cards for the other Pair.}\)
The fifth cards is chosen from the other 44 cards.
\(\displaystyle \text{Hence, there are: }\:78\cdot6\cdot6\cdot44 \:=\:123,552\text{ ways to get Two Pairs.}\)
\(\displaystyle P(\text{Two Pairs}) \;=\;\frac{123,552}{2,598,960} \;\approx\;0.04754\)
e) One pair (one pair of equal face value plus three cards of different values)
Ans: 0.42257
There are 13 choices for the value of the Pair.
\(\displaystyle \text{There are: }\,_4C_2 \,=\,6\) ways to get the 2 cards of that Pair.
The last 3 cards are chosen from the other 48, 44, 40 cards.
. . Again, the order is not important, so we divide by \(\displaystyle 3! = 6.\)
\(\displaystyle \text{Hence, there are: }\:13\cdot6\cdot\frac{48\cdot44\cdot40}{6} \:=\:1,098,240\text{ ways to One Pair.}\)
\(\displaystyle P(\text{One Pair}) \;=\;\frac{1,098,240}{2,598,960} \;\approx\;0.42257\)