Two Simultaneous Cumulative Hypergeometric Distributions
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Suppose we have a standard 52 card deck from which we draw five cards. What are the chances of drawing one or more Aces and one or more Kings? How do I calculate this?
I know that we can calculate the chance of drawing one or more Ace by using cumulative hypergeometric distribution (Population 52, Subpopulation 4, Sample Size 5). This gives us a result of 34.116%.
How do I then find the odds of drawing one or more kings as well?
I mistakenly thought I could "set aside" the Ace, reducing the Population and Sample Size by 1, then calculate the chance for drawing one or more kings out of the rest (28%). Multiplying the two would find the total odds. However, the results off this - 9.765% - is well under the 10% that I continuously get out of a game simulator (multiple tries of 10 million hands).
So then, how do I calculate the above question, and - more directly - what is the correct way to calculate two different cumulative hypergeometric distributions that occur simultaneously?
probability probability-distributions hypergeometric-function card-games
asked Aug 10 at 22:17
Courtney Stucker
31
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up vote
0
down vote
favorite
Suppose we have a standard 52 card deck from which we draw five cards. What are the chances of drawing one or more Aces and one or more Kings? How do I calculate this?
I know that we can calculate the chance of drawing one or more Ace by using cumulative hypergeometric distribution (Population 52, Subpopulation 4, Sample Size 5). This gives us a result of 34.116%.
How do I then find the odds of drawing one or more kings as well?
I mistakenly thought I could "set aside" the Ace, reducing the Population and Sample Size by 1, then calculate the chance for drawing one or more kings out of the rest (28%). Multiplying the two would find the total odds. However, the results off this - 9.765% - is well under the 10% that I continuously get out of a game simulator (multiple tries of 10 million hands).
So then, how do I calculate the above question, and - more directly - what is the correct way to calculate two different cumulative hypergeometric distributions that occur simultaneously?
probability probability-distributions hypergeometric-function card-games
asked Aug 10 at 22:17
Courtney Stucker
31
You can use the multivariate hypergeometric distribution.
– JMoravitz
Aug 10 at 22:20
1
"Multiplying the two would find total odds..." Remember, $Pr(Acap B)=Pr(A)times Pr(B)$ is true only when $A$ and $B$ are independent events. These are certainly not independent.
– JMoravitz
Aug 10 at 22:22
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose we have a standard 52 card deck from which we draw five cards. What are the chances of drawing one or more Aces and one or more Kings? How do I calculate this?
I know that we can calculate the chance of drawing one or more Ace by using cumulative hypergeometric distribution (Population 52, Subpopulation 4, Sample Size 5). This gives us a result of 34.116%.
How do I then find the odds of drawing one or more kings as well?
I mistakenly thought I could "set aside" the Ace, reducing the Population and Sample Size by 1, then calculate the chance for drawing one or more kings out of the rest (28%). Multiplying the two would find the total odds. However, the results off this - 9.765% - is well under the 10% that I continuously get out of a game simulator (multiple tries of 10 million hands).
So then, how do I calculate the above question, and - more directly - what is the correct way to calculate two different cumulative hypergeometric distributions that occur simultaneously?
probability probability-distributions hypergeometric-function card-games
asked Aug 10 at 22:17
Courtney Stucker
31
Suppose we have a standard 52 card deck from which we draw five cards. What are the chances of drawing one or more Aces and one or more Kings? How do I calculate this?
I know that we can calculate the chance of drawing one or more Ace by using cumulative hypergeometric distribution (Population 52, Subpopulation 4, Sample Size 5). This gives us a result of 34.116%.
How do I then find the odds of drawing one or more kings as well?
I mistakenly thought I could "set aside" the Ace, reducing the Population and Sample Size by 1, then calculate the chance for drawing one or more kings out of the rest (28%). Multiplying the two would find the total odds. However, the results off this - 9.765% - is well under the 10% that I continuously get out of a game simulator (multiple tries of 10 million hands).
So then, how do I calculate the above question, and - more directly - what is the correct way to calculate two different cumulative hypergeometric distributions that occur simultaneously?
probability probability-distributions hypergeometric-function card-games
asked Aug 10 at 22:17
Courtney Stucker
31
asked Aug 10 at 22:17
Courtney Stucker
31
asked Aug 10 at 22:17
Courtney Stucker
31
asked Aug 10 at 22:17
Courtney Stucker
31
31
You can use the multivariate hypergeometric distribution.
– JMoravitz
Aug 10 at 22:20
1
"Multiplying the two would find total odds..." Remember, $Pr(Acap B)=Pr(A)times Pr(B)$ is true only when $A$ and $B$ are independent events. These are certainly not independent.
– JMoravitz
Aug 10 at 22:22
add a comment |Â
You can use the multivariate hypergeometric distribution.
– JMoravitz
Aug 10 at 22:20
1
"Multiplying the two would find total odds..." Remember, $Pr(Acap B)=Pr(A)times Pr(B)$ is true only when $A$ and $B$ are independent events. These are certainly not independent.
– JMoravitz
Aug 10 at 22:22
You can use the multivariate hypergeometric distribution.
– JMoravitz
Aug 10 at 22:20
You can use the multivariate hypergeometric distribution.
– JMoravitz
Aug 10 at 22:20
1
1
"Multiplying the two would find total odds..." Remember, $Pr(Acap B)=Pr(A)times Pr(B)$ is true only when $A$ and $B$ are independent events. These are certainly not independent.
– JMoravitz
Aug 10 at 22:22
"Multiplying the two would find total odds..." Remember, $Pr(Acap B)=Pr(A)times Pr(B)$ is true only when $A$ and $B$ are independent events. These are certainly not independent.
– JMoravitz
Aug 10 at 22:22
add a comment |Â
1 Answer
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Let $A$ represent the event that you have at least one ace in your hand.
Let $B$ represent the event that you have at least one king in your hand.
You ask how to calculate $Pr(Acap B)$.
To do this, let us try to calculate the complementary event and apply some common identities.
$Pr(Acap B)=1-Pr((Acap B)^c)=1-Pr(A^ccup B^c)=1-Pr(A^c)-Pr(B^c)+Pr(A^ccap B^c)$
Each of these terms at the end are relatively easy to calculate once you recognize what they represent. $Pr(A^c)$ represents the probability that there are no aces in your hand. $Pr(A^ccap B^c)$ represents the probability that there are no aces and there are no kings in your hand, etc...
Running the calculations, we arrive at a final answer of:
$$Pr(Acap B)=1-dfracbinom485binom525-dfracbinom485binom525+dfracbinom445binom525approx 0.1001785$$
answered Aug 10 at 22:36
JMoravitz
44.4k33481
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Let $A$ represent the event that you have at least one ace in your hand.
Let $B$ represent the event that you have at least one king in your hand.
You ask how to calculate $Pr(Acap B)$.
To do this, let us try to calculate the complementary event and apply some common identities.
$Pr(Acap B)=1-Pr((Acap B)^c)=1-Pr(A^ccup B^c)=1-Pr(A^c)-Pr(B^c)+Pr(A^ccap B^c)$
Each of these terms at the end are relatively easy to calculate once you recognize what they represent. $Pr(A^c)$ represents the probability that there are no aces in your hand. $Pr(A^ccap B^c)$ represents the probability that there are no aces and there are no kings in your hand, etc...
Running the calculations, we arrive at a final answer of:
$$Pr(Acap B)=1-dfracbinom485binom525-dfracbinom485binom525+dfracbinom445binom525approx 0.1001785$$
answered Aug 10 at 22:36
JMoravitz
44.4k33481
add a comment |Â
up vote
0
down vote
accepted
Let $A$ represent the event that you have at least one ace in your hand.
Let $B$ represent the event that you have at least one king in your hand.
You ask how to calculate $Pr(Acap B)$.
To do this, let us try to calculate the complementary event and apply some common identities.
$Pr(Acap B)=1-Pr((Acap B)^c)=1-Pr(A^ccup B^c)=1-Pr(A^c)-Pr(B^c)+Pr(A^ccap B^c)$
Each of these terms at the end are relatively easy to calculate once you recognize what they represent. $Pr(A^c)$ represents the probability that there are no aces in your hand. $Pr(A^ccap B^c)$ represents the probability that there are no aces and there are no kings in your hand, etc...
Running the calculations, we arrive at a final answer of:
$$Pr(Acap B)=1-dfracbinom485binom525-dfracbinom485binom525+dfracbinom445binom525approx 0.1001785$$
answered Aug 10 at 22:36
JMoravitz
44.4k33481
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let $A$ represent the event that you have at least one ace in your hand.
Let $B$ represent the event that you have at least one king in your hand.
You ask how to calculate $Pr(Acap B)$.
To do this, let us try to calculate the complementary event and apply some common identities.
$Pr(Acap B)=1-Pr((Acap B)^c)=1-Pr(A^ccup B^c)=1-Pr(A^c)-Pr(B^c)+Pr(A^ccap B^c)$
Each of these terms at the end are relatively easy to calculate once you recognize what they represent. $Pr(A^c)$ represents the probability that there are no aces in your hand. $Pr(A^ccap B^c)$ represents the probability that there are no aces and there are no kings in your hand, etc...
Running the calculations, we arrive at a final answer of:
$$Pr(Acap B)=1-dfracbinom485binom525-dfracbinom485binom525+dfracbinom445binom525approx 0.1001785$$
answered Aug 10 at 22:36
JMoravitz
44.4k33481
Let $A$ represent the event that you have at least one ace in your hand.
Let $B$ represent the event that you have at least one king in your hand.
You ask how to calculate $Pr(Acap B)$.
To do this, let us try to calculate the complementary event and apply some common identities.
$Pr(Acap B)=1-Pr((Acap B)^c)=1-Pr(A^ccup B^c)=1-Pr(A^c)-Pr(B^c)+Pr(A^ccap B^c)$
Each of these terms at the end are relatively easy to calculate once you recognize what they represent. $Pr(A^c)$ represents the probability that there are no aces in your hand. $Pr(A^ccap B^c)$ represents the probability that there are no aces and there are no kings in your hand, etc...
Running the calculations, we arrive at a final answer of:
$$Pr(Acap B)=1-dfracbinom485binom525-dfracbinom485binom525+dfracbinom445binom525approx 0.1001785$$
answered Aug 10 at 22:36
JMoravitz
44.4k33481
answered Aug 10 at 22:36
JMoravitz
44.4k33481
answered Aug 10 at 22:36
JMoravitz
44.4k33481
answered Aug 10 at 22:36
JMoravitz
44.4k33481
44.4k33481
add a comment |Â
add a comment |Â
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You can use the multivariate hypergeometric distribution.
– JMoravitz
Aug 10 at 22:20
1
"Multiplying the two would find total odds..." Remember, $Pr(Acap B)=Pr(A)times Pr(B)$ is true only when $A$ and $B$ are independent events. These are certainly not independent.
– JMoravitz
Aug 10 at 22:22