What is the probability that a five card poker hand contains at most 1 ace?
What is the probability that a five card poker hand contains at most 1 ace?
I know that if it is at least $1$:
$$frac{C(4,1) cdot C(51,4)}{C(52,5)}$$
I also know that if it is exactly $1$, then it is:
$$frac{C(4,1) cdot C(48,4)}{C(52,5)}$$
What would at most be:
$$1 - frac{C(4,1) cdot C(51,4)}{C(52,5)}$$
and if so what is non complement representation?
combinatorics combinations
edited Nov 27 at 12:38
N. F. Taussig
43.5k93355
asked Nov 27 at 8:34
anon6789
12
add a comment |
What is the probability that a five card poker hand contains at most 1 ace?
I know that if it is at least $1$:
$$frac{C(4,1) cdot C(51,4)}{C(52,5)}$$
I also know that if it is exactly $1$, then it is:
$$frac{C(4,1) cdot C(48,4)}{C(52,5)}$$
What would at most be:
$$1 - frac{C(4,1) cdot C(51,4)}{C(52,5)}$$
and if so what is non complement representation?
combinatorics combinations
edited Nov 27 at 12:38
N. F. Taussig
43.5k93355
asked Nov 27 at 8:34
anon6789
12
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 27 at 12:30
I think your first expression (for at least one Ace) is wrong. How did you calculate it?
– TonyK
Nov 27 at 12:43
add a comment |
What is the probability that a five card poker hand contains at most 1 ace?
I know that if it is at least $1$:
$$frac{C(4,1) cdot C(51,4)}{C(52,5)}$$
I also know that if it is exactly $1$, then it is:
$$frac{C(4,1) cdot C(48,4)}{C(52,5)}$$
What would at most be:
$$1 - frac{C(4,1) cdot C(51,4)}{C(52,5)}$$
and if so what is non complement representation?
combinatorics combinations
edited Nov 27 at 12:38
N. F. Taussig
43.5k93355
asked Nov 27 at 8:34
anon6789
12
What is the probability that a five card poker hand contains at most 1 ace?
I know that if it is at least $1$:
$$frac{C(4,1) cdot C(51,4)}{C(52,5)}$$
I also know that if it is exactly $1$, then it is:
$$frac{C(4,1) cdot C(48,4)}{C(52,5)}$$
What would at most be:
$$1 - frac{C(4,1) cdot C(51,4)}{C(52,5)}$$
and if so what is non complement representation?
combinatorics combinations
combinatorics combinations
edited Nov 27 at 12:38
N. F. Taussig
43.5k93355
asked Nov 27 at 8:34
anon6789
12
edited Nov 27 at 12:38
N. F. Taussig
43.5k93355
asked Nov 27 at 8:34
anon6789
12
edited Nov 27 at 12:38
N. F. Taussig
43.5k93355
edited Nov 27 at 12:38
N. F. Taussig
43.5k93355
edited Nov 27 at 12:38
N. F. Taussig
43.5k93355
43.5k93355
asked Nov 27 at 8:34
anon6789
12
asked Nov 27 at 8:34
anon6789
12
asked Nov 27 at 8:34
anon6789
12
12
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 27 at 12:30
I think your first expression (for at least one Ace) is wrong. How did you calculate it?
– TonyK
Nov 27 at 12:43
add a comment |
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 27 at 12:30
I think your first expression (for at least one Ace) is wrong. How did you calculate it?
– TonyK
Nov 27 at 12:43
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 27 at 12:30
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 27 at 12:30
I think your first expression (for at least one Ace) is wrong. How did you calculate it?
– TonyK
Nov 27 at 12:43
I think your first expression (for at least one Ace) is wrong. How did you calculate it?
– TonyK
Nov 27 at 12:43
add a comment |
1 Answer
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If a hand has at most one ace, it has either no aces or one ace.
The number of five-card hands with no aces is
$$binom{4}{0}binom{48}{5} = binom{48}{5}$$
since all five cards must be chosen from the $52 - 4 = 48$ cards that are not aces.
The number of five-card hands with exactly one ace is
$$binom{4}{1}binom{48}{4}$$
since one of the four aces and four of the other $48$ cards in the deck must be selected.
Since there are
$$binom{52}{5}$$
possible five-card hands, the probability of selecting at most one ace is
$$Pr(text{at most one ace}) = frac{dbinom{48}{5} + dbinom{4}{1}binom{48}{4}}{dbinom{52}{5}}$$
What errors did you make?
The number of hands with at least one ace is found by subtracting the number of hands with no aces from the total number of hands, which is
$$binom{52}{5} - binom{48}{5}$$
Alternatively, you could add the number of hands with exactly $k$ aces, where $1 leq k leq 5$.
$$sum_{k = 1}^{4} binom{4}{k}binom{48}{5 - k} = binom{4}{1}binom{48}{4} + binom{4}{2}binom{48}{3} + binom{4}{3}binom{48}{2} + binom{4}{4}binom{48}{1}$$
Also, at most $1$ is the complement of at least $2$.
answered Nov 27 at 12:38
N. F. Taussig
43.5k93355
add a comment |
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1 Answer
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1 Answer
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If a hand has at most one ace, it has either no aces or one ace.
The number of five-card hands with no aces is
$$binom{4}{0}binom{48}{5} = binom{48}{5}$$
since all five cards must be chosen from the $52 - 4 = 48$ cards that are not aces.
The number of five-card hands with exactly one ace is
$$binom{4}{1}binom{48}{4}$$
since one of the four aces and four of the other $48$ cards in the deck must be selected.
Since there are
$$binom{52}{5}$$
possible five-card hands, the probability of selecting at most one ace is
$$Pr(text{at most one ace}) = frac{dbinom{48}{5} + dbinom{4}{1}binom{48}{4}}{dbinom{52}{5}}$$
What errors did you make?
The number of hands with at least one ace is found by subtracting the number of hands with no aces from the total number of hands, which is
$$binom{52}{5} - binom{48}{5}$$
Alternatively, you could add the number of hands with exactly $k$ aces, where $1 leq k leq 5$.
$$sum_{k = 1}^{4} binom{4}{k}binom{48}{5 - k} = binom{4}{1}binom{48}{4} + binom{4}{2}binom{48}{3} + binom{4}{3}binom{48}{2} + binom{4}{4}binom{48}{1}$$
Also, at most $1$ is the complement of at least $2$.
answered Nov 27 at 12:38
N. F. Taussig
43.5k93355
add a comment |
If a hand has at most one ace, it has either no aces or one ace.
The number of five-card hands with no aces is
$$binom{4}{0}binom{48}{5} = binom{48}{5}$$
since all five cards must be chosen from the $52 - 4 = 48$ cards that are not aces.
The number of five-card hands with exactly one ace is
$$binom{4}{1}binom{48}{4}$$
since one of the four aces and four of the other $48$ cards in the deck must be selected.
Since there are
$$binom{52}{5}$$
possible five-card hands, the probability of selecting at most one ace is
$$Pr(text{at most one ace}) = frac{dbinom{48}{5} + dbinom{4}{1}binom{48}{4}}{dbinom{52}{5}}$$
What errors did you make?
The number of hands with at least one ace is found by subtracting the number of hands with no aces from the total number of hands, which is
$$binom{52}{5} - binom{48}{5}$$
Alternatively, you could add the number of hands with exactly $k$ aces, where $1 leq k leq 5$.
$$sum_{k = 1}^{4} binom{4}{k}binom{48}{5 - k} = binom{4}{1}binom{48}{4} + binom{4}{2}binom{48}{3} + binom{4}{3}binom{48}{2} + binom{4}{4}binom{48}{1}$$
Also, at most $1$ is the complement of at least $2$.
answered Nov 27 at 12:38
N. F. Taussig
43.5k93355
add a comment |
If a hand has at most one ace, it has either no aces or one ace.
The number of five-card hands with no aces is
$$binom{4}{0}binom{48}{5} = binom{48}{5}$$
since all five cards must be chosen from the $52 - 4 = 48$ cards that are not aces.
The number of five-card hands with exactly one ace is
$$binom{4}{1}binom{48}{4}$$
since one of the four aces and four of the other $48$ cards in the deck must be selected.
Since there are
$$binom{52}{5}$$
possible five-card hands, the probability of selecting at most one ace is
$$Pr(text{at most one ace}) = frac{dbinom{48}{5} + dbinom{4}{1}binom{48}{4}}{dbinom{52}{5}}$$
What errors did you make?
The number of hands with at least one ace is found by subtracting the number of hands with no aces from the total number of hands, which is
$$binom{52}{5} - binom{48}{5}$$
Alternatively, you could add the number of hands with exactly $k$ aces, where $1 leq k leq 5$.
$$sum_{k = 1}^{4} binom{4}{k}binom{48}{5 - k} = binom{4}{1}binom{48}{4} + binom{4}{2}binom{48}{3} + binom{4}{3}binom{48}{2} + binom{4}{4}binom{48}{1}$$
Also, at most $1$ is the complement of at least $2$.
answered Nov 27 at 12:38
N. F. Taussig
43.5k93355
If a hand has at most one ace, it has either no aces or one ace.
The number of five-card hands with no aces is
$$binom{4}{0}binom{48}{5} = binom{48}{5}$$
since all five cards must be chosen from the $52 - 4 = 48$ cards that are not aces.
The number of five-card hands with exactly one ace is
$$binom{4}{1}binom{48}{4}$$
since one of the four aces and four of the other $48$ cards in the deck must be selected.
Since there are
$$binom{52}{5}$$
possible five-card hands, the probability of selecting at most one ace is
$$Pr(text{at most one ace}) = frac{dbinom{48}{5} + dbinom{4}{1}binom{48}{4}}{dbinom{52}{5}}$$
What errors did you make?
The number of hands with at least one ace is found by subtracting the number of hands with no aces from the total number of hands, which is
$$binom{52}{5} - binom{48}{5}$$
Alternatively, you could add the number of hands with exactly $k$ aces, where $1 leq k leq 5$.
$$sum_{k = 1}^{4} binom{4}{k}binom{48}{5 - k} = binom{4}{1}binom{48}{4} + binom{4}{2}binom{48}{3} + binom{4}{3}binom{48}{2} + binom{4}{4}binom{48}{1}$$
Also, at most $1$ is the complement of at least $2$.
answered Nov 27 at 12:38
N. F. Taussig
43.5k93355
answered Nov 27 at 12:38
N. F. Taussig
43.5k93355
answered Nov 27 at 12:38
N. F. Taussig
43.5k93355
answered Nov 27 at 12:38
N. F. Taussig
43.5k93355
43.5k93355
add a comment |
add a comment |
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Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 27 at 12:30
I think your first expression (for at least one Ace) is wrong. How did you calculate it?
– TonyK
Nov 27 at 12:43