Mixing
5 cards are chosen from a standard deck. What is the probability that we get all four aces, plus the king of spades?
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We have $binom525$ ways of doing this. This will be our denominator.
We want to select all 4 aces, there are there are exactly $binom44$ ways of doing this.
Now we have selected 4 cards, and we need to select one more. That one card has to be the king of spades. Out of the 48 remaining cards, only one of them is our wanted one. This can be chosen in $binom481$ ways.
So is the answer:
$$dfracbinom44binom481binom525=dfrac154145?$$
Is this correct?
If we look at it another way:
Then then our probability for the aces and specific king are $(4/52)times (3/51) times (2/50) times (1/49) times (1/48)$ which is a completely different here.
Which is the correct approach?
probability statistics discrete-mathematics
asked 2 hours ago
K Split X
3,862827
add a comment |Â
up vote
2
down vote
favorite
We have $binom525$ ways of doing this. This will be our denominator.
We want to select all 4 aces, there are there are exactly $binom44$ ways of doing this.
Now we have selected 4 cards, and we need to select one more. That one card has to be the king of spades. Out of the 48 remaining cards, only one of them is our wanted one. This can be chosen in $binom481$ ways.
So is the answer:
$$dfracbinom44binom481binom525=dfrac154145?$$
Is this correct?
If we look at it another way:
Then then our probability for the aces and specific king are $(4/52)times (3/51) times (2/50) times (1/49) times (1/48)$ which is a completely different here.
Which is the correct approach?
probability statistics discrete-mathematics
asked 2 hours ago
K Split X
3,862827
numerator should be 1 if you want a specific combination
– Nong
2 hours ago
So the second one is correct?
– K Split X
2 hours ago
No, it seems like you're permuting those four aces.
– Nong
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
We have $binom525$ ways of doing this. This will be our denominator.
We want to select all 4 aces, there are there are exactly $binom44$ ways of doing this.
Now we have selected 4 cards, and we need to select one more. That one card has to be the king of spades. Out of the 48 remaining cards, only one of them is our wanted one. This can be chosen in $binom481$ ways.
So is the answer:
$$dfracbinom44binom481binom525=dfrac154145?$$
Is this correct?
If we look at it another way:
Then then our probability for the aces and specific king are $(4/52)times (3/51) times (2/50) times (1/49) times (1/48)$ which is a completely different here.
Which is the correct approach?
probability statistics discrete-mathematics
asked 2 hours ago
K Split X
3,862827
We have $binom525$ ways of doing this. This will be our denominator.
We want to select all 4 aces, there are there are exactly $binom44$ ways of doing this.
Now we have selected 4 cards, and we need to select one more. That one card has to be the king of spades. Out of the 48 remaining cards, only one of them is our wanted one. This can be chosen in $binom481$ ways.
So is the answer:
$$dfracbinom44binom481binom525=dfrac154145?$$
Is this correct?
If we look at it another way:
Then then our probability for the aces and specific king are $(4/52)times (3/51) times (2/50) times (1/49) times (1/48)$ which is a completely different here.
Which is the correct approach?
probability statistics discrete-mathematics
probability statistics discrete-mathematics
asked 2 hours ago
K Split X
3,862827
asked 2 hours ago
K Split X
3,862827
asked 2 hours ago
K Split X
3,862827
asked 2 hours ago
K Split X
3,862827
asked 2 hours ago
K Split X
3,862827
3,862827
numerator should be 1 if you want a specific combination
– Nong
2 hours ago
So the second one is correct?
– K Split X
2 hours ago
No, it seems like you're permuting those four aces.
– Nong
2 hours ago
add a comment |Â
numerator should be 1 if you want a specific combination
– Nong
2 hours ago
So the second one is correct?
– K Split X
2 hours ago
No, it seems like you're permuting those four aces.
– Nong
2 hours ago
numerator should be 1 if you want a specific combination
– Nong
2 hours ago
numerator should be 1 if you want a specific combination
– Nong
2 hours ago
So the second one is correct?
– K Split X
2 hours ago
So the second one is correct?
– K Split X
2 hours ago
No, it seems like you're permuting those four aces.
– Nong
2 hours ago
No, it seems like you're permuting those four aces.
– Nong
2 hours ago
add a comment |Â
2 Answers
2
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up vote
4
down vote
Your first computation gets the chance you get four aces and any other card. The $48 choose 1$ factor is selecting the other card. If you want specifically the king of spades it should be $1$, so the chance is $48$ times less. Your second computation gets the chance you draw the four aces first in some order, then draw the king of spades. You should multiply by $5$ for the number of positions the king can be in. Neither is correct for the question in the title.
answered 2 hours ago
Ross Millikan
279k23189355
add a comment |Â
up vote
1
down vote
Assuming you just want those 5 cards in any order:
Since you have specified all 5 specific cards that you want, you don't need to consider the aces separately from the king. The probability of selecting these 5 cards is then
(Probability choosing any one of the 5) $times$ (Probability of choosing one of the remaining 4) $times$ ..., i.e. $$frac552 times frac451 times frac350 times frac249 times frac148 = frac154145 times frac148,$$ just to confirm what Ross Millikan said.
If, instead, you want the 4 aces before the king, we have $$frac452 times frac351 times frac250 times frac149 times frac148.$$
answered 2 hours ago
combinatoricky
614
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Your first computation gets the chance you get four aces and any other card. The $48 choose 1$ factor is selecting the other card. If you want specifically the king of spades it should be $1$, so the chance is $48$ times less. Your second computation gets the chance you draw the four aces first in some order, then draw the king of spades. You should multiply by $5$ for the number of positions the king can be in. Neither is correct for the question in the title.
answered 2 hours ago
Ross Millikan
279k23189355
add a comment |Â
up vote
4
down vote
Your first computation gets the chance you get four aces and any other card. The $48 choose 1$ factor is selecting the other card. If you want specifically the king of spades it should be $1$, so the chance is $48$ times less. Your second computation gets the chance you draw the four aces first in some order, then draw the king of spades. You should multiply by $5$ for the number of positions the king can be in. Neither is correct for the question in the title.
answered 2 hours ago
Ross Millikan
279k23189355
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Your first computation gets the chance you get four aces and any other card. The $48 choose 1$ factor is selecting the other card. If you want specifically the king of spades it should be $1$, so the chance is $48$ times less. Your second computation gets the chance you draw the four aces first in some order, then draw the king of spades. You should multiply by $5$ for the number of positions the king can be in. Neither is correct for the question in the title.
answered 2 hours ago
Ross Millikan
279k23189355
Your first computation gets the chance you get four aces and any other card. The $48 choose 1$ factor is selecting the other card. If you want specifically the king of spades it should be $1$, so the chance is $48$ times less. Your second computation gets the chance you draw the four aces first in some order, then draw the king of spades. You should multiply by $5$ for the number of positions the king can be in. Neither is correct for the question in the title.
answered 2 hours ago
Ross Millikan
279k23189355
answered 2 hours ago
Ross Millikan
279k23189355
answered 2 hours ago
Ross Millikan
279k23189355
answered 2 hours ago
Ross Millikan
279k23189355
279k23189355
add a comment |Â
add a comment |Â
up vote
1
down vote
Assuming you just want those 5 cards in any order:
Since you have specified all 5 specific cards that you want, you don't need to consider the aces separately from the king. The probability of selecting these 5 cards is then
(Probability choosing any one of the 5) $times$ (Probability of choosing one of the remaining 4) $times$ ..., i.e. $$frac552 times frac451 times frac350 times frac249 times frac148 = frac154145 times frac148,$$ just to confirm what Ross Millikan said.
If, instead, you want the 4 aces before the king, we have $$frac452 times frac351 times frac250 times frac149 times frac148.$$
answered 2 hours ago
combinatoricky
614
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
Assuming you just want those 5 cards in any order:
Since you have specified all 5 specific cards that you want, you don't need to consider the aces separately from the king. The probability of selecting these 5 cards is then
(Probability choosing any one of the 5) $times$ (Probability of choosing one of the remaining 4) $times$ ..., i.e. $$frac552 times frac451 times frac350 times frac249 times frac148 = frac154145 times frac148,$$ just to confirm what Ross Millikan said.
If, instead, you want the 4 aces before the king, we have $$frac452 times frac351 times frac250 times frac149 times frac148.$$
answered 2 hours ago
combinatoricky
614
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Assuming you just want those 5 cards in any order:
Since you have specified all 5 specific cards that you want, you don't need to consider the aces separately from the king. The probability of selecting these 5 cards is then
(Probability choosing any one of the 5) $times$ (Probability of choosing one of the remaining 4) $times$ ..., i.e. $$frac552 times frac451 times frac350 times frac249 times frac148 = frac154145 times frac148,$$ just to confirm what Ross Millikan said.
If, instead, you want the 4 aces before the king, we have $$frac452 times frac351 times frac250 times frac149 times frac148.$$
answered 2 hours ago
combinatoricky
614
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Assuming you just want those 5 cards in any order:
Since you have specified all 5 specific cards that you want, you don't need to consider the aces separately from the king. The probability of selecting these 5 cards is then
(Probability choosing any one of the 5) $times$ (Probability of choosing one of the remaining 4) $times$ ..., i.e. $$frac552 times frac451 times frac350 times frac249 times frac148 = frac154145 times frac148,$$ just to confirm what Ross Millikan said.
If, instead, you want the 4 aces before the king, we have $$frac452 times frac351 times frac250 times frac149 times frac148.$$
answered 2 hours ago
combinatoricky
614
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
answered 2 hours ago
combinatoricky
614
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
combinatoricky
614
answered 2 hours ago
combinatoricky
614
614
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
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numerator should be 1 if you want a specific combination
– Nong
2 hours ago
So the second one is correct?
– K Split X
2 hours ago
No, it seems like you're permuting those four aces.
– Nong
2 hours ago