From a deck of 52 cards, extract 10. In how many combinations do you get at least one ace?
If from a deck of 52 cards, I extract 10. In how many combinations do you get at least one ace?
I have come up with two possible answers, but I don't know which one is the right one and why.
So one is $$4{51 choose 9}$$
the reasoning is that first I extract an ace from the four that there are, and then I have ${51 choose 9}$ combinations for the other 9 cards.
The second is
$${52 choose 10} - {48 choose 10}$$ reasoning that there are ${52 choose 10}$ total combinations of ten cards and I subtract ${48 choose 10}$ combinations without any aces.
So from other questions it seems that the first one is the correct answer, but why would the second one be wrong?
Thanks for the answers.
combinatorics combinations
edited Nov 27 '18 at 23:56
N. F. Taussig
43.5k93355
asked Nov 27 '18 at 16:45
Miguel Vázquez Caraballo
41
|
show 2 more comments
If from a deck of 52 cards, I extract 10. In how many combinations do you get at least one ace?
I have come up with two possible answers, but I don't know which one is the right one and why.
So one is $$4{51 choose 9}$$
the reasoning is that first I extract an ace from the four that there are, and then I have ${51 choose 9}$ combinations for the other 9 cards.
The second is
$${52 choose 10} - {48 choose 10}$$ reasoning that there are ${52 choose 10}$ total combinations of ten cards and I subtract ${48 choose 10}$ combinations without any aces.
So from other questions it seems that the first one is the correct answer, but why would the second one be wrong?
Thanks for the answers.
combinatorics combinations
edited Nov 27 '18 at 23:56
N. F. Taussig
43.5k93355
asked Nov 27 '18 at 16:45
Miguel Vázquez Caraballo
41
3
Your first method counts hands with more than one ace multiple times.
– lulu
Nov 27 '18 at 16:49
@lulu That's important as a point, but the question wants to count all hands with at least one ace. There is still overcounting going on though.
– amWhy
Nov 27 '18 at 16:50
@amWhy lulu is pointing out that certain hands are counted too many times.
– Théophile
Nov 27 '18 at 16:53
2
The second method is correct. It is the best way for the general question. - to obtain probability of at least one, calculate the probability of none and subtract from $1$.
– herb steinberg
Nov 27 '18 at 17:00
2
@amWhy Not sure I see your point. if ${X_i}_{i=1}^8$ denote specific non-aces, then the OP's method counts $Aspadesuit, Aheartsuit, {X_i}_{i=1}^8$ as one case, and the equivalent reordered hand $Aheartsuit, Aspadesuit, {X_i}_{i=1}^8$ as another. I understand the OP is counting hands containing at least one ace.
– lulu
Nov 27 '18 at 17:01
|
show 2 more comments
If from a deck of 52 cards, I extract 10. In how many combinations do you get at least one ace?
I have come up with two possible answers, but I don't know which one is the right one and why.
So one is $$4{51 choose 9}$$
the reasoning is that first I extract an ace from the four that there are, and then I have ${51 choose 9}$ combinations for the other 9 cards.
The second is
$${52 choose 10} - {48 choose 10}$$ reasoning that there are ${52 choose 10}$ total combinations of ten cards and I subtract ${48 choose 10}$ combinations without any aces.
So from other questions it seems that the first one is the correct answer, but why would the second one be wrong?
Thanks for the answers.
combinatorics combinations
edited Nov 27 '18 at 23:56
N. F. Taussig
43.5k93355
asked Nov 27 '18 at 16:45
Miguel Vázquez Caraballo
41
If from a deck of 52 cards, I extract 10. In how many combinations do you get at least one ace?
I have come up with two possible answers, but I don't know which one is the right one and why.
So one is $$4{51 choose 9}$$
the reasoning is that first I extract an ace from the four that there are, and then I have ${51 choose 9}$ combinations for the other 9 cards.
The second is
$${52 choose 10} - {48 choose 10}$$ reasoning that there are ${52 choose 10}$ total combinations of ten cards and I subtract ${48 choose 10}$ combinations without any aces.
So from other questions it seems that the first one is the correct answer, but why would the second one be wrong?
Thanks for the answers.
combinatorics combinations
combinatorics combinations
edited Nov 27 '18 at 23:56
N. F. Taussig
43.5k93355
asked Nov 27 '18 at 16:45
Miguel Vázquez Caraballo
41
edited Nov 27 '18 at 23:56
N. F. Taussig
43.5k93355
asked Nov 27 '18 at 16:45
Miguel Vázquez Caraballo
41
edited Nov 27 '18 at 23:56
N. F. Taussig
43.5k93355
edited Nov 27 '18 at 23:56
N. F. Taussig
43.5k93355
edited Nov 27 '18 at 23:56
N. F. Taussig
43.5k93355
43.5k93355
asked Nov 27 '18 at 16:45
Miguel Vázquez Caraballo
41
asked Nov 27 '18 at 16:45
Miguel Vázquez Caraballo
41
asked Nov 27 '18 at 16:45
Miguel Vázquez Caraballo
41
41
3
Your first method counts hands with more than one ace multiple times.
– lulu
Nov 27 '18 at 16:49
@lulu That's important as a point, but the question wants to count all hands with at least one ace. There is still overcounting going on though.
– amWhy
Nov 27 '18 at 16:50
@amWhy lulu is pointing out that certain hands are counted too many times.
– Théophile
Nov 27 '18 at 16:53
2
The second method is correct. It is the best way for the general question. - to obtain probability of at least one, calculate the probability of none and subtract from $1$.
– herb steinberg
Nov 27 '18 at 17:00
2
@amWhy Not sure I see your point. if ${X_i}_{i=1}^8$ denote specific non-aces, then the OP's method counts $Aspadesuit, Aheartsuit, {X_i}_{i=1}^8$ as one case, and the equivalent reordered hand $Aheartsuit, Aspadesuit, {X_i}_{i=1}^8$ as another. I understand the OP is counting hands containing at least one ace.
– lulu
Nov 27 '18 at 17:01
|
show 2 more comments
3
Your first method counts hands with more than one ace multiple times.
– lulu
Nov 27 '18 at 16:49
@lulu That's important as a point, but the question wants to count all hands with at least one ace. There is still overcounting going on though.
– amWhy
Nov 27 '18 at 16:50
@amWhy lulu is pointing out that certain hands are counted too many times.
– Théophile
Nov 27 '18 at 16:53
2
The second method is correct. It is the best way for the general question. - to obtain probability of at least one, calculate the probability of none and subtract from $1$.
– herb steinberg
Nov 27 '18 at 17:00
2
@amWhy Not sure I see your point. if ${X_i}_{i=1}^8$ denote specific non-aces, then the OP's method counts $Aspadesuit, Aheartsuit, {X_i}_{i=1}^8$ as one case, and the equivalent reordered hand $Aheartsuit, Aspadesuit, {X_i}_{i=1}^8$ as another. I understand the OP is counting hands containing at least one ace.
– lulu
Nov 27 '18 at 17:01
3
3
Your first method counts hands with more than one ace multiple times.
– lulu
Nov 27 '18 at 16:49
Your first method counts hands with more than one ace multiple times.
– lulu
Nov 27 '18 at 16:49
@lulu That's important as a point, but the question wants to count all hands with at least one ace. There is still overcounting going on though.
– amWhy
Nov 27 '18 at 16:50
@lulu That's important as a point, but the question wants to count all hands with at least one ace. There is still overcounting going on though.
– amWhy
Nov 27 '18 at 16:50
@amWhy lulu is pointing out that certain hands are counted too many times.
– Théophile
Nov 27 '18 at 16:53
@amWhy lulu is pointing out that certain hands are counted too many times.
– Théophile
Nov 27 '18 at 16:53
2
2
The second method is correct. It is the best way for the general question. - to obtain probability of at least one, calculate the probability of none and subtract from $1$.
– herb steinberg
Nov 27 '18 at 17:00
The second method is correct. It is the best way for the general question. - to obtain probability of at least one, calculate the probability of none and subtract from $1$.
– herb steinberg
Nov 27 '18 at 17:00
2
2
@amWhy Not sure I see your point. if ${X_i}_{i=1}^8$ denote specific non-aces, then the OP's method counts $Aspadesuit, Aheartsuit, {X_i}_{i=1}^8$ as one case, and the equivalent reordered hand $Aheartsuit, Aspadesuit, {X_i}_{i=1}^8$ as another. I understand the OP is counting hands containing at least one ace.
– lulu
Nov 27 '18 at 17:01
@amWhy Not sure I see your point. if ${X_i}_{i=1}^8$ denote specific non-aces, then the OP's method counts $Aspadesuit, Aheartsuit, {X_i}_{i=1}^8$ as one case, and the equivalent reordered hand $Aheartsuit, Aspadesuit, {X_i}_{i=1}^8$ as another. I understand the OP is counting hands containing at least one ace.
– lulu
Nov 27 '18 at 17:01
|
show 2 more comments
2 Answers
2
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The first approach needs to be modified to ${4choose1}{48choose 9}+{4choose2}{48choose 8}+{4choose 3}{48choose 7}+{4choose 4}{48choose 6}$, for instance.
The second is correct.
answered Nov 27 '18 at 23:19
Chris Custer
10.8k3724
add a comment |
To see why your first answer is not correct, assume you have a deck of 3 with 2 aces and you'll draw 2 cards with at least one ace. Call the cards $A_1,A_2,B$
Based on your logic, you compute $2 {2 choose 1}=4$ combinations. However, clearly you have only $3$. This is due to double counting ${A_1,A_2}$
answered Nov 27 '18 at 17:10
karakfa
1,973811
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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The first approach needs to be modified to ${4choose1}{48choose 9}+{4choose2}{48choose 8}+{4choose 3}{48choose 7}+{4choose 4}{48choose 6}$, for instance.
The second is correct.
answered Nov 27 '18 at 23:19
Chris Custer
10.8k3724
add a comment |
The first approach needs to be modified to ${4choose1}{48choose 9}+{4choose2}{48choose 8}+{4choose 3}{48choose 7}+{4choose 4}{48choose 6}$, for instance.
The second is correct.
answered Nov 27 '18 at 23:19
Chris Custer
10.8k3724
add a comment |
The first approach needs to be modified to ${4choose1}{48choose 9}+{4choose2}{48choose 8}+{4choose 3}{48choose 7}+{4choose 4}{48choose 6}$, for instance.
The second is correct.
answered Nov 27 '18 at 23:19
Chris Custer
10.8k3724
The first approach needs to be modified to ${4choose1}{48choose 9}+{4choose2}{48choose 8}+{4choose 3}{48choose 7}+{4choose 4}{48choose 6}$, for instance.
The second is correct.
answered Nov 27 '18 at 23:19
Chris Custer
10.8k3724
answered Nov 27 '18 at 23:19
Chris Custer
10.8k3724
answered Nov 27 '18 at 23:19
Chris Custer
10.8k3724
answered Nov 27 '18 at 23:19
Chris Custer
10.8k3724
10.8k3724
add a comment |
add a comment |
To see why your first answer is not correct, assume you have a deck of 3 with 2 aces and you'll draw 2 cards with at least one ace. Call the cards $A_1,A_2,B$
Based on your logic, you compute $2 {2 choose 1}=4$ combinations. However, clearly you have only $3$. This is due to double counting ${A_1,A_2}$
answered Nov 27 '18 at 17:10
karakfa
1,973811
add a comment |
To see why your first answer is not correct, assume you have a deck of 3 with 2 aces and you'll draw 2 cards with at least one ace. Call the cards $A_1,A_2,B$
Based on your logic, you compute $2 {2 choose 1}=4$ combinations. However, clearly you have only $3$. This is due to double counting ${A_1,A_2}$
answered Nov 27 '18 at 17:10
karakfa
1,973811
add a comment |
To see why your first answer is not correct, assume you have a deck of 3 with 2 aces and you'll draw 2 cards with at least one ace. Call the cards $A_1,A_2,B$
Based on your logic, you compute $2 {2 choose 1}=4$ combinations. However, clearly you have only $3$. This is due to double counting ${A_1,A_2}$
answered Nov 27 '18 at 17:10
karakfa
1,973811
To see why your first answer is not correct, assume you have a deck of 3 with 2 aces and you'll draw 2 cards with at least one ace. Call the cards $A_1,A_2,B$
Based on your logic, you compute $2 {2 choose 1}=4$ combinations. However, clearly you have only $3$. This is due to double counting ${A_1,A_2}$
answered Nov 27 '18 at 17:10
karakfa
1,973811
answered Nov 27 '18 at 17:10
karakfa
1,973811
answered Nov 27 '18 at 17:10
karakfa
1,973811
answered Nov 27 '18 at 17:10
karakfa
1,973811
1,973811
add a comment |
add a comment |
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3
Your first method counts hands with more than one ace multiple times.
– lulu
Nov 27 '18 at 16:49
@lulu That's important as a point, but the question wants to count all hands with at least one ace. There is still overcounting going on though.
– amWhy
Nov 27 '18 at 16:50
@amWhy lulu is pointing out that certain hands are counted too many times.
– Théophile
Nov 27 '18 at 16:53
2
The second method is correct. It is the best way for the general question. - to obtain probability of at least one, calculate the probability of none and subtract from $1$.
– herb steinberg
Nov 27 '18 at 17:00
2
@amWhy Not sure I see your point. if ${X_i}_{i=1}^8$ denote specific non-aces, then the OP's method counts $Aspadesuit, Aheartsuit, {X_i}_{i=1}^8$ as one case, and the equivalent reordered hand $Aheartsuit, Aspadesuit, {X_i}_{i=1}^8$ as another. I understand the OP is counting hands containing at least one ace.
– lulu
Nov 27 '18 at 17:01