Probability of picking 3 playing cards that belong to different suits
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If you have 52 playing cards with 4 different suits (13 spades, 13 clubs, 13 hearts, and 13 diamonds), what is the probability of picking 3 random cards that are all from a different suit?
I tried going about this by writing down all 52 cards and I found out that, if you were to pick a second card, the probability of the second card being from a different suit is 39/51, since there are 51 cards left, with 12 of them being from the same suit as the first card you picked (51-12=39).
However, when I tried to do the same thing with a third card, I got 26/50, since there are 50 cards left, and, assuming the first 2 cards you picked are from a different suit, the probability of the third card being from a different suit would be 50 cards - 12 cards of the same suit * 2 different suits =26 out of 50.
Unfortunately, I realized that it's also possible for the second card you pick to belong to the same suit as the first one, so, in that case, there would also be 50 cards left, and the probability is not the same since there would also be 50 cards left, but 39 cards would belong to a different suit, so I got 39/50.
After that, I realized that I had just managed to confuse myself more than I should have and couldn't really figure out where to go from there. Where did I go wrong and what is the proper way to solve this problem?
probability random
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If you have 52 playing cards with 4 different suits (13 spades, 13 clubs, 13 hearts, and 13 diamonds), what is the probability of picking 3 random cards that are all from a different suit?
I tried going about this by writing down all 52 cards and I found out that, if you were to pick a second card, the probability of the second card being from a different suit is 39/51, since there are 51 cards left, with 12 of them being from the same suit as the first card you picked (51-12=39).
However, when I tried to do the same thing with a third card, I got 26/50, since there are 50 cards left, and, assuming the first 2 cards you picked are from a different suit, the probability of the third card being from a different suit would be 50 cards - 12 cards of the same suit * 2 different suits =26 out of 50.
Unfortunately, I realized that it's also possible for the second card you pick to belong to the same suit as the first one, so, in that case, there would also be 50 cards left, and the probability is not the same since there would also be 50 cards left, but 39 cards would belong to a different suit, so I got 39/50.
After that, I realized that I had just managed to confuse myself more than I should have and couldn't really figure out where to go from there. Where did I go wrong and what is the proper way to solve this problem?
probability random
"I realized that it's also possible for the second card you pick to belong to the same suit as the first one" -- With respect this to this comment, it brings up a question. The question you seem to be asking in the opening parts of the question seems to imply that you want all three to be from different suits (i.e. Spade/Spade/Heart is not valid). Yet the quote seems to imply this is okay. This is an important nuance to settle for probably everyone's sake.
– Eevee Trainer
Nov 16 at 5:16
Sorry about that. You're right. Spade/Spade/Heart would not be a valid combination in this scenario. I nonetheless mentioned it, thinking that it could somehow affect the probability of the third card belonging to a different suit. I didn't mean to imply that it actually was a valid combination. Which is why I also said that I just confused myself even more, since this would contradict the initial statement.
– John Smith
Nov 16 at 5:20
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If you have 52 playing cards with 4 different suits (13 spades, 13 clubs, 13 hearts, and 13 diamonds), what is the probability of picking 3 random cards that are all from a different suit?
I tried going about this by writing down all 52 cards and I found out that, if you were to pick a second card, the probability of the second card being from a different suit is 39/51, since there are 51 cards left, with 12 of them being from the same suit as the first card you picked (51-12=39).
However, when I tried to do the same thing with a third card, I got 26/50, since there are 50 cards left, and, assuming the first 2 cards you picked are from a different suit, the probability of the third card being from a different suit would be 50 cards - 12 cards of the same suit * 2 different suits =26 out of 50.
Unfortunately, I realized that it's also possible for the second card you pick to belong to the same suit as the first one, so, in that case, there would also be 50 cards left, and the probability is not the same since there would also be 50 cards left, but 39 cards would belong to a different suit, so I got 39/50.
After that, I realized that I had just managed to confuse myself more than I should have and couldn't really figure out where to go from there. Where did I go wrong and what is the proper way to solve this problem?
probability random
If you have 52 playing cards with 4 different suits (13 spades, 13 clubs, 13 hearts, and 13 diamonds), what is the probability of picking 3 random cards that are all from a different suit?
I tried going about this by writing down all 52 cards and I found out that, if you were to pick a second card, the probability of the second card being from a different suit is 39/51, since there are 51 cards left, with 12 of them being from the same suit as the first card you picked (51-12=39).
However, when I tried to do the same thing with a third card, I got 26/50, since there are 50 cards left, and, assuming the first 2 cards you picked are from a different suit, the probability of the third card being from a different suit would be 50 cards - 12 cards of the same suit * 2 different suits =26 out of 50.
Unfortunately, I realized that it's also possible for the second card you pick to belong to the same suit as the first one, so, in that case, there would also be 50 cards left, and the probability is not the same since there would also be 50 cards left, but 39 cards would belong to a different suit, so I got 39/50.
After that, I realized that I had just managed to confuse myself more than I should have and couldn't really figure out where to go from there. Where did I go wrong and what is the proper way to solve this problem?
probability random
probability random
asked Nov 16 at 5:13
John Smith
1
1
"I realized that it's also possible for the second card you pick to belong to the same suit as the first one" -- With respect this to this comment, it brings up a question. The question you seem to be asking in the opening parts of the question seems to imply that you want all three to be from different suits (i.e. Spade/Spade/Heart is not valid). Yet the quote seems to imply this is okay. This is an important nuance to settle for probably everyone's sake.
– Eevee Trainer
Nov 16 at 5:16
Sorry about that. You're right. Spade/Spade/Heart would not be a valid combination in this scenario. I nonetheless mentioned it, thinking that it could somehow affect the probability of the third card belonging to a different suit. I didn't mean to imply that it actually was a valid combination. Which is why I also said that I just confused myself even more, since this would contradict the initial statement.
– John Smith
Nov 16 at 5:20
add a comment |
"I realized that it's also possible for the second card you pick to belong to the same suit as the first one" -- With respect this to this comment, it brings up a question. The question you seem to be asking in the opening parts of the question seems to imply that you want all three to be from different suits (i.e. Spade/Spade/Heart is not valid). Yet the quote seems to imply this is okay. This is an important nuance to settle for probably everyone's sake.
– Eevee Trainer
Nov 16 at 5:16
Sorry about that. You're right. Spade/Spade/Heart would not be a valid combination in this scenario. I nonetheless mentioned it, thinking that it could somehow affect the probability of the third card belonging to a different suit. I didn't mean to imply that it actually was a valid combination. Which is why I also said that I just confused myself even more, since this would contradict the initial statement.
– John Smith
Nov 16 at 5:20
"I realized that it's also possible for the second card you pick to belong to the same suit as the first one" -- With respect this to this comment, it brings up a question. The question you seem to be asking in the opening parts of the question seems to imply that you want all three to be from different suits (i.e. Spade/Spade/Heart is not valid). Yet the quote seems to imply this is okay. This is an important nuance to settle for probably everyone's sake.
– Eevee Trainer
Nov 16 at 5:16
"I realized that it's also possible for the second card you pick to belong to the same suit as the first one" -- With respect this to this comment, it brings up a question. The question you seem to be asking in the opening parts of the question seems to imply that you want all three to be from different suits (i.e. Spade/Spade/Heart is not valid). Yet the quote seems to imply this is okay. This is an important nuance to settle for probably everyone's sake.
– Eevee Trainer
Nov 16 at 5:16
Sorry about that. You're right. Spade/Spade/Heart would not be a valid combination in this scenario. I nonetheless mentioned it, thinking that it could somehow affect the probability of the third card belonging to a different suit. I didn't mean to imply that it actually was a valid combination. Which is why I also said that I just confused myself even more, since this would contradict the initial statement.
– John Smith
Nov 16 at 5:20
Sorry about that. You're right. Spade/Spade/Heart would not be a valid combination in this scenario. I nonetheless mentioned it, thinking that it could somehow affect the probability of the third card belonging to a different suit. I didn't mean to imply that it actually was a valid combination. Which is why I also said that I just confused myself even more, since this would contradict the initial statement.
– John Smith
Nov 16 at 5:20
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2 Answers
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You are correct that your chance to get an acceptable second card is $frac {39}{51}$ and that assuming the second is acceptable the chance for the third to be acceptable is $frac {26}{50}$. You don't need to worry about the chance the third is acceptable if the second is not because you have already failed. Now you can use the multiplication principle to get the probability of success
$$frac {39}{51} cdot frac {26}{50}=frac {169}{425}$$
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There are 4 suits from which you choose 3, C(4,3)=4. There are 4 ways of choosing which suits to represent. For each of the 3 chosen suits, there are 13 options from which to choose one representative, C(13,1)=13. So there are $4*13^3$ ways of selecting these cards.
Overall, the number of ways of choosing 3 cards from a group of 52 is :
$C(52,3)=frac{52!}{3!49!}=frac{52*51*50}{6}=26*17*50$
So the probability of choosing 3 cards each of different suits is : number of ways of selecting 3 cards of different suits/ total number of ways of picking 3 cards out of 52.
ans = $frac{4*13^3}{26*17*50}=frac{4*13^2}{2*17*50}=frac{169}{425}$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You are correct that your chance to get an acceptable second card is $frac {39}{51}$ and that assuming the second is acceptable the chance for the third to be acceptable is $frac {26}{50}$. You don't need to worry about the chance the third is acceptable if the second is not because you have already failed. Now you can use the multiplication principle to get the probability of success
$$frac {39}{51} cdot frac {26}{50}=frac {169}{425}$$
add a comment |
up vote
1
down vote
You are correct that your chance to get an acceptable second card is $frac {39}{51}$ and that assuming the second is acceptable the chance for the third to be acceptable is $frac {26}{50}$. You don't need to worry about the chance the third is acceptable if the second is not because you have already failed. Now you can use the multiplication principle to get the probability of success
$$frac {39}{51} cdot frac {26}{50}=frac {169}{425}$$
add a comment |
up vote
1
down vote
up vote
1
down vote
You are correct that your chance to get an acceptable second card is $frac {39}{51}$ and that assuming the second is acceptable the chance for the third to be acceptable is $frac {26}{50}$. You don't need to worry about the chance the third is acceptable if the second is not because you have already failed. Now you can use the multiplication principle to get the probability of success
$$frac {39}{51} cdot frac {26}{50}=frac {169}{425}$$
You are correct that your chance to get an acceptable second card is $frac {39}{51}$ and that assuming the second is acceptable the chance for the third to be acceptable is $frac {26}{50}$. You don't need to worry about the chance the third is acceptable if the second is not because you have already failed. Now you can use the multiplication principle to get the probability of success
$$frac {39}{51} cdot frac {26}{50}=frac {169}{425}$$
answered Nov 16 at 5:58
Ross Millikan
287k23195364
287k23195364
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There are 4 suits from which you choose 3, C(4,3)=4. There are 4 ways of choosing which suits to represent. For each of the 3 chosen suits, there are 13 options from which to choose one representative, C(13,1)=13. So there are $4*13^3$ ways of selecting these cards.
Overall, the number of ways of choosing 3 cards from a group of 52 is :
$C(52,3)=frac{52!}{3!49!}=frac{52*51*50}{6}=26*17*50$
So the probability of choosing 3 cards each of different suits is : number of ways of selecting 3 cards of different suits/ total number of ways of picking 3 cards out of 52.
ans = $frac{4*13^3}{26*17*50}=frac{4*13^2}{2*17*50}=frac{169}{425}$
add a comment |
up vote
0
down vote
There are 4 suits from which you choose 3, C(4,3)=4. There are 4 ways of choosing which suits to represent. For each of the 3 chosen suits, there are 13 options from which to choose one representative, C(13,1)=13. So there are $4*13^3$ ways of selecting these cards.
Overall, the number of ways of choosing 3 cards from a group of 52 is :
$C(52,3)=frac{52!}{3!49!}=frac{52*51*50}{6}=26*17*50$
So the probability of choosing 3 cards each of different suits is : number of ways of selecting 3 cards of different suits/ total number of ways of picking 3 cards out of 52.
ans = $frac{4*13^3}{26*17*50}=frac{4*13^2}{2*17*50}=frac{169}{425}$
add a comment |
up vote
0
down vote
up vote
0
down vote
There are 4 suits from which you choose 3, C(4,3)=4. There are 4 ways of choosing which suits to represent. For each of the 3 chosen suits, there are 13 options from which to choose one representative, C(13,1)=13. So there are $4*13^3$ ways of selecting these cards.
Overall, the number of ways of choosing 3 cards from a group of 52 is :
$C(52,3)=frac{52!}{3!49!}=frac{52*51*50}{6}=26*17*50$
So the probability of choosing 3 cards each of different suits is : number of ways of selecting 3 cards of different suits/ total number of ways of picking 3 cards out of 52.
ans = $frac{4*13^3}{26*17*50}=frac{4*13^2}{2*17*50}=frac{169}{425}$
There are 4 suits from which you choose 3, C(4,3)=4. There are 4 ways of choosing which suits to represent. For each of the 3 chosen suits, there are 13 options from which to choose one representative, C(13,1)=13. So there are $4*13^3$ ways of selecting these cards.
Overall, the number of ways of choosing 3 cards from a group of 52 is :
$C(52,3)=frac{52!}{3!49!}=frac{52*51*50}{6}=26*17*50$
So the probability of choosing 3 cards each of different suits is : number of ways of selecting 3 cards of different suits/ total number of ways of picking 3 cards out of 52.
ans = $frac{4*13^3}{26*17*50}=frac{4*13^2}{2*17*50}=frac{169}{425}$
answered Nov 16 at 7:21
TurlocTheRed
751210
751210
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"I realized that it's also possible for the second card you pick to belong to the same suit as the first one" -- With respect this to this comment, it brings up a question. The question you seem to be asking in the opening parts of the question seems to imply that you want all three to be from different suits (i.e. Spade/Spade/Heart is not valid). Yet the quote seems to imply this is okay. This is an important nuance to settle for probably everyone's sake.
– Eevee Trainer
Nov 16 at 5:16
Sorry about that. You're right. Spade/Spade/Heart would not be a valid combination in this scenario. I nonetheless mentioned it, thinking that it could somehow affect the probability of the third card belonging to a different suit. I didn't mean to imply that it actually was a valid combination. Which is why I also said that I just confused myself even more, since this would contradict the initial statement.
– John Smith
Nov 16 at 5:20