What is the probability that only one of the cards will have the matching suit?
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A group of 4 friends are playing a card game using a standard deck of 52 cards. Each friend receives one card from the deck, and the next card is flipped up. Only one of the friend has a card that matches the suit of the flipped up card. What is the probability that only one person is able to have a suit that matches the suit of the card that's flipped up?
My work (it's most likely wrong):
Let's say that the turned up card is a hearts. There are 13 hearts in a deck of hearts.
13*39*38*37= 712,842
Only one of the cards is a hearts (13). The rest of the cards are not hearts (39, 38, 37).
The probability that one player has the matching suit -> 13/712,842.
The number is too small, and I feel that I did something wrong.
probability combinatorics permutations combinations card-games
edited Dec 3 '18 at 12:20
N. F. Taussig
43.9k93355
asked Dec 3 '18 at 11:43
Lauren PabloLauren Pablo
11
$endgroup$
add a comment |
$begingroup$
A group of 4 friends are playing a card game using a standard deck of 52 cards. Each friend receives one card from the deck, and the next card is flipped up. Only one of the friend has a card that matches the suit of the flipped up card. What is the probability that only one person is able to have a suit that matches the suit of the card that's flipped up?
My work (it's most likely wrong):
Let's say that the turned up card is a hearts. There are 13 hearts in a deck of hearts.
13*39*38*37= 712,842
Only one of the cards is a hearts (13). The rest of the cards are not hearts (39, 38, 37).
The probability that one player has the matching suit -> 13/712,842.
The number is too small, and I feel that I did something wrong.
probability combinatorics permutations combinations card-games
edited Dec 3 '18 at 12:20
N. F. Taussig
43.9k93355
asked Dec 3 '18 at 11:43
Lauren PabloLauren Pablo
11
$endgroup$
add a comment |
$begingroup$
A group of 4 friends are playing a card game using a standard deck of 52 cards. Each friend receives one card from the deck, and the next card is flipped up. Only one of the friend has a card that matches the suit of the flipped up card. What is the probability that only one person is able to have a suit that matches the suit of the card that's flipped up?
My work (it's most likely wrong):
Let's say that the turned up card is a hearts. There are 13 hearts in a deck of hearts.
13*39*38*37= 712,842
Only one of the cards is a hearts (13). The rest of the cards are not hearts (39, 38, 37).
The probability that one player has the matching suit -> 13/712,842.
The number is too small, and I feel that I did something wrong.
probability combinatorics permutations combinations card-games
edited Dec 3 '18 at 12:20
N. F. Taussig
43.9k93355
asked Dec 3 '18 at 11:43
Lauren PabloLauren Pablo
11
$endgroup$
A group of 4 friends are playing a card game using a standard deck of 52 cards. Each friend receives one card from the deck, and the next card is flipped up. Only one of the friend has a card that matches the suit of the flipped up card. What is the probability that only one person is able to have a suit that matches the suit of the card that's flipped up?
My work (it's most likely wrong):
Let's say that the turned up card is a hearts. There are 13 hearts in a deck of hearts.
13*39*38*37= 712,842
Only one of the cards is a hearts (13). The rest of the cards are not hearts (39, 38, 37).
The probability that one player has the matching suit -> 13/712,842.
The number is too small, and I feel that I did something wrong.
probability combinatorics permutations combinations card-games
probability combinatorics permutations combinations card-games
edited Dec 3 '18 at 12:20
N. F. Taussig
43.9k93355
asked Dec 3 '18 at 11:43
Lauren PabloLauren Pablo
11
edited Dec 3 '18 at 12:20
N. F. Taussig
43.9k93355
asked Dec 3 '18 at 11:43
Lauren PabloLauren Pablo
11
edited Dec 3 '18 at 12:20
N. F. Taussig
43.9k93355
edited Dec 3 '18 at 12:20
N. F. Taussig
43.9k93355
edited Dec 3 '18 at 12:20
N. F. Taussig
43.9k93355
43.9k93355
asked Dec 3 '18 at 11:43
Lauren PabloLauren Pablo
11
asked Dec 3 '18 at 11:43
Lauren PabloLauren Pablo
11
asked Dec 3 '18 at 11:43
Lauren PabloLauren Pablo
11
11
add a comment |
add a comment |
1 Answer
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$begingroup$
Three things:
With the card being flipped one of the $13$ hearts, the player whose card is a heart as well must have one of the $12$ remaining hearts, not $13$. So you must use a factor of $12$ not $13$
There are $4$ possible players whose card is a heart, so you need to have a factor of $4$ in there.
Most importantly, you should not be dividing by the factors of $12$, $39$, $38$, and $37$: these are the possible target combinations that go in the numerator. The denominator are all the possible ways to get $4$ cards to the $4$ players, so there you get something like $51$, $50$, $49$, and $48$
OK, using these hints, try again!
answered Dec 3 '18 at 12:17
Bram28Bram28
60.6k44590
$endgroup$
$begingroup$
@N.F.Taussig Right, thanks! :P
$endgroup$
– Bram28
Dec 3 '18 at 12:22
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Since one card is placed on the table, there are $51$ cards left from which the four players can each select a card.
$endgroup$
– N. F. Taussig
Dec 3 '18 at 12:27
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Thanks! Would it be instead 51, 50, 49, and 48 since one of the cards is already used up?
$endgroup$
– Lauren Pablo
Dec 3 '18 at 12:28
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@laurenpablo exactly!
$endgroup$
– Bram28
Dec 3 '18 at 13:01
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Three things:
With the card being flipped one of the $13$ hearts, the player whose card is a heart as well must have one of the $12$ remaining hearts, not $13$. So you must use a factor of $12$ not $13$
There are $4$ possible players whose card is a heart, so you need to have a factor of $4$ in there.
Most importantly, you should not be dividing by the factors of $12$, $39$, $38$, and $37$: these are the possible target combinations that go in the numerator. The denominator are all the possible ways to get $4$ cards to the $4$ players, so there you get something like $51$, $50$, $49$, and $48$
OK, using these hints, try again!
answered Dec 3 '18 at 12:17
Bram28Bram28
60.6k44590
$endgroup$
$begingroup$
@N.F.Taussig Right, thanks! :P
$endgroup$
– Bram28
Dec 3 '18 at 12:22
$begingroup$
Since one card is placed on the table, there are $51$ cards left from which the four players can each select a card.
$endgroup$
– N. F. Taussig
Dec 3 '18 at 12:27
$begingroup$
Thanks! Would it be instead 51, 50, 49, and 48 since one of the cards is already used up?
$endgroup$
– Lauren Pablo
Dec 3 '18 at 12:28
$begingroup$
@laurenpablo exactly!
$endgroup$
– Bram28
Dec 3 '18 at 13:01
add a comment |
$begingroup$
Three things:
With the card being flipped one of the $13$ hearts, the player whose card is a heart as well must have one of the $12$ remaining hearts, not $13$. So you must use a factor of $12$ not $13$
There are $4$ possible players whose card is a heart, so you need to have a factor of $4$ in there.
Most importantly, you should not be dividing by the factors of $12$, $39$, $38$, and $37$: these are the possible target combinations that go in the numerator. The denominator are all the possible ways to get $4$ cards to the $4$ players, so there you get something like $51$, $50$, $49$, and $48$
OK, using these hints, try again!
answered Dec 3 '18 at 12:17
Bram28Bram28
60.6k44590
$endgroup$
$begingroup$
@N.F.Taussig Right, thanks! :P
$endgroup$
– Bram28
Dec 3 '18 at 12:22
$begingroup$
Since one card is placed on the table, there are $51$ cards left from which the four players can each select a card.
$endgroup$
– N. F. Taussig
Dec 3 '18 at 12:27
$begingroup$
Thanks! Would it be instead 51, 50, 49, and 48 since one of the cards is already used up?
$endgroup$
– Lauren Pablo
Dec 3 '18 at 12:28
$begingroup$
@laurenpablo exactly!
$endgroup$
– Bram28
Dec 3 '18 at 13:01
add a comment |
$begingroup$
Three things:
With the card being flipped one of the $13$ hearts, the player whose card is a heart as well must have one of the $12$ remaining hearts, not $13$. So you must use a factor of $12$ not $13$
There are $4$ possible players whose card is a heart, so you need to have a factor of $4$ in there.
Most importantly, you should not be dividing by the factors of $12$, $39$, $38$, and $37$: these are the possible target combinations that go in the numerator. The denominator are all the possible ways to get $4$ cards to the $4$ players, so there you get something like $51$, $50$, $49$, and $48$
OK, using these hints, try again!
answered Dec 3 '18 at 12:17
Bram28Bram28
60.6k44590
$endgroup$
Three things:
With the card being flipped one of the $13$ hearts, the player whose card is a heart as well must have one of the $12$ remaining hearts, not $13$. So you must use a factor of $12$ not $13$
There are $4$ possible players whose card is a heart, so you need to have a factor of $4$ in there.
Most importantly, you should not be dividing by the factors of $12$, $39$, $38$, and $37$: these are the possible target combinations that go in the numerator. The denominator are all the possible ways to get $4$ cards to the $4$ players, so there you get something like $51$, $50$, $49$, and $48$
OK, using these hints, try again!
answered Dec 3 '18 at 12:17
Bram28Bram28
60.6k44590
edited Dec 3 '18 at 13:02
answered Dec 3 '18 at 12:17
Bram28Bram28
60.6k44590
answered Dec 3 '18 at 12:17
Bram28Bram28
60.6k44590
answered Dec 3 '18 at 12:17
Bram28Bram28
60.6k44590
60.6k44590
$begingroup$
@N.F.Taussig Right, thanks! :P
$endgroup$
– Bram28
Dec 3 '18 at 12:22
$begingroup$
Since one card is placed on the table, there are $51$ cards left from which the four players can each select a card.
$endgroup$
– N. F. Taussig
Dec 3 '18 at 12:27
$begingroup$
Thanks! Would it be instead 51, 50, 49, and 48 since one of the cards is already used up?
$endgroup$
– Lauren Pablo
Dec 3 '18 at 12:28
$begingroup$
@laurenpablo exactly!
$endgroup$
– Bram28
Dec 3 '18 at 13:01
add a comment |
$begingroup$
@N.F.Taussig Right, thanks! :P
$endgroup$
– Bram28
Dec 3 '18 at 12:22
$begingroup$
Since one card is placed on the table, there are $51$ cards left from which the four players can each select a card.
$endgroup$
– N. F. Taussig
Dec 3 '18 at 12:27
$begingroup$
Thanks! Would it be instead 51, 50, 49, and 48 since one of the cards is already used up?
$endgroup$
– Lauren Pablo
Dec 3 '18 at 12:28
$begingroup$
@laurenpablo exactly!
$endgroup$
– Bram28
Dec 3 '18 at 13:01
$begingroup$
@N.F.Taussig Right, thanks! :P
$endgroup$
– Bram28
Dec 3 '18 at 12:22
$begingroup$
@N.F.Taussig Right, thanks! :P
$endgroup$
– Bram28
Dec 3 '18 at 12:22
$begingroup$
Since one card is placed on the table, there are $51$ cards left from which the four players can each select a card.
$endgroup$
– N. F. Taussig
Dec 3 '18 at 12:27
$begingroup$
Since one card is placed on the table, there are $51$ cards left from which the four players can each select a card.
$endgroup$
– N. F. Taussig
Dec 3 '18 at 12:27
$begingroup$
Thanks! Would it be instead 51, 50, 49, and 48 since one of the cards is already used up?
$endgroup$
– Lauren Pablo
Dec 3 '18 at 12:28
$begingroup$
Thanks! Would it be instead 51, 50, 49, and 48 since one of the cards is already used up?
$endgroup$
– Lauren Pablo
Dec 3 '18 at 12:28
$begingroup$
@laurenpablo exactly!
$endgroup$
– Bram28
Dec 3 '18 at 13:01
$begingroup$
@laurenpablo exactly!
$endgroup$
– Bram28
Dec 3 '18 at 13:01
add a comment |
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