6 cards are distributed between 3 people among other cards, what is the probability that someone will have at...
I play a card game called "Whist". In this game, all four players get 13 cards each, and you have to estimate how many tricks you will win with one trump color.
Here is my scenario:
In my hand, I have 7 hearts : Ace, King, Jack, 10, ... (the rest doesn't matter).
My question is : what is the probability that I will win those 7 cards if the trump is Heart? The only way for me to lose the third trick is if somebody has 3 hearts or more AND the queen. This is because if the player who has the Queen has only one or two cards he will lose it when I play my Ace and my King (people have to follow your color).
So my question is:
I know there are 6 remaining hearts in the game, they are separated into 3 players, what is the probability that one player has 3 cards or more containing the Queen?
I tried to solve it doing a probability tree but it quickly became complicated so I wonder which formula we can use.
Thanks in advance for your response!
(PS: I know that if for example I have Ace, King, Jack, 10, 8, etc, I could lose one trick if someone has 5 hearts and the 9 of hearts but I think if I am not mistaken that the probability of this is quite low (correct me if I am wrong))
probability combinatorics card-games
asked Sep 22 '17 at 9:46
Ricola
133
add a comment |
I play a card game called "Whist". In this game, all four players get 13 cards each, and you have to estimate how many tricks you will win with one trump color.
Here is my scenario:
In my hand, I have 7 hearts : Ace, King, Jack, 10, ... (the rest doesn't matter).
My question is : what is the probability that I will win those 7 cards if the trump is Heart? The only way for me to lose the third trick is if somebody has 3 hearts or more AND the queen. This is because if the player who has the Queen has only one or two cards he will lose it when I play my Ace and my King (people have to follow your color).
So my question is:
I know there are 6 remaining hearts in the game, they are separated into 3 players, what is the probability that one player has 3 cards or more containing the Queen?
I tried to solve it doing a probability tree but it quickly became complicated so I wonder which formula we can use.
Thanks in advance for your response!
(PS: I know that if for example I have Ace, King, Jack, 10, 8, etc, I could lose one trick if someone has 5 hearts and the 9 of hearts but I think if I am not mistaken that the probability of this is quite low (correct me if I am wrong))
probability combinatorics card-games
asked Sep 22 '17 at 9:46
Ricola
133
Slightly simplification : You only have to conider the player having the queen of hearts. You need the probability that this player has at least two more hearts. We need the hypergeometric distribution to determine the probability.
– Peter
Sep 22 '17 at 9:53
I've always wanted to play Whist. Want to play in person, though, and don't have players. :(
– Wildcard
Sep 22 '17 at 10:02
add a comment |
I play a card game called "Whist". In this game, all four players get 13 cards each, and you have to estimate how many tricks you will win with one trump color.
Here is my scenario:
In my hand, I have 7 hearts : Ace, King, Jack, 10, ... (the rest doesn't matter).
My question is : what is the probability that I will win those 7 cards if the trump is Heart? The only way for me to lose the third trick is if somebody has 3 hearts or more AND the queen. This is because if the player who has the Queen has only one or two cards he will lose it when I play my Ace and my King (people have to follow your color).
So my question is:
I know there are 6 remaining hearts in the game, they are separated into 3 players, what is the probability that one player has 3 cards or more containing the Queen?
I tried to solve it doing a probability tree but it quickly became complicated so I wonder which formula we can use.
Thanks in advance for your response!
(PS: I know that if for example I have Ace, King, Jack, 10, 8, etc, I could lose one trick if someone has 5 hearts and the 9 of hearts but I think if I am not mistaken that the probability of this is quite low (correct me if I am wrong))
probability combinatorics card-games
asked Sep 22 '17 at 9:46
Ricola
133
I play a card game called "Whist". In this game, all four players get 13 cards each, and you have to estimate how many tricks you will win with one trump color.
Here is my scenario:
In my hand, I have 7 hearts : Ace, King, Jack, 10, ... (the rest doesn't matter).
My question is : what is the probability that I will win those 7 cards if the trump is Heart? The only way for me to lose the third trick is if somebody has 3 hearts or more AND the queen. This is because if the player who has the Queen has only one or two cards he will lose it when I play my Ace and my King (people have to follow your color).
So my question is:
I know there are 6 remaining hearts in the game, they are separated into 3 players, what is the probability that one player has 3 cards or more containing the Queen?
I tried to solve it doing a probability tree but it quickly became complicated so I wonder which formula we can use.
Thanks in advance for your response!
(PS: I know that if for example I have Ace, King, Jack, 10, 8, etc, I could lose one trick if someone has 5 hearts and the 9 of hearts but I think if I am not mistaken that the probability of this is quite low (correct me if I am wrong))
probability combinatorics card-games
probability combinatorics card-games
asked Sep 22 '17 at 9:46
Ricola
133
asked Sep 22 '17 at 9:46
Ricola
133
edited Nov 28 at 13:35
asked Sep 22 '17 at 9:46
Ricola
133
asked Sep 22 '17 at 9:46
Ricola
133
asked Sep 22 '17 at 9:46
Ricola
133
133
Slightly simplification : You only have to conider the player having the queen of hearts. You need the probability that this player has at least two more hearts. We need the hypergeometric distribution to determine the probability.
– Peter
Sep 22 '17 at 9:53
I've always wanted to play Whist. Want to play in person, though, and don't have players. :(
– Wildcard
Sep 22 '17 at 10:02
add a comment |
Slightly simplification : You only have to conider the player having the queen of hearts. You need the probability that this player has at least two more hearts. We need the hypergeometric distribution to determine the probability.
– Peter
Sep 22 '17 at 9:53
I've always wanted to play Whist. Want to play in person, though, and don't have players. :(
– Wildcard
Sep 22 '17 at 10:02
Slightly simplification : You only have to conider the player having the queen of hearts. You need the probability that this player has at least two more hearts. We need the hypergeometric distribution to determine the probability.
– Peter
Sep 22 '17 at 9:53
Slightly simplification : You only have to conider the player having the queen of hearts. You need the probability that this player has at least two more hearts. We need the hypergeometric distribution to determine the probability.
– Peter
Sep 22 '17 at 9:53
I've always wanted to play Whist. Want to play in person, though, and don't have players. :(
– Wildcard
Sep 22 '17 at 10:02
I've always wanted to play Whist. Want to play in person, though, and don't have players. :(
– Wildcard
Sep 22 '17 at 10:02
add a comment |
1 Answer
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One of the $3$ opponents has the queen of hearts.
The question is: what is the probability that this opponent has at least $2$ other hearts.
$$sum_{k=2}^5frac{binom{12}{k}binom{26}{5-k}}{binom{38}{5}}$$
Here we are practizing hypergeometric distribution and term $frac{binom{12}{k}binom{26}{5-k}}{binom{38}{5}}$ is the probability that the opponent in possession of the queen of hearts received exactly $k$ other hearts.
Observe that this opponent received $12$ cards next to the queen of hearts and the other two opponents together received $26$ cars. Among these cards there are $5$ hearts.
You think of the opponent in possession of the queen of hearts as drawing $12$ balls out of an urn containing $5$ blue balls and $33$ red balls. Then what is the probability that he draws at least $2$ blue balls.
Here balls are cards and blue balls are hearts.
answered Sep 22 '17 at 9:58
drhab
96.6k544127
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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One of the $3$ opponents has the queen of hearts.
The question is: what is the probability that this opponent has at least $2$ other hearts.
$$sum_{k=2}^5frac{binom{12}{k}binom{26}{5-k}}{binom{38}{5}}$$
Here we are practizing hypergeometric distribution and term $frac{binom{12}{k}binom{26}{5-k}}{binom{38}{5}}$ is the probability that the opponent in possession of the queen of hearts received exactly $k$ other hearts.
Observe that this opponent received $12$ cards next to the queen of hearts and the other two opponents together received $26$ cars. Among these cards there are $5$ hearts.
You think of the opponent in possession of the queen of hearts as drawing $12$ balls out of an urn containing $5$ blue balls and $33$ red balls. Then what is the probability that he draws at least $2$ blue balls.
Here balls are cards and blue balls are hearts.
answered Sep 22 '17 at 9:58
drhab
96.6k544127
add a comment |
One of the $3$ opponents has the queen of hearts.
The question is: what is the probability that this opponent has at least $2$ other hearts.
$$sum_{k=2}^5frac{binom{12}{k}binom{26}{5-k}}{binom{38}{5}}$$
Here we are practizing hypergeometric distribution and term $frac{binom{12}{k}binom{26}{5-k}}{binom{38}{5}}$ is the probability that the opponent in possession of the queen of hearts received exactly $k$ other hearts.
Observe that this opponent received $12$ cards next to the queen of hearts and the other two opponents together received $26$ cars. Among these cards there are $5$ hearts.
You think of the opponent in possession of the queen of hearts as drawing $12$ balls out of an urn containing $5$ blue balls and $33$ red balls. Then what is the probability that he draws at least $2$ blue balls.
Here balls are cards and blue balls are hearts.
answered Sep 22 '17 at 9:58
drhab
96.6k544127
add a comment |
One of the $3$ opponents has the queen of hearts.
The question is: what is the probability that this opponent has at least $2$ other hearts.
$$sum_{k=2}^5frac{binom{12}{k}binom{26}{5-k}}{binom{38}{5}}$$
Here we are practizing hypergeometric distribution and term $frac{binom{12}{k}binom{26}{5-k}}{binom{38}{5}}$ is the probability that the opponent in possession of the queen of hearts received exactly $k$ other hearts.
Observe that this opponent received $12$ cards next to the queen of hearts and the other two opponents together received $26$ cars. Among these cards there are $5$ hearts.
You think of the opponent in possession of the queen of hearts as drawing $12$ balls out of an urn containing $5$ blue balls and $33$ red balls. Then what is the probability that he draws at least $2$ blue balls.
Here balls are cards and blue balls are hearts.
answered Sep 22 '17 at 9:58
drhab
96.6k544127
One of the $3$ opponents has the queen of hearts.
The question is: what is the probability that this opponent has at least $2$ other hearts.
$$sum_{k=2}^5frac{binom{12}{k}binom{26}{5-k}}{binom{38}{5}}$$
Here we are practizing hypergeometric distribution and term $frac{binom{12}{k}binom{26}{5-k}}{binom{38}{5}}$ is the probability that the opponent in possession of the queen of hearts received exactly $k$ other hearts.
Observe that this opponent received $12$ cards next to the queen of hearts and the other two opponents together received $26$ cars. Among these cards there are $5$ hearts.
You think of the opponent in possession of the queen of hearts as drawing $12$ balls out of an urn containing $5$ blue balls and $33$ red balls. Then what is the probability that he draws at least $2$ blue balls.
Here balls are cards and blue balls are hearts.
answered Sep 22 '17 at 9:58
drhab
96.6k544127
edited Sep 22 '17 at 10:13
answered Sep 22 '17 at 9:58
drhab
96.6k544127
answered Sep 22 '17 at 9:58
drhab
96.6k544127
answered Sep 22 '17 at 9:58
drhab
96.6k544127
96.6k544127
add a comment |
add a comment |
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Slightly simplification : You only have to conider the player having the queen of hearts. You need the probability that this player has at least two more hearts. We need the hypergeometric distribution to determine the probability.
– Peter
Sep 22 '17 at 9:53
I've always wanted to play Whist. Want to play in person, though, and don't have players. :(
– Wildcard
Sep 22 '17 at 10:02