Mixing
In a bridge game, North and South have 9 spades between them. Find the probability that either East or West...
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I am a complete novice in playing card games and I am unable to solve problems of this type . However I have tried to solve it as shown below:
Firstly we select $9$ cards out of $13$ spades and $17$ non spade cards out of $39$ non spade cards. These make a total of $26$ cards. Now we try to divide these $26$ cards between North and South. Now for each of these possible cases we try to choose $13$ non spade cards from $22$ non spade cards for either of East or West . Two such ways shall be possible i.e. either for East or for West. So the total number of favourable outcomes(m):
13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
Total number of possible outcomes(n):
(52!)/(13!)^4
Required probability $=frac{m}{n}approx7.036337939times10^{-3}$
But the answer given in book is not matching.( It is given as $=frac{11}{115}$ , I am not sure whether the answer given is book is wrong or that of mine is wrong.)
Please help me out.
probability
asked Jan 25 at 17:27
Abhishek GhoshAbhishek Ghosh
878
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add a comment |
$begingroup$
I am a complete novice in playing card games and I am unable to solve problems of this type . However I have tried to solve it as shown below:
Firstly we select $9$ cards out of $13$ spades and $17$ non spade cards out of $39$ non spade cards. These make a total of $26$ cards. Now we try to divide these $26$ cards between North and South. Now for each of these possible cases we try to choose $13$ non spade cards from $22$ non spade cards for either of East or West . Two such ways shall be possible i.e. either for East or for West. So the total number of favourable outcomes(m):
13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
Total number of possible outcomes(n):
(52!)/(13!)^4
Required probability $=frac{m}{n}approx7.036337939times10^{-3}$
But the answer given in book is not matching.( It is given as $=frac{11}{115}$ , I am not sure whether the answer given is book is wrong or that of mine is wrong.)
Please help me out.
probability
asked Jan 25 at 17:27
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
$begingroup$
You answered your own, similar question on this here: math.stackexchange.com/questions/3087310/… There are four players in Bridge: North and South (partners) and East and West (partners). There are $13$ spades. $9$ of them are held in either the North or South hand and you need to work out what the probability is that one of the remaining two players holds no spades (or equivalently, all four remaining spades)
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– postmortes
Jan 25 at 17:40
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I think your denominator is too big. It shoul count, not all possible distributions of the $52$ cards, just those in which North-South have $9$ spades.
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– bof
Jan 25 at 18:51
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Do you mean to say conditional probability?
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– Abhishek Ghosh
Jan 25 at 18:52
$begingroup$
Thank @bof for helping me out with the problem
$endgroup$
– Abhishek Ghosh
Jan 25 at 19:22
add a comment |
$begingroup$
I am a complete novice in playing card games and I am unable to solve problems of this type . However I have tried to solve it as shown below:
Firstly we select $9$ cards out of $13$ spades and $17$ non spade cards out of $39$ non spade cards. These make a total of $26$ cards. Now we try to divide these $26$ cards between North and South. Now for each of these possible cases we try to choose $13$ non spade cards from $22$ non spade cards for either of East or West . Two such ways shall be possible i.e. either for East or for West. So the total number of favourable outcomes(m):
13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
Total number of possible outcomes(n):
(52!)/(13!)^4
Required probability $=frac{m}{n}approx7.036337939times10^{-3}$
But the answer given in book is not matching.( It is given as $=frac{11}{115}$ , I am not sure whether the answer given is book is wrong or that of mine is wrong.)
Please help me out.
probability
asked Jan 25 at 17:27
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
I am a complete novice in playing card games and I am unable to solve problems of this type . However I have tried to solve it as shown below:
Firstly we select $9$ cards out of $13$ spades and $17$ non spade cards out of $39$ non spade cards. These make a total of $26$ cards. Now we try to divide these $26$ cards between North and South. Now for each of these possible cases we try to choose $13$ non spade cards from $22$ non spade cards for either of East or West . Two such ways shall be possible i.e. either for East or for West. So the total number of favourable outcomes(m):
13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
Total number of possible outcomes(n):
(52!)/(13!)^4
Required probability $=frac{m}{n}approx7.036337939times10^{-3}$
But the answer given in book is not matching.( It is given as $=frac{11}{115}$ , I am not sure whether the answer given is book is wrong or that of mine is wrong.)
Please help me out.
probability
probability
asked Jan 25 at 17:27
Abhishek GhoshAbhishek Ghosh
878
asked Jan 25 at 17:27
Abhishek GhoshAbhishek Ghosh
878
edited Jan 25 at 18:44
Abhishek Ghosh
asked Jan 25 at 17:27
Abhishek GhoshAbhishek Ghosh
878
asked Jan 25 at 17:27
Abhishek GhoshAbhishek Ghosh
878
asked Jan 25 at 17:27
Abhishek GhoshAbhishek Ghosh
878
878
$begingroup$
You answered your own, similar question on this here: math.stackexchange.com/questions/3087310/… There are four players in Bridge: North and South (partners) and East and West (partners). There are $13$ spades. $9$ of them are held in either the North or South hand and you need to work out what the probability is that one of the remaining two players holds no spades (or equivalently, all four remaining spades)
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– postmortes
Jan 25 at 17:40
$begingroup$
I think your denominator is too big. It shoul count, not all possible distributions of the $52$ cards, just those in which North-South have $9$ spades.
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– bof
Jan 25 at 18:51
$begingroup$
Do you mean to say conditional probability?
$endgroup$
– Abhishek Ghosh
Jan 25 at 18:52
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Thank @bof for helping me out with the problem
$endgroup$
– Abhishek Ghosh
Jan 25 at 19:22
add a comment |
$begingroup$
You answered your own, similar question on this here: math.stackexchange.com/questions/3087310/… There are four players in Bridge: North and South (partners) and East and West (partners). There are $13$ spades. $9$ of them are held in either the North or South hand and you need to work out what the probability is that one of the remaining two players holds no spades (or equivalently, all four remaining spades)
$endgroup$
– postmortes
Jan 25 at 17:40
$begingroup$
I think your denominator is too big. It shoul count, not all possible distributions of the $52$ cards, just those in which North-South have $9$ spades.
$endgroup$
– bof
Jan 25 at 18:51
$begingroup$
Do you mean to say conditional probability?
$endgroup$
– Abhishek Ghosh
Jan 25 at 18:52
$begingroup$
Thank @bof for helping me out with the problem
$endgroup$
– Abhishek Ghosh
Jan 25 at 19:22
$begingroup$
You answered your own, similar question on this here: math.stackexchange.com/questions/3087310/… There are four players in Bridge: North and South (partners) and East and West (partners). There are $13$ spades. $9$ of them are held in either the North or South hand and you need to work out what the probability is that one of the remaining two players holds no spades (or equivalently, all four remaining spades)
$endgroup$
– postmortes
Jan 25 at 17:40
$begingroup$
You answered your own, similar question on this here: math.stackexchange.com/questions/3087310/… There are four players in Bridge: North and South (partners) and East and West (partners). There are $13$ spades. $9$ of them are held in either the North or South hand and you need to work out what the probability is that one of the remaining two players holds no spades (or equivalently, all four remaining spades)
$endgroup$
– postmortes
Jan 25 at 17:40
$begingroup$
I think your denominator is too big. It shoul count, not all possible distributions of the $52$ cards, just those in which North-South have $9$ spades.
$endgroup$
– bof
Jan 25 at 18:51
$begingroup$
I think your denominator is too big. It shoul count, not all possible distributions of the $52$ cards, just those in which North-South have $9$ spades.
$endgroup$
– bof
Jan 25 at 18:51
$begingroup$
Do you mean to say conditional probability?
$endgroup$
– Abhishek Ghosh
Jan 25 at 18:52
$begingroup$
Do you mean to say conditional probability?
$endgroup$
– Abhishek Ghosh
Jan 25 at 18:52
$begingroup$
Thank @bof for helping me out with the problem
$endgroup$
– Abhishek Ghosh
Jan 25 at 19:22
$begingroup$
Thank @bof for helping me out with the problem
$endgroup$
– Abhishek Ghosh
Jan 25 at 19:22
add a comment |
4 Answers
4
active
oldest
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We have 4 spades and 22 others to distribute between East and West. Let's suppose we deal the cards out, one by one, with the first 13 going to East and the second 13 going to West. We will consider the order of the cards to be significant, so this can be done in $26!$ ways, all of which are equally likely.
We want to count the number of sequences in which East or West gets all 4 spades. There are $2$ ways to choose East/West, then there are $binom{13}{4}$ ways to choose the locations of the spades within the chosen hand, $4!$ ways to sequence the spades, and $22!$ ways to sequence all the remaining cards. So all together, there are
$$2 times binom{13}{4} times 4! times 22!$$
ways to sequence the cards in which either East or West gets all the spades.
So the probability that either East or West gets all the spades is
$$frac{2 times binom{13}{4} times 4! times 22!}{26!} = frac{11}{115}$$
answered Jan 25 at 21:47
awkwardawkward
6,56511025
$endgroup$
add a comment |
$begingroup$
In your attempt, you tried to work out the probability that North and South have $9$ spades and that one of East or West has none.
But the problem says we are supposed to assume North and South have $9$ spades between them:
In a bridge game, North and South have 9 spades between them.
We don’t have to be concerned about how likely this event is; we know it happened, because the problem statement says so.
So the cards in the hands of North and South can be considered “already dealt,” and we only have to consider one thing: given that the remaining $26$ cards are divided $13$ to East and $13$ to West, what is the probability that one of them gets no spades?
You correctly found that there are $binom{22}{13}$ ways for East to get no spades. You correctly multiply by $2$ to include the case in which West gets no spades.
The denominator is the number of ways to divide the $26$ cards, $binom{26}{13}.$
So the probability comes out to
$$frac{2 binom{22}{13}}{binom{26}{13}}
= frac{11}{115}.
$$
answered Jan 26 at 14:29
David KDavid K
55.2k344120
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add a comment |
$begingroup$
Let having $9$ spades between North and South be event A and having no spades between East and West is event B.
So the required probability is P(B|A).
Now reducing the sample space as regards "$9$ spades are already present between North and South" so
So the favourable outcomes are corresponding to getting no spades in East or West.
Firstly we select $9$ cards out of $13$ spades and $17$ non spade cards out of $39$ non spade cards. These make a total of $26$ cards. Now we try to divide these $26$ cards between North and South. Now for each of these possible cases we try to choose $13$ non spade cards from $22$ non spade cards for either of East or West . Two such ways shall be possible i.e. either for East or for West. So the total number of favourable outcomes(n(AB)):
13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
Total number of possible outcomes when A has occurred already(n(A)):
13C9 ×39C17×((26!)/(13!)^2)× 26!/(13!)^2
Required probability $=frac{P(AB)}{P(A)}$
$=frac{n(AB)}{n(A)}$
$=frac{13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
}{13C9 ×39C17×((26!)/(13!)^2)× 26!/(13!)^2}$
$=frac{22C13×2!}{26!/(13!)^2}$
$=frac{11}{115}$
answered Jan 25 at 19:21
Abhishek GhoshAbhishek Ghosh
878
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add a comment |
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Since this problem asks only about spades, there is no reason to say anything about "non-spade cards". If North and South, together, have 9 spades then East and West, together, have 13- 9= 4 spades. The probability any one of those 4 spades is in East's hand is 1/2. The probability all 4 are in East's hand is $(1/2)^4= 1/16$. Similarly the probability all 4 are in West's hand is $(1/2)^4= 1/16$. The probability that "either East or West has no spades" which is the same as the probability "either East has all 4 remaining spades or West has all 4 remaining spades" is 1/16+ 1/16= 2/16= 1/8.
answered Jan 25 at 19:58
user247327user247327
11.5k1516
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How did you get that one of those 4 spades to be with East has probability 1/2 ? And also the answer given is not 1/8. Please explain me further. I am confused a bit.
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– Abhishek Ghosh
Jan 25 at 20:04
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Is there any reason why a spade, that is not in North and South's hands, would be more likely to be in East's than Wests' or vice-versa? If not then the probability is 1/2-1/2.
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– user247327
Jan 25 at 20:08
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How ever you also need to take into consideration that 9 spade are distributed between the north and south in terms of probability as this probability is not a sure event and as per dependence and multiplication theorem this probability shall also affect the value found out by you i hope
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– Abhishek Ghosh
Jan 25 at 20:13
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Assuming the K and Q of spades are not held by North or South, then the probability that West holds the K is $1/2$, and the probability that West holds the Q iis $1/2$, but the probability that West holds both the K and the Q is less than $1/4$; the events are not independent.
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– bof
Jan 25 at 23:17
add a comment |
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4 Answers
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4 Answers
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$begingroup$
We have 4 spades and 22 others to distribute between East and West. Let's suppose we deal the cards out, one by one, with the first 13 going to East and the second 13 going to West. We will consider the order of the cards to be significant, so this can be done in $26!$ ways, all of which are equally likely.
We want to count the number of sequences in which East or West gets all 4 spades. There are $2$ ways to choose East/West, then there are $binom{13}{4}$ ways to choose the locations of the spades within the chosen hand, $4!$ ways to sequence the spades, and $22!$ ways to sequence all the remaining cards. So all together, there are
$$2 times binom{13}{4} times 4! times 22!$$
ways to sequence the cards in which either East or West gets all the spades.
So the probability that either East or West gets all the spades is
$$frac{2 times binom{13}{4} times 4! times 22!}{26!} = frac{11}{115}$$
answered Jan 25 at 21:47
awkwardawkward
6,56511025
$endgroup$
add a comment |
$begingroup$
We have 4 spades and 22 others to distribute between East and West. Let's suppose we deal the cards out, one by one, with the first 13 going to East and the second 13 going to West. We will consider the order of the cards to be significant, so this can be done in $26!$ ways, all of which are equally likely.
We want to count the number of sequences in which East or West gets all 4 spades. There are $2$ ways to choose East/West, then there are $binom{13}{4}$ ways to choose the locations of the spades within the chosen hand, $4!$ ways to sequence the spades, and $22!$ ways to sequence all the remaining cards. So all together, there are
$$2 times binom{13}{4} times 4! times 22!$$
ways to sequence the cards in which either East or West gets all the spades.
So the probability that either East or West gets all the spades is
$$frac{2 times binom{13}{4} times 4! times 22!}{26!} = frac{11}{115}$$
answered Jan 25 at 21:47
awkwardawkward
6,56511025
$endgroup$
add a comment |
$begingroup$
We have 4 spades and 22 others to distribute between East and West. Let's suppose we deal the cards out, one by one, with the first 13 going to East and the second 13 going to West. We will consider the order of the cards to be significant, so this can be done in $26!$ ways, all of which are equally likely.
We want to count the number of sequences in which East or West gets all 4 spades. There are $2$ ways to choose East/West, then there are $binom{13}{4}$ ways to choose the locations of the spades within the chosen hand, $4!$ ways to sequence the spades, and $22!$ ways to sequence all the remaining cards. So all together, there are
$$2 times binom{13}{4} times 4! times 22!$$
ways to sequence the cards in which either East or West gets all the spades.
So the probability that either East or West gets all the spades is
$$frac{2 times binom{13}{4} times 4! times 22!}{26!} = frac{11}{115}$$
answered Jan 25 at 21:47
awkwardawkward
6,56511025
$endgroup$
We have 4 spades and 22 others to distribute between East and West. Let's suppose we deal the cards out, one by one, with the first 13 going to East and the second 13 going to West. We will consider the order of the cards to be significant, so this can be done in $26!$ ways, all of which are equally likely.
We want to count the number of sequences in which East or West gets all 4 spades. There are $2$ ways to choose East/West, then there are $binom{13}{4}$ ways to choose the locations of the spades within the chosen hand, $4!$ ways to sequence the spades, and $22!$ ways to sequence all the remaining cards. So all together, there are
$$2 times binom{13}{4} times 4! times 22!$$
ways to sequence the cards in which either East or West gets all the spades.
So the probability that either East or West gets all the spades is
$$frac{2 times binom{13}{4} times 4! times 22!}{26!} = frac{11}{115}$$
answered Jan 25 at 21:47
awkwardawkward
6,56511025
edited Jan 26 at 14:16
answered Jan 25 at 21:47
awkwardawkward
6,56511025
answered Jan 25 at 21:47
awkwardawkward
6,56511025
answered Jan 25 at 21:47
awkwardawkward
6,56511025
6,56511025
add a comment |
add a comment |
$begingroup$
In your attempt, you tried to work out the probability that North and South have $9$ spades and that one of East or West has none.
But the problem says we are supposed to assume North and South have $9$ spades between them:
In a bridge game, North and South have 9 spades between them.
We don’t have to be concerned about how likely this event is; we know it happened, because the problem statement says so.
So the cards in the hands of North and South can be considered “already dealt,” and we only have to consider one thing: given that the remaining $26$ cards are divided $13$ to East and $13$ to West, what is the probability that one of them gets no spades?
You correctly found that there are $binom{22}{13}$ ways for East to get no spades. You correctly multiply by $2$ to include the case in which West gets no spades.
The denominator is the number of ways to divide the $26$ cards, $binom{26}{13}.$
So the probability comes out to
$$frac{2 binom{22}{13}}{binom{26}{13}}
= frac{11}{115}.
$$
answered Jan 26 at 14:29
David KDavid K
55.2k344120
$endgroup$
add a comment |
$begingroup$
In your attempt, you tried to work out the probability that North and South have $9$ spades and that one of East or West has none.
But the problem says we are supposed to assume North and South have $9$ spades between them:
In a bridge game, North and South have 9 spades between them.
We don’t have to be concerned about how likely this event is; we know it happened, because the problem statement says so.
So the cards in the hands of North and South can be considered “already dealt,” and we only have to consider one thing: given that the remaining $26$ cards are divided $13$ to East and $13$ to West, what is the probability that one of them gets no spades?
You correctly found that there are $binom{22}{13}$ ways for East to get no spades. You correctly multiply by $2$ to include the case in which West gets no spades.
The denominator is the number of ways to divide the $26$ cards, $binom{26}{13}.$
So the probability comes out to
$$frac{2 binom{22}{13}}{binom{26}{13}}
= frac{11}{115}.
$$
answered Jan 26 at 14:29
David KDavid K
55.2k344120
$endgroup$
add a comment |
$begingroup$
In your attempt, you tried to work out the probability that North and South have $9$ spades and that one of East or West has none.
But the problem says we are supposed to assume North and South have $9$ spades between them:
In a bridge game, North and South have 9 spades between them.
We don’t have to be concerned about how likely this event is; we know it happened, because the problem statement says so.
So the cards in the hands of North and South can be considered “already dealt,” and we only have to consider one thing: given that the remaining $26$ cards are divided $13$ to East and $13$ to West, what is the probability that one of them gets no spades?
You correctly found that there are $binom{22}{13}$ ways for East to get no spades. You correctly multiply by $2$ to include the case in which West gets no spades.
The denominator is the number of ways to divide the $26$ cards, $binom{26}{13}.$
So the probability comes out to
$$frac{2 binom{22}{13}}{binom{26}{13}}
= frac{11}{115}.
$$
answered Jan 26 at 14:29
David KDavid K
55.2k344120
$endgroup$
In your attempt, you tried to work out the probability that North and South have $9$ spades and that one of East or West has none.
But the problem says we are supposed to assume North and South have $9$ spades between them:
In a bridge game, North and South have 9 spades between them.
We don’t have to be concerned about how likely this event is; we know it happened, because the problem statement says so.
So the cards in the hands of North and South can be considered “already dealt,” and we only have to consider one thing: given that the remaining $26$ cards are divided $13$ to East and $13$ to West, what is the probability that one of them gets no spades?
You correctly found that there are $binom{22}{13}$ ways for East to get no spades. You correctly multiply by $2$ to include the case in which West gets no spades.
The denominator is the number of ways to divide the $26$ cards, $binom{26}{13}.$
So the probability comes out to
$$frac{2 binom{22}{13}}{binom{26}{13}}
= frac{11}{115}.
$$
answered Jan 26 at 14:29
David KDavid K
55.2k344120
edited Jan 26 at 14:36
answered Jan 26 at 14:29
David KDavid K
55.2k344120
answered Jan 26 at 14:29
David KDavid K
55.2k344120
answered Jan 26 at 14:29
David KDavid K
55.2k344120
55.2k344120
add a comment |
add a comment |
$begingroup$
Let having $9$ spades between North and South be event A and having no spades between East and West is event B.
So the required probability is P(B|A).
Now reducing the sample space as regards "$9$ spades are already present between North and South" so
So the favourable outcomes are corresponding to getting no spades in East or West.
Firstly we select $9$ cards out of $13$ spades and $17$ non spade cards out of $39$ non spade cards. These make a total of $26$ cards. Now we try to divide these $26$ cards between North and South. Now for each of these possible cases we try to choose $13$ non spade cards from $22$ non spade cards for either of East or West . Two such ways shall be possible i.e. either for East or for West. So the total number of favourable outcomes(n(AB)):
13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
Total number of possible outcomes when A has occurred already(n(A)):
13C9 ×39C17×((26!)/(13!)^2)× 26!/(13!)^2
Required probability $=frac{P(AB)}{P(A)}$
$=frac{n(AB)}{n(A)}$
$=frac{13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
}{13C9 ×39C17×((26!)/(13!)^2)× 26!/(13!)^2}$
$=frac{22C13×2!}{26!/(13!)^2}$
$=frac{11}{115}$
answered Jan 25 at 19:21
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
add a comment |
$begingroup$
Let having $9$ spades between North and South be event A and having no spades between East and West is event B.
So the required probability is P(B|A).
Now reducing the sample space as regards "$9$ spades are already present between North and South" so
So the favourable outcomes are corresponding to getting no spades in East or West.
Firstly we select $9$ cards out of $13$ spades and $17$ non spade cards out of $39$ non spade cards. These make a total of $26$ cards. Now we try to divide these $26$ cards between North and South. Now for each of these possible cases we try to choose $13$ non spade cards from $22$ non spade cards for either of East or West . Two such ways shall be possible i.e. either for East or for West. So the total number of favourable outcomes(n(AB)):
13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
Total number of possible outcomes when A has occurred already(n(A)):
13C9 ×39C17×((26!)/(13!)^2)× 26!/(13!)^2
Required probability $=frac{P(AB)}{P(A)}$
$=frac{n(AB)}{n(A)}$
$=frac{13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
}{13C9 ×39C17×((26!)/(13!)^2)× 26!/(13!)^2}$
$=frac{22C13×2!}{26!/(13!)^2}$
$=frac{11}{115}$
answered Jan 25 at 19:21
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
add a comment |
$begingroup$
Let having $9$ spades between North and South be event A and having no spades between East and West is event B.
So the required probability is P(B|A).
Now reducing the sample space as regards "$9$ spades are already present between North and South" so
So the favourable outcomes are corresponding to getting no spades in East or West.
Firstly we select $9$ cards out of $13$ spades and $17$ non spade cards out of $39$ non spade cards. These make a total of $26$ cards. Now we try to divide these $26$ cards between North and South. Now for each of these possible cases we try to choose $13$ non spade cards from $22$ non spade cards for either of East or West . Two such ways shall be possible i.e. either for East or for West. So the total number of favourable outcomes(n(AB)):
13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
Total number of possible outcomes when A has occurred already(n(A)):
13C9 ×39C17×((26!)/(13!)^2)× 26!/(13!)^2
Required probability $=frac{P(AB)}{P(A)}$
$=frac{n(AB)}{n(A)}$
$=frac{13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
}{13C9 ×39C17×((26!)/(13!)^2)× 26!/(13!)^2}$
$=frac{22C13×2!}{26!/(13!)^2}$
$=frac{11}{115}$
answered Jan 25 at 19:21
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
Let having $9$ spades between North and South be event A and having no spades between East and West is event B.
So the required probability is P(B|A).
Now reducing the sample space as regards "$9$ spades are already present between North and South" so
So the favourable outcomes are corresponding to getting no spades in East or West.
Firstly we select $9$ cards out of $13$ spades and $17$ non spade cards out of $39$ non spade cards. These make a total of $26$ cards. Now we try to divide these $26$ cards between North and South. Now for each of these possible cases we try to choose $13$ non spade cards from $22$ non spade cards for either of East or West . Two such ways shall be possible i.e. either for East or for West. So the total number of favourable outcomes(n(AB)):
13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
Total number of possible outcomes when A has occurred already(n(A)):
13C9 ×39C17×((26!)/(13!)^2)× 26!/(13!)^2
Required probability $=frac{P(AB)}{P(A)}$
$=frac{n(AB)}{n(A)}$
$=frac{13C9 ×39C17×((26!)/(13!)^2)×22C13×2!
}{13C9 ×39C17×((26!)/(13!)^2)× 26!/(13!)^2}$
$=frac{22C13×2!}{26!/(13!)^2}$
$=frac{11}{115}$
answered Jan 25 at 19:21
Abhishek GhoshAbhishek Ghosh
878
answered Jan 25 at 19:21
Abhishek GhoshAbhishek Ghosh
878
answered Jan 25 at 19:21
Abhishek GhoshAbhishek Ghosh
878
answered Jan 25 at 19:21
Abhishek GhoshAbhishek Ghosh
878
878
add a comment |
add a comment |
$begingroup$
Since this problem asks only about spades, there is no reason to say anything about "non-spade cards". If North and South, together, have 9 spades then East and West, together, have 13- 9= 4 spades. The probability any one of those 4 spades is in East's hand is 1/2. The probability all 4 are in East's hand is $(1/2)^4= 1/16$. Similarly the probability all 4 are in West's hand is $(1/2)^4= 1/16$. The probability that "either East or West has no spades" which is the same as the probability "either East has all 4 remaining spades or West has all 4 remaining spades" is 1/16+ 1/16= 2/16= 1/8.
answered Jan 25 at 19:58
user247327user247327
11.5k1516
$endgroup$
$begingroup$
How did you get that one of those 4 spades to be with East has probability 1/2 ? And also the answer given is not 1/8. Please explain me further. I am confused a bit.
$endgroup$
– Abhishek Ghosh
Jan 25 at 20:04
$begingroup$
Is there any reason why a spade, that is not in North and South's hands, would be more likely to be in East's than Wests' or vice-versa? If not then the probability is 1/2-1/2.
$endgroup$
– user247327
Jan 25 at 20:08
$begingroup$
How ever you also need to take into consideration that 9 spade are distributed between the north and south in terms of probability as this probability is not a sure event and as per dependence and multiplication theorem this probability shall also affect the value found out by you i hope
$endgroup$
– Abhishek Ghosh
Jan 25 at 20:13
$begingroup$
Assuming the K and Q of spades are not held by North or South, then the probability that West holds the K is $1/2$, and the probability that West holds the Q iis $1/2$, but the probability that West holds both the K and the Q is less than $1/4$; the events are not independent.
$endgroup$
– bof
Jan 25 at 23:17
add a comment |
$begingroup$
Since this problem asks only about spades, there is no reason to say anything about "non-spade cards". If North and South, together, have 9 spades then East and West, together, have 13- 9= 4 spades. The probability any one of those 4 spades is in East's hand is 1/2. The probability all 4 are in East's hand is $(1/2)^4= 1/16$. Similarly the probability all 4 are in West's hand is $(1/2)^4= 1/16$. The probability that "either East or West has no spades" which is the same as the probability "either East has all 4 remaining spades or West has all 4 remaining spades" is 1/16+ 1/16= 2/16= 1/8.
answered Jan 25 at 19:58
user247327user247327
11.5k1516
$endgroup$
$begingroup$
How did you get that one of those 4 spades to be with East has probability 1/2 ? And also the answer given is not 1/8. Please explain me further. I am confused a bit.
$endgroup$
– Abhishek Ghosh
Jan 25 at 20:04
$begingroup$
Is there any reason why a spade, that is not in North and South's hands, would be more likely to be in East's than Wests' or vice-versa? If not then the probability is 1/2-1/2.
$endgroup$
– user247327
Jan 25 at 20:08
$begingroup$
How ever you also need to take into consideration that 9 spade are distributed between the north and south in terms of probability as this probability is not a sure event and as per dependence and multiplication theorem this probability shall also affect the value found out by you i hope
$endgroup$
– Abhishek Ghosh
Jan 25 at 20:13
$begingroup$
Assuming the K and Q of spades are not held by North or South, then the probability that West holds the K is $1/2$, and the probability that West holds the Q iis $1/2$, but the probability that West holds both the K and the Q is less than $1/4$; the events are not independent.
$endgroup$
– bof
Jan 25 at 23:17
add a comment |
$begingroup$
Since this problem asks only about spades, there is no reason to say anything about "non-spade cards". If North and South, together, have 9 spades then East and West, together, have 13- 9= 4 spades. The probability any one of those 4 spades is in East's hand is 1/2. The probability all 4 are in East's hand is $(1/2)^4= 1/16$. Similarly the probability all 4 are in West's hand is $(1/2)^4= 1/16$. The probability that "either East or West has no spades" which is the same as the probability "either East has all 4 remaining spades or West has all 4 remaining spades" is 1/16+ 1/16= 2/16= 1/8.
answered Jan 25 at 19:58
user247327user247327
11.5k1516
$endgroup$
Since this problem asks only about spades, there is no reason to say anything about "non-spade cards". If North and South, together, have 9 spades then East and West, together, have 13- 9= 4 spades. The probability any one of those 4 spades is in East's hand is 1/2. The probability all 4 are in East's hand is $(1/2)^4= 1/16$. Similarly the probability all 4 are in West's hand is $(1/2)^4= 1/16$. The probability that "either East or West has no spades" which is the same as the probability "either East has all 4 remaining spades or West has all 4 remaining spades" is 1/16+ 1/16= 2/16= 1/8.
answered Jan 25 at 19:58
user247327user247327
11.5k1516
answered Jan 25 at 19:58
user247327user247327
11.5k1516
answered Jan 25 at 19:58
user247327user247327
11.5k1516
answered Jan 25 at 19:58
user247327user247327
11.5k1516
11.5k1516
$begingroup$
How did you get that one of those 4 spades to be with East has probability 1/2 ? And also the answer given is not 1/8. Please explain me further. I am confused a bit.
$endgroup$
– Abhishek Ghosh
Jan 25 at 20:04
$begingroup$
Is there any reason why a spade, that is not in North and South's hands, would be more likely to be in East's than Wests' or vice-versa? If not then the probability is 1/2-1/2.
$endgroup$
– user247327
Jan 25 at 20:08
$begingroup$
How ever you also need to take into consideration that 9 spade are distributed between the north and south in terms of probability as this probability is not a sure event and as per dependence and multiplication theorem this probability shall also affect the value found out by you i hope
$endgroup$
– Abhishek Ghosh
Jan 25 at 20:13
$begingroup$
Assuming the K and Q of spades are not held by North or South, then the probability that West holds the K is $1/2$, and the probability that West holds the Q iis $1/2$, but the probability that West holds both the K and the Q is less than $1/4$; the events are not independent.
$endgroup$
– bof
Jan 25 at 23:17
add a comment |
$begingroup$
How did you get that one of those 4 spades to be with East has probability 1/2 ? And also the answer given is not 1/8. Please explain me further. I am confused a bit.
$endgroup$
– Abhishek Ghosh
Jan 25 at 20:04
$begingroup$
Is there any reason why a spade, that is not in North and South's hands, would be more likely to be in East's than Wests' or vice-versa? If not then the probability is 1/2-1/2.
$endgroup$
– user247327
Jan 25 at 20:08
$begingroup$
How ever you also need to take into consideration that 9 spade are distributed between the north and south in terms of probability as this probability is not a sure event and as per dependence and multiplication theorem this probability shall also affect the value found out by you i hope
$endgroup$
– Abhishek Ghosh
Jan 25 at 20:13
$begingroup$
Assuming the K and Q of spades are not held by North or South, then the probability that West holds the K is $1/2$, and the probability that West holds the Q iis $1/2$, but the probability that West holds both the K and the Q is less than $1/4$; the events are not independent.
$endgroup$
– bof
Jan 25 at 23:17
$begingroup$
How did you get that one of those 4 spades to be with East has probability 1/2 ? And also the answer given is not 1/8. Please explain me further. I am confused a bit.
$endgroup$
– Abhishek Ghosh
Jan 25 at 20:04
$begingroup$
How did you get that one of those 4 spades to be with East has probability 1/2 ? And also the answer given is not 1/8. Please explain me further. I am confused a bit.
$endgroup$
– Abhishek Ghosh
Jan 25 at 20:04
$begingroup$
Is there any reason why a spade, that is not in North and South's hands, would be more likely to be in East's than Wests' or vice-versa? If not then the probability is 1/2-1/2.
$endgroup$
– user247327
Jan 25 at 20:08
$begingroup$
Is there any reason why a spade, that is not in North and South's hands, would be more likely to be in East's than Wests' or vice-versa? If not then the probability is 1/2-1/2.
$endgroup$
– user247327
Jan 25 at 20:08
$begingroup$
How ever you also need to take into consideration that 9 spade are distributed between the north and south in terms of probability as this probability is not a sure event and as per dependence and multiplication theorem this probability shall also affect the value found out by you i hope
$endgroup$
– Abhishek Ghosh
Jan 25 at 20:13
$begingroup$
How ever you also need to take into consideration that 9 spade are distributed between the north and south in terms of probability as this probability is not a sure event and as per dependence and multiplication theorem this probability shall also affect the value found out by you i hope
$endgroup$
– Abhishek Ghosh
Jan 25 at 20:13
$begingroup$
Assuming the K and Q of spades are not held by North or South, then the probability that West holds the K is $1/2$, and the probability that West holds the Q iis $1/2$, but the probability that West holds both the K and the Q is less than $1/4$; the events are not independent.
$endgroup$
– bof
Jan 25 at 23:17
$begingroup$
Assuming the K and Q of spades are not held by North or South, then the probability that West holds the K is $1/2$, and the probability that West holds the Q iis $1/2$, but the probability that West holds both the K and the Q is less than $1/4$; the events are not independent.
$endgroup$
– bof
Jan 25 at 23:17
add a comment |
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$begingroup$
You answered your own, similar question on this here: math.stackexchange.com/questions/3087310/… There are four players in Bridge: North and South (partners) and East and West (partners). There are $13$ spades. $9$ of them are held in either the North or South hand and you need to work out what the probability is that one of the remaining two players holds no spades (or equivalently, all four remaining spades)
$endgroup$
– postmortes
Jan 25 at 17:40
$begingroup$
I think your denominator is too big. It shoul count, not all possible distributions of the $52$ cards, just those in which North-South have $9$ spades.
$endgroup$
– bof
Jan 25 at 18:51
$begingroup$
Do you mean to say conditional probability?
$endgroup$
– Abhishek Ghosh
Jan 25 at 18:52
$begingroup$
Thank @bof for helping me out with the problem
$endgroup$
– Abhishek Ghosh
Jan 25 at 19:22