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Combinatorics bridge card game.
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Bridge is played with four players and an ordinary deck of 52 cards. Each player begins with a hand of 13 cards. In how many ways can a bridge game start?
(Ignore the fact that bridge is played in partnership)
Here is my attempt: For the first player I have $52 choose 13 $. For the second player I have $39choose 13$. Then third player $26choose 13$ and lastly $13choose 13$.
Therefore, I have $52choose 13$$39choose 13$$26choose 13$$13choose 13$ of ways a bridge game can start.
Am I correct with this ? Is there another way to think this questions?
combinatorics combinations
edited Jul 22 at 21:08
N. F. Taussig
38.2k93053
asked Jul 22 at 17:39
Ling Min Hao
31118
add a comment |Â
up vote
0
down vote
favorite
Bridge is played with four players and an ordinary deck of 52 cards. Each player begins with a hand of 13 cards. In how many ways can a bridge game start?
(Ignore the fact that bridge is played in partnership)
Here is my attempt: For the first player I have $52 choose 13 $. For the second player I have $39choose 13$. Then third player $26choose 13$ and lastly $13choose 13$.
Therefore, I have $52choose 13$$39choose 13$$26choose 13$$13choose 13$ of ways a bridge game can start.
Am I correct with this ? Is there another way to think this questions?
combinatorics combinations
edited Jul 22 at 21:08
N. F. Taussig
38.2k93053
asked Jul 22 at 17:39
Ling Min Hao
31118
3
This is right for four distinct players (table positions), which is the correct way to look at bridge; who bids when is important. I'm not quite sure what the question means when it specifies ignoring partnership.
– Joffan
Jul 22 at 17:52
1
An equivalent solution is that the number of ways is a multinomial coefficient, $binom5213 ; 13 ; 13 ; 13 = 52!/(13!)^4$
– awkward
Jul 22 at 18:46
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Bridge is played with four players and an ordinary deck of 52 cards. Each player begins with a hand of 13 cards. In how many ways can a bridge game start?
(Ignore the fact that bridge is played in partnership)
Here is my attempt: For the first player I have $52 choose 13 $. For the second player I have $39choose 13$. Then third player $26choose 13$ and lastly $13choose 13$.
Therefore, I have $52choose 13$$39choose 13$$26choose 13$$13choose 13$ of ways a bridge game can start.
Am I correct with this ? Is there another way to think this questions?
combinatorics combinations
edited Jul 22 at 21:08
N. F. Taussig
38.2k93053
asked Jul 22 at 17:39
Ling Min Hao
31118
Bridge is played with four players and an ordinary deck of 52 cards. Each player begins with a hand of 13 cards. In how many ways can a bridge game start?
(Ignore the fact that bridge is played in partnership)
Here is my attempt: For the first player I have $52 choose 13 $. For the second player I have $39choose 13$. Then third player $26choose 13$ and lastly $13choose 13$.
Therefore, I have $52choose 13$$39choose 13$$26choose 13$$13choose 13$ of ways a bridge game can start.
Am I correct with this ? Is there another way to think this questions?
combinatorics combinations
edited Jul 22 at 21:08
N. F. Taussig
38.2k93053
asked Jul 22 at 17:39
Ling Min Hao
31118
edited Jul 22 at 21:08
N. F. Taussig
38.2k93053
edited Jul 22 at 21:08
N. F. Taussig
38.2k93053
edited Jul 22 at 21:08
N. F. Taussig
38.2k93053
38.2k93053
asked Jul 22 at 17:39
Ling Min Hao
31118
asked Jul 22 at 17:39
Ling Min Hao
31118
asked Jul 22 at 17:39
Ling Min Hao
31118
31118
3
This is right for four distinct players (table positions), which is the correct way to look at bridge; who bids when is important. I'm not quite sure what the question means when it specifies ignoring partnership.
– Joffan
Jul 22 at 17:52
1
An equivalent solution is that the number of ways is a multinomial coefficient, $binom5213 ; 13 ; 13 ; 13 = 52!/(13!)^4$
– awkward
Jul 22 at 18:46
add a comment |Â
3
This is right for four distinct players (table positions), which is the correct way to look at bridge; who bids when is important. I'm not quite sure what the question means when it specifies ignoring partnership.
– Joffan
Jul 22 at 17:52
1
An equivalent solution is that the number of ways is a multinomial coefficient, $binom5213 ; 13 ; 13 ; 13 = 52!/(13!)^4$
– awkward
Jul 22 at 18:46
3
3
This is right for four distinct players (table positions), which is the correct way to look at bridge; who bids when is important. I'm not quite sure what the question means when it specifies ignoring partnership.
– Joffan
Jul 22 at 17:52
This is right for four distinct players (table positions), which is the correct way to look at bridge; who bids when is important. I'm not quite sure what the question means when it specifies ignoring partnership.
– Joffan
Jul 22 at 17:52
1
1
An equivalent solution is that the number of ways is a multinomial coefficient, $binom5213 ; 13 ; 13 ; 13 = 52!/(13!)^4$
– awkward
Jul 22 at 18:46
An equivalent solution is that the number of ways is a multinomial coefficient, $binom5213 ; 13 ; 13 ; 13 = 52!/(13!)^4$
– awkward
Jul 22 at 18:46
add a comment |Â
1 Answer
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I agree with the comment that it's unclear what the prescription to ignore the fact that bridge is played in partnership means. You've correctly calculated the number of ways a bridge game can start if you regard the players as distinct, which is the appropriate way to look at the situation in bridge.
To illustrate how the result can depend on what you regard as different situations, consider the game of Getaway, where the player who holds the ace of spades starts the game. If you want to know how many different situations you might face as a player, the result would again be the same (assuming four players are playing), since each player is in a distinct position relative to you. However, if you're not one of the players and you don't care about their identities and just want to know how many inequivalent situations might arise in this game, you need to take into account the cyclical symmetry among the players. The order of the players still matters, so there's no full permutational symmetry among them, but it doesn't matter whether you number them e.g. $1234$ or $2341$. Thus you'd have to divide your result by $4$ to get the number of inequivalent starting positions. This doesn't apply to bridge, though, since in bridge the dealer opens the auction, so there are distinguished first, second, third and fourth players. Thus, in the case of bridge, your result is right no matter whether you look from the position of a distinguished player or from outside without regard for the identity of the players. Since this is true with or without partnerships, it's not clear what the prescription to ignore the partnerships means.
answered Jul 23 at 4:13
joriki
164k10180328
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I agree with the comment that it's unclear what the prescription to ignore the fact that bridge is played in partnership means. You've correctly calculated the number of ways a bridge game can start if you regard the players as distinct, which is the appropriate way to look at the situation in bridge.
To illustrate how the result can depend on what you regard as different situations, consider the game of Getaway, where the player who holds the ace of spades starts the game. If you want to know how many different situations you might face as a player, the result would again be the same (assuming four players are playing), since each player is in a distinct position relative to you. However, if you're not one of the players and you don't care about their identities and just want to know how many inequivalent situations might arise in this game, you need to take into account the cyclical symmetry among the players. The order of the players still matters, so there's no full permutational symmetry among them, but it doesn't matter whether you number them e.g. $1234$ or $2341$. Thus you'd have to divide your result by $4$ to get the number of inequivalent starting positions. This doesn't apply to bridge, though, since in bridge the dealer opens the auction, so there are distinguished first, second, third and fourth players. Thus, in the case of bridge, your result is right no matter whether you look from the position of a distinguished player or from outside without regard for the identity of the players. Since this is true with or without partnerships, it's not clear what the prescription to ignore the partnerships means.
answered Jul 23 at 4:13
joriki
164k10180328
add a comment |Â
up vote
0
down vote
I agree with the comment that it's unclear what the prescription to ignore the fact that bridge is played in partnership means. You've correctly calculated the number of ways a bridge game can start if you regard the players as distinct, which is the appropriate way to look at the situation in bridge.
To illustrate how the result can depend on what you regard as different situations, consider the game of Getaway, where the player who holds the ace of spades starts the game. If you want to know how many different situations you might face as a player, the result would again be the same (assuming four players are playing), since each player is in a distinct position relative to you. However, if you're not one of the players and you don't care about their identities and just want to know how many inequivalent situations might arise in this game, you need to take into account the cyclical symmetry among the players. The order of the players still matters, so there's no full permutational symmetry among them, but it doesn't matter whether you number them e.g. $1234$ or $2341$. Thus you'd have to divide your result by $4$ to get the number of inequivalent starting positions. This doesn't apply to bridge, though, since in bridge the dealer opens the auction, so there are distinguished first, second, third and fourth players. Thus, in the case of bridge, your result is right no matter whether you look from the position of a distinguished player or from outside without regard for the identity of the players. Since this is true with or without partnerships, it's not clear what the prescription to ignore the partnerships means.
answered Jul 23 at 4:13
joriki
164k10180328
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I agree with the comment that it's unclear what the prescription to ignore the fact that bridge is played in partnership means. You've correctly calculated the number of ways a bridge game can start if you regard the players as distinct, which is the appropriate way to look at the situation in bridge.
To illustrate how the result can depend on what you regard as different situations, consider the game of Getaway, where the player who holds the ace of spades starts the game. If you want to know how many different situations you might face as a player, the result would again be the same (assuming four players are playing), since each player is in a distinct position relative to you. However, if you're not one of the players and you don't care about their identities and just want to know how many inequivalent situations might arise in this game, you need to take into account the cyclical symmetry among the players. The order of the players still matters, so there's no full permutational symmetry among them, but it doesn't matter whether you number them e.g. $1234$ or $2341$. Thus you'd have to divide your result by $4$ to get the number of inequivalent starting positions. This doesn't apply to bridge, though, since in bridge the dealer opens the auction, so there are distinguished first, second, third and fourth players. Thus, in the case of bridge, your result is right no matter whether you look from the position of a distinguished player or from outside without regard for the identity of the players. Since this is true with or without partnerships, it's not clear what the prescription to ignore the partnerships means.
answered Jul 23 at 4:13
joriki
164k10180328
I agree with the comment that it's unclear what the prescription to ignore the fact that bridge is played in partnership means. You've correctly calculated the number of ways a bridge game can start if you regard the players as distinct, which is the appropriate way to look at the situation in bridge.
To illustrate how the result can depend on what you regard as different situations, consider the game of Getaway, where the player who holds the ace of spades starts the game. If you want to know how many different situations you might face as a player, the result would again be the same (assuming four players are playing), since each player is in a distinct position relative to you. However, if you're not one of the players and you don't care about their identities and just want to know how many inequivalent situations might arise in this game, you need to take into account the cyclical symmetry among the players. The order of the players still matters, so there's no full permutational symmetry among them, but it doesn't matter whether you number them e.g. $1234$ or $2341$. Thus you'd have to divide your result by $4$ to get the number of inequivalent starting positions. This doesn't apply to bridge, though, since in bridge the dealer opens the auction, so there are distinguished first, second, third and fourth players. Thus, in the case of bridge, your result is right no matter whether you look from the position of a distinguished player or from outside without regard for the identity of the players. Since this is true with or without partnerships, it's not clear what the prescription to ignore the partnerships means.
answered Jul 23 at 4:13
joriki
164k10180328
answered Jul 23 at 4:13
joriki
164k10180328
answered Jul 23 at 4:13
joriki
164k10180328
answered Jul 23 at 4:13
joriki
164k10180328
164k10180328
add a comment |Â
add a comment |Â
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3
This is right for four distinct players (table positions), which is the correct way to look at bridge; who bids when is important. I'm not quite sure what the question means when it specifies ignoring partnership.
– Joffan
Jul 22 at 17:52
1
An equivalent solution is that the number of ways is a multinomial coefficient, $binom5213 ; 13 ; 13 ; 13 = 52!/(13!)^4$
– awkward
Jul 22 at 18:46