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Are the two item selection turn lottery methods identical?
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Owners of a land plot build a building together, in which there are many apartments. Every owner has the right to own a certain number of apartments. The sum of the rights of all owners is equal to the number of apartments in the building. Let's take for example a building with five apartments, and two owners, Reuven and Shimon, so that Reuven has the right to three apartments and Shimon has the right to two apartments.
The owners choose the apartments in a certain order and the order is determined in a lottery. We want the lottery to reflect an advantage for those who have rights to more apartments. In the example above we want the probability of Reuven choosing the first apartment to be 60%, and Shimon's 40%.
Two methods are suggested taken to decide the order of selection:
In the Naming System we put notes in a hat. The number of notes is the same as the number of apartments - in our case five notes. The name of each owner is recorded on the same number of notes as the number of apartments to which he has the right: Reuven's name on three notes, and Shimon's name on two. After registering the names and mixing the notes, they begin to take them out, note by note, and read aloud the name on each note, Reuven or Shimon. The order of the names called is the order of choosing the apartments.
Using the Numbers method, they also put notes in the hat. The number of notes is again the number of apartments - again five notes. However, this time a number is written on each note - on the first note we write 1, on the second 2, and so on until on the fifth note we write 5. Now, each owner takes out of the hat X notes, where X is the number of apartments he is entitled to. Reuven will take out three notes, and Shimon will take out two. Each will select apartments on the turns whose numbers appear on the notes he has taken out of the hat. If, for example, Reuven has taken out the notes marked by the numbers 1, 3 and 4, then Reuven will choose the first apartment. The, on the second turn, Shimon's will choose an apartment. Then Reuven will select apartments on the third and fourth turns, and finally Shimon will receive the fifth apartment - the one left unselected.
Are both methods equivalent?
probability lotteries
edited Jul 24 at 8:48
N. F. Taussig
38.2k93053
asked Jul 24 at 8:26
Avi
1688
add a comment |Â
up vote
2
down vote
favorite
Owners of a land plot build a building together, in which there are many apartments. Every owner has the right to own a certain number of apartments. The sum of the rights of all owners is equal to the number of apartments in the building. Let's take for example a building with five apartments, and two owners, Reuven and Shimon, so that Reuven has the right to three apartments and Shimon has the right to two apartments.
The owners choose the apartments in a certain order and the order is determined in a lottery. We want the lottery to reflect an advantage for those who have rights to more apartments. In the example above we want the probability of Reuven choosing the first apartment to be 60%, and Shimon's 40%.
Two methods are suggested taken to decide the order of selection:
In the Naming System we put notes in a hat. The number of notes is the same as the number of apartments - in our case five notes. The name of each owner is recorded on the same number of notes as the number of apartments to which he has the right: Reuven's name on three notes, and Shimon's name on two. After registering the names and mixing the notes, they begin to take them out, note by note, and read aloud the name on each note, Reuven or Shimon. The order of the names called is the order of choosing the apartments.
Using the Numbers method, they also put notes in the hat. The number of notes is again the number of apartments - again five notes. However, this time a number is written on each note - on the first note we write 1, on the second 2, and so on until on the fifth note we write 5. Now, each owner takes out of the hat X notes, where X is the number of apartments he is entitled to. Reuven will take out three notes, and Shimon will take out two. Each will select apartments on the turns whose numbers appear on the notes he has taken out of the hat. If, for example, Reuven has taken out the notes marked by the numbers 1, 3 and 4, then Reuven will choose the first apartment. The, on the second turn, Shimon's will choose an apartment. Then Reuven will select apartments on the third and fourth turns, and finally Shimon will receive the fifth apartment - the one left unselected.
Are both methods equivalent?
probability lotteries
edited Jul 24 at 8:48
N. F. Taussig
38.2k93053
asked Jul 24 at 8:26
Avi
1688
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Owners of a land plot build a building together, in which there are many apartments. Every owner has the right to own a certain number of apartments. The sum of the rights of all owners is equal to the number of apartments in the building. Let's take for example a building with five apartments, and two owners, Reuven and Shimon, so that Reuven has the right to three apartments and Shimon has the right to two apartments.
The owners choose the apartments in a certain order and the order is determined in a lottery. We want the lottery to reflect an advantage for those who have rights to more apartments. In the example above we want the probability of Reuven choosing the first apartment to be 60%, and Shimon's 40%.
Two methods are suggested taken to decide the order of selection:
In the Naming System we put notes in a hat. The number of notes is the same as the number of apartments - in our case five notes. The name of each owner is recorded on the same number of notes as the number of apartments to which he has the right: Reuven's name on three notes, and Shimon's name on two. After registering the names and mixing the notes, they begin to take them out, note by note, and read aloud the name on each note, Reuven or Shimon. The order of the names called is the order of choosing the apartments.
Using the Numbers method, they also put notes in the hat. The number of notes is again the number of apartments - again five notes. However, this time a number is written on each note - on the first note we write 1, on the second 2, and so on until on the fifth note we write 5. Now, each owner takes out of the hat X notes, where X is the number of apartments he is entitled to. Reuven will take out three notes, and Shimon will take out two. Each will select apartments on the turns whose numbers appear on the notes he has taken out of the hat. If, for example, Reuven has taken out the notes marked by the numbers 1, 3 and 4, then Reuven will choose the first apartment. The, on the second turn, Shimon's will choose an apartment. Then Reuven will select apartments on the third and fourth turns, and finally Shimon will receive the fifth apartment - the one left unselected.
Are both methods equivalent?
probability lotteries
edited Jul 24 at 8:48
N. F. Taussig
38.2k93053
asked Jul 24 at 8:26
Avi
1688
Owners of a land plot build a building together, in which there are many apartments. Every owner has the right to own a certain number of apartments. The sum of the rights of all owners is equal to the number of apartments in the building. Let's take for example a building with five apartments, and two owners, Reuven and Shimon, so that Reuven has the right to three apartments and Shimon has the right to two apartments.
The owners choose the apartments in a certain order and the order is determined in a lottery. We want the lottery to reflect an advantage for those who have rights to more apartments. In the example above we want the probability of Reuven choosing the first apartment to be 60%, and Shimon's 40%.
Two methods are suggested taken to decide the order of selection:
In the Naming System we put notes in a hat. The number of notes is the same as the number of apartments - in our case five notes. The name of each owner is recorded on the same number of notes as the number of apartments to which he has the right: Reuven's name on three notes, and Shimon's name on two. After registering the names and mixing the notes, they begin to take them out, note by note, and read aloud the name on each note, Reuven or Shimon. The order of the names called is the order of choosing the apartments.
Using the Numbers method, they also put notes in the hat. The number of notes is again the number of apartments - again five notes. However, this time a number is written on each note - on the first note we write 1, on the second 2, and so on until on the fifth note we write 5. Now, each owner takes out of the hat X notes, where X is the number of apartments he is entitled to. Reuven will take out three notes, and Shimon will take out two. Each will select apartments on the turns whose numbers appear on the notes he has taken out of the hat. If, for example, Reuven has taken out the notes marked by the numbers 1, 3 and 4, then Reuven will choose the first apartment. The, on the second turn, Shimon's will choose an apartment. Then Reuven will select apartments on the third and fourth turns, and finally Shimon will receive the fifth apartment - the one left unselected.
Are both methods equivalent?
probability lotteries
edited Jul 24 at 8:48
N. F. Taussig
38.2k93053
asked Jul 24 at 8:26
Avi
1688
edited Jul 24 at 8:48
N. F. Taussig
38.2k93053
edited Jul 24 at 8:48
N. F. Taussig
38.2k93053
edited Jul 24 at 8:48
N. F. Taussig
38.2k93053
38.2k93053
asked Jul 24 at 8:26
Avi
1688
asked Jul 24 at 8:26
Avi
1688
asked Jul 24 at 8:26
Avi
1688
1688
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
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Imagine that Reuben and Shimon bring their children $R_1$, $R_2$, $R_3$ and $S_1$, $S_2$ along to help in the lottery. In the first protocol the five notes carry their names, and in the second protocol they individually choose a ticket from the hat.
I hope you agree that in both protocols a random pairing between the sets $$C:=R_1,R_2,R_3,S_1,S_2quadrm andquad[5]:=1,2,3,4,5 ,$$ in other words: a random bijective map $pi:>Cto [5]$ has been selected, with $pi$ uniformly distributed over the $5!=120$ possible such maps. By symmetry the set $pibigl(S_1,S_2bigr)$ can be any of the $5choose2$ two element subsets of $[5]$ with equal probability, and this under both protocols.
It follows that Shimon (as well as Reuben) has no reason to prefer one of the protocols over the other.
answered Jul 24 at 12:04
Christian Blatter
163k7107306
OMG. From my math edu 30 yeas ago I recall that :-> means a relation, and denotes a set. I also recall that () denote selection without ordering. Figuring out the rest for you answer will take a few more readings :) Thanks!
– Avi
Jul 24 at 12:09
OK. I don't get it. First, if by pairing you mean selecting one item from C and another from [5], then there are IMHO 25 such pairs: <R1,1> to <R1,5>, then <R2,1> to <R2,5>... then <S2,1> to <S2,5>. Obviously, I don't get it. :(
– Avi
Jul 24 at 19:01
@Avi For each of the five ways you can assign a number to $R_1$, there are four ways to assign one of the remaining numbers to $R_2$, three ways to assign one of the remaining numbers to $R_3$, two ways to assign one of the remaining numbers to $S_1$, and one way to assign the remaining number to $S_2$, giving $5! = 5 cdot 4 cdot 3 cdot 2 cdot 1$ possible assignments.
– N. F. Taussig
Jul 24 at 20:08
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Imagine that Reuben and Shimon bring their children $R_1$, $R_2$, $R_3$ and $S_1$, $S_2$ along to help in the lottery. In the first protocol the five notes carry their names, and in the second protocol they individually choose a ticket from the hat.
I hope you agree that in both protocols a random pairing between the sets $$C:=R_1,R_2,R_3,S_1,S_2quadrm andquad[5]:=1,2,3,4,5 ,$$ in other words: a random bijective map $pi:>Cto [5]$ has been selected, with $pi$ uniformly distributed over the $5!=120$ possible such maps. By symmetry the set $pibigl(S_1,S_2bigr)$ can be any of the $5choose2$ two element subsets of $[5]$ with equal probability, and this under both protocols.
It follows that Shimon (as well as Reuben) has no reason to prefer one of the protocols over the other.
answered Jul 24 at 12:04
Christian Blatter
163k7107306
OMG. From my math edu 30 yeas ago I recall that :-> means a relation, and denotes a set. I also recall that () denote selection without ordering. Figuring out the rest for you answer will take a few more readings :) Thanks!
– Avi
Jul 24 at 12:09
OK. I don't get it. First, if by pairing you mean selecting one item from C and another from [5], then there are IMHO 25 such pairs: <R1,1> to <R1,5>, then <R2,1> to <R2,5>... then <S2,1> to <S2,5>. Obviously, I don't get it. :(
– Avi
Jul 24 at 19:01
@Avi For each of the five ways you can assign a number to $R_1$, there are four ways to assign one of the remaining numbers to $R_2$, three ways to assign one of the remaining numbers to $R_3$, two ways to assign one of the remaining numbers to $S_1$, and one way to assign the remaining number to $S_2$, giving $5! = 5 cdot 4 cdot 3 cdot 2 cdot 1$ possible assignments.
– N. F. Taussig
Jul 24 at 20:08
add a comment |Â
up vote
1
down vote
Imagine that Reuben and Shimon bring their children $R_1$, $R_2$, $R_3$ and $S_1$, $S_2$ along to help in the lottery. In the first protocol the five notes carry their names, and in the second protocol they individually choose a ticket from the hat.
I hope you agree that in both protocols a random pairing between the sets $$C:=R_1,R_2,R_3,S_1,S_2quadrm andquad[5]:=1,2,3,4,5 ,$$ in other words: a random bijective map $pi:>Cto [5]$ has been selected, with $pi$ uniformly distributed over the $5!=120$ possible such maps. By symmetry the set $pibigl(S_1,S_2bigr)$ can be any of the $5choose2$ two element subsets of $[5]$ with equal probability, and this under both protocols.
It follows that Shimon (as well as Reuben) has no reason to prefer one of the protocols over the other.
answered Jul 24 at 12:04
Christian Blatter
163k7107306
OMG. From my math edu 30 yeas ago I recall that :-> means a relation, and denotes a set. I also recall that () denote selection without ordering. Figuring out the rest for you answer will take a few more readings :) Thanks!
– Avi
Jul 24 at 12:09
OK. I don't get it. First, if by pairing you mean selecting one item from C and another from [5], then there are IMHO 25 such pairs: <R1,1> to <R1,5>, then <R2,1> to <R2,5>... then <S2,1> to <S2,5>. Obviously, I don't get it. :(
– Avi
Jul 24 at 19:01
@Avi For each of the five ways you can assign a number to $R_1$, there are four ways to assign one of the remaining numbers to $R_2$, three ways to assign one of the remaining numbers to $R_3$, two ways to assign one of the remaining numbers to $S_1$, and one way to assign the remaining number to $S_2$, giving $5! = 5 cdot 4 cdot 3 cdot 2 cdot 1$ possible assignments.
– N. F. Taussig
Jul 24 at 20:08
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Imagine that Reuben and Shimon bring their children $R_1$, $R_2$, $R_3$ and $S_1$, $S_2$ along to help in the lottery. In the first protocol the five notes carry their names, and in the second protocol they individually choose a ticket from the hat.
I hope you agree that in both protocols a random pairing between the sets $$C:=R_1,R_2,R_3,S_1,S_2quadrm andquad[5]:=1,2,3,4,5 ,$$ in other words: a random bijective map $pi:>Cto [5]$ has been selected, with $pi$ uniformly distributed over the $5!=120$ possible such maps. By symmetry the set $pibigl(S_1,S_2bigr)$ can be any of the $5choose2$ two element subsets of $[5]$ with equal probability, and this under both protocols.
It follows that Shimon (as well as Reuben) has no reason to prefer one of the protocols over the other.
answered Jul 24 at 12:04
Christian Blatter
163k7107306
Imagine that Reuben and Shimon bring their children $R_1$, $R_2$, $R_3$ and $S_1$, $S_2$ along to help in the lottery. In the first protocol the five notes carry their names, and in the second protocol they individually choose a ticket from the hat.
I hope you agree that in both protocols a random pairing between the sets $$C:=R_1,R_2,R_3,S_1,S_2quadrm andquad[5]:=1,2,3,4,5 ,$$ in other words: a random bijective map $pi:>Cto [5]$ has been selected, with $pi$ uniformly distributed over the $5!=120$ possible such maps. By symmetry the set $pibigl(S_1,S_2bigr)$ can be any of the $5choose2$ two element subsets of $[5]$ with equal probability, and this under both protocols.
It follows that Shimon (as well as Reuben) has no reason to prefer one of the protocols over the other.
answered Jul 24 at 12:04
Christian Blatter
163k7107306
edited Jul 24 at 20:02
answered Jul 24 at 12:04
Christian Blatter
163k7107306
answered Jul 24 at 12:04
Christian Blatter
163k7107306
answered Jul 24 at 12:04
Christian Blatter
163k7107306
163k7107306
OMG. From my math edu 30 yeas ago I recall that :-> means a relation, and denotes a set. I also recall that () denote selection without ordering. Figuring out the rest for you answer will take a few more readings :) Thanks!
– Avi
Jul 24 at 12:09
OK. I don't get it. First, if by pairing you mean selecting one item from C and another from [5], then there are IMHO 25 such pairs: <R1,1> to <R1,5>, then <R2,1> to <R2,5>... then <S2,1> to <S2,5>. Obviously, I don't get it. :(
– Avi
Jul 24 at 19:01
@Avi For each of the five ways you can assign a number to $R_1$, there are four ways to assign one of the remaining numbers to $R_2$, three ways to assign one of the remaining numbers to $R_3$, two ways to assign one of the remaining numbers to $S_1$, and one way to assign the remaining number to $S_2$, giving $5! = 5 cdot 4 cdot 3 cdot 2 cdot 1$ possible assignments.
– N. F. Taussig
Jul 24 at 20:08
add a comment |Â
OMG. From my math edu 30 yeas ago I recall that :-> means a relation, and denotes a set. I also recall that () denote selection without ordering. Figuring out the rest for you answer will take a few more readings :) Thanks!
– Avi
Jul 24 at 12:09
OK. I don't get it. First, if by pairing you mean selecting one item from C and another from [5], then there are IMHO 25 such pairs: <R1,1> to <R1,5>, then <R2,1> to <R2,5>... then <S2,1> to <S2,5>. Obviously, I don't get it. :(
– Avi
Jul 24 at 19:01
@Avi For each of the five ways you can assign a number to $R_1$, there are four ways to assign one of the remaining numbers to $R_2$, three ways to assign one of the remaining numbers to $R_3$, two ways to assign one of the remaining numbers to $S_1$, and one way to assign the remaining number to $S_2$, giving $5! = 5 cdot 4 cdot 3 cdot 2 cdot 1$ possible assignments.
– N. F. Taussig
Jul 24 at 20:08
OMG. From my math edu 30 yeas ago I recall that :-> means a relation, and denotes a set. I also recall that () denote selection without ordering. Figuring out the rest for you answer will take a few more readings :) Thanks!
– Avi
Jul 24 at 12:09
OMG. From my math edu 30 yeas ago I recall that :-> means a relation, and denotes a set. I also recall that () denote selection without ordering. Figuring out the rest for you answer will take a few more readings :) Thanks!
– Avi
Jul 24 at 12:09
OK. I don't get it. First, if by pairing you mean selecting one item from C and another from [5], then there are IMHO 25 such pairs: <R1,1> to <R1,5>, then <R2,1> to <R2,5>... then <S2,1> to <S2,5>. Obviously, I don't get it. :(
– Avi
Jul 24 at 19:01
OK. I don't get it. First, if by pairing you mean selecting one item from C and another from [5], then there are IMHO 25 such pairs: <R1,1> to <R1,5>, then <R2,1> to <R2,5>... then <S2,1> to <S2,5>. Obviously, I don't get it. :(
– Avi
Jul 24 at 19:01
@Avi For each of the five ways you can assign a number to $R_1$, there are four ways to assign one of the remaining numbers to $R_2$, three ways to assign one of the remaining numbers to $R_3$, two ways to assign one of the remaining numbers to $S_1$, and one way to assign the remaining number to $S_2$, giving $5! = 5 cdot 4 cdot 3 cdot 2 cdot 1$ possible assignments.
– N. F. Taussig
Jul 24 at 20:08
@Avi For each of the five ways you can assign a number to $R_1$, there are four ways to assign one of the remaining numbers to $R_2$, three ways to assign one of the remaining numbers to $R_3$, two ways to assign one of the remaining numbers to $S_1$, and one way to assign the remaining number to $S_2$, giving $5! = 5 cdot 4 cdot 3 cdot 2 cdot 1$ possible assignments.
– N. F. Taussig
Jul 24 at 20:08
add a comment |Â
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