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Is it true that $sum_j=k^2k 2k choose jbig(frac12big)^2k-frac12=frac12Big(2k choose kbig(frac12big)^2kBig)$?
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On the first page of Lionel Penrose's 1946 paper "The Elementary Statistics of Majority Voting," Penrose claims that "the amount by which [an individual voter's] chance of winning exceeds one half" (call this quantity $A$) is equal to "half the likelihood of a situation in which an individual vote can be decisive--that is to say, a situation in which the remaining votes are equally divided upon the issue at stake" (call this quantity $B$).
Suppose an individual Bob has a given preference on a vote between two options/candidates. Based on the discussion in the paragraph quoted above, it seems like what Penrose means by quantity $A$ is the difference between $frac12$ and the probability that enough other voters will agree with Bob for his choice to win a majority vote, assuming that the probability any other voter agrees with Bob is $frac12$. Hence, if the number of voters is $2k+1$, it seems that what Penrose means by what we call quantity $A$ is $sum_j=k^2k 2k choose jbig(frac12big)^2k-frac12$.
Similarly, it seems that what Penrose means by quantity $B$ is one half times the probability that exactly one half of all voters other than Bob will agree with him, assuming each other voter has a $frac12$ chance of agreeing with Bob. So if there are $2k+1$ voters, it seems that what Penrose has in mind for what we call quantity $B$ is $frac12Big(2k choose kbig(frac12big)^2kBig)$.
So it seems that Penrose is claiming that for all integers $kge 1$ we have $$sum_j=k^2k 2k choose jbig(frac12big)^2k-frac12=frac12Big(2k choose kbig(frac12big)^2kBig).$$
But I'm wondering, how can one prove this identity? And is it even true? It seems to hold for the cases $k=1$ and $k=2$, but I haven't yet checked any others.
probability combinatorics statistics binomial-coefficients binomial-distribution
asked Jul 22 at 18:48
Rasputin
38619
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On the first page of Lionel Penrose's 1946 paper "The Elementary Statistics of Majority Voting," Penrose claims that "the amount by which [an individual voter's] chance of winning exceeds one half" (call this quantity $A$) is equal to "half the likelihood of a situation in which an individual vote can be decisive--that is to say, a situation in which the remaining votes are equally divided upon the issue at stake" (call this quantity $B$).
Suppose an individual Bob has a given preference on a vote between two options/candidates. Based on the discussion in the paragraph quoted above, it seems like what Penrose means by quantity $A$ is the difference between $frac12$ and the probability that enough other voters will agree with Bob for his choice to win a majority vote, assuming that the probability any other voter agrees with Bob is $frac12$. Hence, if the number of voters is $2k+1$, it seems that what Penrose means by what we call quantity $A$ is $sum_j=k^2k 2k choose jbig(frac12big)^2k-frac12$.
Similarly, it seems that what Penrose means by quantity $B$ is one half times the probability that exactly one half of all voters other than Bob will agree with him, assuming each other voter has a $frac12$ chance of agreeing with Bob. So if there are $2k+1$ voters, it seems that what Penrose has in mind for what we call quantity $B$ is $frac12Big(2k choose kbig(frac12big)^2kBig)$.
So it seems that Penrose is claiming that for all integers $kge 1$ we have $$sum_j=k^2k 2k choose jbig(frac12big)^2k-frac12=frac12Big(2k choose kbig(frac12big)^2kBig).$$
But I'm wondering, how can one prove this identity? And is it even true? It seems to hold for the cases $k=1$ and $k=2$, but I haven't yet checked any others.
probability combinatorics statistics binomial-coefficients binomial-distribution
asked Jul 22 at 18:48
Rasputin
38619
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
On the first page of Lionel Penrose's 1946 paper "The Elementary Statistics of Majority Voting," Penrose claims that "the amount by which [an individual voter's] chance of winning exceeds one half" (call this quantity $A$) is equal to "half the likelihood of a situation in which an individual vote can be decisive--that is to say, a situation in which the remaining votes are equally divided upon the issue at stake" (call this quantity $B$).
Suppose an individual Bob has a given preference on a vote between two options/candidates. Based on the discussion in the paragraph quoted above, it seems like what Penrose means by quantity $A$ is the difference between $frac12$ and the probability that enough other voters will agree with Bob for his choice to win a majority vote, assuming that the probability any other voter agrees with Bob is $frac12$. Hence, if the number of voters is $2k+1$, it seems that what Penrose means by what we call quantity $A$ is $sum_j=k^2k 2k choose jbig(frac12big)^2k-frac12$.
Similarly, it seems that what Penrose means by quantity $B$ is one half times the probability that exactly one half of all voters other than Bob will agree with him, assuming each other voter has a $frac12$ chance of agreeing with Bob. So if there are $2k+1$ voters, it seems that what Penrose has in mind for what we call quantity $B$ is $frac12Big(2k choose kbig(frac12big)^2kBig)$.
So it seems that Penrose is claiming that for all integers $kge 1$ we have $$sum_j=k^2k 2k choose jbig(frac12big)^2k-frac12=frac12Big(2k choose kbig(frac12big)^2kBig).$$
But I'm wondering, how can one prove this identity? And is it even true? It seems to hold for the cases $k=1$ and $k=2$, but I haven't yet checked any others.
probability combinatorics statistics binomial-coefficients binomial-distribution
asked Jul 22 at 18:48
Rasputin
38619
On the first page of Lionel Penrose's 1946 paper "The Elementary Statistics of Majority Voting," Penrose claims that "the amount by which [an individual voter's] chance of winning exceeds one half" (call this quantity $A$) is equal to "half the likelihood of a situation in which an individual vote can be decisive--that is to say, a situation in which the remaining votes are equally divided upon the issue at stake" (call this quantity $B$).
Suppose an individual Bob has a given preference on a vote between two options/candidates. Based on the discussion in the paragraph quoted above, it seems like what Penrose means by quantity $A$ is the difference between $frac12$ and the probability that enough other voters will agree with Bob for his choice to win a majority vote, assuming that the probability any other voter agrees with Bob is $frac12$. Hence, if the number of voters is $2k+1$, it seems that what Penrose means by what we call quantity $A$ is $sum_j=k^2k 2k choose jbig(frac12big)^2k-frac12$.
Similarly, it seems that what Penrose means by quantity $B$ is one half times the probability that exactly one half of all voters other than Bob will agree with him, assuming each other voter has a $frac12$ chance of agreeing with Bob. So if there are $2k+1$ voters, it seems that what Penrose has in mind for what we call quantity $B$ is $frac12Big(2k choose kbig(frac12big)^2kBig)$.
So it seems that Penrose is claiming that for all integers $kge 1$ we have $$sum_j=k^2k 2k choose jbig(frac12big)^2k-frac12=frac12Big(2k choose kbig(frac12big)^2kBig).$$
But I'm wondering, how can one prove this identity? And is it even true? It seems to hold for the cases $k=1$ and $k=2$, but I haven't yet checked any others.
probability combinatorics statistics binomial-coefficients binomial-distribution
asked Jul 22 at 18:48
Rasputin
38619
asked Jul 22 at 18:48
Rasputin
38619
asked Jul 22 at 18:48
Rasputin
38619
asked Jul 22 at 18:48
Rasputin
38619
38619
add a comment |Â
add a comment |Â
1 Answer
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Take out the factor $(1/2)^2k$ and repeat the terms in reverse (we will have the middle term twice)
begineqnarray*
sum_j=k^2k binom2kj left( frac12 right)^2k &=& left( frac12 right)^2k+1 left( sum_j=0^2k binom2kj + binom2kk right) \
&=& left( frac12 right)^2k+1 left( 2^2k + binom2kkright) \
&=& frac12 + binom2kk left(frac12 right)^2k
endeqnarray*
answered Jul 22 at 19:01
Donald Splutterwit
21.3k21243
1
begineqnarray* sum_j=k^2k binom2kj &=& frac12 left( sum_j=k^2k binom2kj + sum_j=k^0 binom2kj right) \ endeqnarray* begineqnarray* &=& frac12 left(binom2kk +sum_j=0^2k binom2kj right) \ endeqnarray*
– Donald Splutterwit
Jul 22 at 19:06
This looks a little bit weird, I guess there is an error somewhere in your math mode ^^'. I just realized that my first question was nonsense
– mrtaurho
Jul 22 at 19:08
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
accepted
Take out the factor $(1/2)^2k$ and repeat the terms in reverse (we will have the middle term twice)
begineqnarray*
sum_j=k^2k binom2kj left( frac12 right)^2k &=& left( frac12 right)^2k+1 left( sum_j=0^2k binom2kj + binom2kk right) \
&=& left( frac12 right)^2k+1 left( 2^2k + binom2kkright) \
&=& frac12 + binom2kk left(frac12 right)^2k
endeqnarray*
answered Jul 22 at 19:01
Donald Splutterwit
21.3k21243
1
begineqnarray* sum_j=k^2k binom2kj &=& frac12 left( sum_j=k^2k binom2kj + sum_j=k^0 binom2kj right) \ endeqnarray* begineqnarray* &=& frac12 left(binom2kk +sum_j=0^2k binom2kj right) \ endeqnarray*
– Donald Splutterwit
Jul 22 at 19:06
This looks a little bit weird, I guess there is an error somewhere in your math mode ^^'. I just realized that my first question was nonsense
– mrtaurho
Jul 22 at 19:08
add a comment |Â
up vote
1
down vote
accepted
Take out the factor $(1/2)^2k$ and repeat the terms in reverse (we will have the middle term twice)
begineqnarray*
sum_j=k^2k binom2kj left( frac12 right)^2k &=& left( frac12 right)^2k+1 left( sum_j=0^2k binom2kj + binom2kk right) \
&=& left( frac12 right)^2k+1 left( 2^2k + binom2kkright) \
&=& frac12 + binom2kk left(frac12 right)^2k
endeqnarray*
answered Jul 22 at 19:01
Donald Splutterwit
21.3k21243
1
begineqnarray* sum_j=k^2k binom2kj &=& frac12 left( sum_j=k^2k binom2kj + sum_j=k^0 binom2kj right) \ endeqnarray* begineqnarray* &=& frac12 left(binom2kk +sum_j=0^2k binom2kj right) \ endeqnarray*
– Donald Splutterwit
Jul 22 at 19:06
This looks a little bit weird, I guess there is an error somewhere in your math mode ^^'. I just realized that my first question was nonsense
– mrtaurho
Jul 22 at 19:08
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Take out the factor $(1/2)^2k$ and repeat the terms in reverse (we will have the middle term twice)
begineqnarray*
sum_j=k^2k binom2kj left( frac12 right)^2k &=& left( frac12 right)^2k+1 left( sum_j=0^2k binom2kj + binom2kk right) \
&=& left( frac12 right)^2k+1 left( 2^2k + binom2kkright) \
&=& frac12 + binom2kk left(frac12 right)^2k
endeqnarray*
answered Jul 22 at 19:01
Donald Splutterwit
21.3k21243
Take out the factor $(1/2)^2k$ and repeat the terms in reverse (we will have the middle term twice)
begineqnarray*
sum_j=k^2k binom2kj left( frac12 right)^2k &=& left( frac12 right)^2k+1 left( sum_j=0^2k binom2kj + binom2kk right) \
&=& left( frac12 right)^2k+1 left( 2^2k + binom2kkright) \
&=& frac12 + binom2kk left(frac12 right)^2k
endeqnarray*
answered Jul 22 at 19:01
Donald Splutterwit
21.3k21243
answered Jul 22 at 19:01
Donald Splutterwit
21.3k21243
answered Jul 22 at 19:01
Donald Splutterwit
21.3k21243
answered Jul 22 at 19:01
Donald Splutterwit
21.3k21243
21.3k21243
1
begineqnarray* sum_j=k^2k binom2kj &=& frac12 left( sum_j=k^2k binom2kj + sum_j=k^0 binom2kj right) \ endeqnarray* begineqnarray* &=& frac12 left(binom2kk +sum_j=0^2k binom2kj right) \ endeqnarray*
– Donald Splutterwit
Jul 22 at 19:06
This looks a little bit weird, I guess there is an error somewhere in your math mode ^^'. I just realized that my first question was nonsense
– mrtaurho
Jul 22 at 19:08
add a comment |Â
1
begineqnarray* sum_j=k^2k binom2kj &=& frac12 left( sum_j=k^2k binom2kj + sum_j=k^0 binom2kj right) \ endeqnarray* begineqnarray* &=& frac12 left(binom2kk +sum_j=0^2k binom2kj right) \ endeqnarray*
– Donald Splutterwit
Jul 22 at 19:06
This looks a little bit weird, I guess there is an error somewhere in your math mode ^^'. I just realized that my first question was nonsense
– mrtaurho
Jul 22 at 19:08
1
1
begineqnarray* sum_j=k^2k binom2kj &=& frac12 left( sum_j=k^2k binom2kj + sum_j=k^0 binom2kj right) \ endeqnarray* begineqnarray* &=& frac12 left(binom2kk +sum_j=0^2k binom2kj right) \ endeqnarray*
– Donald Splutterwit
Jul 22 at 19:06
begineqnarray* sum_j=k^2k binom2kj &=& frac12 left( sum_j=k^2k binom2kj + sum_j=k^0 binom2kj right) \ endeqnarray* begineqnarray* &=& frac12 left(binom2kk +sum_j=0^2k binom2kj right) \ endeqnarray*
– Donald Splutterwit
Jul 22 at 19:06
This looks a little bit weird, I guess there is an error somewhere in your math mode ^^'. I just realized that my first question was nonsense
– mrtaurho
Jul 22 at 19:08
This looks a little bit weird, I guess there is an error somewhere in your math mode ^^'. I just realized that my first question was nonsense
– mrtaurho
Jul 22 at 19:08
add a comment |Â
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