Given a random permutation of 1 to N, what is probability to get $N-1$ with a certain strategy?
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Given a random permutation of $1$ to $N$, let the sequence be $a_1,a_2,cdots,a_N$.
Erase the first $k$ items, and find out the item (let it be $a_I$) which is first item greater than the fist $k$ items.
I already know that $$P(a_I = N)=fracknsum_i=k+1^Nfrac1i-1.$$
But what is the probability of $a_I=N-1$?
Furthermore, what is the expected value of $a_I$?
probability combinatorics permutations
asked Aug 10 at 17:17
Charles Bao
746919
add a comment |Â
up vote
2
down vote
favorite
Given a random permutation of $1$ to $N$, let the sequence be $a_1,a_2,cdots,a_N$.
Erase the first $k$ items, and find out the item (let it be $a_I$) which is first item greater than the fist $k$ items.
I already know that $$P(a_I = N)=fracknsum_i=k+1^Nfrac1i-1.$$
But what is the probability of $a_I=N-1$?
Furthermore, what is the expected value of $a_I$?
probability combinatorics permutations
asked Aug 10 at 17:17
Charles Bao
746919
I'm assuming this is related somehow to the secretary problem/sultan's dowry problem/etc.?
– Brian Tung
Aug 10 at 17:21
2
I think you should use $a_I$ where $I$ is a random index. Of course, you would also need to define what happens if number $N$ lies in the first $k$ items, since then there would be no "first item greater than [all of?] the first $k$ items."
– Michael
Aug 10 at 17:35
Also, your expression for $P(a_i=N)$ isn't a probability; e.g., if $k=frac n2$ and $n$ is sufficiently large, you're summing several terms near $frac12$, so the result is greater than $1$.
– joriki
Aug 10 at 19:02
It seems you've edited the question to address one of three issues but not the other two?
– joriki
Aug 11 at 6:55
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given a random permutation of $1$ to $N$, let the sequence be $a_1,a_2,cdots,a_N$.
Erase the first $k$ items, and find out the item (let it be $a_I$) which is first item greater than the fist $k$ items.
I already know that $$P(a_I = N)=fracknsum_i=k+1^Nfrac1i-1.$$
But what is the probability of $a_I=N-1$?
Furthermore, what is the expected value of $a_I$?
probability combinatorics permutations
asked Aug 10 at 17:17
Charles Bao
746919
Given a random permutation of $1$ to $N$, let the sequence be $a_1,a_2,cdots,a_N$.
Erase the first $k$ items, and find out the item (let it be $a_I$) which is first item greater than the fist $k$ items.
I already know that $$P(a_I = N)=fracknsum_i=k+1^Nfrac1i-1.$$
But what is the probability of $a_I=N-1$?
Furthermore, what is the expected value of $a_I$?
probability combinatorics permutations
asked Aug 10 at 17:17
Charles Bao
746919
edited Aug 11 at 0:46
asked Aug 10 at 17:17
Charles Bao
746919
asked Aug 10 at 17:17
Charles Bao
746919
asked Aug 10 at 17:17
Charles Bao
746919
746919
I'm assuming this is related somehow to the secretary problem/sultan's dowry problem/etc.?
– Brian Tung
Aug 10 at 17:21
2
I think you should use $a_I$ where $I$ is a random index. Of course, you would also need to define what happens if number $N$ lies in the first $k$ items, since then there would be no "first item greater than [all of?] the first $k$ items."
– Michael
Aug 10 at 17:35
Also, your expression for $P(a_i=N)$ isn't a probability; e.g., if $k=frac n2$ and $n$ is sufficiently large, you're summing several terms near $frac12$, so the result is greater than $1$.
– joriki
Aug 10 at 19:02
It seems you've edited the question to address one of three issues but not the other two?
– joriki
Aug 11 at 6:55
add a comment |Â
I'm assuming this is related somehow to the secretary problem/sultan's dowry problem/etc.?
– Brian Tung
Aug 10 at 17:21
2
I think you should use $a_I$ where $I$ is a random index. Of course, you would also need to define what happens if number $N$ lies in the first $k$ items, since then there would be no "first item greater than [all of?] the first $k$ items."
– Michael
Aug 10 at 17:35
Also, your expression for $P(a_i=N)$ isn't a probability; e.g., if $k=frac n2$ and $n$ is sufficiently large, you're summing several terms near $frac12$, so the result is greater than $1$.
– joriki
Aug 10 at 19:02
It seems you've edited the question to address one of three issues but not the other two?
– joriki
Aug 11 at 6:55
I'm assuming this is related somehow to the secretary problem/sultan's dowry problem/etc.?
– Brian Tung
Aug 10 at 17:21
I'm assuming this is related somehow to the secretary problem/sultan's dowry problem/etc.?
– Brian Tung
Aug 10 at 17:21
2
2
I think you should use $a_I$ where $I$ is a random index. Of course, you would also need to define what happens if number $N$ lies in the first $k$ items, since then there would be no "first item greater than [all of?] the first $k$ items."
– Michael
Aug 10 at 17:35
I think you should use $a_I$ where $I$ is a random index. Of course, you would also need to define what happens if number $N$ lies in the first $k$ items, since then there would be no "first item greater than [all of?] the first $k$ items."
– Michael
Aug 10 at 17:35
Also, your expression for $P(a_i=N)$ isn't a probability; e.g., if $k=frac n2$ and $n$ is sufficiently large, you're summing several terms near $frac12$, so the result is greater than $1$.
– joriki
Aug 10 at 19:02
Also, your expression for $P(a_i=N)$ isn't a probability; e.g., if $k=frac n2$ and $n$ is sufficiently large, you're summing several terms near $frac12$, so the result is greater than $1$.
– joriki
Aug 10 at 19:02
It seems you've edited the question to address one of three issues but not the other two?
– joriki
Aug 11 at 6:55
It seems you've edited the question to address one of three issues but not the other two?
– joriki
Aug 11 at 6:55
add a comment |Â
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I'm assuming this is related somehow to the secretary problem/sultan's dowry problem/etc.?
– Brian Tung
Aug 10 at 17:21
2
I think you should use $a_I$ where $I$ is a random index. Of course, you would also need to define what happens if number $N$ lies in the first $k$ items, since then there would be no "first item greater than [all of?] the first $k$ items."
– Michael
Aug 10 at 17:35
Also, your expression for $P(a_i=N)$ isn't a probability; e.g., if $k=frac n2$ and $n$ is sufficiently large, you're summing several terms near $frac12$, so the result is greater than $1$.
– joriki
Aug 10 at 19:02
It seems you've edited the question to address one of three issues but not the other two?
– joriki
Aug 11 at 6:55