Mixing
Probability of picking a number from a set of unique integers
$begingroup$
Suppose I have a set of $k$ integers such that every number is unique.
Let $A = {1,2,3,4,...,k}$
Now suppose that we rearrange these numbers to a random permutation in the set. I want to find the probability of finding a fixed number $x$ at any position of the set.
Here's my understanding:
The probability of the first number in the set being $x$ is $frac{1}{k}$
The probability of the second number in the set being $x$ is $(1-frac{1}{k})frac{1}{k}$
The probability of the third number being $x$ is $(1-frac{1}{k})^2 frac{1}{k}$
That leads to the probability of the $n^{th}$ term being $x$ to be $(1-frac{1}{k})^{n-1} frac{1}{k}$
Is this correct?
probability combinatorics
asked Jan 23 at 19:38
user2965071user2965071
1627
$endgroup$
add a comment |
$begingroup$
Suppose I have a set of $k$ integers such that every number is unique.
Let $A = {1,2,3,4,...,k}$
Now suppose that we rearrange these numbers to a random permutation in the set. I want to find the probability of finding a fixed number $x$ at any position of the set.
Here's my understanding:
The probability of the first number in the set being $x$ is $frac{1}{k}$
The probability of the second number in the set being $x$ is $(1-frac{1}{k})frac{1}{k}$
The probability of the third number being $x$ is $(1-frac{1}{k})^2 frac{1}{k}$
That leads to the probability of the $n^{th}$ term being $x$ to be $(1-frac{1}{k})^{n-1} frac{1}{k}$
Is this correct?
probability combinatorics
asked Jan 23 at 19:38
user2965071user2965071
1627
$endgroup$
1
$begingroup$
Not quite: your are forgetting about dependence. Given that the first number is not $x$, there are now only $k-1$ equally likely places $x$ can go, so the probability the second number is $x$ is $(1-frac1k)cdot frac1{color{red}{k-1}}=frac1k$.
$endgroup$
– Mike Earnest
Jan 23 at 19:42
1
$begingroup$
Which matches intuition - why should $x$ being in the first place be any more likely than in the second place?
$endgroup$
– bounceback
Jan 23 at 19:43
add a comment |
$begingroup$
Suppose I have a set of $k$ integers such that every number is unique.
Let $A = {1,2,3,4,...,k}$
Now suppose that we rearrange these numbers to a random permutation in the set. I want to find the probability of finding a fixed number $x$ at any position of the set.
Here's my understanding:
The probability of the first number in the set being $x$ is $frac{1}{k}$
The probability of the second number in the set being $x$ is $(1-frac{1}{k})frac{1}{k}$
The probability of the third number being $x$ is $(1-frac{1}{k})^2 frac{1}{k}$
That leads to the probability of the $n^{th}$ term being $x$ to be $(1-frac{1}{k})^{n-1} frac{1}{k}$
Is this correct?
probability combinatorics
asked Jan 23 at 19:38
user2965071user2965071
1627
$endgroup$
Suppose I have a set of $k$ integers such that every number is unique.
Let $A = {1,2,3,4,...,k}$
Now suppose that we rearrange these numbers to a random permutation in the set. I want to find the probability of finding a fixed number $x$ at any position of the set.
Here's my understanding:
The probability of the first number in the set being $x$ is $frac{1}{k}$
The probability of the second number in the set being $x$ is $(1-frac{1}{k})frac{1}{k}$
The probability of the third number being $x$ is $(1-frac{1}{k})^2 frac{1}{k}$
That leads to the probability of the $n^{th}$ term being $x$ to be $(1-frac{1}{k})^{n-1} frac{1}{k}$
Is this correct?
probability combinatorics
probability combinatorics
asked Jan 23 at 19:38
user2965071user2965071
1627
asked Jan 23 at 19:38
user2965071user2965071
1627
asked Jan 23 at 19:38
user2965071user2965071
1627
asked Jan 23 at 19:38
user2965071user2965071
1627
asked Jan 23 at 19:38
user2965071user2965071
1627
1627
1
$begingroup$
Not quite: your are forgetting about dependence. Given that the first number is not $x$, there are now only $k-1$ equally likely places $x$ can go, so the probability the second number is $x$ is $(1-frac1k)cdot frac1{color{red}{k-1}}=frac1k$.
$endgroup$
– Mike Earnest
Jan 23 at 19:42
1
$begingroup$
Which matches intuition - why should $x$ being in the first place be any more likely than in the second place?
$endgroup$
– bounceback
Jan 23 at 19:43
add a comment |
1
$begingroup$
Not quite: your are forgetting about dependence. Given that the first number is not $x$, there are now only $k-1$ equally likely places $x$ can go, so the probability the second number is $x$ is $(1-frac1k)cdot frac1{color{red}{k-1}}=frac1k$.
$endgroup$
– Mike Earnest
Jan 23 at 19:42
1
$begingroup$
Which matches intuition - why should $x$ being in the first place be any more likely than in the second place?
$endgroup$
– bounceback
Jan 23 at 19:43
1
1
$begingroup$
Not quite: your are forgetting about dependence. Given that the first number is not $x$, there are now only $k-1$ equally likely places $x$ can go, so the probability the second number is $x$ is $(1-frac1k)cdot frac1{color{red}{k-1}}=frac1k$.
$endgroup$
– Mike Earnest
Jan 23 at 19:42
$begingroup$
Not quite: your are forgetting about dependence. Given that the first number is not $x$, there are now only $k-1$ equally likely places $x$ can go, so the probability the second number is $x$ is $(1-frac1k)cdot frac1{color{red}{k-1}}=frac1k$.
$endgroup$
– Mike Earnest
Jan 23 at 19:42
1
1
$begingroup$
Which matches intuition - why should $x$ being in the first place be any more likely than in the second place?
$endgroup$
– bounceback
Jan 23 at 19:43
$begingroup$
Which matches intuition - why should $x$ being in the first place be any more likely than in the second place?
$endgroup$
– bounceback
Jan 23 at 19:43
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
If you want to know the probability of finding the fixed number $x$ in $n_{th}$ position you should proceed by first finding the total number of favorable cases and dividing by the total number of possible cases. The number of favorable cases can be obtained by fixing $x$ in the $n_{th}$ position and considering all permutations of the remaining $k-1$ numbers.
$P(x is in n_{th} position) = frac{(k-1)!}{k!}=frac{1}{k}$
Also note that you may consider $n$ to be any position, the result will remain the same.
answered Jan 23 at 19:47
s0ulr3aper07s0ulr3aper07
566111
$endgroup$
add a comment |
$begingroup$
Well, no. The probability that the second number is $x$ is
$(1 - frac{1}{k})frac{1}{k-1} = frac{k-1}{k}frac{1}{k-1} = frac{1}{k}$.
You forgot that, once you know that the first number is not $x$, you only have $k-1$ remaining, not $k$, so the probability that the first of the rest is $x$ is $frac{1}{k-1}$.
It's also not a coincidence that the answer is $frac{1}{k}$. The situation is symmetric, so the probability that $x$ is in any one position is the same.
answered Jan 23 at 19:43
Todor MarkovTodor Markov
2,420412
$endgroup$
add a comment |
$begingroup$
Unless I am misunderstanding what you're trying to figure out, that's not correct.
It sounds like you're trying to answer the question:
First, fix some number for $x$, say $17$. Now you want to know:
What is the probability that the first number in the set is $17$?
What is the probability that the second number in the set is $17$?
etc.
Well, they all have a probability of $frac{1}{k}$ of being $17$
This should be immediately intutive, since the probability should be the same for each of the positions. But, if you use your approach of calculating it, you find it as well:
$$P(x_2 = 17)= P(x_2=17|x_1not = 17)cdot P(x_1 not = 17) = frac{1}{kcolor{red}{-1}}cdot frac{k-1}{k}=frac{1}{k}$$
answered Jan 23 at 19:46
Bram28Bram28
63.7k44793
$endgroup$
add a comment |
$begingroup$
Hint:
You are wrong at the second place in the set.
$P(x=2)=P(xneq1 )cdot P(x=2|xneq1)=(1-frac{1}{k})cdot frac{1}{k-1}=frac{k-1}{k}cdot frac{1}{k-1}=frac{1}{k} $
This is because you already know it is not in the first place.
Now try to look for the third one, and then the fourth..
And you can ask yourself, why should the first place will have more chances the the others to find x in it?
answered Jan 23 at 19:46
ShaqShaq
3049
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084957%2fprobability-of-picking-a-number-from-a-set-of-unique-integers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you want to know the probability of finding the fixed number $x$ in $n_{th}$ position you should proceed by first finding the total number of favorable cases and dividing by the total number of possible cases. The number of favorable cases can be obtained by fixing $x$ in the $n_{th}$ position and considering all permutations of the remaining $k-1$ numbers.
$P(x is in n_{th} position) = frac{(k-1)!}{k!}=frac{1}{k}$
Also note that you may consider $n$ to be any position, the result will remain the same.
answered Jan 23 at 19:47
s0ulr3aper07s0ulr3aper07
566111
$endgroup$
add a comment |
$begingroup$
If you want to know the probability of finding the fixed number $x$ in $n_{th}$ position you should proceed by first finding the total number of favorable cases and dividing by the total number of possible cases. The number of favorable cases can be obtained by fixing $x$ in the $n_{th}$ position and considering all permutations of the remaining $k-1$ numbers.
$P(x is in n_{th} position) = frac{(k-1)!}{k!}=frac{1}{k}$
Also note that you may consider $n$ to be any position, the result will remain the same.
answered Jan 23 at 19:47
s0ulr3aper07s0ulr3aper07
566111
$endgroup$
add a comment |
$begingroup$
If you want to know the probability of finding the fixed number $x$ in $n_{th}$ position you should proceed by first finding the total number of favorable cases and dividing by the total number of possible cases. The number of favorable cases can be obtained by fixing $x$ in the $n_{th}$ position and considering all permutations of the remaining $k-1$ numbers.
$P(x is in n_{th} position) = frac{(k-1)!}{k!}=frac{1}{k}$
Also note that you may consider $n$ to be any position, the result will remain the same.
answered Jan 23 at 19:47
s0ulr3aper07s0ulr3aper07
566111
$endgroup$
If you want to know the probability of finding the fixed number $x$ in $n_{th}$ position you should proceed by first finding the total number of favorable cases and dividing by the total number of possible cases. The number of favorable cases can be obtained by fixing $x$ in the $n_{th}$ position and considering all permutations of the remaining $k-1$ numbers.
$P(x is in n_{th} position) = frac{(k-1)!}{k!}=frac{1}{k}$
Also note that you may consider $n$ to be any position, the result will remain the same.
answered Jan 23 at 19:47
s0ulr3aper07s0ulr3aper07
566111
answered Jan 23 at 19:47
s0ulr3aper07s0ulr3aper07
566111
answered Jan 23 at 19:47
s0ulr3aper07s0ulr3aper07
566111
answered Jan 23 at 19:47
s0ulr3aper07s0ulr3aper07
566111
566111
add a comment |
add a comment |
$begingroup$
Well, no. The probability that the second number is $x$ is
$(1 - frac{1}{k})frac{1}{k-1} = frac{k-1}{k}frac{1}{k-1} = frac{1}{k}$.
You forgot that, once you know that the first number is not $x$, you only have $k-1$ remaining, not $k$, so the probability that the first of the rest is $x$ is $frac{1}{k-1}$.
It's also not a coincidence that the answer is $frac{1}{k}$. The situation is symmetric, so the probability that $x$ is in any one position is the same.
answered Jan 23 at 19:43
Todor MarkovTodor Markov
2,420412
$endgroup$
add a comment |
$begingroup$
Well, no. The probability that the second number is $x$ is
$(1 - frac{1}{k})frac{1}{k-1} = frac{k-1}{k}frac{1}{k-1} = frac{1}{k}$.
You forgot that, once you know that the first number is not $x$, you only have $k-1$ remaining, not $k$, so the probability that the first of the rest is $x$ is $frac{1}{k-1}$.
It's also not a coincidence that the answer is $frac{1}{k}$. The situation is symmetric, so the probability that $x$ is in any one position is the same.
answered Jan 23 at 19:43
Todor MarkovTodor Markov
2,420412
$endgroup$
add a comment |
$begingroup$
Well, no. The probability that the second number is $x$ is
$(1 - frac{1}{k})frac{1}{k-1} = frac{k-1}{k}frac{1}{k-1} = frac{1}{k}$.
You forgot that, once you know that the first number is not $x$, you only have $k-1$ remaining, not $k$, so the probability that the first of the rest is $x$ is $frac{1}{k-1}$.
It's also not a coincidence that the answer is $frac{1}{k}$. The situation is symmetric, so the probability that $x$ is in any one position is the same.
answered Jan 23 at 19:43
Todor MarkovTodor Markov
2,420412
$endgroup$
Well, no. The probability that the second number is $x$ is
$(1 - frac{1}{k})frac{1}{k-1} = frac{k-1}{k}frac{1}{k-1} = frac{1}{k}$.
You forgot that, once you know that the first number is not $x$, you only have $k-1$ remaining, not $k$, so the probability that the first of the rest is $x$ is $frac{1}{k-1}$.
It's also not a coincidence that the answer is $frac{1}{k}$. The situation is symmetric, so the probability that $x$ is in any one position is the same.
answered Jan 23 at 19:43
Todor MarkovTodor Markov
2,420412
answered Jan 23 at 19:43
Todor MarkovTodor Markov
2,420412
answered Jan 23 at 19:43
Todor MarkovTodor Markov
2,420412
answered Jan 23 at 19:43
Todor MarkovTodor Markov
2,420412
2,420412
add a comment |
add a comment |
$begingroup$
Unless I am misunderstanding what you're trying to figure out, that's not correct.
It sounds like you're trying to answer the question:
First, fix some number for $x$, say $17$. Now you want to know:
What is the probability that the first number in the set is $17$?
What is the probability that the second number in the set is $17$?
etc.
Well, they all have a probability of $frac{1}{k}$ of being $17$
This should be immediately intutive, since the probability should be the same for each of the positions. But, if you use your approach of calculating it, you find it as well:
$$P(x_2 = 17)= P(x_2=17|x_1not = 17)cdot P(x_1 not = 17) = frac{1}{kcolor{red}{-1}}cdot frac{k-1}{k}=frac{1}{k}$$
answered Jan 23 at 19:46
Bram28Bram28
63.7k44793
$endgroup$
add a comment |
$begingroup$
Unless I am misunderstanding what you're trying to figure out, that's not correct.
It sounds like you're trying to answer the question:
First, fix some number for $x$, say $17$. Now you want to know:
What is the probability that the first number in the set is $17$?
What is the probability that the second number in the set is $17$?
etc.
Well, they all have a probability of $frac{1}{k}$ of being $17$
This should be immediately intutive, since the probability should be the same for each of the positions. But, if you use your approach of calculating it, you find it as well:
$$P(x_2 = 17)= P(x_2=17|x_1not = 17)cdot P(x_1 not = 17) = frac{1}{kcolor{red}{-1}}cdot frac{k-1}{k}=frac{1}{k}$$
answered Jan 23 at 19:46
Bram28Bram28
63.7k44793
$endgroup$
add a comment |
$begingroup$
Unless I am misunderstanding what you're trying to figure out, that's not correct.
It sounds like you're trying to answer the question:
First, fix some number for $x$, say $17$. Now you want to know:
What is the probability that the first number in the set is $17$?
What is the probability that the second number in the set is $17$?
etc.
Well, they all have a probability of $frac{1}{k}$ of being $17$
This should be immediately intutive, since the probability should be the same for each of the positions. But, if you use your approach of calculating it, you find it as well:
$$P(x_2 = 17)= P(x_2=17|x_1not = 17)cdot P(x_1 not = 17) = frac{1}{kcolor{red}{-1}}cdot frac{k-1}{k}=frac{1}{k}$$
answered Jan 23 at 19:46
Bram28Bram28
63.7k44793
$endgroup$
Unless I am misunderstanding what you're trying to figure out, that's not correct.
It sounds like you're trying to answer the question:
First, fix some number for $x$, say $17$. Now you want to know:
What is the probability that the first number in the set is $17$?
What is the probability that the second number in the set is $17$?
etc.
Well, they all have a probability of $frac{1}{k}$ of being $17$
This should be immediately intutive, since the probability should be the same for each of the positions. But, if you use your approach of calculating it, you find it as well:
$$P(x_2 = 17)= P(x_2=17|x_1not = 17)cdot P(x_1 not = 17) = frac{1}{kcolor{red}{-1}}cdot frac{k-1}{k}=frac{1}{k}$$
answered Jan 23 at 19:46
Bram28Bram28
63.7k44793
answered Jan 23 at 19:46
Bram28Bram28
63.7k44793
answered Jan 23 at 19:46
Bram28Bram28
63.7k44793
answered Jan 23 at 19:46
Bram28Bram28
63.7k44793
63.7k44793
add a comment |
add a comment |
$begingroup$
Hint:
You are wrong at the second place in the set.
$P(x=2)=P(xneq1 )cdot P(x=2|xneq1)=(1-frac{1}{k})cdot frac{1}{k-1}=frac{k-1}{k}cdot frac{1}{k-1}=frac{1}{k} $
This is because you already know it is not in the first place.
Now try to look for the third one, and then the fourth..
And you can ask yourself, why should the first place will have more chances the the others to find x in it?
answered Jan 23 at 19:46
ShaqShaq
3049
$endgroup$
add a comment |
$begingroup$
Hint:
You are wrong at the second place in the set.
$P(x=2)=P(xneq1 )cdot P(x=2|xneq1)=(1-frac{1}{k})cdot frac{1}{k-1}=frac{k-1}{k}cdot frac{1}{k-1}=frac{1}{k} $
This is because you already know it is not in the first place.
Now try to look for the third one, and then the fourth..
And you can ask yourself, why should the first place will have more chances the the others to find x in it?
answered Jan 23 at 19:46
ShaqShaq
3049
$endgroup$
add a comment |
$begingroup$
Hint:
You are wrong at the second place in the set.
$P(x=2)=P(xneq1 )cdot P(x=2|xneq1)=(1-frac{1}{k})cdot frac{1}{k-1}=frac{k-1}{k}cdot frac{1}{k-1}=frac{1}{k} $
This is because you already know it is not in the first place.
Now try to look for the third one, and then the fourth..
And you can ask yourself, why should the first place will have more chances the the others to find x in it?
answered Jan 23 at 19:46
ShaqShaq
3049
$endgroup$
Hint:
You are wrong at the second place in the set.
$P(x=2)=P(xneq1 )cdot P(x=2|xneq1)=(1-frac{1}{k})cdot frac{1}{k-1}=frac{k-1}{k}cdot frac{1}{k-1}=frac{1}{k} $
This is because you already know it is not in the first place.
Now try to look for the third one, and then the fourth..
And you can ask yourself, why should the first place will have more chances the the others to find x in it?
answered Jan 23 at 19:46
ShaqShaq
3049
answered Jan 23 at 19:46
ShaqShaq
3049
answered Jan 23 at 19:46
ShaqShaq
3049
answered Jan 23 at 19:46
ShaqShaq
3049
3049
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084957%2fprobability-of-picking-a-number-from-a-set-of-unique-integers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Not quite: your are forgetting about dependence. Given that the first number is not $x$, there are now only $k-1$ equally likely places $x$ can go, so the probability the second number is $x$ is $(1-frac1k)cdot frac1{color{red}{k-1}}=frac1k$.
$endgroup$
– Mike Earnest
Jan 23 at 19:42
1
$begingroup$
Which matches intuition - why should $x$ being in the first place be any more likely than in the second place?
$endgroup$
– bounceback
Jan 23 at 19:43