How many containers contain at least one item?
$begingroup$
I have X containers with Y slots each. The containers are not unique.
I also have n items that are to be placed in any of the X * Y slots at random. n is always less than X * Y.
How would I find the number of containers with at least one item that is statistically most likely? Is this possible to determine theoretically?
statistics probability-distributions
asked Dec 5 '18 at 22:43
Edward CaseyEdward Casey
83
$endgroup$
add a comment |
$begingroup$
I have X containers with Y slots each. The containers are not unique.
I also have n items that are to be placed in any of the X * Y slots at random. n is always less than X * Y.
How would I find the number of containers with at least one item that is statistically most likely? Is this possible to determine theoretically?
statistics probability-distributions
asked Dec 5 '18 at 22:43
Edward CaseyEdward Casey
83
$endgroup$
$begingroup$
So you can only place one item per slot? When you go to place an item, do you pick a slot at random from among the empty ones, or pick a container at random and put the item in it? In the latter case, what if the container is full?
$endgroup$
– Ross Millikan
Dec 5 '18 at 23:03
$begingroup$
@RossMillikan You can only place one item per slot. When you place an item, you can pick any empty slot at random. All empty slots across all containers should have equal probability of getting picked. If a container is full, it can hold no more items, since there are no more slots.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:11
add a comment |
$begingroup$
I have X containers with Y slots each. The containers are not unique.
I also have n items that are to be placed in any of the X * Y slots at random. n is always less than X * Y.
How would I find the number of containers with at least one item that is statistically most likely? Is this possible to determine theoretically?
statistics probability-distributions
asked Dec 5 '18 at 22:43
Edward CaseyEdward Casey
83
$endgroup$
I have X containers with Y slots each. The containers are not unique.
I also have n items that are to be placed in any of the X * Y slots at random. n is always less than X * Y.
How would I find the number of containers with at least one item that is statistically most likely? Is this possible to determine theoretically?
statistics probability-distributions
statistics probability-distributions
asked Dec 5 '18 at 22:43
Edward CaseyEdward Casey
83
asked Dec 5 '18 at 22:43
Edward CaseyEdward Casey
83
edited Dec 5 '18 at 23:16
Edward Casey
asked Dec 5 '18 at 22:43
Edward CaseyEdward Casey
83
asked Dec 5 '18 at 22:43
Edward CaseyEdward Casey
83
asked Dec 5 '18 at 22:43
Edward CaseyEdward Casey
83
83
$begingroup$
So you can only place one item per slot? When you go to place an item, do you pick a slot at random from among the empty ones, or pick a container at random and put the item in it? In the latter case, what if the container is full?
$endgroup$
– Ross Millikan
Dec 5 '18 at 23:03
$begingroup$
@RossMillikan You can only place one item per slot. When you place an item, you can pick any empty slot at random. All empty slots across all containers should have equal probability of getting picked. If a container is full, it can hold no more items, since there are no more slots.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:11
add a comment |
$begingroup$
So you can only place one item per slot? When you go to place an item, do you pick a slot at random from among the empty ones, or pick a container at random and put the item in it? In the latter case, what if the container is full?
$endgroup$
– Ross Millikan
Dec 5 '18 at 23:03
$begingroup$
@RossMillikan You can only place one item per slot. When you place an item, you can pick any empty slot at random. All empty slots across all containers should have equal probability of getting picked. If a container is full, it can hold no more items, since there are no more slots.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:11
$begingroup$
So you can only place one item per slot? When you go to place an item, do you pick a slot at random from among the empty ones, or pick a container at random and put the item in it? In the latter case, what if the container is full?
$endgroup$
– Ross Millikan
Dec 5 '18 at 23:03
$begingroup$
So you can only place one item per slot? When you go to place an item, do you pick a slot at random from among the empty ones, or pick a container at random and put the item in it? In the latter case, what if the container is full?
$endgroup$
– Ross Millikan
Dec 5 '18 at 23:03
$begingroup$
@RossMillikan You can only place one item per slot. When you place an item, you can pick any empty slot at random. All empty slots across all containers should have equal probability of getting picked. If a container is full, it can hold no more items, since there are no more slots.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:11
$begingroup$
@RossMillikan You can only place one item per slot. When you place an item, you can pick any empty slot at random. All empty slots across all containers should have equal probability of getting picked. If a container is full, it can hold no more items, since there are no more slots.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I suspect this may be quite difficult in general, though for $Y=1$ the answer is $n$
But a related question is the expected number of containers with exactly one item (i.e. the mean rather than the mode): that would be $Xdfrac{{Y choose 1}{XY-Y choose n-1}}{XY choose n} = dfrac{n(XY-Y)!(XY-n)! }{(XY-Y-n+1)!(XY-1)!}$
You then edited the question to at least one item. Here the expected number is $Xleft(1-dfrac{XY-Y choose n}{XY choose n}right)$
answered Dec 5 '18 at 23:13
HenryHenry
98.5k476163
$endgroup$
$begingroup$
I swear I mess up every time I post to stack exchange. The question is supposed to be how many containers with at least one item is statistically most likely. I reread the question half a dozen times before I posted and still didn't catch my mistake.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:22
$begingroup$
@EdwardCasey Many least one questions are easier to calculate if you approach them as an exactly zero question and then adjust the answer, as in my final "$1-$"
$endgroup$
– Henry
Dec 5 '18 at 23:30
$begingroup$
Yes, I did learn that trick back in college, but it's been so long since I did probabilities that I legitimately didn't know how to structure the rest of the expression.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:39
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%2f3027761%2fhow-many-containers-contain-at-least-one-item%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I suspect this may be quite difficult in general, though for $Y=1$ the answer is $n$
But a related question is the expected number of containers with exactly one item (i.e. the mean rather than the mode): that would be $Xdfrac{{Y choose 1}{XY-Y choose n-1}}{XY choose n} = dfrac{n(XY-Y)!(XY-n)! }{(XY-Y-n+1)!(XY-1)!}$
You then edited the question to at least one item. Here the expected number is $Xleft(1-dfrac{XY-Y choose n}{XY choose n}right)$
answered Dec 5 '18 at 23:13
HenryHenry
98.5k476163
$endgroup$
$begingroup$
I swear I mess up every time I post to stack exchange. The question is supposed to be how many containers with at least one item is statistically most likely. I reread the question half a dozen times before I posted and still didn't catch my mistake.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:22
$begingroup$
@EdwardCasey Many least one questions are easier to calculate if you approach them as an exactly zero question and then adjust the answer, as in my final "$1-$"
$endgroup$
– Henry
Dec 5 '18 at 23:30
$begingroup$
Yes, I did learn that trick back in college, but it's been so long since I did probabilities that I legitimately didn't know how to structure the rest of the expression.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:39
add a comment |
$begingroup$
I suspect this may be quite difficult in general, though for $Y=1$ the answer is $n$
But a related question is the expected number of containers with exactly one item (i.e. the mean rather than the mode): that would be $Xdfrac{{Y choose 1}{XY-Y choose n-1}}{XY choose n} = dfrac{n(XY-Y)!(XY-n)! }{(XY-Y-n+1)!(XY-1)!}$
You then edited the question to at least one item. Here the expected number is $Xleft(1-dfrac{XY-Y choose n}{XY choose n}right)$
answered Dec 5 '18 at 23:13
HenryHenry
98.5k476163
$endgroup$
$begingroup$
I swear I mess up every time I post to stack exchange. The question is supposed to be how many containers with at least one item is statistically most likely. I reread the question half a dozen times before I posted and still didn't catch my mistake.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:22
$begingroup$
@EdwardCasey Many least one questions are easier to calculate if you approach them as an exactly zero question and then adjust the answer, as in my final "$1-$"
$endgroup$
– Henry
Dec 5 '18 at 23:30
$begingroup$
Yes, I did learn that trick back in college, but it's been so long since I did probabilities that I legitimately didn't know how to structure the rest of the expression.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:39
add a comment |
$begingroup$
I suspect this may be quite difficult in general, though for $Y=1$ the answer is $n$
But a related question is the expected number of containers with exactly one item (i.e. the mean rather than the mode): that would be $Xdfrac{{Y choose 1}{XY-Y choose n-1}}{XY choose n} = dfrac{n(XY-Y)!(XY-n)! }{(XY-Y-n+1)!(XY-1)!}$
You then edited the question to at least one item. Here the expected number is $Xleft(1-dfrac{XY-Y choose n}{XY choose n}right)$
answered Dec 5 '18 at 23:13
HenryHenry
98.5k476163
$endgroup$
I suspect this may be quite difficult in general, though for $Y=1$ the answer is $n$
But a related question is the expected number of containers with exactly one item (i.e. the mean rather than the mode): that would be $Xdfrac{{Y choose 1}{XY-Y choose n-1}}{XY choose n} = dfrac{n(XY-Y)!(XY-n)! }{(XY-Y-n+1)!(XY-1)!}$
You then edited the question to at least one item. Here the expected number is $Xleft(1-dfrac{XY-Y choose n}{XY choose n}right)$
answered Dec 5 '18 at 23:13
HenryHenry
98.5k476163
edited Dec 5 '18 at 23:21
answered Dec 5 '18 at 23:13
HenryHenry
98.5k476163
answered Dec 5 '18 at 23:13
HenryHenry
98.5k476163
answered Dec 5 '18 at 23:13
HenryHenry
98.5k476163
98.5k476163
$begingroup$
I swear I mess up every time I post to stack exchange. The question is supposed to be how many containers with at least one item is statistically most likely. I reread the question half a dozen times before I posted and still didn't catch my mistake.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:22
$begingroup$
@EdwardCasey Many least one questions are easier to calculate if you approach them as an exactly zero question and then adjust the answer, as in my final "$1-$"
$endgroup$
– Henry
Dec 5 '18 at 23:30
$begingroup$
Yes, I did learn that trick back in college, but it's been so long since I did probabilities that I legitimately didn't know how to structure the rest of the expression.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:39
add a comment |
$begingroup$
I swear I mess up every time I post to stack exchange. The question is supposed to be how many containers with at least one item is statistically most likely. I reread the question half a dozen times before I posted and still didn't catch my mistake.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:22
$begingroup$
@EdwardCasey Many least one questions are easier to calculate if you approach them as an exactly zero question and then adjust the answer, as in my final "$1-$"
$endgroup$
– Henry
Dec 5 '18 at 23:30
$begingroup$
Yes, I did learn that trick back in college, but it's been so long since I did probabilities that I legitimately didn't know how to structure the rest of the expression.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:39
$begingroup$
I swear I mess up every time I post to stack exchange. The question is supposed to be how many containers with at least one item is statistically most likely. I reread the question half a dozen times before I posted and still didn't catch my mistake.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:22
$begingroup$
I swear I mess up every time I post to stack exchange. The question is supposed to be how many containers with at least one item is statistically most likely. I reread the question half a dozen times before I posted and still didn't catch my mistake.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:22
$begingroup$
@EdwardCasey Many least one questions are easier to calculate if you approach them as an exactly zero question and then adjust the answer, as in my final "$1-$"
$endgroup$
– Henry
Dec 5 '18 at 23:30
$begingroup$
@EdwardCasey Many least one questions are easier to calculate if you approach them as an exactly zero question and then adjust the answer, as in my final "$1-$"
$endgroup$
– Henry
Dec 5 '18 at 23:30
$begingroup$
Yes, I did learn that trick back in college, but it's been so long since I did probabilities that I legitimately didn't know how to structure the rest of the expression.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:39
$begingroup$
Yes, I did learn that trick back in college, but it's been so long since I did probabilities that I legitimately didn't know how to structure the rest of the expression.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:39
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%2f3027761%2fhow-many-containers-contain-at-least-one-item%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
$begingroup$
So you can only place one item per slot? When you go to place an item, do you pick a slot at random from among the empty ones, or pick a container at random and put the item in it? In the latter case, what if the container is full?
$endgroup$
– Ross Millikan
Dec 5 '18 at 23:03
$begingroup$
@RossMillikan You can only place one item per slot. When you place an item, you can pick any empty slot at random. All empty slots across all containers should have equal probability of getting picked. If a container is full, it can hold no more items, since there are no more slots.
$endgroup$
– Edward Casey
Dec 5 '18 at 23:11