How many containers contain at least one item?












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 23:16







Edward Casey

















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




















  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















0












$begingroup$

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 $Xleft(1-dfrac{XY-Y choose n}{XY choose n}right)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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













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
});


}
});














draft saved

draft discarded


















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









0












$begingroup$

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 $Xleft(1-dfrac{XY-Y choose n}{XY choose n}right)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















0












$begingroup$

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 $Xleft(1-dfrac{XY-Y choose n}{XY choose n}right)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












0








0





$begingroup$

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 $Xleft(1-dfrac{XY-Y choose n}{XY choose n}right)$






share|cite|improve this answer











$endgroup$



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 $Xleft(1-dfrac{XY-Y choose n}{XY choose n}right)$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 23:21

























answered Dec 5 '18 at 23:13









HenryHenry

99.5k479165




99.5k479165












  • $begingroup$
    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










  • $begingroup$
    @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




















draft saved

draft discarded




















































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.




draft saved


draft discarded














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





















































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







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more