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
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$begingroup$
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
99.5k479165
$endgroup$
$begingroup$
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 |
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1 Answer
1
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oldest
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1 Answer
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oldest
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$begingroup$
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
99.5k479165
$endgroup$
$begingroup$
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
99.5k479165
$endgroup$
$begingroup$
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
99.5k479165
$endgroup$
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
99.5k479165
edited Dec 5 '18 at 23:21
answered Dec 5 '18 at 23:13
HenryHenry
99.5k479165
answered Dec 5 '18 at 23:13
HenryHenry
99.5k479165
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
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 |
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$begingroup$
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