Mixing
Probability of 4 balls in one of 80 bins after 26 throws [closed]
Clash Royale CLAN TAG#URR8PPP
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This relates to a stock problem:
I have a product with 80 sizes, the stock level of each size is 3, I forecast 26 customers per week so what is the probability I will run out of stock of at least one size ie. have an order for at east 4 of at least one size.
From this I would like to be able to try various combinations of stock level and forecast customers.
probability balls-in-bins
edited Jul 26 at 15:56
joriki
164k10180328
asked Jul 26 at 15:20
Beefos
61
closed as off-topic by amWhy, Siong Thye Goh, Isaac Browne, Lord Shark the Unknown, Xander Henderson Jul 27 at 12:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Siong Thye Goh, Isaac Browne, Lord Shark the Unknown, Xander Henderson
add a comment |Â
up vote
1
down vote
favorite
This relates to a stock problem:
I have a product with 80 sizes, the stock level of each size is 3, I forecast 26 customers per week so what is the probability I will run out of stock of at least one size ie. have an order for at east 4 of at least one size.
From this I would like to be able to try various combinations of stock level and forecast customers.
probability balls-in-bins
edited Jul 26 at 15:56
joriki
164k10180328
asked Jul 26 at 15:20
Beefos
61
closed as off-topic by amWhy, Siong Thye Goh, Isaac Browne, Lord Shark the Unknown, Xander Henderson Jul 27 at 12:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Siong Thye Goh, Isaac Browne, Lord Shark the Unknown, Xander Henderson
@saulspatz: as I understand it, the stock contains $80 cdot 3$ items, and $26$ are sold per week
– G Cab
Jul 26 at 15:40
@GCab You're right, I read it backwards!
– saulspatz
Jul 26 at 15:41
2
Is a customer equally likely to want any of the $80$ items? Aren't some more popular than others? One would have to know the relative likelihoods.
– saulspatz
Jul 26 at 15:43
Are you assuming that each customer only buys 1 item, or should it read "I forecast sales of 26 items per week"? Also as @saulspatz says, this will vary with what the items are. If you're selling 10 different styles of t-shirt, in 8 different sizes each, then you are more likely to sell out of Medium size shirts than you are to sell out of XXXXL.
– Chronocidal
Jul 26 at 15:50
assuming "$n$ customer /w = items sold /w", and assuming equal probability per size, then your problem is equivalent to the number of ways to compose a word, of lenght $n$ (26), from the alphabet $1, cdots,r$ (80), that does not have any character repeated more than $m$ times (3).
– G Cab
Jul 26 at 16:05
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This relates to a stock problem:
I have a product with 80 sizes, the stock level of each size is 3, I forecast 26 customers per week so what is the probability I will run out of stock of at least one size ie. have an order for at east 4 of at least one size.
From this I would like to be able to try various combinations of stock level and forecast customers.
probability balls-in-bins
edited Jul 26 at 15:56
joriki
164k10180328
asked Jul 26 at 15:20
Beefos
61
This relates to a stock problem:
I have a product with 80 sizes, the stock level of each size is 3, I forecast 26 customers per week so what is the probability I will run out of stock of at least one size ie. have an order for at east 4 of at least one size.
From this I would like to be able to try various combinations of stock level and forecast customers.
probability balls-in-bins
edited Jul 26 at 15:56
joriki
164k10180328
asked Jul 26 at 15:20
Beefos
61
edited Jul 26 at 15:56
joriki
164k10180328
edited Jul 26 at 15:56
joriki
164k10180328
edited Jul 26 at 15:56
joriki
164k10180328
164k10180328
asked Jul 26 at 15:20
Beefos
61
asked Jul 26 at 15:20
Beefos
61
asked Jul 26 at 15:20
Beefos
61
61
closed as off-topic by amWhy, Siong Thye Goh, Isaac Browne, Lord Shark the Unknown, Xander Henderson Jul 27 at 12:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Siong Thye Goh, Isaac Browne, Lord Shark the Unknown, Xander Henderson
closed as off-topic by amWhy, Siong Thye Goh, Isaac Browne, Lord Shark the Unknown, Xander Henderson Jul 27 at 12:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Siong Thye Goh, Isaac Browne, Lord Shark the Unknown, Xander Henderson
@saulspatz: as I understand it, the stock contains $80 cdot 3$ items, and $26$ are sold per week
– G Cab
Jul 26 at 15:40
@GCab You're right, I read it backwards!
– saulspatz
Jul 26 at 15:41
2
Is a customer equally likely to want any of the $80$ items? Aren't some more popular than others? One would have to know the relative likelihoods.
– saulspatz
Jul 26 at 15:43
Are you assuming that each customer only buys 1 item, or should it read "I forecast sales of 26 items per week"? Also as @saulspatz says, this will vary with what the items are. If you're selling 10 different styles of t-shirt, in 8 different sizes each, then you are more likely to sell out of Medium size shirts than you are to sell out of XXXXL.
– Chronocidal
Jul 26 at 15:50
assuming "$n$ customer /w = items sold /w", and assuming equal probability per size, then your problem is equivalent to the number of ways to compose a word, of lenght $n$ (26), from the alphabet $1, cdots,r$ (80), that does not have any character repeated more than $m$ times (3).
– G Cab
Jul 26 at 16:05
add a comment |Â
@saulspatz: as I understand it, the stock contains $80 cdot 3$ items, and $26$ are sold per week
– G Cab
Jul 26 at 15:40
@GCab You're right, I read it backwards!
– saulspatz
Jul 26 at 15:41
2
Is a customer equally likely to want any of the $80$ items? Aren't some more popular than others? One would have to know the relative likelihoods.
– saulspatz
Jul 26 at 15:43
Are you assuming that each customer only buys 1 item, or should it read "I forecast sales of 26 items per week"? Also as @saulspatz says, this will vary with what the items are. If you're selling 10 different styles of t-shirt, in 8 different sizes each, then you are more likely to sell out of Medium size shirts than you are to sell out of XXXXL.
– Chronocidal
Jul 26 at 15:50
assuming "$n$ customer /w = items sold /w", and assuming equal probability per size, then your problem is equivalent to the number of ways to compose a word, of lenght $n$ (26), from the alphabet $1, cdots,r$ (80), that does not have any character repeated more than $m$ times (3).
– G Cab
Jul 26 at 16:05
@saulspatz: as I understand it, the stock contains $80 cdot 3$ items, and $26$ are sold per week
– G Cab
Jul 26 at 15:40
@saulspatz: as I understand it, the stock contains $80 cdot 3$ items, and $26$ are sold per week
– G Cab
Jul 26 at 15:40
@GCab You're right, I read it backwards!
– saulspatz
Jul 26 at 15:41
@GCab You're right, I read it backwards!
– saulspatz
Jul 26 at 15:41
2
2
Is a customer equally likely to want any of the $80$ items? Aren't some more popular than others? One would have to know the relative likelihoods.
– saulspatz
Jul 26 at 15:43
Is a customer equally likely to want any of the $80$ items? Aren't some more popular than others? One would have to know the relative likelihoods.
– saulspatz
Jul 26 at 15:43
Are you assuming that each customer only buys 1 item, or should it read "I forecast sales of 26 items per week"? Also as @saulspatz says, this will vary with what the items are. If you're selling 10 different styles of t-shirt, in 8 different sizes each, then you are more likely to sell out of Medium size shirts than you are to sell out of XXXXL.
– Chronocidal
Jul 26 at 15:50
Are you assuming that each customer only buys 1 item, or should it read "I forecast sales of 26 items per week"? Also as @saulspatz says, this will vary with what the items are. If you're selling 10 different styles of t-shirt, in 8 different sizes each, then you are more likely to sell out of Medium size shirts than you are to sell out of XXXXL.
– Chronocidal
Jul 26 at 15:50
assuming "$n$ customer /w = items sold /w", and assuming equal probability per size, then your problem is equivalent to the number of ways to compose a word, of lenght $n$ (26), from the alphabet $1, cdots,r$ (80), that does not have any character repeated more than $m$ times (3).
– G Cab
Jul 26 at 16:05
assuming "$n$ customer /w = items sold /w", and assuming equal probability per size, then your problem is equivalent to the number of ways to compose a word, of lenght $n$ (26), from the alphabet $1, cdots,r$ (80), that does not have any character repeated more than $m$ times (3).
– G Cab
Jul 26 at 16:05
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
For practical purposes, the probability that this happens for more than one size is so low that you can just take the expected number of sold-out sizes as an excellent approximation of the probability that at least one size is sold out. With $n=80$, $m=26$, $k=4$, this is
begineqnarray*
nleft(1-n^-msum_j=0^k-1binom mj(n-1)^m-jright)
&=&
80left(1-frac79^26+26cdot79^25+binom262cdot79^24+binom263cdot79^2380^26right)
\
&approx&
0.0234;.
endeqnarray*
If you want to know the exact probability, you can find it using inclusion–exclusion , but the calculation would be tedious since you'd have to sum over many possibilities for each of the sold-out sizes. Another way to find the exact probability would be to model this as a Markov chain, where the states would be characterized by the number of sizes that have $0$, $1$, $2$ and $3$ items left; the number of states would be the number $binom80+33=91881$ of ways to distribute the $80$ sizes over the $4$ stock values, so this could easily be modeled exactly on a computer.
answered Jul 26 at 16:17
joriki
164k10180328
This looks reasonable to me, but I get a wildly different result. Please take a look at my answer.
– saulspatz
Jul 26 at 16:47
Simulation shows my answer is completely wrong. I'm going to re-think it.
– saulspatz
Jul 26 at 16:57
BTW, the simulations confirm your solution.
– saulspatz
Jul 26 at 17:04
@saulspatz: The problem is that you're treating the balls as indistinguishable and thus significantly boosting the probability for them to be in the same bin. E.g. if there were $2$ bins, you'd make it equally likely that half the balls are in each bin as that they're all in one bin.
– joriki
Jul 26 at 17:06
1
Oh yes, I see now. I'm counting all the distributions as equiprobable. I really do have my head screwed on wrong!
– saulspatz
Jul 26 at 17:40
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
For practical purposes, the probability that this happens for more than one size is so low that you can just take the expected number of sold-out sizes as an excellent approximation of the probability that at least one size is sold out. With $n=80$, $m=26$, $k=4$, this is
begineqnarray*
nleft(1-n^-msum_j=0^k-1binom mj(n-1)^m-jright)
&=&
80left(1-frac79^26+26cdot79^25+binom262cdot79^24+binom263cdot79^2380^26right)
\
&approx&
0.0234;.
endeqnarray*
If you want to know the exact probability, you can find it using inclusion–exclusion , but the calculation would be tedious since you'd have to sum over many possibilities for each of the sold-out sizes. Another way to find the exact probability would be to model this as a Markov chain, where the states would be characterized by the number of sizes that have $0$, $1$, $2$ and $3$ items left; the number of states would be the number $binom80+33=91881$ of ways to distribute the $80$ sizes over the $4$ stock values, so this could easily be modeled exactly on a computer.
answered Jul 26 at 16:17
joriki
164k10180328
This looks reasonable to me, but I get a wildly different result. Please take a look at my answer.
– saulspatz
Jul 26 at 16:47
Simulation shows my answer is completely wrong. I'm going to re-think it.
– saulspatz
Jul 26 at 16:57
BTW, the simulations confirm your solution.
– saulspatz
Jul 26 at 17:04
@saulspatz: The problem is that you're treating the balls as indistinguishable and thus significantly boosting the probability for them to be in the same bin. E.g. if there were $2$ bins, you'd make it equally likely that half the balls are in each bin as that they're all in one bin.
– joriki
Jul 26 at 17:06
1
Oh yes, I see now. I'm counting all the distributions as equiprobable. I really do have my head screwed on wrong!
– saulspatz
Jul 26 at 17:40
 |Â
show 5 more comments
up vote
2
down vote
For practical purposes, the probability that this happens for more than one size is so low that you can just take the expected number of sold-out sizes as an excellent approximation of the probability that at least one size is sold out. With $n=80$, $m=26$, $k=4$, this is
begineqnarray*
nleft(1-n^-msum_j=0^k-1binom mj(n-1)^m-jright)
&=&
80left(1-frac79^26+26cdot79^25+binom262cdot79^24+binom263cdot79^2380^26right)
\
&approx&
0.0234;.
endeqnarray*
If you want to know the exact probability, you can find it using inclusion–exclusion , but the calculation would be tedious since you'd have to sum over many possibilities for each of the sold-out sizes. Another way to find the exact probability would be to model this as a Markov chain, where the states would be characterized by the number of sizes that have $0$, $1$, $2$ and $3$ items left; the number of states would be the number $binom80+33=91881$ of ways to distribute the $80$ sizes over the $4$ stock values, so this could easily be modeled exactly on a computer.
answered Jul 26 at 16:17
joriki
164k10180328
This looks reasonable to me, but I get a wildly different result. Please take a look at my answer.
– saulspatz
Jul 26 at 16:47
Simulation shows my answer is completely wrong. I'm going to re-think it.
– saulspatz
Jul 26 at 16:57
BTW, the simulations confirm your solution.
– saulspatz
Jul 26 at 17:04
@saulspatz: The problem is that you're treating the balls as indistinguishable and thus significantly boosting the probability for them to be in the same bin. E.g. if there were $2$ bins, you'd make it equally likely that half the balls are in each bin as that they're all in one bin.
– joriki
Jul 26 at 17:06
1
Oh yes, I see now. I'm counting all the distributions as equiprobable. I really do have my head screwed on wrong!
– saulspatz
Jul 26 at 17:40
 |Â
show 5 more comments
up vote
2
down vote
up vote
2
down vote
For practical purposes, the probability that this happens for more than one size is so low that you can just take the expected number of sold-out sizes as an excellent approximation of the probability that at least one size is sold out. With $n=80$, $m=26$, $k=4$, this is
begineqnarray*
nleft(1-n^-msum_j=0^k-1binom mj(n-1)^m-jright)
&=&
80left(1-frac79^26+26cdot79^25+binom262cdot79^24+binom263cdot79^2380^26right)
\
&approx&
0.0234;.
endeqnarray*
If you want to know the exact probability, you can find it using inclusion–exclusion , but the calculation would be tedious since you'd have to sum over many possibilities for each of the sold-out sizes. Another way to find the exact probability would be to model this as a Markov chain, where the states would be characterized by the number of sizes that have $0$, $1$, $2$ and $3$ items left; the number of states would be the number $binom80+33=91881$ of ways to distribute the $80$ sizes over the $4$ stock values, so this could easily be modeled exactly on a computer.
answered Jul 26 at 16:17
joriki
164k10180328
For practical purposes, the probability that this happens for more than one size is so low that you can just take the expected number of sold-out sizes as an excellent approximation of the probability that at least one size is sold out. With $n=80$, $m=26$, $k=4$, this is
begineqnarray*
nleft(1-n^-msum_j=0^k-1binom mj(n-1)^m-jright)
&=&
80left(1-frac79^26+26cdot79^25+binom262cdot79^24+binom263cdot79^2380^26right)
\
&approx&
0.0234;.
endeqnarray*
If you want to know the exact probability, you can find it using inclusion–exclusion , but the calculation would be tedious since you'd have to sum over many possibilities for each of the sold-out sizes. Another way to find the exact probability would be to model this as a Markov chain, where the states would be characterized by the number of sizes that have $0$, $1$, $2$ and $3$ items left; the number of states would be the number $binom80+33=91881$ of ways to distribute the $80$ sizes over the $4$ stock values, so this could easily be modeled exactly on a computer.
answered Jul 26 at 16:17
joriki
164k10180328
answered Jul 26 at 16:17
joriki
164k10180328
answered Jul 26 at 16:17
joriki
164k10180328
answered Jul 26 at 16:17
joriki
164k10180328
164k10180328
This looks reasonable to me, but I get a wildly different result. Please take a look at my answer.
– saulspatz
Jul 26 at 16:47
Simulation shows my answer is completely wrong. I'm going to re-think it.
– saulspatz
Jul 26 at 16:57
BTW, the simulations confirm your solution.
– saulspatz
Jul 26 at 17:04
@saulspatz: The problem is that you're treating the balls as indistinguishable and thus significantly boosting the probability for them to be in the same bin. E.g. if there were $2$ bins, you'd make it equally likely that half the balls are in each bin as that they're all in one bin.
– joriki
Jul 26 at 17:06
1
Oh yes, I see now. I'm counting all the distributions as equiprobable. I really do have my head screwed on wrong!
– saulspatz
Jul 26 at 17:40
 |Â
show 5 more comments
This looks reasonable to me, but I get a wildly different result. Please take a look at my answer.
– saulspatz
Jul 26 at 16:47
Simulation shows my answer is completely wrong. I'm going to re-think it.
– saulspatz
Jul 26 at 16:57
BTW, the simulations confirm your solution.
– saulspatz
Jul 26 at 17:04
@saulspatz: The problem is that you're treating the balls as indistinguishable and thus significantly boosting the probability for them to be in the same bin. E.g. if there were $2$ bins, you'd make it equally likely that half the balls are in each bin as that they're all in one bin.
– joriki
Jul 26 at 17:06
1
Oh yes, I see now. I'm counting all the distributions as equiprobable. I really do have my head screwed on wrong!
– saulspatz
Jul 26 at 17:40
This looks reasonable to me, but I get a wildly different result. Please take a look at my answer.
– saulspatz
Jul 26 at 16:47
This looks reasonable to me, but I get a wildly different result. Please take a look at my answer.
– saulspatz
Jul 26 at 16:47
Simulation shows my answer is completely wrong. I'm going to re-think it.
– saulspatz
Jul 26 at 16:57
Simulation shows my answer is completely wrong. I'm going to re-think it.
– saulspatz
Jul 26 at 16:57
BTW, the simulations confirm your solution.
– saulspatz
Jul 26 at 17:04
BTW, the simulations confirm your solution.
– saulspatz
Jul 26 at 17:04
@saulspatz: The problem is that you're treating the balls as indistinguishable and thus significantly boosting the probability for them to be in the same bin. E.g. if there were $2$ bins, you'd make it equally likely that half the balls are in each bin as that they're all in one bin.
– joriki
Jul 26 at 17:06
@saulspatz: The problem is that you're treating the balls as indistinguishable and thus significantly boosting the probability for them to be in the same bin. E.g. if there were $2$ bins, you'd make it equally likely that half the balls are in each bin as that they're all in one bin.
– joriki
Jul 26 at 17:06
1
1
Oh yes, I see now. I'm counting all the distributions as equiprobable. I really do have my head screwed on wrong!
– saulspatz
Jul 26 at 17:40
Oh yes, I see now. I'm counting all the distributions as equiprobable. I really do have my head screwed on wrong!
– saulspatz
Jul 26 at 17:40
 |Â
show 5 more comments
@saulspatz: as I understand it, the stock contains $80 cdot 3$ items, and $26$ are sold per week
– G Cab
Jul 26 at 15:40
@GCab You're right, I read it backwards!
– saulspatz
Jul 26 at 15:41
2
Is a customer equally likely to want any of the $80$ items? Aren't some more popular than others? One would have to know the relative likelihoods.
– saulspatz
Jul 26 at 15:43
Are you assuming that each customer only buys 1 item, or should it read "I forecast sales of 26 items per week"? Also as @saulspatz says, this will vary with what the items are. If you're selling 10 different styles of t-shirt, in 8 different sizes each, then you are more likely to sell out of Medium size shirts than you are to sell out of XXXXL.
– Chronocidal
Jul 26 at 15:50
assuming "$n$ customer /w = items sold /w", and assuming equal probability per size, then your problem is equivalent to the number of ways to compose a word, of lenght $n$ (26), from the alphabet $1, cdots,r$ (80), that does not have any character repeated more than $m$ times (3).
– G Cab
Jul 26 at 16:05