Probability of two of each object
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So if drawing from a pot at random where when an object is drawn there is another one added, so if I draw a red ball a red ball is added so probability remains the same. How can I find the average amount of draws to get 2 or 3 of each object, if all objects have an even chance to be drawn?
probability probability-theory
asked Dec 5 '18 at 0:21
James WilderJames Wilder
31
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add a comment |
$begingroup$
So if drawing from a pot at random where when an object is drawn there is another one added, so if I draw a red ball a red ball is added so probability remains the same. How can I find the average amount of draws to get 2 or 3 of each object, if all objects have an even chance to be drawn?
probability probability-theory
asked Dec 5 '18 at 0:21
James WilderJames Wilder
31
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$begingroup$
Are you looking for the multinomial distribution?
$endgroup$
– Antoni Parellada
Dec 5 '18 at 0:39
add a comment |
$begingroup$
So if drawing from a pot at random where when an object is drawn there is another one added, so if I draw a red ball a red ball is added so probability remains the same. How can I find the average amount of draws to get 2 or 3 of each object, if all objects have an even chance to be drawn?
probability probability-theory
asked Dec 5 '18 at 0:21
James WilderJames Wilder
31
$endgroup$
So if drawing from a pot at random where when an object is drawn there is another one added, so if I draw a red ball a red ball is added so probability remains the same. How can I find the average amount of draws to get 2 or 3 of each object, if all objects have an even chance to be drawn?
probability probability-theory
probability probability-theory
asked Dec 5 '18 at 0:21
James WilderJames Wilder
31
asked Dec 5 '18 at 0:21
James WilderJames Wilder
31
asked Dec 5 '18 at 0:21
James WilderJames Wilder
31
asked Dec 5 '18 at 0:21
James WilderJames Wilder
31
asked Dec 5 '18 at 0:21
James WilderJames Wilder
31
31
$begingroup$
Are you looking for the multinomial distribution?
$endgroup$
– Antoni Parellada
Dec 5 '18 at 0:39
add a comment |
$begingroup$
Are you looking for the multinomial distribution?
$endgroup$
– Antoni Parellada
Dec 5 '18 at 0:39
$begingroup$
Are you looking for the multinomial distribution?
$endgroup$
– Antoni Parellada
Dec 5 '18 at 0:39
$begingroup$
Are you looking for the multinomial distribution?
$endgroup$
– Antoni Parellada
Dec 5 '18 at 0:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is called sampling with replacement. For an exact answer, you need to know how many objects you have. For a formula, you need the "Double Dixie Cup Formula". Check out Numberphile's video about it.
Let n = the number of objects to be drawn and d = the number of times you want to draw them. Then the expected number of draws is
$$ncdotint_0^inftybigg[1-bigg(1-frac{sum_{i=0}^{d-1}frac{x^i}{i!}}{e^x}bigg)^nbigg]dx$$
answered Dec 5 '18 at 1:01
David DiazDavid Diaz
979420
$endgroup$
1
$begingroup$
Thank you so much. I do know how many objects I am doing this on but I am doing this a couple of times, but the formula was what I was looking for.
$endgroup$
– James Wilder
Dec 5 '18 at 2:05
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is called sampling with replacement. For an exact answer, you need to know how many objects you have. For a formula, you need the "Double Dixie Cup Formula". Check out Numberphile's video about it.
Let n = the number of objects to be drawn and d = the number of times you want to draw them. Then the expected number of draws is
$$ncdotint_0^inftybigg[1-bigg(1-frac{sum_{i=0}^{d-1}frac{x^i}{i!}}{e^x}bigg)^nbigg]dx$$
answered Dec 5 '18 at 1:01
David DiazDavid Diaz
979420
$endgroup$
1
$begingroup$
Thank you so much. I do know how many objects I am doing this on but I am doing this a couple of times, but the formula was what I was looking for.
$endgroup$
– James Wilder
Dec 5 '18 at 2:05
add a comment |
$begingroup$
This is called sampling with replacement. For an exact answer, you need to know how many objects you have. For a formula, you need the "Double Dixie Cup Formula". Check out Numberphile's video about it.
Let n = the number of objects to be drawn and d = the number of times you want to draw them. Then the expected number of draws is
$$ncdotint_0^inftybigg[1-bigg(1-frac{sum_{i=0}^{d-1}frac{x^i}{i!}}{e^x}bigg)^nbigg]dx$$
answered Dec 5 '18 at 1:01
David DiazDavid Diaz
979420
$endgroup$
1
$begingroup$
Thank you so much. I do know how many objects I am doing this on but I am doing this a couple of times, but the formula was what I was looking for.
$endgroup$
– James Wilder
Dec 5 '18 at 2:05
add a comment |
$begingroup$
This is called sampling with replacement. For an exact answer, you need to know how many objects you have. For a formula, you need the "Double Dixie Cup Formula". Check out Numberphile's video about it.
Let n = the number of objects to be drawn and d = the number of times you want to draw them. Then the expected number of draws is
$$ncdotint_0^inftybigg[1-bigg(1-frac{sum_{i=0}^{d-1}frac{x^i}{i!}}{e^x}bigg)^nbigg]dx$$
answered Dec 5 '18 at 1:01
David DiazDavid Diaz
979420
$endgroup$
This is called sampling with replacement. For an exact answer, you need to know how many objects you have. For a formula, you need the "Double Dixie Cup Formula". Check out Numberphile's video about it.
Let n = the number of objects to be drawn and d = the number of times you want to draw them. Then the expected number of draws is
$$ncdotint_0^inftybigg[1-bigg(1-frac{sum_{i=0}^{d-1}frac{x^i}{i!}}{e^x}bigg)^nbigg]dx$$
answered Dec 5 '18 at 1:01
David DiazDavid Diaz
979420
answered Dec 5 '18 at 1:01
David DiazDavid Diaz
979420
answered Dec 5 '18 at 1:01
David DiazDavid Diaz
979420
answered Dec 5 '18 at 1:01
David DiazDavid Diaz
979420
979420
1
$begingroup$
Thank you so much. I do know how many objects I am doing this on but I am doing this a couple of times, but the formula was what I was looking for.
$endgroup$
– James Wilder
Dec 5 '18 at 2:05
add a comment |
1
$begingroup$
Thank you so much. I do know how many objects I am doing this on but I am doing this a couple of times, but the formula was what I was looking for.
$endgroup$
– James Wilder
Dec 5 '18 at 2:05
1
1
$begingroup$
Thank you so much. I do know how many objects I am doing this on but I am doing this a couple of times, but the formula was what I was looking for.
$endgroup$
– James Wilder
Dec 5 '18 at 2:05
$begingroup$
Thank you so much. I do know how many objects I am doing this on but I am doing this a couple of times, but the formula was what I was looking for.
$endgroup$
– James Wilder
Dec 5 '18 at 2:05
add a comment |
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$begingroup$
Are you looking for the multinomial distribution?
$endgroup$
– Antoni Parellada
Dec 5 '18 at 0:39