Observing all possible values of a discrete random variable with the given probability
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Suppose there's a true random number generator which can output an integer value in [0..N). How many times does it have to be invoked for it to generate every possible value [0..N) at least once, with the given probability P?
For example, for coin clips, there's a 50% probability of seeing both heads and tails in two rounds, 75% probability in three rounds etc. I wonder how to generalize this for a system of N possible outcomes and any given probability P?
probability combinatorics
asked Dec 10 '18 at 1:04
dragonrootdragonroot
1363
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add a comment |
$begingroup$
Suppose there's a true random number generator which can output an integer value in [0..N). How many times does it have to be invoked for it to generate every possible value [0..N) at least once, with the given probability P?
For example, for coin clips, there's a 50% probability of seeing both heads and tails in two rounds, 75% probability in three rounds etc. I wonder how to generalize this for a system of N possible outcomes and any given probability P?
probability combinatorics
asked Dec 10 '18 at 1:04
dragonrootdragonroot
1363
$endgroup$
1
$begingroup$
This situation is called the "coupon collector" problem with $n$ coupons. For some problems you get a more insightful answer by changing the question: If you ask the expected time to get all coupons the answer is $(n)(1 + frac{1}{2} + frac{1}{3} + ...+frac{1}{n})$. Or, the probability of getting all of them in exactly $n$ picks is $frac{n}{n}frac{n-1}{n}frac{n-2}{n}...frac{1}{n}$.
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– Michael
Dec 10 '18 at 1:42
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The probability $p_n$ to generate every possible value at least once in $n$ outputs of the TRNG generating $N$ equiprobable possible values, is $$p_n=sum_{k=0}^{N-1}(-1)^k{Nchoose k}left(1-frac kNright)^n$$ Thus, for $N$ fixed, when $ntoinfty$, $$1-p_nsim Nleft(1-frac1Nright)^n$$
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– Did
Dec 10 '18 at 3:02
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Thanks! Consider posting as an answer so I could accept.
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– dragonroot
Dec 10 '18 at 3:11
add a comment |
$begingroup$
Suppose there's a true random number generator which can output an integer value in [0..N). How many times does it have to be invoked for it to generate every possible value [0..N) at least once, with the given probability P?
For example, for coin clips, there's a 50% probability of seeing both heads and tails in two rounds, 75% probability in three rounds etc. I wonder how to generalize this for a system of N possible outcomes and any given probability P?
probability combinatorics
asked Dec 10 '18 at 1:04
dragonrootdragonroot
1363
$endgroup$
Suppose there's a true random number generator which can output an integer value in [0..N). How many times does it have to be invoked for it to generate every possible value [0..N) at least once, with the given probability P?
For example, for coin clips, there's a 50% probability of seeing both heads and tails in two rounds, 75% probability in three rounds etc. I wonder how to generalize this for a system of N possible outcomes and any given probability P?
probability combinatorics
probability combinatorics
asked Dec 10 '18 at 1:04
dragonrootdragonroot
1363
asked Dec 10 '18 at 1:04
dragonrootdragonroot
1363
asked Dec 10 '18 at 1:04
dragonrootdragonroot
1363
asked Dec 10 '18 at 1:04
dragonrootdragonroot
1363
asked Dec 10 '18 at 1:04
dragonrootdragonroot
1363
1363
1
$begingroup$
This situation is called the "coupon collector" problem with $n$ coupons. For some problems you get a more insightful answer by changing the question: If you ask the expected time to get all coupons the answer is $(n)(1 + frac{1}{2} + frac{1}{3} + ...+frac{1}{n})$. Or, the probability of getting all of them in exactly $n$ picks is $frac{n}{n}frac{n-1}{n}frac{n-2}{n}...frac{1}{n}$.
$endgroup$
– Michael
Dec 10 '18 at 1:42
$begingroup$
The probability $p_n$ to generate every possible value at least once in $n$ outputs of the TRNG generating $N$ equiprobable possible values, is $$p_n=sum_{k=0}^{N-1}(-1)^k{Nchoose k}left(1-frac kNright)^n$$ Thus, for $N$ fixed, when $ntoinfty$, $$1-p_nsim Nleft(1-frac1Nright)^n$$
$endgroup$
– Did
Dec 10 '18 at 3:02
$begingroup$
Thanks! Consider posting as an answer so I could accept.
$endgroup$
– dragonroot
Dec 10 '18 at 3:11
add a comment |
1
$begingroup$
This situation is called the "coupon collector" problem with $n$ coupons. For some problems you get a more insightful answer by changing the question: If you ask the expected time to get all coupons the answer is $(n)(1 + frac{1}{2} + frac{1}{3} + ...+frac{1}{n})$. Or, the probability of getting all of them in exactly $n$ picks is $frac{n}{n}frac{n-1}{n}frac{n-2}{n}...frac{1}{n}$.
$endgroup$
– Michael
Dec 10 '18 at 1:42
$begingroup$
The probability $p_n$ to generate every possible value at least once in $n$ outputs of the TRNG generating $N$ equiprobable possible values, is $$p_n=sum_{k=0}^{N-1}(-1)^k{Nchoose k}left(1-frac kNright)^n$$ Thus, for $N$ fixed, when $ntoinfty$, $$1-p_nsim Nleft(1-frac1Nright)^n$$
$endgroup$
– Did
Dec 10 '18 at 3:02
$begingroup$
Thanks! Consider posting as an answer so I could accept.
$endgroup$
– dragonroot
Dec 10 '18 at 3:11
1
1
$begingroup$
This situation is called the "coupon collector" problem with $n$ coupons. For some problems you get a more insightful answer by changing the question: If you ask the expected time to get all coupons the answer is $(n)(1 + frac{1}{2} + frac{1}{3} + ...+frac{1}{n})$. Or, the probability of getting all of them in exactly $n$ picks is $frac{n}{n}frac{n-1}{n}frac{n-2}{n}...frac{1}{n}$.
$endgroup$
– Michael
Dec 10 '18 at 1:42
$begingroup$
This situation is called the "coupon collector" problem with $n$ coupons. For some problems you get a more insightful answer by changing the question: If you ask the expected time to get all coupons the answer is $(n)(1 + frac{1}{2} + frac{1}{3} + ...+frac{1}{n})$. Or, the probability of getting all of them in exactly $n$ picks is $frac{n}{n}frac{n-1}{n}frac{n-2}{n}...frac{1}{n}$.
$endgroup$
– Michael
Dec 10 '18 at 1:42
$begingroup$
The probability $p_n$ to generate every possible value at least once in $n$ outputs of the TRNG generating $N$ equiprobable possible values, is $$p_n=sum_{k=0}^{N-1}(-1)^k{Nchoose k}left(1-frac kNright)^n$$ Thus, for $N$ fixed, when $ntoinfty$, $$1-p_nsim Nleft(1-frac1Nright)^n$$
$endgroup$
– Did
Dec 10 '18 at 3:02
$begingroup$
The probability $p_n$ to generate every possible value at least once in $n$ outputs of the TRNG generating $N$ equiprobable possible values, is $$p_n=sum_{k=0}^{N-1}(-1)^k{Nchoose k}left(1-frac kNright)^n$$ Thus, for $N$ fixed, when $ntoinfty$, $$1-p_nsim Nleft(1-frac1Nright)^n$$
$endgroup$
– Did
Dec 10 '18 at 3:02
$begingroup$
Thanks! Consider posting as an answer so I could accept.
$endgroup$
– dragonroot
Dec 10 '18 at 3:11
$begingroup$
Thanks! Consider posting as an answer so I could accept.
$endgroup$
– dragonroot
Dec 10 '18 at 3:11
add a comment |
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$begingroup$
This situation is called the "coupon collector" problem with $n$ coupons. For some problems you get a more insightful answer by changing the question: If you ask the expected time to get all coupons the answer is $(n)(1 + frac{1}{2} + frac{1}{3} + ...+frac{1}{n})$. Or, the probability of getting all of them in exactly $n$ picks is $frac{n}{n}frac{n-1}{n}frac{n-2}{n}...frac{1}{n}$.
$endgroup$
– Michael
Dec 10 '18 at 1:42
$begingroup$
The probability $p_n$ to generate every possible value at least once in $n$ outputs of the TRNG generating $N$ equiprobable possible values, is $$p_n=sum_{k=0}^{N-1}(-1)^k{Nchoose k}left(1-frac kNright)^n$$ Thus, for $N$ fixed, when $ntoinfty$, $$1-p_nsim Nleft(1-frac1Nright)^n$$
$endgroup$
– Did
Dec 10 '18 at 3:02
$begingroup$
Thanks! Consider posting as an answer so I could accept.
$endgroup$
– dragonroot
Dec 10 '18 at 3:11