If i have a $20$ digit number, what are the chances I get the same randomly generated number?
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I've gotten a bit mixed up here, if you have a number comprised of $20$ digits and need to find the possibility of a randomly generated number ending up the same, do I consider each digit independent? I keep thinking the answer should be $frac{1}{10^{20}}$ but that seems wrong somehow?
probability discrete-mathematics
edited Dec 19 '18 at 5:58
Brahadeesh
6,46942363
asked Dec 19 '18 at 4:48
ThetaTheta
11
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add a comment |
$begingroup$
I've gotten a bit mixed up here, if you have a number comprised of $20$ digits and need to find the possibility of a randomly generated number ending up the same, do I consider each digit independent? I keep thinking the answer should be $frac{1}{10^{20}}$ but that seems wrong somehow?
probability discrete-mathematics
edited Dec 19 '18 at 5:58
Brahadeesh
6,46942363
asked Dec 19 '18 at 4:48
ThetaTheta
11
$endgroup$
$begingroup$
Presumably you want to know if given a twenty digit number $n$, what is the probability that another twenty digit number is equal to $n$? Provided you don't mind $00000000000000000001$ being a twenty digit number then I believe you are correct.
$endgroup$
– Robert Thingum
Dec 19 '18 at 5:16
add a comment |
$begingroup$
I've gotten a bit mixed up here, if you have a number comprised of $20$ digits and need to find the possibility of a randomly generated number ending up the same, do I consider each digit independent? I keep thinking the answer should be $frac{1}{10^{20}}$ but that seems wrong somehow?
probability discrete-mathematics
edited Dec 19 '18 at 5:58
Brahadeesh
6,46942363
asked Dec 19 '18 at 4:48
ThetaTheta
11
$endgroup$
I've gotten a bit mixed up here, if you have a number comprised of $20$ digits and need to find the possibility of a randomly generated number ending up the same, do I consider each digit independent? I keep thinking the answer should be $frac{1}{10^{20}}$ but that seems wrong somehow?
probability discrete-mathematics
probability discrete-mathematics
edited Dec 19 '18 at 5:58
Brahadeesh
6,46942363
asked Dec 19 '18 at 4:48
ThetaTheta
11
edited Dec 19 '18 at 5:58
Brahadeesh
6,46942363
asked Dec 19 '18 at 4:48
ThetaTheta
11
edited Dec 19 '18 at 5:58
Brahadeesh
6,46942363
edited Dec 19 '18 at 5:58
Brahadeesh
6,46942363
edited Dec 19 '18 at 5:58
Brahadeesh
6,46942363
6,46942363
asked Dec 19 '18 at 4:48
ThetaTheta
11
asked Dec 19 '18 at 4:48
ThetaTheta
11
asked Dec 19 '18 at 4:48
ThetaTheta
11
11
$begingroup$
Presumably you want to know if given a twenty digit number $n$, what is the probability that another twenty digit number is equal to $n$? Provided you don't mind $00000000000000000001$ being a twenty digit number then I believe you are correct.
$endgroup$
– Robert Thingum
Dec 19 '18 at 5:16
add a comment |
$begingroup$
Presumably you want to know if given a twenty digit number $n$, what is the probability that another twenty digit number is equal to $n$? Provided you don't mind $00000000000000000001$ being a twenty digit number then I believe you are correct.
$endgroup$
– Robert Thingum
Dec 19 '18 at 5:16
$begingroup$
Presumably you want to know if given a twenty digit number $n$, what is the probability that another twenty digit number is equal to $n$? Provided you don't mind $00000000000000000001$ being a twenty digit number then I believe you are correct.
$endgroup$
– Robert Thingum
Dec 19 '18 at 5:16
$begingroup$
Presumably you want to know if given a twenty digit number $n$, what is the probability that another twenty digit number is equal to $n$? Provided you don't mind $00000000000000000001$ being a twenty digit number then I believe you are correct.
$endgroup$
– Robert Thingum
Dec 19 '18 at 5:16
add a comment |
1 Answer
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$begingroup$
If the first digit can't be $0$, then we have $$frac1{9cdot 10^{19}}$$
If the first digit can be $0$, then we have $$frac1{10^{20}}.$$
We can assume independence of digit, otherwise, it won't be uniformly distributed.
answered Dec 19 '18 at 4:54
Siong Thye GohSiong Thye Goh
102k1466118
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
If the first digit can't be $0$, then we have $$frac1{9cdot 10^{19}}$$
If the first digit can be $0$, then we have $$frac1{10^{20}}.$$
We can assume independence of digit, otherwise, it won't be uniformly distributed.
answered Dec 19 '18 at 4:54
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
add a comment |
$begingroup$
If the first digit can't be $0$, then we have $$frac1{9cdot 10^{19}}$$
If the first digit can be $0$, then we have $$frac1{10^{20}}.$$
We can assume independence of digit, otherwise, it won't be uniformly distributed.
answered Dec 19 '18 at 4:54
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
add a comment |
$begingroup$
If the first digit can't be $0$, then we have $$frac1{9cdot 10^{19}}$$
If the first digit can be $0$, then we have $$frac1{10^{20}}.$$
We can assume independence of digit, otherwise, it won't be uniformly distributed.
answered Dec 19 '18 at 4:54
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
If the first digit can't be $0$, then we have $$frac1{9cdot 10^{19}}$$
If the first digit can be $0$, then we have $$frac1{10^{20}}.$$
We can assume independence of digit, otherwise, it won't be uniformly distributed.
answered Dec 19 '18 at 4:54
Siong Thye GohSiong Thye Goh
102k1466118
answered Dec 19 '18 at 4:54
Siong Thye GohSiong Thye Goh
102k1466118
answered Dec 19 '18 at 4:54
Siong Thye GohSiong Thye Goh
102k1466118
answered Dec 19 '18 at 4:54
Siong Thye GohSiong Thye Goh
102k1466118
102k1466118
add a comment |
add a comment |
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$begingroup$
Presumably you want to know if given a twenty digit number $n$, what is the probability that another twenty digit number is equal to $n$? Provided you don't mind $00000000000000000001$ being a twenty digit number then I believe you are correct.
$endgroup$
– Robert Thingum
Dec 19 '18 at 5:16