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Expected number of rolls of die until gcd/lcm exceeds certain value?
$begingroup$
I was presented with the following problem:
What is the expected number of rolls of a $n$-sided die until
i) the lcm of the rolls exceeds $k$?
ii) the gcd of the rolls exceeds $k$?
Is this something that can be analytically done, or must this be done numerically?
As a hint, I am told that for $n=10$ and $k=2018$, the EV for part i) converges to $18.8$, but that seems oddly low.
Could someone prod me the right direction? Thanks!
probability statistics conditional-expectation expected-value
asked Feb 1 at 17:17
user107224user107224
475314
$endgroup$
add a comment |
$begingroup$
I was presented with the following problem:
What is the expected number of rolls of a $n$-sided die until
i) the lcm of the rolls exceeds $k$?
ii) the gcd of the rolls exceeds $k$?
Is this something that can be analytically done, or must this be done numerically?
As a hint, I am told that for $n=10$ and $k=2018$, the EV for part i) converges to $18.8$, but that seems oddly low.
Could someone prod me the right direction? Thanks!
probability statistics conditional-expectation expected-value
asked Feb 1 at 17:17
user107224user107224
475314
$endgroup$
1
$begingroup$
The answer in your example is not particularly low. If you wanted the expected number of rolls to see all ten faces then this would be about $29.3$. But here you only need to see faces $7$, $8$, $9$ and either of $5$ or $10$ to get a lowest common multiple above $2018$ (or indeed of $1261$ or more, since $2560$ is the only possibility)
$endgroup$
– Henry
Feb 1 at 18:30
$begingroup$
@Henry I can see that now; how do you prove it analytically though?
$endgroup$
– user107224
Feb 1 at 22:47
add a comment |
$begingroup$
I was presented with the following problem:
What is the expected number of rolls of a $n$-sided die until
i) the lcm of the rolls exceeds $k$?
ii) the gcd of the rolls exceeds $k$?
Is this something that can be analytically done, or must this be done numerically?
As a hint, I am told that for $n=10$ and $k=2018$, the EV for part i) converges to $18.8$, but that seems oddly low.
Could someone prod me the right direction? Thanks!
probability statistics conditional-expectation expected-value
asked Feb 1 at 17:17
user107224user107224
475314
$endgroup$
I was presented with the following problem:
What is the expected number of rolls of a $n$-sided die until
i) the lcm of the rolls exceeds $k$?
ii) the gcd of the rolls exceeds $k$?
Is this something that can be analytically done, or must this be done numerically?
As a hint, I am told that for $n=10$ and $k=2018$, the EV for part i) converges to $18.8$, but that seems oddly low.
Could someone prod me the right direction? Thanks!
probability statistics conditional-expectation expected-value
probability statistics conditional-expectation expected-value
asked Feb 1 at 17:17
user107224user107224
475314
asked Feb 1 at 17:17
user107224user107224
475314
edited Feb 2 at 8:43
user107224
asked Feb 1 at 17:17
user107224user107224
475314
asked Feb 1 at 17:17
user107224user107224
475314
asked Feb 1 at 17:17
user107224user107224
475314
475314
1
$begingroup$
The answer in your example is not particularly low. If you wanted the expected number of rolls to see all ten faces then this would be about $29.3$. But here you only need to see faces $7$, $8$, $9$ and either of $5$ or $10$ to get a lowest common multiple above $2018$ (or indeed of $1261$ or more, since $2560$ is the only possibility)
$endgroup$
– Henry
Feb 1 at 18:30
$begingroup$
@Henry I can see that now; how do you prove it analytically though?
$endgroup$
– user107224
Feb 1 at 22:47
add a comment |
1
$begingroup$
The answer in your example is not particularly low. If you wanted the expected number of rolls to see all ten faces then this would be about $29.3$. But here you only need to see faces $7$, $8$, $9$ and either of $5$ or $10$ to get a lowest common multiple above $2018$ (or indeed of $1261$ or more, since $2560$ is the only possibility)
$endgroup$
– Henry
Feb 1 at 18:30
$begingroup$
@Henry I can see that now; how do you prove it analytically though?
$endgroup$
– user107224
Feb 1 at 22:47
1
1
$begingroup$
The answer in your example is not particularly low. If you wanted the expected number of rolls to see all ten faces then this would be about $29.3$. But here you only need to see faces $7$, $8$, $9$ and either of $5$ or $10$ to get a lowest common multiple above $2018$ (or indeed of $1261$ or more, since $2560$ is the only possibility)
$endgroup$
– Henry
Feb 1 at 18:30
$begingroup$
The answer in your example is not particularly low. If you wanted the expected number of rolls to see all ten faces then this would be about $29.3$. But here you only need to see faces $7$, $8$, $9$ and either of $5$ or $10$ to get a lowest common multiple above $2018$ (or indeed of $1261$ or more, since $2560$ is the only possibility)
$endgroup$
– Henry
Feb 1 at 18:30
$begingroup$
@Henry I can see that now; how do you prove it analytically though?
$endgroup$
– user107224
Feb 1 at 22:47
$begingroup$
@Henry I can see that now; how do you prove it analytically though?
$endgroup$
– user107224
Feb 1 at 22:47
add a comment |
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1
$begingroup$
The answer in your example is not particularly low. If you wanted the expected number of rolls to see all ten faces then this would be about $29.3$. But here you only need to see faces $7$, $8$, $9$ and either of $5$ or $10$ to get a lowest common multiple above $2018$ (or indeed of $1261$ or more, since $2560$ is the only possibility)
$endgroup$
– Henry
Feb 1 at 18:30
$begingroup$
@Henry I can see that now; how do you prove it analytically though?
$endgroup$
– user107224
Feb 1 at 22:47