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How many tries to get a closer number to a target when $X$ others numbers are present? [closed]
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I have a target number in a $2^64$ space.
How many random tries do I have to do to be the closest to this target number if there is $X$ other random number?
probability discrete-mathematics
asked Jul 28 at 17:34
benoit jego
31
closed as off-topic by amWhy, Jyrki Lahtonen, Xander Henderson, Leucippus, Adrian Keister Jul 29 at 0:48
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, Xander Henderson, Adrian Keister
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up vote
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down vote
favorite
I have a target number in a $2^64$ space.
How many random tries do I have to do to be the closest to this target number if there is $X$ other random number?
probability discrete-mathematics
asked Jul 28 at 17:34
benoit jego
31
closed as off-topic by amWhy, Jyrki Lahtonen, Xander Henderson, Leucippus, Adrian Keister Jul 29 at 0:48
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, Xander Henderson, Adrian Keister
Well, it's random, so you never know :-). Do you want to reach a particular probability to be the closest? Also, how is the target number chosen? Is it also random? And are all these random numbers drawn uniformly from all integers in a contiguous range of $2^64$ integers?
– joriki
Jul 28 at 18:04
Yeah everything is random, drawn uniformly. I would like to know the formula to know how many tries on average it would take to be the closest
– benoit jego
Jul 28 at 18:23
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I have a target number in a $2^64$ space.
How many random tries do I have to do to be the closest to this target number if there is $X$ other random number?
probability discrete-mathematics
asked Jul 28 at 17:34
benoit jego
31
I have a target number in a $2^64$ space.
How many random tries do I have to do to be the closest to this target number if there is $X$ other random number?
probability discrete-mathematics
asked Jul 28 at 17:34
benoit jego
31
edited Jul 28 at 18:21
asked Jul 28 at 17:34
benoit jego
31
asked Jul 28 at 17:34
benoit jego
31
asked Jul 28 at 17:34
benoit jego
31
31
closed as off-topic by amWhy, Jyrki Lahtonen, Xander Henderson, Leucippus, Adrian Keister Jul 29 at 0:48
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, Xander Henderson, Adrian Keister
closed as off-topic by amWhy, Jyrki Lahtonen, Xander Henderson, Leucippus, Adrian Keister Jul 29 at 0:48
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, Xander Henderson, Adrian Keister
Well, it's random, so you never know :-). Do you want to reach a particular probability to be the closest? Also, how is the target number chosen? Is it also random? And are all these random numbers drawn uniformly from all integers in a contiguous range of $2^64$ integers?
– joriki
Jul 28 at 18:04
Yeah everything is random, drawn uniformly. I would like to know the formula to know how many tries on average it would take to be the closest
– benoit jego
Jul 28 at 18:23
add a comment |Â
Well, it's random, so you never know :-). Do you want to reach a particular probability to be the closest? Also, how is the target number chosen? Is it also random? And are all these random numbers drawn uniformly from all integers in a contiguous range of $2^64$ integers?
– joriki
Jul 28 at 18:04
Yeah everything is random, drawn uniformly. I would like to know the formula to know how many tries on average it would take to be the closest
– benoit jego
Jul 28 at 18:23
Well, it's random, so you never know :-). Do you want to reach a particular probability to be the closest? Also, how is the target number chosen? Is it also random? And are all these random numbers drawn uniformly from all integers in a contiguous range of $2^64$ integers?
– joriki
Jul 28 at 18:04
Well, it's random, so you never know :-). Do you want to reach a particular probability to be the closest? Also, how is the target number chosen? Is it also random? And are all these random numbers drawn uniformly from all integers in a contiguous range of $2^64$ integers?
– joriki
Jul 28 at 18:04
Yeah everything is random, drawn uniformly. I would like to know the formula to know how many tries on average it would take to be the closest
– benoit jego
Jul 28 at 18:23
Yeah everything is random, drawn uniformly. I would like to know the formula to know how many tries on average it would take to be the closest
– benoit jego
Jul 28 at 18:23
add a comment |Â
1 Answer
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Let $N$ be the number of tries it takes for you to be closest. You want the expected value of $N$. After $n$ tries, your probability of being the closest is $n/(X+n)$. Thus the expected value of $N$ is
begineqnarray*
mathsf E[N]
&=&sum_n=1^inftymathsf P(Nge n)\
&=&sum_n=1^inftyfrac XX+n\
&=&infty;.
endeqnarray*
Thus, perhaps somewhat counterintuitively, the expected number of tries turns out to be infinite.
I glossed over the very low probability of ties; in order to deal with that, you'd have to specify what happens in case of ties – both ties between guesses and coincidences between a guess and the target. If “closest†is taken to mean “closer than any other numberâ€, then the expected number of tries is infinite simply because there's a non-zero ($2^-64$) probability that one of the other numbers coincides with the target and you can never be closer.
answered Jul 28 at 18:34
joriki
164k10179328
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Let $N$ be the number of tries it takes for you to be closest. You want the expected value of $N$. After $n$ tries, your probability of being the closest is $n/(X+n)$. Thus the expected value of $N$ is
begineqnarray*
mathsf E[N]
&=&sum_n=1^inftymathsf P(Nge n)\
&=&sum_n=1^inftyfrac XX+n\
&=&infty;.
endeqnarray*
Thus, perhaps somewhat counterintuitively, the expected number of tries turns out to be infinite.
I glossed over the very low probability of ties; in order to deal with that, you'd have to specify what happens in case of ties – both ties between guesses and coincidences between a guess and the target. If “closest†is taken to mean “closer than any other numberâ€, then the expected number of tries is infinite simply because there's a non-zero ($2^-64$) probability that one of the other numbers coincides with the target and you can never be closer.
answered Jul 28 at 18:34
joriki
164k10179328
add a comment |Â
up vote
0
down vote
accepted
Let $N$ be the number of tries it takes for you to be closest. You want the expected value of $N$. After $n$ tries, your probability of being the closest is $n/(X+n)$. Thus the expected value of $N$ is
begineqnarray*
mathsf E[N]
&=&sum_n=1^inftymathsf P(Nge n)\
&=&sum_n=1^inftyfrac XX+n\
&=&infty;.
endeqnarray*
Thus, perhaps somewhat counterintuitively, the expected number of tries turns out to be infinite.
I glossed over the very low probability of ties; in order to deal with that, you'd have to specify what happens in case of ties – both ties between guesses and coincidences between a guess and the target. If “closest†is taken to mean “closer than any other numberâ€, then the expected number of tries is infinite simply because there's a non-zero ($2^-64$) probability that one of the other numbers coincides with the target and you can never be closer.
answered Jul 28 at 18:34
joriki
164k10179328
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let $N$ be the number of tries it takes for you to be closest. You want the expected value of $N$. After $n$ tries, your probability of being the closest is $n/(X+n)$. Thus the expected value of $N$ is
begineqnarray*
mathsf E[N]
&=&sum_n=1^inftymathsf P(Nge n)\
&=&sum_n=1^inftyfrac XX+n\
&=&infty;.
endeqnarray*
Thus, perhaps somewhat counterintuitively, the expected number of tries turns out to be infinite.
I glossed over the very low probability of ties; in order to deal with that, you'd have to specify what happens in case of ties – both ties between guesses and coincidences between a guess and the target. If “closest†is taken to mean “closer than any other numberâ€, then the expected number of tries is infinite simply because there's a non-zero ($2^-64$) probability that one of the other numbers coincides with the target and you can never be closer.
answered Jul 28 at 18:34
joriki
164k10179328
Let $N$ be the number of tries it takes for you to be closest. You want the expected value of $N$. After $n$ tries, your probability of being the closest is $n/(X+n)$. Thus the expected value of $N$ is
begineqnarray*
mathsf E[N]
&=&sum_n=1^inftymathsf P(Nge n)\
&=&sum_n=1^inftyfrac XX+n\
&=&infty;.
endeqnarray*
Thus, perhaps somewhat counterintuitively, the expected number of tries turns out to be infinite.
I glossed over the very low probability of ties; in order to deal with that, you'd have to specify what happens in case of ties – both ties between guesses and coincidences between a guess and the target. If “closest†is taken to mean “closer than any other numberâ€, then the expected number of tries is infinite simply because there's a non-zero ($2^-64$) probability that one of the other numbers coincides with the target and you can never be closer.
answered Jul 28 at 18:34
joriki
164k10179328
answered Jul 28 at 18:34
joriki
164k10179328
answered Jul 28 at 18:34
joriki
164k10179328
answered Jul 28 at 18:34
joriki
164k10179328
164k10179328
add a comment |Â
add a comment |Â
Well, it's random, so you never know :-). Do you want to reach a particular probability to be the closest? Also, how is the target number chosen? Is it also random? And are all these random numbers drawn uniformly from all integers in a contiguous range of $2^64$ integers?
– joriki
Jul 28 at 18:04
Yeah everything is random, drawn uniformly. I would like to know the formula to know how many tries on average it would take to be the closest
– benoit jego
Jul 28 at 18:23