How to calculate the times to generate number for “generate number from 0 to 1 until it is greater than X...
$begingroup$
for example, I'm trying to test the average count for generate a number from 0 to 1 which is greater than X, let says 0.75, the program that I use (in javascript) to test it 100000 times:
let avgCount=0;
for(let i=0;i<100000;i++){
let count=0;
while(Math.random()<0.75){
count++;
}
avgCount+=count;
}
document.write(avgCount/100000);
I found it roughly needs to generate 3 numbers until it becomes greater than 0.75.
And I found if X is 0.25, it becomes roughly 0.33 (I guess it is 1/3)
the relationship between X and avgCount (average count) I guess by editing the value of X is:
0.25 --- 1/3
0.5 --- 1
0.75 ---3
0.875 --- 7
->1 --- ->infinity
My question is, how to find the relationship between X and avgCount without programming testing (e.g.:by formula)? And what is the principle of the formula, if any?
probability
asked Dec 14 '18 at 2:38
mmmaaammmaaa
1061
$endgroup$
add a comment |
$begingroup$
for example, I'm trying to test the average count for generate a number from 0 to 1 which is greater than X, let says 0.75, the program that I use (in javascript) to test it 100000 times:
let avgCount=0;
for(let i=0;i<100000;i++){
let count=0;
while(Math.random()<0.75){
count++;
}
avgCount+=count;
}
document.write(avgCount/100000);
I found it roughly needs to generate 3 numbers until it becomes greater than 0.75.
And I found if X is 0.25, it becomes roughly 0.33 (I guess it is 1/3)
the relationship between X and avgCount (average count) I guess by editing the value of X is:
0.25 --- 1/3
0.5 --- 1
0.75 ---3
0.875 --- 7
->1 --- ->infinity
My question is, how to find the relationship between X and avgCount without programming testing (e.g.:by formula)? And what is the principle of the formula, if any?
probability
asked Dec 14 '18 at 2:38
mmmaaammmaaa
1061
$endgroup$
add a comment |
$begingroup$
for example, I'm trying to test the average count for generate a number from 0 to 1 which is greater than X, let says 0.75, the program that I use (in javascript) to test it 100000 times:
let avgCount=0;
for(let i=0;i<100000;i++){
let count=0;
while(Math.random()<0.75){
count++;
}
avgCount+=count;
}
document.write(avgCount/100000);
I found it roughly needs to generate 3 numbers until it becomes greater than 0.75.
And I found if X is 0.25, it becomes roughly 0.33 (I guess it is 1/3)
the relationship between X and avgCount (average count) I guess by editing the value of X is:
0.25 --- 1/3
0.5 --- 1
0.75 ---3
0.875 --- 7
->1 --- ->infinity
My question is, how to find the relationship between X and avgCount without programming testing (e.g.:by formula)? And what is the principle of the formula, if any?
probability
asked Dec 14 '18 at 2:38
mmmaaammmaaa
1061
$endgroup$
for example, I'm trying to test the average count for generate a number from 0 to 1 which is greater than X, let says 0.75, the program that I use (in javascript) to test it 100000 times:
let avgCount=0;
for(let i=0;i<100000;i++){
let count=0;
while(Math.random()<0.75){
count++;
}
avgCount+=count;
}
document.write(avgCount/100000);
I found it roughly needs to generate 3 numbers until it becomes greater than 0.75.
And I found if X is 0.25, it becomes roughly 0.33 (I guess it is 1/3)
the relationship between X and avgCount (average count) I guess by editing the value of X is:
0.25 --- 1/3
0.5 --- 1
0.75 ---3
0.875 --- 7
->1 --- ->infinity
My question is, how to find the relationship between X and avgCount without programming testing (e.g.:by formula)? And what is the principle of the formula, if any?
probability
probability
asked Dec 14 '18 at 2:38
mmmaaammmaaa
1061
asked Dec 14 '18 at 2:38
mmmaaammmaaa
1061
asked Dec 14 '18 at 2:38
mmmaaammmaaa
1061
asked Dec 14 '18 at 2:38
mmmaaammmaaa
1061
asked Dec 14 '18 at 2:38
mmmaaammmaaa
1061
1061
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The probability of succeeding in each attempt is $p=1-x$.
The number of tries until a success follows a geometric distribution whose mean is $1/p$ -- but that nice formula also counts the final attempt that succeeds, whereas your program doesn't.
Correcting for this, we get that what your program ought to find is the relation
$$ x quadmapstoquad frac{1}{1-x} - 1 = frac{x}{1-x} $$
which agrees exactly with your observations/guesses.
answered Dec 14 '18 at 2:56
Henning MakholmHenning Makholm
240k17305542
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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
$begingroup$
The probability of succeeding in each attempt is $p=1-x$.
The number of tries until a success follows a geometric distribution whose mean is $1/p$ -- but that nice formula also counts the final attempt that succeeds, whereas your program doesn't.
Correcting for this, we get that what your program ought to find is the relation
$$ x quadmapstoquad frac{1}{1-x} - 1 = frac{x}{1-x} $$
which agrees exactly with your observations/guesses.
answered Dec 14 '18 at 2:56
Henning MakholmHenning Makholm
240k17305542
$endgroup$
add a comment |
$begingroup$
The probability of succeeding in each attempt is $p=1-x$.
The number of tries until a success follows a geometric distribution whose mean is $1/p$ -- but that nice formula also counts the final attempt that succeeds, whereas your program doesn't.
Correcting for this, we get that what your program ought to find is the relation
$$ x quadmapstoquad frac{1}{1-x} - 1 = frac{x}{1-x} $$
which agrees exactly with your observations/guesses.
answered Dec 14 '18 at 2:56
Henning MakholmHenning Makholm
240k17305542
$endgroup$
add a comment |
$begingroup$
The probability of succeeding in each attempt is $p=1-x$.
The number of tries until a success follows a geometric distribution whose mean is $1/p$ -- but that nice formula also counts the final attempt that succeeds, whereas your program doesn't.
Correcting for this, we get that what your program ought to find is the relation
$$ x quadmapstoquad frac{1}{1-x} - 1 = frac{x}{1-x} $$
which agrees exactly with your observations/guesses.
answered Dec 14 '18 at 2:56
Henning MakholmHenning Makholm
240k17305542
$endgroup$
The probability of succeeding in each attempt is $p=1-x$.
The number of tries until a success follows a geometric distribution whose mean is $1/p$ -- but that nice formula also counts the final attempt that succeeds, whereas your program doesn't.
Correcting for this, we get that what your program ought to find is the relation
$$ x quadmapstoquad frac{1}{1-x} - 1 = frac{x}{1-x} $$
which agrees exactly with your observations/guesses.
answered Dec 14 '18 at 2:56
Henning MakholmHenning Makholm
240k17305542
answered Dec 14 '18 at 2:56
Henning MakholmHenning Makholm
240k17305542
answered Dec 14 '18 at 2:56
Henning MakholmHenning Makholm
240k17305542
answered Dec 14 '18 at 2:56
Henning MakholmHenning Makholm
240k17305542
240k17305542
add a comment |
add a comment |
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