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getting 2 random numbers to add up to less than number n

Is this number random?Easy way to generate random numbers?Pseudo Random Number Generation on the GPUCompressing random numbersWhat is the statistical difference (if any) between these two methods of generating an n-digit random number?What constitutes randomness?generate random number from exponential distribution with random right truncationFast method to pick unique random numbers?Is the sequence of non repeating random numbers still random?What does “less random” mean?

0

$begingroup$

So i'm trying to code a game for context. I'd like to be able to get 2 random numbers(a, b) that will add up to less then number n.

Here's my initial approach.

I get a random number between 0 and n.

Lets says n = 100 in an example. And my first random number is 78.

so i would take the remainder, in this case 22 and find a random number between 0 and 22. And then it would spit out some random number.

I can't seem to think of an approach where the second number gets as much "randomness" opportunity as the first number, due to the range being reduced.

What are some more approaches to getting 2 random numbers that must be less then sum n?

algebra-precalculus statistics random

share|cite|improve this question

edited Mar 29 at 1:34

J. W. Tanner

4,4711320

asked Mar 29 at 0:09

Andrew KimAndrew Kim

1033

New contributor

Andrew Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i guess i could also find 2 random numbers from 0 and n / 2.
  $endgroup$
  – Andrew Kim
  Mar 29 at 0:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so sorry, edited!
  $endgroup$
  – Andrew Kim
  Mar 29 at 0:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Definitely not the way to go... highly non-random. Your method never yield $94$ and $2$, for example, even though it should.
  $endgroup$
  – David G. Stork
  Mar 29 at 0:44
  

  
  
  
  

add a comment | 

0

$begingroup$

So i'm trying to code a game for context. I'd like to be able to get 2 random numbers(a, b) that will add up to less then number n.

Here's my initial approach.

I get a random number between 0 and n.

Lets says n = 100 in an example. And my first random number is 78.

so i would take the remainder, in this case 22 and find a random number between 0 and 22. And then it would spit out some random number.

I can't seem to think of an approach where the second number gets as much "randomness" opportunity as the first number, due to the range being reduced.

What are some more approaches to getting 2 random numbers that must be less then sum n?

algebra-precalculus statistics random

share|cite|improve this question

edited Mar 29 at 1:34

J. W. Tanner

4,4711320

asked Mar 29 at 0:09

Andrew KimAndrew Kim

1033

New contributor

Andrew Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i guess i could also find 2 random numbers from 0 and n / 2.
  $endgroup$
  – Andrew Kim
  Mar 29 at 0:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so sorry, edited!
  $endgroup$
  – Andrew Kim
  Mar 29 at 0:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Definitely not the way to go... highly non-random. Your method never yield $94$ and $2$, for example, even though it should.
  $endgroup$
  – David G. Stork
  Mar 29 at 0:44
  

  
  
  
  

add a comment | 

$begingroup$

So i'm trying to code a game for context. I'd like to be able to get 2 random numbers(a, b) that will add up to less then number n.

Here's my initial approach.

I get a random number between 0 and n.

Lets says n = 100 in an example. And my first random number is 78.

so i would take the remainder, in this case 22 and find a random number between 0 and 22. And then it would spit out some random number.

I can't seem to think of an approach where the second number gets as much "randomness" opportunity as the first number, due to the range being reduced.

What are some more approaches to getting 2 random numbers that must be less then sum n?

algebra-precalculus statistics random

share|cite|improve this question

edited Mar 29 at 1:34

J. W. Tanner

4,4711320

asked Mar 29 at 0:09

Andrew KimAndrew Kim

1033

New contributor

Andrew Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 29 at 1:34

J. W. Tanner

4,4711320

asked Mar 29 at 0:09

Andrew KimAndrew Kim

1033

New contributor

Andrew Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 29 at 1:34

J. W. Tanner

4,4711320

asked Mar 29 at 0:09

Andrew KimAndrew Kim

1033

New contributor

Andrew Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 29 at 1:34

J. W. Tanner

4,4711320

edited Mar 29 at 1:34

J. W. Tanner

4,4711320

asked Mar 29 at 0:09

Andrew KimAndrew Kim

1033

New contributor

Andrew Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 29 at 0:09

Andrew KimAndrew Kim

1033

2 Answers
2

active

oldest

votes

1

$begingroup$

This method selects two random nonnegative integers $a$ and $b$ so that $a+b< n$. This method is perfect in the sense that every possible ordered pair $(a,b)$ is equally likely. This means that $a$ and $b$ will have the exact same degree of randomness.

• Choose an integer $x$ randomly in the range $1$ to $n+1$.



• 
  

  Choose an integer $y$ randomly in the range $1$ to $n$.

  
  
  
  • If $yge x$, replace $y$ with $y+1$.
  
  



• Let $a=min(x,y)-1$, and let $b=max(x,y)-min(x,y)-1$.

share|cite|improve this answer

answered Mar 29 at 0:32

Mike EarnestMike Earnest

26.9k22152

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i don't know enough math to understand the other answer unfortunately, I like this answer because I can easily translate it to code. And it seems to be working. Thanks so much!
  $endgroup$
  – Andrew Kim
  Mar 29 at 0:56
  

  
  
  
  

add a comment | 

3

$begingroup$

We might as well take $n=1$. You can multiply the numbers that come from this by $n$ to get the range you want.

You are trying to sample the triangle $a ge 0, b ge 0, a+b le 1$ uniformly. This triangle is half of the square $0 le a le 1, 0 le b le 1$, which is easy to sample-just draw $a$ and $b$ uniformly from $[0,1]$. Now if $a+b gt 1$, we can fold over the diagonal. Instead of $a,b$, use $1-a,1-b$ for your variables. The sum is $2-a-b$, which is less than $1$ because $a+b gt 1$.

$a$ and $b$ have the same distribution, which is a triangle with density $2$ at $0$ falling linearly to $0$ at $1$.

share|cite|improve this answer

edited Mar 29 at 0:28

answered Mar 29 at 0:20

Ross MillikanRoss Millikan

301k24200375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how does that compare to the approach in my comment? are they both equally random?
  $endgroup$
  – Andrew Kim
  Mar 29 at 0:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  whoops, sorry, saw your comment, less than is what i meant =(
  $endgroup$
  – Andrew Kim
  Mar 29 at 0:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have replaced the answer with one that does what you want.
  $endgroup$
  – Ross Millikan
  Mar 29 at 0:29
  

  
  
  
  

add a comment | 

Your Answer

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1

$begingroup$

This method selects two random nonnegative integers $a$ and $b$ so that $a+b< n$. This method is perfect in the sense that every possible ordered pair $(a,b)$ is equally likely. This means that $a$ and $b$ will have the exact same degree of randomness.

• Choose an integer $x$ randomly in the range $1$ to $n+1$.



• 
  

  Choose an integer $y$ randomly in the range $1$ to $n$.

  
  
  
  • If $yge x$, replace $y$ with $y+1$.
  
  



• Let $a=min(x,y)-1$, and let $b=max(x,y)-min(x,y)-1$.

share|cite|improve this answer

answered Mar 29 at 0:32

Mike EarnestMike Earnest

26.9k22152

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i don't know enough math to understand the other answer unfortunately, I like this answer because I can easily translate it to code. And it seems to be working. Thanks so much!
  $endgroup$
  – Andrew Kim
  Mar 29 at 0:56
  

  
  
  
  

add a comment | 

1

$begingroup$

This method selects two random nonnegative integers $a$ and $b$ so that $a+b< n$. This method is perfect in the sense that every possible ordered pair $(a,b)$ is equally likely. This means that $a$ and $b$ will have the exact same degree of randomness.

• Choose an integer $x$ randomly in the range $1$ to $n+1$.



• 
  

  Choose an integer $y$ randomly in the range $1$ to $n$.

  
  
  
  • If $yge x$, replace $y$ with $y+1$.
  
  



• Let $a=min(x,y)-1$, and let $b=max(x,y)-min(x,y)-1$.

share|cite|improve this answer

answered Mar 29 at 0:32

Mike EarnestMike Earnest

26.9k22152

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i don't know enough math to understand the other answer unfortunately, I like this answer because I can easily translate it to code. And it seems to be working. Thanks so much!
  $endgroup$
  – Andrew Kim
  Mar 29 at 0:56
  

  
  
  
  

add a comment | 

$begingroup$

This method selects two random nonnegative integers $a$ and $b$ so that $a+b< n$. This method is perfect in the sense that every possible ordered pair $(a,b)$ is equally likely. This means that $a$ and $b$ will have the exact same degree of randomness.

• Choose an integer $x$ randomly in the range $1$ to $n+1$.



• 
  

  Choose an integer $y$ randomly in the range $1$ to $n$.

  
  
  
  • If $yge x$, replace $y$ with $y+1$.
  
  



• Let $a=min(x,y)-1$, and let $b=max(x,y)-min(x,y)-1$.

share|cite|improve this answer

answered Mar 29 at 0:32

Mike EarnestMike Earnest

26.9k22152

$endgroup$

share|cite|improve this answer

answered Mar 29 at 0:32

Mike EarnestMike Earnest

26.9k22152

answered Mar 29 at 0:32

Mike EarnestMike Earnest

26.9k22152

answered Mar 29 at 0:32

Mike EarnestMike Earnest

26.9k22152

3

$begingroup$

We might as well take $n=1$. You can multiply the numbers that come from this by $n$ to get the range you want.

You are trying to sample the triangle $a ge 0, b ge 0, a+b le 1$ uniformly. This triangle is half of the square $0 le a le 1, 0 le b le 1$, which is easy to sample-just draw $a$ and $b$ uniformly from $[0,1]$. Now if $a+b gt 1$, we can fold over the diagonal. Instead of $a,b$, use $1-a,1-b$ for your variables. The sum is $2-a-b$, which is less than $1$ because $a+b gt 1$.

$a$ and $b$ have the same distribution, which is a triangle with density $2$ at $0$ falling linearly to $0$ at $1$.

share|cite|improve this answer

edited Mar 29 at 0:28

answered Mar 29 at 0:20

Ross MillikanRoss Millikan

301k24200375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how does that compare to the approach in my comment? are they both equally random?
  $endgroup$
  – Andrew Kim
  Mar 29 at 0:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  whoops, sorry, saw your comment, less than is what i meant =(
  $endgroup$
  – Andrew Kim
  Mar 29 at 0:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have replaced the answer with one that does what you want.
  $endgroup$
  – Ross Millikan
  Mar 29 at 0:29
  

  
  
  
  

add a comment | 

3

$begingroup$

We might as well take $n=1$. You can multiply the numbers that come from this by $n$ to get the range you want.

You are trying to sample the triangle $a ge 0, b ge 0, a+b le 1$ uniformly. This triangle is half of the square $0 le a le 1, 0 le b le 1$, which is easy to sample-just draw $a$ and $b$ uniformly from $[0,1]$. Now if $a+b gt 1$, we can fold over the diagonal. Instead of $a,b$, use $1-a,1-b$ for your variables. The sum is $2-a-b$, which is less than $1$ because $a+b gt 1$.

$a$ and $b$ have the same distribution, which is a triangle with density $2$ at $0$ falling linearly to $0$ at $1$.

share|cite|improve this answer

edited Mar 29 at 0:28

answered Mar 29 at 0:20

Ross MillikanRoss Millikan

301k24200375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how does that compare to the approach in my comment? are they both equally random?
  $endgroup$
  – Andrew Kim
  Mar 29 at 0:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  whoops, sorry, saw your comment, less than is what i meant =(
  $endgroup$
  – Andrew Kim
  Mar 29 at 0:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have replaced the answer with one that does what you want.
  $endgroup$
  – Ross Millikan
  Mar 29 at 0:29
  

  
  
  
  

add a comment | 

$begingroup$

We might as well take $n=1$. You can multiply the numbers that come from this by $n$ to get the range you want.

You are trying to sample the triangle $a ge 0, b ge 0, a+b le 1$ uniformly. This triangle is half of the square $0 le a le 1, 0 le b le 1$, which is easy to sample-just draw $a$ and $b$ uniformly from $[0,1]$. Now if $a+b gt 1$, we can fold over the diagonal. Instead of $a,b$, use $1-a,1-b$ for your variables. The sum is $2-a-b$, which is less than $1$ because $a+b gt 1$.

$a$ and $b$ have the same distribution, which is a triangle with density $2$ at $0$ falling linearly to $0$ at $1$.

share|cite|improve this answer

edited Mar 29 at 0:28

answered Mar 29 at 0:20

Ross MillikanRoss Millikan

301k24200375

$endgroup$

share|cite|improve this answer

edited Mar 29 at 0:28

answered Mar 29 at 0:20

Ross MillikanRoss Millikan

301k24200375

answered Mar 29 at 0:20

Ross MillikanRoss Millikan

301k24200375

answered Mar 29 at 0:20

Ross MillikanRoss Millikan

301k24200375