Pólya urn flip and rollRoll Dungeons and Dragons diceBuild a program to analyze coin flip sequence choicesSimple single-player board game, expected score distributionSplit, flip and recombine integersYour Base to 1-2-3-Tribonacci to Binary back to Your BaseFlip a coin for me!Flip rows and columnsWhen Fibonacci meets the QueensBack to the Basics of MathSolve a matrix equation by Jacobi's method (Revised)
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Dice rolling probability game
Pólya urn flip and roll
Roll Dungeons and Dragons diceBuild a program to analyze coin flip sequence choicesSimple single-player board game, expected score distributionSplit, flip and recombine integersYour Base to 1-2-3-Tribonacci to Binary back to Your BaseFlip a coin for me!Flip rows and columnsWhen Fibonacci meets the QueensBack to the Basics of MathSolve a matrix equation by Jacobi's method (Revised)
$begingroup$
Problem statement
Pólya is playing about with his urn again and he wants you to help him calculate some probabilities.
In this urn experiment Pólya has an urn which initially contains 1 red and 1 blue bead.
For every iteration, he reaches in and retrieves a bead, then inspects the colour and places the bead back in the urn.
He then flips a fair coin, if the coin lands heads he will insert a fair 6 sided die roll amount of the same coloured bead into the urn, if it lands tails he will remove half the number of the same colored bead from the urn (Using integer division - so if the number of beads of the selected colour is odd he will remove (c-1)/2
where c is the number of beads of that colour)
Given an integer n ≥ 0 and a decimal r > 0, give the probability to 2 decimal places that the ratio between the colours of beads after n iterations is greater than or equal to r in the shortest number of bytes.
An example set of iterations:
Let (x, y) define the urn such that it contains x red beads and y blue beads.
Iteration Urn Ratio
0 (1,1) 1
1 (5,1) 5 //Red bead retrieved, coin flip heads, die roll 4
2 (5,1) 5 //Blue bead retrieved, coin flip tails
3 (3,1) 3 //Red bead retrieved, coin flip tails
4 (3,4) 1.333... //Blue bead retrieved, coin flip heads, die roll 3
As can be seen the Ratio r is always ≥ 1 (so it's the greater of red or blue divided by the lesser)
Test cases:
Let F(n, r) define application of the function for n iterations and a ratio of r
F(0,5) = 0.00
F(1,2) = 0.50
F(1,3) = 0.42
F(5,5) = 0.28
F(10,4) = 0.31
F(40,6.25) = 0.14
This is code golf, so the shortest solution in bytes wins.
code-golf probability-theory
New contributor
$endgroup$
add a comment |
$begingroup$
Problem statement
Pólya is playing about with his urn again and he wants you to help him calculate some probabilities.
In this urn experiment Pólya has an urn which initially contains 1 red and 1 blue bead.
For every iteration, he reaches in and retrieves a bead, then inspects the colour and places the bead back in the urn.
He then flips a fair coin, if the coin lands heads he will insert a fair 6 sided die roll amount of the same coloured bead into the urn, if it lands tails he will remove half the number of the same colored bead from the urn (Using integer division - so if the number of beads of the selected colour is odd he will remove (c-1)/2
where c is the number of beads of that colour)
Given an integer n ≥ 0 and a decimal r > 0, give the probability to 2 decimal places that the ratio between the colours of beads after n iterations is greater than or equal to r in the shortest number of bytes.
An example set of iterations:
Let (x, y) define the urn such that it contains x red beads and y blue beads.
Iteration Urn Ratio
0 (1,1) 1
1 (5,1) 5 //Red bead retrieved, coin flip heads, die roll 4
2 (5,1) 5 //Blue bead retrieved, coin flip tails
3 (3,1) 3 //Red bead retrieved, coin flip tails
4 (3,4) 1.333... //Blue bead retrieved, coin flip heads, die roll 3
As can be seen the Ratio r is always ≥ 1 (so it's the greater of red or blue divided by the lesser)
Test cases:
Let F(n, r) define application of the function for n iterations and a ratio of r
F(0,5) = 0.00
F(1,2) = 0.50
F(1,3) = 0.42
F(5,5) = 0.28
F(10,4) = 0.31
F(40,6.25) = 0.14
This is code golf, so the shortest solution in bytes wins.
code-golf probability-theory
New contributor
$endgroup$
$begingroup$
I feel like there is a formula for this...
$endgroup$
– Embodiment of Ignorance
9 hours ago
$begingroup$
Something to do with beta binomials maybe, but it might be longer to write that out
$endgroup$
– Expired Data
9 hours ago
$begingroup$
depends on the language; R and Mathematica might be able to do it efficiently.
$endgroup$
– Giuseppe
6 hours ago
add a comment |
$begingroup$
Problem statement
Pólya is playing about with his urn again and he wants you to help him calculate some probabilities.
In this urn experiment Pólya has an urn which initially contains 1 red and 1 blue bead.
For every iteration, he reaches in and retrieves a bead, then inspects the colour and places the bead back in the urn.
He then flips a fair coin, if the coin lands heads he will insert a fair 6 sided die roll amount of the same coloured bead into the urn, if it lands tails he will remove half the number of the same colored bead from the urn (Using integer division - so if the number of beads of the selected colour is odd he will remove (c-1)/2
where c is the number of beads of that colour)
Given an integer n ≥ 0 and a decimal r > 0, give the probability to 2 decimal places that the ratio between the colours of beads after n iterations is greater than or equal to r in the shortest number of bytes.
An example set of iterations:
Let (x, y) define the urn such that it contains x red beads and y blue beads.
Iteration Urn Ratio
0 (1,1) 1
1 (5,1) 5 //Red bead retrieved, coin flip heads, die roll 4
2 (5,1) 5 //Blue bead retrieved, coin flip tails
3 (3,1) 3 //Red bead retrieved, coin flip tails
4 (3,4) 1.333... //Blue bead retrieved, coin flip heads, die roll 3
As can be seen the Ratio r is always ≥ 1 (so it's the greater of red or blue divided by the lesser)
Test cases:
Let F(n, r) define application of the function for n iterations and a ratio of r
F(0,5) = 0.00
F(1,2) = 0.50
F(1,3) = 0.42
F(5,5) = 0.28
F(10,4) = 0.31
F(40,6.25) = 0.14
This is code golf, so the shortest solution in bytes wins.
code-golf probability-theory
New contributor
$endgroup$
Problem statement
Pólya is playing about with his urn again and he wants you to help him calculate some probabilities.
In this urn experiment Pólya has an urn which initially contains 1 red and 1 blue bead.
For every iteration, he reaches in and retrieves a bead, then inspects the colour and places the bead back in the urn.
He then flips a fair coin, if the coin lands heads he will insert a fair 6 sided die roll amount of the same coloured bead into the urn, if it lands tails he will remove half the number of the same colored bead from the urn (Using integer division - so if the number of beads of the selected colour is odd he will remove (c-1)/2
where c is the number of beads of that colour)
Given an integer n ≥ 0 and a decimal r > 0, give the probability to 2 decimal places that the ratio between the colours of beads after n iterations is greater than or equal to r in the shortest number of bytes.
An example set of iterations:
Let (x, y) define the urn such that it contains x red beads and y blue beads.
Iteration Urn Ratio
0 (1,1) 1
1 (5,1) 5 //Red bead retrieved, coin flip heads, die roll 4
2 (5,1) 5 //Blue bead retrieved, coin flip tails
3 (3,1) 3 //Red bead retrieved, coin flip tails
4 (3,4) 1.333... //Blue bead retrieved, coin flip heads, die roll 3
As can be seen the Ratio r is always ≥ 1 (so it's the greater of red or blue divided by the lesser)
Test cases:
Let F(n, r) define application of the function for n iterations and a ratio of r
F(0,5) = 0.00
F(1,2) = 0.50
F(1,3) = 0.42
F(5,5) = 0.28
F(10,4) = 0.31
F(40,6.25) = 0.14
This is code golf, so the shortest solution in bytes wins.
code-golf probability-theory
code-golf probability-theory
New contributor
New contributor
edited 11 hours ago
Expired Data
New contributor
asked 12 hours ago
Expired DataExpired Data
2015
2015
New contributor
New contributor
$begingroup$
I feel like there is a formula for this...
$endgroup$
– Embodiment of Ignorance
9 hours ago
$begingroup$
Something to do with beta binomials maybe, but it might be longer to write that out
$endgroup$
– Expired Data
9 hours ago
$begingroup$
depends on the language; R and Mathematica might be able to do it efficiently.
$endgroup$
– Giuseppe
6 hours ago
add a comment |
$begingroup$
I feel like there is a formula for this...
$endgroup$
– Embodiment of Ignorance
9 hours ago
$begingroup$
Something to do with beta binomials maybe, but it might be longer to write that out
$endgroup$
– Expired Data
9 hours ago
$begingroup$
depends on the language; R and Mathematica might be able to do it efficiently.
$endgroup$
– Giuseppe
6 hours ago
$begingroup$
I feel like there is a formula for this...
$endgroup$
– Embodiment of Ignorance
9 hours ago
$begingroup$
I feel like there is a formula for this...
$endgroup$
– Embodiment of Ignorance
9 hours ago
$begingroup$
Something to do with beta binomials maybe, but it might be longer to write that out
$endgroup$
– Expired Data
9 hours ago
$begingroup$
Something to do with beta binomials maybe, but it might be longer to write that out
$endgroup$
– Expired Data
9 hours ago
$begingroup$
depends on the language; R and Mathematica might be able to do it efficiently.
$endgroup$
– Giuseppe
6 hours ago
$begingroup$
depends on the language; R and Mathematica might be able to do it efficiently.
$endgroup$
– Giuseppe
6 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
JavaScript (ES7), 145 139 135 132 129 bytes
Takes input as (r)(n)
. This is a naive solution that actually performs the entire simulation.
r=>g=(n,B=!(s=0),R=1,h=d=>++d<7?h(d,[0,d].map(b=>g(n,b?-~B>>1:B,b?R:-~R>>1)&g(n,B+b,R+d-b))):s/24**-~n)=>n--?h``:s+=B/R>=r|R/B>=r
Try it online!
Too slow for the last 2 test cases.
$endgroup$
$begingroup$
I really like this answer, I found that in order to solve the later test cases I needed to add code to merge the same ratio probabilities. So I'm not surprised it's too slow
$endgroup$
– Expired Data
6 hours ago
add a comment |
Your Answer
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1 Answer
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$begingroup$
JavaScript (ES7), 145 139 135 132 129 bytes
Takes input as (r)(n)
. This is a naive solution that actually performs the entire simulation.
r=>g=(n,B=!(s=0),R=1,h=d=>++d<7?h(d,[0,d].map(b=>g(n,b?-~B>>1:B,b?R:-~R>>1)&g(n,B+b,R+d-b))):s/24**-~n)=>n--?h``:s+=B/R>=r|R/B>=r
Try it online!
Too slow for the last 2 test cases.
$endgroup$
$begingroup$
I really like this answer, I found that in order to solve the later test cases I needed to add code to merge the same ratio probabilities. So I'm not surprised it's too slow
$endgroup$
– Expired Data
6 hours ago
add a comment |
$begingroup$
JavaScript (ES7), 145 139 135 132 129 bytes
Takes input as (r)(n)
. This is a naive solution that actually performs the entire simulation.
r=>g=(n,B=!(s=0),R=1,h=d=>++d<7?h(d,[0,d].map(b=>g(n,b?-~B>>1:B,b?R:-~R>>1)&g(n,B+b,R+d-b))):s/24**-~n)=>n--?h``:s+=B/R>=r|R/B>=r
Try it online!
Too slow for the last 2 test cases.
$endgroup$
$begingroup$
I really like this answer, I found that in order to solve the later test cases I needed to add code to merge the same ratio probabilities. So I'm not surprised it's too slow
$endgroup$
– Expired Data
6 hours ago
add a comment |
$begingroup$
JavaScript (ES7), 145 139 135 132 129 bytes
Takes input as (r)(n)
. This is a naive solution that actually performs the entire simulation.
r=>g=(n,B=!(s=0),R=1,h=d=>++d<7?h(d,[0,d].map(b=>g(n,b?-~B>>1:B,b?R:-~R>>1)&g(n,B+b,R+d-b))):s/24**-~n)=>n--?h``:s+=B/R>=r|R/B>=r
Try it online!
Too slow for the last 2 test cases.
$endgroup$
JavaScript (ES7), 145 139 135 132 129 bytes
Takes input as (r)(n)
. This is a naive solution that actually performs the entire simulation.
r=>g=(n,B=!(s=0),R=1,h=d=>++d<7?h(d,[0,d].map(b=>g(n,b?-~B>>1:B,b?R:-~R>>1)&g(n,B+b,R+d-b))):s/24**-~n)=>n--?h``:s+=B/R>=r|R/B>=r
Try it online!
Too slow for the last 2 test cases.
edited 3 hours ago
answered 10 hours ago
ArnauldArnauld
78.7k795327
78.7k795327
$begingroup$
I really like this answer, I found that in order to solve the later test cases I needed to add code to merge the same ratio probabilities. So I'm not surprised it's too slow
$endgroup$
– Expired Data
6 hours ago
add a comment |
$begingroup$
I really like this answer, I found that in order to solve the later test cases I needed to add code to merge the same ratio probabilities. So I'm not surprised it's too slow
$endgroup$
– Expired Data
6 hours ago
$begingroup$
I really like this answer, I found that in order to solve the later test cases I needed to add code to merge the same ratio probabilities. So I'm not surprised it's too slow
$endgroup$
– Expired Data
6 hours ago
$begingroup$
I really like this answer, I found that in order to solve the later test cases I needed to add code to merge the same ratio probabilities. So I'm not surprised it's too slow
$endgroup$
– Expired Data
6 hours ago
add a comment |
Expired Data is a new contributor. Be nice, and check out our Code of Conduct.
Expired Data is a new contributor. Be nice, and check out our Code of Conduct.
Expired Data is a new contributor. Be nice, and check out our Code of Conduct.
Expired Data is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I feel like there is a formula for this...
$endgroup$
– Embodiment of Ignorance
9 hours ago
$begingroup$
Something to do with beta binomials maybe, but it might be longer to write that out
$endgroup$
– Expired Data
9 hours ago
$begingroup$
depends on the language; R and Mathematica might be able to do it efficiently.
$endgroup$
– Giuseppe
6 hours ago