Pólya urn flip and roll
$begingroup$
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
asked 5 hours ago
Expired DataExpired Data
1914
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Expired Data is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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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
asked 5 hours ago
Expired DataExpired Data
1914
New contributor
Expired Data 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 feel like there is a formula for this...
$endgroup$
– Embodiment of Ignorance
2 hours ago
$begingroup$
Something to do with beta binomials maybe, but it might be longer to write that out
$endgroup$
– Expired Data
2 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
asked 5 hours ago
Expired DataExpired Data
1914
New contributor
Expired Data is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
asked 5 hours ago
Expired DataExpired Data
1914
New contributor
Expired Data is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Expired DataExpired Data
1914
New contributor
Expired Data is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Expired Data
asked 5 hours ago
Expired DataExpired Data
1914
New contributor
Expired Data is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Expired DataExpired Data
1914
asked 5 hours ago
Expired DataExpired Data
1914
1914
New contributor
Expired Data is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Expired Data is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Expired Data is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I feel like there is a formula for this...
$endgroup$
– Embodiment of Ignorance
2 hours ago
$begingroup$
Something to do with beta binomials maybe, but it might be longer to write that out
$endgroup$
– Expired Data
2 hours ago
add a comment |
$begingroup$
I feel like there is a formula for this...
$endgroup$
– Embodiment of Ignorance
2 hours ago
$begingroup$
Something to do with beta binomials maybe, but it might be longer to write that out
$endgroup$
– Expired Data
2 hours ago
$begingroup$
I feel like there is a formula for this...
$endgroup$
– Embodiment of Ignorance
2 hours ago
$begingroup$
I feel like there is a formula for this...
$endgroup$
– Embodiment of Ignorance
2 hours ago
$begingroup$
Something to do with beta binomials maybe, but it might be longer to write that out
$endgroup$
– Expired Data
2 hours ago
$begingroup$
Something to do with beta binomials maybe, but it might be longer to write that out
$endgroup$
– Expired Data
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
JavaScript (ES6), 145 139 135 132 bytes
Takes input as (r)(n). This is a naive solution that actually performs the entire simulation.
r=>g=(n,B=!(k=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/k)=>n--?h``:s+=(k++,B>R?B/R:R/B)>=r
Try it online!
Too slow for the last 2 test cases.
answered 3 hours ago
ArnauldArnauld
78.7k795327
$endgroup$
add a comment |
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1 Answer
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$begingroup$
JavaScript (ES6), 145 139 135 132 bytes
Takes input as (r)(n). This is a naive solution that actually performs the entire simulation.
r=>g=(n,B=!(k=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/k)=>n--?h``:s+=(k++,B>R?B/R:R/B)>=r
Try it online!
Too slow for the last 2 test cases.
answered 3 hours ago
ArnauldArnauld
78.7k795327
$endgroup$
add a comment |
$begingroup$
JavaScript (ES6), 145 139 135 132 bytes
Takes input as (r)(n). This is a naive solution that actually performs the entire simulation.
r=>g=(n,B=!(k=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/k)=>n--?h``:s+=(k++,B>R?B/R:R/B)>=r
Try it online!
Too slow for the last 2 test cases.
answered 3 hours ago
ArnauldArnauld
78.7k795327
$endgroup$
add a comment |
$begingroup$
JavaScript (ES6), 145 139 135 132 bytes
Takes input as (r)(n). This is a naive solution that actually performs the entire simulation.
r=>g=(n,B=!(k=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/k)=>n--?h``:s+=(k++,B>R?B/R:R/B)>=r
Try it online!
Too slow for the last 2 test cases.
answered 3 hours ago
ArnauldArnauld
78.7k795327
$endgroup$
JavaScript (ES6), 145 139 135 132 bytes
Takes input as (r)(n). This is a naive solution that actually performs the entire simulation.
r=>g=(n,B=!(k=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/k)=>n--?h``:s+=(k++,B>R?B/R:R/B)>=r
Try it online!
Too slow for the last 2 test cases.
answered 3 hours ago
ArnauldArnauld
78.7k795327
edited 2 hours ago
answered 3 hours ago
ArnauldArnauld
78.7k795327
answered 3 hours ago
ArnauldArnauld
78.7k795327
answered 3 hours ago
ArnauldArnauld
78.7k795327
78.7k795327
add a comment |
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
2 hours ago
$begingroup$
Something to do with beta binomials maybe, but it might be longer to write that out
$endgroup$
– Expired Data
2 hours ago