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)













10












$begingroup$


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.










share|improve this question









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
    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















10












$begingroup$


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.










share|improve this question









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
    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













10












10








10


2



$begingroup$


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.










share|improve this question









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






share|improve this question









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.











share|improve this question









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.









share|improve this question




share|improve this question








edited 11 hours ago







Expired Data













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 12 hours ago









Expired DataExpired Data

2015




2015




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
    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










  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















5












$begingroup$

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.






share|improve this answer











$endgroup$












  • $begingroup$
    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










Your Answer





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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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.






share|improve this answer











$endgroup$












  • $begingroup$
    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















5












$begingroup$

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.






share|improve this answer











$endgroup$












  • $begingroup$
    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













5












5








5





$begingroup$

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.






share|improve this answer











$endgroup$



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.







share|improve this answer














share|improve this answer



share|improve this answer








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
















  • $begingroup$
    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










Expired Data is a new contributor. Be nice, and check out our Code of Conduct.









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Expired Data is a new contributor. Be nice, and check out our Code of Conduct.











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