Skip to main content
MathOverflow

Return to Answer

added 1386 characters in body
Source Link
Aaron Meyerowitz
Aaron Meyerowitz
  • 30.1k
  • 1
  • 48
  • 104

edit I indeed was mistaken. The suggested answer is correct as far as I checked (n=10). It is easy enough to set up a system of equations for the expectations and compute. Here are the results to 6, the notation should be clear enough:

$$p_{12}=3,p_{111}=4 $$

$$p_{13}=11/2,p_{22}=7,p_{112}=8,p_{1111 }=9 $$

$$p_{14}={\frac {25}{3}},p_{23}={\frac {35}{3}} ,p_{1^23}={\frac {38}{3}},p_{12^2}=14,p_{1^32}=15,p_{1^5}=16 $$

` $p_{15}={\frac {137}{12}},p_{24}={\frac {101}{ 6}},p_{3^2}={\frac {37}{2}},p_{1^24}={\frac {107}{6}},p_{123} ={\frac {83}{4}}, p_{1^33}={\frac {87}{4}},p_{2^3}=22 ,p_{1^22^2}=23,p_{1^52}=24,p_{1^6}=25 $

AT least up to n=10, If two ones are replaced by a 2, the expected number of moves goes down by 1.

later in reply to a comment, here are results for 10. If anyone wants more, ask me. They are just kind of bulky at least the way I have them now. I think the notation should be clear.

[[1, 10], 81], [[1, 8], [2, 1], 80], [[1, 6], [2, 2], 79], [[1, 4], [2, 3], 78], [[1, 2], [2, 4], 77], [[2, 5], 76], [[1, 7], [3, 1], 623/8], [[1, 5], [2, 1], [3, 1], 615/8], [[1, 3], [2, 2], [3, 1], 607/8], [[1, 1], [2, 3], [3, 1], 599/8], [[1, 4], [3, 2], 299/4], [[1, 2], [2, 1], [3, 2], 295/4], [[2, 2], [3, 2], 291/4], [[1, 1], [3, 3], 573/8], [[1, 6], [4, 1], 2085/28], [[1, 4], [2, 1], [4, 1], 2057/28], [[1, 2], [2, 2], [4, 1], 2029/28], [[2, 3], [4, 1], 2001/28], [[1, 3], [3, 1], [4, 1], 3995/56], [[1, 1], [2, 1], [3, 1], [4, 1], 3939/56], [[3, 2], [4, 1], 955/14], [[1, 2], [4, 2], 951/14], [[2, 1], [4, 2], 937/14], [[1, 5], [5, 1], 3895/56], [[1, 3], [2, 1], [5, 1], 3839/56], [[1, 1], [2, 2], [5, 1], 3783/56], [[1, 2], [3, 1], [5, 1], 465/7], [[2, 1], [3, 1], [5, 1], 458/7], [[1, 1], [4, 1], [5, 1], 3529/56], [[5, 2], 1627/28], [[1, 4], [6, 1], 4399/70], [[1, 2], [2, 1], [6, 1], 4329/70], [[2, 2], [6, 1], 4259/70], [[1, 1], [3, 1], [6, 1], 16721/280], [[4, 1], [6, 1], 7883/140], [[1, 3], [7, 1], 15087/280], [[1, 1], [2, 1], [7, 1], 14807/280], [[3, 1], [7, 1], 3553/70], [[1, 2], [8, 1], 5869/140], [[2, 1], [8, 1], 5729/140], [[1, 1], [9, 1], 7129/280], [[10, 1], 0]

edit I indeed was mistaken. The suggested answer is correct as far as I checked (n=10). It is easy enough to set up a system of equations for the expectations and compute. Here are the results to 6, the notation should be clear enough:

$$p_{12}=3,p_{111}=4 $$

$$p_{13}=11/2,p_{22}=7,p_{112}=8,p_{1111 }=9 $$

$$p_{14}={\frac {25}{3}},p_{23}={\frac {35}{3}} ,p_{1^23}={\frac {38}{3}},p_{12^2}=14,p_{1^32}=15,p_{1^5}=16 $$

` $p_{15}={\frac {137}{12}},p_{24}={\frac {101}{ 6}},p_{3^2}={\frac {37}{2}},p_{1^24}={\frac {107}{6}},p_{123} ={\frac {83}{4}}, p_{1^33}={\frac {87}{4}},p_{2^3}=22 ,p_{1^22^2}=23,p_{1^52}=24,p_{1^6}=25 $

AT least up to n=10, If two ones are replaced by a 2, the expected number of moves goes down by 1.

edit I indeed was mistaken. The suggested answer is correct as far as I checked (n=10). It is easy enough to set up a system of equations for the expectations and compute. Here are the results to 6, the notation should be clear enough:

$$p_{12}=3,p_{111}=4 $$

$$p_{13}=11/2,p_{22}=7,p_{112}=8,p_{1111 }=9 $$

$$p_{14}={\frac {25}{3}},p_{23}={\frac {35}{3}} ,p_{1^23}={\frac {38}{3}},p_{12^2}=14,p_{1^32}=15,p_{1^5}=16 $$

` $p_{15}={\frac {137}{12}},p_{24}={\frac {101}{ 6}},p_{3^2}={\frac {37}{2}},p_{1^24}={\frac {107}{6}},p_{123} ={\frac {83}{4}}, p_{1^33}={\frac {87}{4}},p_{2^3}=22 ,p_{1^22^2}=23,p_{1^52}=24,p_{1^6}=25 $

AT least up to n=10, If two ones are replaced by a 2, the expected number of moves goes down by 1.

later in reply to a comment, here are results for 10. If anyone wants more, ask me. They are just kind of bulky at least the way I have them now. I think the notation should be clear.

[[1, 10], 81], [[1, 8], [2, 1], 80], [[1, 6], [2, 2], 79], [[1, 4], [2, 3], 78], [[1, 2], [2, 4], 77], [[2, 5], 76], [[1, 7], [3, 1], 623/8], [[1, 5], [2, 1], [3, 1], 615/8], [[1, 3], [2, 2], [3, 1], 607/8], [[1, 1], [2, 3], [3, 1], 599/8], [[1, 4], [3, 2], 299/4], [[1, 2], [2, 1], [3, 2], 295/4], [[2, 2], [3, 2], 291/4], [[1, 1], [3, 3], 573/8], [[1, 6], [4, 1], 2085/28], [[1, 4], [2, 1], [4, 1], 2057/28], [[1, 2], [2, 2], [4, 1], 2029/28], [[2, 3], [4, 1], 2001/28], [[1, 3], [3, 1], [4, 1], 3995/56], [[1, 1], [2, 1], [3, 1], [4, 1], 3939/56], [[3, 2], [4, 1], 955/14], [[1, 2], [4, 2], 951/14], [[2, 1], [4, 2], 937/14], [[1, 5], [5, 1], 3895/56], [[1, 3], [2, 1], [5, 1], 3839/56], [[1, 1], [2, 2], [5, 1], 3783/56], [[1, 2], [3, 1], [5, 1], 465/7], [[2, 1], [3, 1], [5, 1], 458/7], [[1, 1], [4, 1], [5, 1], 3529/56], [[5, 2], 1627/28], [[1, 4], [6, 1], 4399/70], [[1, 2], [2, 1], [6, 1], 4329/70], [[2, 2], [6, 1], 4259/70], [[1, 1], [3, 1], [6, 1], 16721/280], [[4, 1], [6, 1], 7883/140], [[1, 3], [7, 1], 15087/280], [[1, 1], [2, 1], [7, 1], 14807/280], [[3, 1], [7, 1], 3553/70], [[1, 2], [8, 1], 5869/140], [[2, 1], [8, 1], 5729/140], [[1, 1], [9, 1], 7129/280], [[10, 1], 0]

took out mistaken answer and put in some computations
Source Link
Aaron Meyerowitz
Aaron Meyerowitz
  • 30.1k
  • 1
  • 48
  • 104

edit I think that your hintindeed was mistaken. The suggested answer is wrongcorrect as far as I checked (n=10). Before all colors are the same there will beIt is easy enough to set up a stage with 2 colorssystem of equations for the expectations and compute. ThatHere are the results to 6, the notation should be easyclear enough:

$$p_{12}=3,p_{111}=4 $$

$$p_{13}=11/2,p_{22}=7,p_{112}=8,p_{1111 }=9 $$

$$p_{14}={\frac {25}{3}},p_{23}={\frac {35}{3}} ,p_{1^23}={\frac {38}{3}},p_{12^2}=14,p_{1^32}=15,p_{1^5}=16 $$

` $p_{15}={\frac {137}{12}},p_{24}={\frac {101}{ 6}},p_{3^2}={\frac {37}{2}},p_{1^24}={\frac {107}{6}},p_{123} ={\frac {83}{4}}, p_{1^33}={\frac {87}{4}},p_{2^3}=22 ,p_{1^22^2}=23,p_{1^52}=24,p_{1^6}=25 $

AT least up to analyze. In the case n=3 you arrive at 2 colors after one step and stay there for a while until alln=10, If two ones are replaced by a 2, the same. The expected total number of turns is 1+3/2moves goes down by 1.

I think that your hint is wrong. Before all colors are the same there will be a stage with 2 colors. That should be easy to analyze. In the case n=3 you arrive at 2 colors after one step and stay there for a while until all are the same. The expected total number of turns is 1+3/2.

edit I indeed was mistaken. The suggested answer is correct as far as I checked (n=10). It is easy enough to set up a system of equations for the expectations and compute. Here are the results to 6, the notation should be clear enough:

$$p_{12}=3,p_{111}=4 $$

$$p_{13}=11/2,p_{22}=7,p_{112}=8,p_{1111 }=9 $$

$$p_{14}={\frac {25}{3}},p_{23}={\frac {35}{3}} ,p_{1^23}={\frac {38}{3}},p_{12^2}=14,p_{1^32}=15,p_{1^5}=16 $$

` $p_{15}={\frac {137}{12}},p_{24}={\frac {101}{ 6}},p_{3^2}={\frac {37}{2}},p_{1^24}={\frac {107}{6}},p_{123} ={\frac {83}{4}}, p_{1^33}={\frac {87}{4}},p_{2^3}=22 ,p_{1^22^2}=23,p_{1^52}=24,p_{1^6}=25 $

AT least up to n=10, If two ones are replaced by a 2, the expected number of moves goes down by 1.

Source Link
Aaron Meyerowitz
Aaron Meyerowitz
  • 30.1k
  • 1
  • 48
  • 104

I think that your hint is wrong. Before all colors are the same there will be a stage with 2 colors. That should be easy to analyze. In the case n=3 you arrive at 2 colors after one step and stay there for a while until all are the same. The expected total number of turns is 1+3/2.