TITLE: Another colored balls puzzle (part II)
QUESTION [8 upvotes]: The same colleague as in Another colored balls puzzle asked me the following variant which she called "part II".
Imagine you have $n$ balls in a bag that are colored from $1$ to $n$.  At each turn you take one ball uniformly at random from the bag and then a second uniformly at random from the remaining balls which have a different color to the first. You then color one ball the color of the other and put them both back in the bag. 
What is the expected number of turns before all the balls have the same color if:

You always paint the first ball the color of the second? Or...
You always paint the second ball the color of the first?

My intuition is that in the first case you are likely to be decreasing the number of balls with a  commonly occurring color  and in the second you are likely to be increasing the number of balls with a commonly occurring color. Hence the time to get all colors the same will be much less in the second case than the first.
Simulations.
For rule 1 I get the following approximate expected values for $n = 2 \dots 6$.
$1$, $4$, $10.33$, $22.48$, $45.23$.
For rule 2 I get the following approximate expected values for $n = 2 \dots 6$.
$1$, $2.500$, $4.416$, $6.698$. $9.310$.
To get some feeling for how the problem scales I also tried some other values up to $n=100$. For rule 1 and $n=100$ my simulation didn't get to all balls being equal even once in the time I gave it. However for $n=10$ I get about $618$ and for $n=20$ I get very roughly $520,000$.    For rule 2 I get a mean of roughly $865$ for $n=100$.  My blind guess is now that the mean for rule is 1 is at least exponential (in fact it looks a little like $2^{n-1}$) but that for rule two it is not quite  $\frac{e}{4} n^{1.55}$ but it may be something similar.

REPLY [2 votes]: For given $n$ it is possible to calculate the expected number of turns 
from an absorbing Markov chain on the integer partitions of $n$.
Each state of the Markov process corresponds to the ordered list of the number of equally colored balls.
The matrix of transition probabilities between the states is easily calculated according to rules 1 or 2, 
e.g.
Removing from position 1, adding to position 2 :  {3,1,1} -> {2,2,1}
    with probability  p = P(choosing position 1 out of all balls) * P(choosing position 2 out of remaining balls) 
                        =3/5 * 1/2 
Given the transition matrix with the unique absorbing state {n} and the initial state {1,1,1,...,1} it is possible to calculate 
the expected number of turns before absorption.
This is done in the  Mathematica program below. Here some results where the first number is n, the second the numerical result, and the third the exact fraction:
Rule 1
{{2, 1., 1}, 

{3, 4., 4}, 

{4, 10.3333, 31/3}, 

{5, 22.4852, 16729/744}, 

{6, 45.2173, 33913/750},

{7, 87.7733, 26707139046097/304274018880}, 

{8, 168.252, 129857255359868261/771803525388385},

{9, 322.292, 4555917617310039296830835441635/14135986803865219963776139264}, 

{10,620.346,822838635777324535445878391148603051494611/1326419190860455039655669536523314862820}}

Only the numerical values for n>10:
{{11, 1202.04}, {12, 2344.58}, {13, 4599.07}, {14, 9062.01}, 
{15, 17916.8}, {16, 35513.4},{17, 70522.1}, {18, 140231.},{19, 279122.}, {20, 555989.}}
Rule 2
{{2, 1., 1}, 

{3, 2.5, 5/2}, 

{4, 4.41667, 53/12}, 

{5, 6.6994, 2251/336}, 

{6, 9.31157, 866023/93005}, 

{7, 12.2249, 1009285097/82560060},

{8, 15.4166, 2246993235815929/145751872750176}, 

{9, 18.868, 2285085765293281062003190373/121108796080566797904702840}, 

{10, 22.5637, 618224636000595187350171250435705332105433100763641/27399140168645771065204844597274355963355735154297}}

Only the numerical values for n>10:
{{11, 26.4901}, {12, 30.6358}, {13, 34.9907}, {14, 39.546}, {15, 
  44.2936}, {16, 49.2266}, {17, 54.3383}, {18, 59.6231}, {19, 
  65.0754}, {20, 70.6904}}
Program:
(*
Defining Rule 1:
  input : lst = Partition of n , e.g. {5,3,2,1} with n=11
        k, m :  Positions to be changed ; remove ball from position k and add to position m
 output : list containing changed partition and probability for this change

 E.g. PartMove[{5, 3, 2, 1}, 1, 3] -> {{4, 3, 3, 1}, 5/33}
*)
PartMove[lst_, k_, m_] :=
 Module[{ll = lst, len = Plus @@ lst, prob},
  prob = (ll[[k]]/len*ll[[m]]/(len - ll[[k]]));
  ll[[k]] -= 1; ll[[m]] += 1;
  If[ll[[k]] == 0, ll = Drop[ll, {k}]]; {Reverse[Sort[ll]], prob}]

(*Alternatively
Defining Rule 2:
  input : lst = Partition of n , e.g. {5,3,2,1} with n=11
        k, m :  Positions to be changed ; add ball to position k and remove from  position m
 output : list containing changed partition and probability for this change

 E.g. PartMove[{5, 3, 2, 1}, 1, 3] ->   {{6, 3, 1, 1}, 5/33}
*)
 PartMove[lst_, k_, m_] :=
 Module[{ll = lst, len = Plus @@ lst, prob},
  prob = (ll[[k]]/len*ll[[m]]/(len - ll[[k]]));
  ll[[k]] += 1; ll[[m]] -= 1;
  If[ll[[m]] == 0, ll = Drop[ll, {m}]]; {Reverse[Sort[ll]], prob}]

  (*
Calculate all possible target partitions for a given partition with respective probabilities:
  input : lst = Partition of n , e.g. {5,3,2,1} with n=11

  output : list containing all possible changed partitions with their  probabilities
 Note: Depends on the chosen rule via PartMove

 E.g. for rule 1:
 Targets[{2, 1, 1}] -> {{{2, 2}, 1/6}, {{3, 1}, 1/3}, {{2, 1, 1}, 1/2}}
*)
  Targets[lst_] := Module[{len = Length[lst], pairs, res},
  If[len <= 1, {{lst, 1}},
   pairs = Select[Tuples[Range[len], 2], #[[1]] != #[[2]] &];
   res = Sort[PartMove[lst, #[[1]], #[[2]]] & /@ pairs];
    {#[[1, 1]], Plus @@ (#[[2]] & /@ #)} & /@ SplitBy[res, First]]]

(*  Define all possible states for chosen n (here n=4) *)

states = IntegerPartitions[4];

(* Define transition matrix PM for Markov chain *)
nn = Length[states];
Clear[PartIndex]; n = 1; (PartIndex[#] = n++) & /@ states;
PM = Table[0, {nn}, {nn}];
Do[(PM[[PartIndex[#[[1]]], k]] = #[[2]]) & /@
  Targets[states[[k]]], {k, 1, nn}]

(* Define submatrix Q for transient state changes *)
Q = (Drop[#, 1] & /@ Drop[PM, 1]);
(* Calculate fundamental matrix NPM of absorbing Markov chain*)
NPM = Inverse[ IdentityMatrix[Length[Q]] - Q];
(*Calculate expected number t of turns when starting from {1,1,..,1} = states[[-1]] *)
t = Plus @@ (#[[-1]] & /@ NPM);
{N[t], t} (*numerical and exact value of t*)