How many even numbers will $99$ dice show if we roll them forever under a certain condition? Consider a six-sided die with numbers from $1$ to $6$. Imagine you have a jar with $99$ of such dice. You throw all dice on the floor randomly. You look at one of the dice on the floor at a time. For each die, you do the following:


*

*If it landed at an even number $(2,4,6)$, you turn the die so that it lands on the number $1$.

*If the die landed on an odd number $(1,3,5)$, you throw the die up in the air, so it can land on any number.
After you finish doing the above for all dice on the floor, you come back to the first die and repeat the entire process again. You keep on doing this until eternity (for a billion years, let’s say). If I come into the room after a billion years, how many dice on the floor will have even numbers up?

OK, its not about Mathematica, but let's make it about using Mathematica to visualize the result, just for fun.
Analytic
The first run produces some number of nE even numbers with probability 
pe0 = PDF[BinomialDistribution[99, 1/2], nE]

Any number from 0 to 99 is possible, justifying the last rule.

Irrespective of the initial distribution, on equilibrium the probability of changing an even number for an odd number must be equal to changing from an odd to an even
Solve[pe == (1 - pe)/2, pe]


{{pe -> 1/3}}


That means in average 1/3 even, 2/3 odd.
Simulation
start = RandomChoice[Range[6], 99];
BarChart[Apply[Labeled, Reverse[Sort@Tally[start], 2], {1}]]




f = Block[
   {rndIndx = RandomInteger[{1, 99}], new},
   new = If[EvenQ@Part[#, rndIndx], 1, RandomInteger[{1, 6}]];
   ReplacePart[#, rndIndx -> new]
   ] &

reslist = NestList[f, start, 100000];
evol = Transpose[(#/Total[#]) &[Part[Sort@Tally[EvenQ[#]], All, 2]] & /@ reslist];
ListLogLinearPlot[evol, Joined -> True, Epilog -> {Line[{{0, 1/3}, {100000, 1/3}}], Line[{{0, 2/3}, {100000, 2/3}}]},
 PlotRange -> {{1, 100000}, {0, 1}},
 Frame -> True
 ]


Analysis
As we iterate there will be random fluctuations continuously changing the number of odd and even numbers. The probability of creating a 1 is bigger than any other number. After a couple thousand iterations most of the numbers are 1
Sort@Tally[Last@reslist]


{{1, 37}, {2, 9}, {3, 10}, {4, 17}, {5, 15}, {6, 11}}


 BarChart[Apply[Labeled, Reverse[%, 2], {1}]]




And the distribution of even and odd numbers is centred around
N[Mean /@ evol]


{0.666522, 0.333478}


or 66.65% Odd, and 33.35% even, very close to 2/3 and 1/3 predicted.
The variation about that average is considerable:
N[StandardDeviation /@ evol]


{0.048476, 0.048476}


so the answer can not be given as a specific number, but as a distribution.
SmoothHistogram[evol, 0.01, Frame -> True, 
 Epilog -> {Line[{{1/3, 0}, {1/3, 9}}], Line[{{2/3, 0}, {2/3, 9}}]}, 
 PlotLabel -> "Probability Distribution", 
 FrameLabel -> {"Proportion", "Probability density"}]




Conclusion
After a not so long time, provided a couple of thousand iterations have pass, with more than 95% probability you will find a pile with roughly between 57% to 76% odd numbers.
Only the average number of odds tends to 2/3 in the limit where the samples goes to infinity. The exact number of a particular instance can not be determined, but the most likely outcome will be 33 even, 66 odds of which a one (1) will be chosen to be rolled again. 

