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?

Using Mathematica:
Clear[a, n, m, even];

For n dice, half are initially even and become odd, the other half are flipped and half of those flipped end up even. So 1/4 are even after the first iteration.
even[1] = n/4;

For each iteration, all of evens become odd and half of the remainder are even after their flip.
even[m_] := even[m] = (n - even[m - 1])/2;

Looking at the tailing end of 100 iterations
Take[Table[even[m], {m, 1, 100}], -3] // N


{0.333333 n, 0.333333 n, 0.333333 n}

This indicates that the limit is tending to n/3.
Clear[even]

even[m_] = 
 a[m] /. RSolve[{a[m] == (n - a[m - 1])/2, a[1] == n/4}, a[m], m][[1]] // 
  ExpandAll


(1/3)*(-1)^(2*m)n + (1/3)(-1)^m*
       2^(-1 - m)*n

Since m is a positive integer, even[m] can be restructured to n/3 (1 + (1/2)*(-1/2)^m)
even[m] == n/3 (1 + (1/2)*(-1/2)^m) //
 Simplify[#, {Element[m, Integers], m > 0}] &


True

Limit[n/3 (1 + (1/2)*(-1/2)^m), m -> Infinity]


n/3

For n = 99, after an infinite number of iterations the number of dice with even numbers up is statistically 33.
