EQUIDISTRIBUTION IN SHARING GAMES CARLOS D’ANDREA (UNIVERSITY OF BARCELONA)

Games 
often provide a good introduction to interesting phenomena in mathematics. In 
this note, we examine three variations of an iterative sharing game played 
around a circular (or not so circular) table. More precisely, for each variation, we 
study the tendency toward equal distribution among the players. In 
the first variation, the players have discrete amounts at 
each step. The second variation removes this restriction, and the third one 
considers an infinitely long table with an infinite number of players.

Games often provide a good introduction to interesting phenomena in mathematics.In this note, we examine three variations of an iterative sharing game played around a circular (or not so circular) table.More precisely, for each variation we study the tendency toward equal distribution among the players.In the first variation, the players have discrete amounts at each step.The second variation removes this restriction, and the third one considers an infinitely long table with an infinite number of players.

Sharing around a table with discrete amounts
Suppose you have n persons seated around a circular table, each having an even number of dimes.Number them from 1 to n counterclockwise, and denote by a i the (even) number of dimes that the i-th person has, i = 1, . . ., n.
The game is played by repeating the same two steps over and over, as follows: first, each person gives one half of her dimes to the person sitting on her right.In symbols, (1) with the convention a 0 = a n .Note that the new a i 's are not all necessarily even.The second step, which allows us to repeat (1), is to add one dime to each odd a i : (2) a i + 1 − (−1) a i 2 → a i , i = 1, . . ., n.
We call this game the "discrete sharing game".Claim 1.By iterating (1) and (2), after a finite number of steps we will have a 1 = a 2 = . . .= a n , i.e. every person ends up with the same number of dimes.
This problem can be found in [1].The proof follows easily from two observations.First, the maximum amount around the table cannot increase, nor can the minimum amount decrease.Second, if the minimum amount is strictly smaller than the maximum amount, then after the next step, either the minimum amount will have increased, or else the number of players having the minimum amount will have decreased.Together, these observations imply that after finitely many steps, the maximum and minimum amounts will coincide.We leave the details of the proof to the reader.Another interesting problem is to determine the number of iterations needed to reach equidistribution in terms of the initial data (see the figures below).An iterative sharing game that tends toward equal distribution of the players' amounts is what we call an "equidistribution process".You may think that in the above example, the equidistribution is due to the fact that the sharing is done by halves.However, Claim 1 generalizes without much effort: suppose all the a i 's are initially multiples of a fixed number k.

Replace (1) with
and (2) with a similar step where you add dimes until you reach the first multiple of k greater than or equal to a i for all i = 1, . . ., n.Then again you will end up with a 1 = a 2 = . . .= a n after a finite number of steps.

Sharing around a table with complex numbers
Suppose there are n persons around a circular table, numbered counterclockwise as before.But this time each of them starts with an initial amount a i , i ≤ 1 ≤ n, that can be any complex number.Let r be a positive real  k above.We want to study the behavior of (1 − r)a i + ra i−1 → a i , i = 1, . . ., n, as the number of iterations goes to infinity.Note that the balancing step (2) does not make sense here anymore, as neither the a i 's are necessarily integers nor r = 1 k for some k ∈ N. We can give a realistic feeling to the game if we restrict the a i 's to real numbers.Then at each step, each person is sharing her wealth (a i > 0) or debt (a i < 0) by giving a portion r of her number to the person on her right.But this restriction is unnecessary, so we will work in the more general setting of complex numbers.In contrast to the discrete version of the last section, we call this the "complex sharing game".For = 0, 1, 2, . .., let a i be the amount that person i has after the -th step.Then a 0 i = a i , and we also extend the definition by setting a i = a j whenever i ≡ j mod n, so that a i makes sense for any integer i.The rule for sharing at each step then yields where the sub-indices are understood modulo n.
, clearly this sharing game with only two people is an equidistribution process.This is particularly trivial if r = 1  2 , in which case both persons will have the same number after just one step.From now on, we assume n > 2. Note also that the sum of the numbers around the table remains constant with each step.Let us denote this constant by S, that is 2.1.The sharing transformation and its eigenvalues.We can model the sharing process (3) with the aid of the C-linear transformation T r : With the notation as above, we see that (a 1 , . . ., a n ) = T r (a 1 , . . ., a n ), where the exponent on T r indicates the number of times that this transformation is composed with itself: In the canonical basis of C n , T r has matrix , which has entries 1 − r along the main diagonal, r on the diagonal immediately below and at the upper right corner, and zero elsewhere.This is a "stochastic" or "Markov" matrix, as its entries are all non-negative and the sum of the entries in each row is equal to 1.For more information on these matrices and their interesting properties, we refer the reader to [2].We will presently prove some of these properties for our matrix A r in order to make this note more self-contained, but the reader should be aware that both Claim 2 and Proposition 1 below hold for any stochastic matrix.
Claim 2.The matrix A r has n distinct eigenvalues Proof.Let f r (t) be the characteristic polynomial of T r .Computing it explicitly, we obtain Let ω 1 , . . ., ω n be the nth distinct complex roots of r n .Then it is easy to see that the n (distinct) roots of f r are If we set ω 1 = r, then we have that λ 1 = 1.For i = 2, . . ., n, we have The inequality is strict, as ω i is not a positive real number if i > 1.
An eigenvector associated to the eigenvalue 1 is v 1 = (1, 1, . . ., 1).In our game, this eigenvector corresponds to the case where every person has the same amount (a i = S/n, i = 1, . . ., n).Clearly, this situation is stable, i.e. does not change after applying T r .Let v 2 , . . ., v n denote eigenvectors for the eigenvalues λ 2 , . . ., λ n respectively.Thus T r (v i ) = λ i v i for each i.Now we are ready for the main result of this section, which is that the complex sharing game also tends toward equal distribution among the n players.Proposition 1.The complex sharing game described by (3) is an equidistribution process.That is, lim →∞ (a 1 , . . ., a n ) = S n , . . ., S n .
Proof.Note that {v 1 , . . ., v n } is a basis for C n consisting of eigenvectors of Therefore the game is indeed an equidistribution process.Note of course that, since the sum of the n numbers remains constant at each step, we must have 2.2.An explicit formula for a i .Recall that we defined a i = a j , and in particular a i = a j , whenever i ≡ j mod n.Let us find a formula, in terms of the initial data a 1 , . . ., a n , for the number that each person will have after any given step of the complex sharing game: Claim 3.For any i ∈ Z and ∈ Z ≥0 , we have Proof.We use induction on , the initial step being obvious.Next, assume the formula holds for all i ∈ Z and for all 0 with ≥ 0 ≥ 0, for a fixed ≥ 0.Then, +1 j r j (1 − r) +1−j a i−j , so the formula holds for + 1 and the proof is complete.
In particular, when each person shares one half of her amount at each step, we have 2 , then a i = 1 2 j=0 j a i−j for all i ∈ Z and ≥ 0. By grouping the binomial coefficients j in congruence classes of j modulo n, we obtain an equivalent formulation of this result (always for r = 1  2 ): where each congruence is understood to be modulo n.If all the initial amounts a 1 , . . ., a n are equal, the sharing process is stable from the beginning, and this formula reduces to the well known fact about the sum of the entries in the -th row of the Pascal triangle, j=0 j = 2 .What is more interesting is that, in light of Proposition 1, lim →∞ a i = S n for all i, regardless of the initial amounts.So given n, if we group the binomial coefficients j by the congruence classes of j modulo n and give each class an arbitrary weight, then as we go further down the rows of the Pascal triangle, the weighed sum of the entries divided by the sum with no weights, 2 , always tends to the average of the weights.More generally, regardless of the value of r, Proposition 1 shows that the sums given in the statement of Claim 3 all tend to the same limit S n as → ∞.

Sharing at an infinite table
Suppose now that, instead of a circular table, we have an infinite number of persons seated on one side of an infinitely long "rectangular" table , their number being unbounded both from the left and the right.Each person has an initial (complex number) amount to be shared in the same way as in the game with the circular table: at each step, each person gives a portion r of her amount to the person on her right.As before, r is a fixed positive real number, 0 < r < 1.We call this version the "infinite sharing game".
The initial data now is a sequence (a i ) i∈Z of complex numbers.We define a i as in (3) for i ∈ Z and ≥ 0. Note that this time there is no "congruence modulo n", as there are infinitely many players, indexed by the integers.Indeed, the rectangular table can be regarded as the limit of circular tables having n → ∞.The recursion from sharing is the same as before: Claim 3 also holds here, the proof is the same.The only difference is that now we do not have a i = a j whenever i ≡ j mod n, the condition that effectively made the game circular (and with n players) before.So we still get (5) This formula shows something that is obvious in the new context, namely that the amount of the person in the i-th place will depend only on the initial amounts of those people seated to her left (and on her own initial amount, of course).So there is no reason to believe that this new game should be an equidistribution process.For instance, if there is a concentration of wealth (by which we mean numbers with large absolute value) in some section of the table, it will never spread to the left.However, as the following proposition shows, if one of the sequences (a i ) ≥0 converges, then the game will tend toward equal distribution.
Proposition 2. If lim →∞ a i exists for some i ∈ Z, then lim →∞ a j exists for all j ∈ Z, and all of these limits coincide.
Proof.Let i ∈ Z be such that L i = lim →∞ a i exists.Then, using (4), we have lim i.e.L i−1 exists and it is equal to L i .By induction, we have then that L j exists and L j = L i for all j ≤ i.Now, using (4) again, (6) for all ≥ 0. Since a i → L i as → ∞, given ε > 0 there exists 0 ∈ N such that |a i − L i | < ε 2 for all ≥ 0 .But then we have that for k ≥ 1, (7) ε 2 thanks to (6), and which, together with (7), shows that L i = lim →∞ a i+1 = L i+1 , and from here we deduce (again by induction) that L j = lim →∞ a j = L i for all j > i.
It would be an interesting problem to characterize those sequences (a i ) i∈Z of initial data that induce an equidistribution process in the infinite rectangular table.Note for instance that if f : Z → C is any periodic function and we set a i = f (i) for i ∈ Z, then the game will be an equidistribution process (this is essentially the complex game around a circular table, where the number of players is the period of f , "lifted" to the infinite rectangular table ).One natural question to ask is this: if the initial data is bounded, must the infinite game be an equidistribution process?Suppose that |a i | ≤ C for all i ∈ Z.Then, since (8) we see from (5) that |a i | ≤ C for all i ∈ Z and ≥ 0. Does this imply that the sequence (a i ) ≥0 converges for some (and hence all) i ∈ Z? The answer is NO, as the following example shows.
Example 1. Suppose r ≤ 1 2 , and set This sequence can be defined recursively by first setting all its values equal to zero for i ≥ 0, then a −1 = 1, and from there on we move to the left two zeroes, then three ones.Every time we end with a list of ones, we put a sequence of zeroes doubling in number the sequence already built so far (starting from a −1 ).And every time we end with a list of zeroes, we put a sequence of ones equal to the number of the sequence already built so far (starting from a −1 ), always moving to the left.So, the sequence looks like this (starting from a −1 on the right): . . ., 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1 With this sequence as the initial data and (5), one can show that for any positive integer k, so the sequence (a −1 ) ≥0 does not converge and, in light of Proposition 2, (a i ) ≥0 does not converge for any i ∈ Z.
An immediate consequence of Proposition 2 is that if there exist j ∈ Z and a ∈ C such that a i = a for all i < j, then the infinite game is an equidistribution process.This is, in fact, a particular case of a more general fact: it turns out that equidistribution follows from the existence of lim i→−∞ a i (and not, as we just saw, from the boundedness of (a i ) i∈Z ).
Proposition 3. Suppose that lim i→−∞ a i = a ∈ C. Then lim →∞ a i = a for every i ∈ Z.
Proof.Consider the sequence (b i ) i∈Z defined as b i = a i − a for all i ∈ Z.For a fixed i, due to (4), it is easy to see that lim →∞ a i = a if and only if lim →∞ b i = 0, so we can assume without loss of generality that lim i→−∞ a i = 0.For a fixed i, since the sequence (a i−j ) j∈N converges, there exists C > 0 such that |a i−j | < C for all j ≥ 1.Let ε > 0 be given.Since a i− → 0, there exists 0 ∈ N such that |a i− | < ε 2 if > 0 .For > 0 , we use (5) to write (9) a i = 0 j=0 j r j (1 − r) −j a i−j + j= 0 +1 j r j (1 − r) −j a i−j .
As for the first summand of (9), we have: which proves that lim →∞ a i = 0.
When the initial amounts are all real, (a i ) i∈Z ⊂ R, one can similarly show that if lim i→−∞ a i = +∞ (resp.−∞), then a i → +∞ (resp.−∞) as → ∞, for every i ∈ Z.Note that the existence of lim i→−∞ a i is sufficient but not necessary for the infinite game to be an equidistribution process.This can be seen from the above example where the initial data take on the values of a periodic function f : Z → C. As mentioned above, it would be interesting to find more explicit necessary and sufficient conditions than the existence of lim →∞ j=0 j r j (1 − r) −j a i−j for some i ∈ Z, which is essentially the characterization we get from Proposition 2 and (5).

Figure 1 .
Figure 1.3-D graph of the discrete sharing game with 10 players and with initial distribution as shown.In this example, equidistribution is achieved at the 16-th iteration.

Figure 2 .
Figure 2. 2-D graph of the game given in Figure 1.This graph does not show how the initial amounts are distributed around the table.

Figure 3 .
Figure 3. 2-D graph of a generalized discrete sharing game with 20 players and k = 3.

Figure 4 .
Figure 4. Evolution of the complex sharing game with 20 players and r = 3 4 .The initial amounts are integers between 0 and 10.

Figure 5 .
Figure 5. Evolution of the complex sharing game with 50 players and r = 1 2 .The initial amounts are integers between 0 and 50.