CC BY
2Contributions to Game Theory and Management, X, 226—232
Blotto Games with Costly Winnings
Irit Nowik1 and Tahl Nowik2
1 Department of Industrial Engineering and Management, Lev Academic Center, P.O.B 16031, Jerusalem 9116001, Israel E-mail: nowikSSj ct.ac. il 2 Department of Mathematics, Bar-Ilan University, Ramat-Gan 5290002, Israel E-mail: tahl@Smath.biu.ac.il URL: www.math.biu.ac.il/ tahl
Abstract We introduce a new variation of the stochastic asymmetric Colonel Blotto game, where the n battles occur as sequential stages of the game, and the winner of each stage needs to spend resources for maintaining his win. The limited resources of the players are thus needed both for increasing the probability of winning and for the maintenance costs. We show that if the initial resources of the players are not too small, then the game has a unique Nash equilibrium, and the given equilibrium strategies guarantee the given expected payoff for each player.
1. Introduction
We present a new n-stage game, which is a variation of the Colonel Blotto game. Each player starts the game with some given resource, and at the beginning of each stage he must decide how much resource to invest in that stage. A player wins the given stage with probability corresponding to the relative investments of the players, and if both players invest 0 then no player wins that stage. The winner of the stage receives a payoff which may differ from stage to stage. Since it is possible that certain stages will not be won by any player, this is not a fixed sum game.
The players' resources from which the investments are taken can be thought of as money, whereas the payoffs should be thought of as a quantity of different nature, such as political gain. The two quantities cannot be interchanged, that is, the payoff cannot be converted into resources for further investment.
The new feature of our game is the following. The winner of each stage is required to spend additional resources on the maintenance of his winning. This is a real life situation, where the winnings are some assets, and resources are required for their maintenance, as in wars, territorial contests among organisms, or in the political arena. The winner of a given stage must put aside all resources that will be required for future maintenance costs of the won asset. Thus, a fixed amount will be deducted from the resources of the winner immediately after winning, which should be thought of as the sum of all future maintenance costs for the given acquired asset.
At each stage the player thus needs to decide how much to invest in the given stage, where winning that stage on one hand leads to the payoff of the given stage, but on the other hand the maintenance cost for the given winning negatively affects the probabilities for future winnings. In the present work we show that if the initial resources of the players are not too small then the game has a unique Nash equilibrium, and each player guarantees the payoff of this Nash equilibrium (Theorem 2.)
As mentioned, our game presents a variation of the well known Colonel Blotto game (Borel, 1921). In Blotto games two players simultaneously distribute forces across several battlefields. At each battlefield, the player that allocates the largest force wins. The Blotto game has been developed and generalized in many directions (see e.g., Borel, 1921; Friedman, 1958; Lake, 1979; Roberson,2006; Hart, 2008; Duffy and Matros, 2015). Two main developments are the "asymmetric" and the "stochastic" models. The asymmetric version allows the payoffs of the battlefields to differ from each other, and in the stochastic model the deterministic rule deciding on the winner is replaced by a probabilistic one, by which the chances of winning a battlefield depends on the size of investment.
The present work adds a new feature which changes the nature of the game, in making the winnings costly. The players thus do not know before hand how much of their resources will be available for investing in winning rather than on maintenance, and so the game cannot be formulated with simultaneous investments, as in the usual Blotto games, but rather must be formulated with sequential stages. At each stage the players need to decide how much to invest in the given stage, based on their remaining available resources and on the future fees and payoffs.
This work was inspired by previous work of the first author with S. Zamir and I. Segev ( Nowik, 2009; Nowik et al., 2012) on a developmental competition that occurs in the nervous system, which we now describe. A muscle is composed of many muscle-fibers. At birth each muscle-fiber is innervated by several motorneurons (MNs) that "compete" to singly innervate it. It has been found that MNs with higher activation-threshold win in more competitions than MNs with lower activation thresholds. In Nowik, 2009 this competitive process is modeled as a multi-stage game between two groups of players: those with lower and those with higher thresholds. At each stage a competition at the most active muscle-fiber is resolved. The strategy of a group is defined as the average activity level of its members and the payoff is defined as the sum of their wins. If a MN wins (i.e., singly innervates) a muscle-fiber, then from that stage on, it must continually devote resources for maintaining this muscle-fiber. Hence the MNs use their resources both for winning competitions and for maintaining previously acquired muscle-fibers. It is proved in Nowik, 2009 that in such circumstances it is advantageous to win in later competitions rather than in earlier ones, since winning at a late stage will encounter less maintenance and thus will negatively affect only the few competitions that were not yet resolved. If ^ is the cost of maintaining a win at each subsequent stage, then in the terminology of the present work, the fee payed by the MNs for winning the kth stage of an n stage game is (n — k
2. The game
The initial data for our game is the following.
1 The number n of stages of the game.
2 Fixed payoffs wk > 0, 1 < k < n, to be received by the winner of the kth stage.
3 The initial resources A, B > 0 of players I,II respectively.
4 Fixed fees ck > 0, 1 < k < n — 1, to be deducted from the resources of the winner after the kth stage.
The rules of the game are as follows. At the kth stage of the game, the two players, which we name PI,PII, each has some remaining resource Ak, Bk, where
Ai = A, Bi = B. PI,PII each needs to decide his investment xk,yk for that stage, respectively, with 0 < xk < Ak — ck, 0 < yk < Bk — ck, and where if Ak < ck then PI may only invest 0, and similarly for PII. These rules ensure that the winner of the given stage will have the resources for paying the given fee ck. The probability for PI,PII of winning this stage is respectively —^— and —^—, where if xk = yk = 0 then no player wins. The resource of the winner of the kth stage is then reduced by an additional ck, that is, if PI wins the kth stage then Ak+i = Ak — xk — ck and Bk+i = Bfc — yfc, and if PII wins then Afc+i = Afc — xfc and Bfc+i = Bfc — yfc — cfc. The role of ck is in determining Ak+i,Bk+i, thus there is no cn. It will however be convenient in the sequel to formally define cn =0. The payoff received by the winner of the kth stage is wk. Since it is possible that no player wins certain stages, this game is not a fixed sum game.
As already mentioned, the resource quantities Ak, Bk ,xk, yk, ck used for the investments and fees are of different nature than that of the payoffs wk. The two quantities cannot be interchanged and should be thought of as having different "units". Note that all expressions below are unit consistent, that is, if say we divide resources by payoff, then such expression has units of r6p°^ges, and may only be added or equated to expressions of the same units.
If A and B are too small in comparison to ci,..., cn-i then the players' strategies are strongly influenced by the possibility of running out of resources before the end of the game. In the present work we analyze the game when A, B are not too small. Namely, we introduce a quantity M depending on ci,..., cn-i and wi,..., wn, and prove that if A, B > M then there is a unique Nash equilibrium for the game, and each player guarantees the value of this Nash equilibrium.
For k = 1,..., n let Wk = Y,n=k W and W = Wi. We now show that if A > M, then if PI always chooses to invest xj~ < -p^^fc (as holds for our strategy an,A,B presented in Definition 1 below), then whatever the random outcomes of the game are, his resources will not run out before the end of the game. We in fact give a specific lower bound on Ak for every k, which will be used repeatedly in the sequel.
Proposition 1. Let
M = W ■ max ( — + V |.
i<fc<n y Wk Wj+i J
If A> M, and if PI plays xk < ^^ for all k, then Ak > for all 1 < k < n. In particular Ak > 0 for all 1 < k < n. And similarly for PII.
Proof. For every KKnwe have £ > f > ^ + J2-=i so
. k-i A v Cj
> ~k
i=i
Thus it is enough to show that > ^ — yh=i wi+l f°r all 1 < A; < n. We show this by induction on k. For k = 1 the sum is empty and we get equality. Assuming
k-i
Ak A \ Cj Wfc -
j=i 1
we get
Afc+i 1 wk Ak \ 1 / Wfc+iAfc
> 777- Ak--777--ck = 777- -777--ck
Wk+1~ Wk+i\ Wk V Wk+i\ Wk
k — 1
\ q A
—- / — — > -—
Wk Wk+1- W ^ Wi+1 Wk+1 W = Wi+1 • 3. Nash equilibrium
We define the following two strategies an,A,B and Tn,A,B for PI,PII respectively. We prove that for A, B > M as given in Proposition 1, this pair of strategies is a unique Nash equilibrium, and these strategies guarantee the given payoffs.
Definition 1. At the kth stage of the game, let
wk Ak Ak ck WkBk Bk ck
ak = —rrz-----— and bk =
Wk Ak + Bk Wk Ak + Bk
where as mentioned, we formally define cn = 0. The strategy an,A,B for PI is the following: At the kth stage PI invests ak if it is allowed by the rules of the game. Otherwise he invest 0. The strategy Tn,A,B for PII is similarly defined with bk.
Recall that ak = 0 is allowed by the rules of the game if 0 < ak < Ak — ck, whereas ak = 0 is always allowed, even when Ak — ck < 0. We interpret the quantities ak, bk as follows. PI first divides his remaining resource Ak to the remaining stages in proportion to the payoff for each remaining stage, which gives -^¡r-Ak. From this he subtracts ck which is the expected fee he will pay for this stage, since
ak — —^—. Note that Wn = wn and formally c„ = 0, so an = An, bn = Bn
ak+bk ~ Ak+Bk - "n - -n '"""""J - "> - "n, "n - "n,
i.e. at the last stage the two players invest all their remaining resources.
Depending on A and B and on the random outcomes of the game, it may be that PI indeed reaches a stage where ak is not allowed. In this regard we make the following definition.
Definition 2. The triple (n, A, B) is Pi-effective if when PI and PII use an,A,B and Tn,A,B, then it is impossible that they reach a stage where ak is not allowed for PI. Similarly PII-effectiveness is defined for PII with bk.
Proposition 2. Let M be as in Proposition 1. If A > M and B is arbitrary, then (n, A, B) is Pi-effective. Furthermore, ak > 0 for all k. And similarly for PII when B > M.
Proof. We need to show that necessarily 0 < ak < Ak — ck for all 1 < k < n. We have ak = ^ - < so by Proposition 1, ^ > % > ^ and
Ak>0,so^> giving ak > 0.
For the inequality ak < Ak — ck we first consider k < n — 1. We have from the proof of Proposition 1 that ^ - ^ > A - £-=i ^ > ^L > 0, so
It >W^7'and so
,, Ak Wk+1Ak wk
ih)Ck^Ck<^r = {l-wk)Ak-
This gives ck - <Ak-^,so ak = ^- ^ < Ak - ck. For k = n
we note that cn = 0 by definition, and Wn = wn, so an = An = An — cn.
c
In general, an inductive characterization of PI-effectiveness will also involve induction regarding PII. But if we assume that B > M, and so by Proposition 2 all bk are known to be allowed and positive, then the notion of PI-effectiveness becomes simpler, and may be characterized inductively as follows. When saying that a triple (n — 1,A',B') is PI-effective, we refer to the n — 1 stage game with fees C2,..., cn-i and payoffs w2,..., wn. Starting with n =1, (1, A, B) is always PI-effective. For n > 2, if ai is not allowed then (n, A, B) is not PI-effective. If ai =0 then it is allowed, and PI surely loses the first stage, and so (n, A, B) is PI-effective iff (n — 1, A, B — bi — ci) is PI-effective. Finally if ai > 0 and it is allowed then (n, A, B) is PI-effective iff both (n—1, A—ai — ci, B—bi) and (n—1, A—ai, B—bi — ci) are PI-effective.
The crucial step in proving Theorem 2 below, on the unique Nash equilibrium and the guaranteed payoffs, is the following Theorem 1. We point out that in Theorem 2 we will assume that A > M, in which case (n, A, B) is PI-effective, by Proposition 2. But here in Theorem 1 we must consider arbitrary A > 0 in order for an induction argument to carry through.
Theorem 1. Given ci,. .., cn-i and wi,. .., wn let M be as in Proposition 1, and assume that B > M and PII plays the strategy rnjA,B. For A > 0, if (n, A, B) is PI-effective, and PI plays according to an,A,B, then his expected payoff is . On the other hand, if (n, A, B) is not PI-effective, or if PI uses a different strategy, then his expected payoff is strictly less than .
Proof. By induction on n. We note that throughout the present proof we do not use the condition B > M directly, but rather only through the statements of Propositions 2 and 1 saying that (n,A,B) is PH-effective, bk > 0 and ck < w^kk for all 1 < k < n, which indeed continue to hold along the induction process.
If A = 0 then ak = 0 for all k, which is the only possible investment, and its payoff is 0 = ^b i so the statement holds. We thus assume from now on that A > 0. For n = 1 we have bi = B. The allowed investment for PI is 0 < s < A with expected payoff j^-gwi = j^W which indeed attains a strict maximum A^B W at s = A = ai.
For n > 2, let s be the investment of PI in the first stage. Assume first that s = 0. In this case PII surely wins the first stage and so following this stage we have A2 = A and B2 = B — bi — ci. The moves for PII dictated by t„ a,B for the remaining n — 1 stages of the game are t„-1ja,B-&i-ci, and so by the induction hypothesis the expected total payoff of PI is at most ■ Since Proposition I holds for
PII, we have ci < ^ < , that is, ^ - > 0, and since A > 0 we get
a\ = — a+b ) > This means that s = 0 a\, so we must verify the strict
inequality < ■ This is readily verified, using A > 0, c\ <
W2 = W-wu and = = ^ +
We now assume s > 0. This is allowed only if A > c1 and 0 < s < A — c1. The moves for PII dictated by t„ a,B for the remaining n — 1 stages of the game are Tn-1,A2,B2. By the induction hypothesis, if PI wins the first stage, which happens with probability -^j- > 0, then his expected payoff in the remaining n — I stages
of the game is at most • Similarly, if he loses the first stage, which
happens with probability > 0, then his expected payoff in the remaining n — 1
stages is at most A+B-s-bi-ci ■ Thus, the expected payoff of PI for the whole n
stage game is at most F (s), where
F(s) = TTT- + TT-s—:—i—r + '
s + A + B — s — b1 — c1 J s + b1 A + B — s — b1 — c1
wifli h- — W1B -Bci with bi - -jy- - :5q7g.
By the induction hypothesis we know furthermore, that in case PI wins the first
stage, he will attain the maximal expected payoff a+b-s-I^ci ™ remaining
stages of the game only if (n — 1, A — s — c1, B — b1) is PI-effective, and he uses
^n-1,A-s-ci,B-6i. Similarly, if he loses the first stage, he will attain the maximal
expected payoff only if (n — 1, A — s, B — b\ — ci) is Pi-effective and
he uses ffn-1jA-s,B-61-c1. If not, then since both alternatives occur with positive
probability, his expected total payoff for the whole n stage game will be strictly less
than F(s).
To analyze F(s), we make a change of variable s = a1 + x, that is, we define F(x) = F{a\ + x) = F— + x). After some manipulations we get:
N AW BW3x2
b (x)
A + B (A + B)(V2(A + B) - W^Wi - Wei + wi(A + B)j
Under this substitution, s = a1 corresponds to x = 0, and the allowed domain 0 < s < A — e1 corresponds to
Ac1 w1A W2A Be1
< x <
- - <L I. \ - -
A + B W " W A + B
Using c\ < one may verify that in the above expression for F the two linear factors appearing in the denominator of the second term are both strictly positive in this domain. It follows that F in the given domain is at most ^b > an<^ this maximal value is attained only for x = 0 (if it is in the domain), which corresponds to s = a1 for the original F. Finally, as mentioned, unless (n — 1,A2,B2) is PI-effective and PI plays an-1jA2,B2, his expected payoff will be strictly less than F(s), and so strictly less than .
We may now prove our main result.
Theorem 2. Given C1,. .., cn-1 and W1,. .., wn, let M be as in Proposition 1, and assume A, B > M. Then the pair of strategies an,A,B , Tn,A,B is a unique Nash equilibrium for the game, with expected total payoffs j^g, a+^b ■ Furthermore, Cn,A,B o,nd, <Jn,A,B guarantee the expected payoffs and .
Proof. Denote a0 = an,A,B and t0 = Tn,A,B, and for any pair of strategies a, t let S1(a, t), S2(a, t) be the expected payoffs of PI, PII respectively. We first prove the second statement of the theorem. Recall that if both players invest 0 in a given stage then there is no winner to that stage. However, if B > M and PII plays t0, then by Proposition 2 we have bk > 0 for all k, and so indeed there is a winner to each stage of the game, and thus the total combined payoff of PI and PII is necessarily W. It thus follows from Theorem 1 that for any strategy a of PI we have s2(cr, r0) = w - si(a, t0) > Similarly, if a > m then S^cto, t) > ^ for all T, establishing the second statement of the theorem.
As to the first statement, Theorem 1 applied to both PI and PII implies that the pair a0, t0 is a Nash equilibrium with the given expected payoffs. To show it is unique we argue as follows. Let a, t be any other Nash equilibrium and assume that a = a0. By Theorem 1 we have Si (a, t0) < S1(a0, t0 ) and since playing t0 guarantees a combined total payoff of W, we have S2 (a, t0 ) = W — S1 (a, t0 ) > W — S1 (a0, t0 ) = S2(a0, t0). Since the pair a, t is a Nash equilibrium we also have S2(a, t) > S2(a, t0), and together we get S2(a, t) > S2(a0,T0). Since S1(a, t) + S2(a, t) < W and S1(a0, t0) + S2(a0, t0) = W, we must have S1(a, t) < S1(a0, t0). Again since a, t is a Nash equilibrium we have S1(a0,T) < S1(a, t) so together S1 (a0,T) < S1(a0,T0) = -^p^r, contradicting the conclusion of the previous paragraph.
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