<p>The idea of a solvable game is probably inspired by the theory of two-player zero-sum games. Such games are known to be solvable; a simple consequence of the <a href="https://en.wikipedia.org/wiki/Minimax_theorem" rel="noreferrer">minimax theorem</a>.</p>
<p>One way to think about solvability is that using equilibrium points as a solution concept does not tell a player how they should behave—for this they would need to know which equilibrium point is played. If the game is solvable, this issue does not arise.</p>
<p>An undesirable feature of equilibrium points in mixed strategies is that a player playing a mixed best response will always be indifferent among all pure strategies they are randomizing over and, consequently, all mixtures of these pure strategies. Nevertheless, equilibrium points may require a player to randomize in a very special way. In a strong solution, all possible best replies must be included and the mentioned problem does not arise.</p>
<p>Finally, the issue with the definition of a subsolution comes from a typo Nash made, presumably while turning his thesis into a paper. The published paper refers to a condition (1) that is actually the definition of an equilibrium point. <a href="https://library.princeton.edu/special-collections/sites/default/files/Non-Cooperative_Games_Nash.pdf" rel="noreferrer">Nash's original thesis</a> also refers to (1) in the same place but (1) refers to something different there.</p>
<p>The correct definition of a subsolution is that it is a maximal set <span class="math-container">$S$</span> of equilibrium points satisfying the condition:
<span class="math-container">$$(t;r_{i})\in S\,\,and\,\,s\in S \implies (s;r_{i})\in S.$$</span></p>

CC BY-SA 4.0
2022-05-19T16:16:00.150Z
2022-05-19T16:16:00.150Z
Michael Greinecker
<p>I am now reading Nash's 1951 paper <a href="https://www.jstor.org/stable/1969529?seq=1" rel="nofollow noreferrer">Non-cooperative games</a> and I am trying to understand the intuition behind solution concepts.</p>
<blockquote>
<p><strong>Solutions(Nash 1951)</strong></p>
<p>A game is <strong>solvable</strong> if its set <span class="math-container">$S$</span> of equilibrium points satisfies the condition for all <span class="math-container">$i$</span>'s <span class="math-container">$$(t;r_{i})\in S\,\,and\,\,s\in S \implies (s;r_{i})\in S\tag{1}$$</span>
Here <span class="math-container">$t$</span> and <span class="math-container">$s$</span> are <span class="math-container">$n$</span>-tuple strategy profile. Additionally, if <span class="math-container">$t=(t_{1},t_{2},\cdots,t_{n})$</span>, then <span class="math-container">$(t;r_{i})=(t_{1},\cdots,t_{i-1},r_{i},t_{i+1},\cdots)$</span>. I.e <span class="math-container">$(t;r_{i})$</span> is the strategy profile with player <span class="math-container">$i$</span>'s strategy now being <span class="math-container">$r_{i}$</span>.</p>
<p>The solution of a solvable game is its set <span class="math-container">$S$</span> of equilibrium points.</p>
<p>A game is <strong>strongly solvable</strong> if it has a solution <span class="math-container">$S$</span> s.t for all <span class="math-container">$i$</span>'s <span class="math-container">$$s\in S\,\,and\,\, p_{i}(s;r_{i})=p_{i}(s)\implies (s;r_{i})\in S\tag{2}$$</span>
Here <span class="math-container">$p_{i}(s)=\max_{r_{i}}p_{i}(s;r_{i})$</span>, where <span class="math-container">$p_{i}$</span> is the payoff for player <span class="math-container">$i$</span>. i.e <span class="math-container">$p_{i}(s)$</span> is the maximal payoff for player <span class="math-container">$i$</span> given the opponents playing <span class="math-container">$s_{-i}$</span>.</p>
<p><strong>Sub-solutions:</strong> If <span class="math-container">$S$</span> is a subset of the set of equilibrium points of a solvable game and for every <span class="math-container">$s \in S$</span>, the condition <span class="math-container">$p_{i}(s)=max_{r_{i}}p_{i}(s;r_{i})$</span> is satisfied, and if <span class="math-container">$S$</span> is maximal relative to this property, we call <span class="math-container">$S$</span> a sub-solution.</p>
<p><strong>Factor sets:</strong> For any sub-solution <span class="math-container">$S$</span> we define the <span class="math-container">$i$</span>-th factor set <span class="math-container">$S_{i}$</span> as the set of all <span class="math-container">$s_{i}$</span> s.t <span class="math-container">$(t;s_{i})\in S$</span> for some <span class="math-container">$t$</span>,</p>
<p><strong>Theorem 3:</strong> A sub-solution <span class="math-container">$S$</span> is the set of all <span class="math-container">$n$</span>-tuples <span class="math-container">$(s_{1},\dots,s_{n})$</span> s.t each <span class="math-container">$s_{i}\in S_{i}$</span> where <span class="math-container">$S_{i}$</span> is the <span class="math-container">$i^{th}$</span> factor set of <span class="math-container">$S$</span>. Geometrically, <span class="math-container">$S$</span> is the product of its factor sets.</p>
</blockquote>
<p>Here are my questions.</p>
<p><strong>1. Why do we need these solution concepts?</strong></p>
<p>I don't quite get the intuition behind these solution concepts. Moreover, since a solution may not exist in a game, it seems that the concept of equilibrium points can be applied to a broader class of games than these solution concepts. Additionally, it has been mentioned in the paper that if a solution exists, then it is unique. Then it seems that we don't need to differentiate between solutions and strong solutions because there can be only one solution (strong solution if it exists) and those conditions (1) and (2) will be trivially satisfied.</p>
<p><strong>2. The difference between sub-solution and equilibrium points</strong></p>
<p>I am a little confused about the difference between sub-solution and equilibrium points. From the definition of sub-solution, it seems that it is exactly the set which contains all the equilibrium points. However, based on Theorem 3, it is the Cartesian product of its factor sets. For example, suppose in a game where the set of all equilibrium points is <span class="math-container">$S=\{(U,L),(D,R)\}$</span>. Then based on the definition, the sub-solution should also be <span class="math-container">$S$</span>. But by theorem <span class="math-container">$3$</span>, the sub-solution should be <span class="math-container">$\{(U,L),(U,R),(D,L),(D,R)\}$</span>.</p>
<p>I really appreciate it if someone can give me some help on these. Thanks in advance!</p>

<game-theory>
Intuition of solution concepts in Non-cooperative games
John
# Intuition of solution concepts in Non-cooperative games

(Asked by: username_0 on 2022-05-19)

I am now reading Nash's 1951 paper [Non-cooperative games](https://www.jstor.org/stable/1969529?seq=1) and I am trying to understand the intuition behind solution concepts.

> **Solutions(Nash 1951)**
>
> A game is **solvable** if its set $S$ of equilibrium points satisfies the condition for all $i$'s $$(t;r\_{i})\in S\,\,and\,\,s\in S \implies (s;r\_{i})\in S\tag{1}$$
> Here $t$ and $s$ are $n$-tuple strategy profile. Additionally, if $t=(t\_{1},t\_{2},\cdots,t\_{n})$, then $(t;r\_{i})=(t\_{1},\cdots,t\_{i-1},r\_{i},t\_{i+1},\cdots)$. I.e $(t;r\_{i})$ is the strategy profile with player $i$'s strategy now being $r\_{i}$.
>
> The solution of a solvable game is its set $S$ of equilibrium points.
>
> A game is **strongly solvable** if it has a solution $S$ s.t for all $i$'s $$s\in S\,\,and\,\, p\_{i}(s;r\_{i})=p\_{i}(s)\implies (s;r\_{i})\in S\tag{2}$$
> Here $p\_{i}(s)=\max\_{r\_{i}}p\_{i}(s;r\_{i})$, where $p\_{i}$ is the payoff for player $i$. i.e $p\_{i}(s)$ is the maximal payoff for player $i$ given the opponents playing $s\_{-i}$.
>
> **Sub-solutions:** If $S$ is a subset of the set of equilibrium points of a solvable game and for every $s \in S$, the condition $p\_{i}(s)=max\_{r\_{i}}p\_{i}(s;r\_{i})$ is satisfied, and if $S$ is maximal relative to this property, we call $S$ a sub-solution.
>
> **Factor sets:** For any sub-solution $S$ we define the $i$-th factor set $S\_{i}$ as the set of all $s\_{i}$ s.t $(t;s\_{i})\in S$ for some $t$,
>
> **Theorem 3:** A sub-solution $S$ is the set of all $n$-tuples $(s\_{1},\dots,s\_{n})$ s.t each $s\_{i}\in S\_{i}$ where $S\_{i}$ is the $i^{th}$ factor set of $S$. Geometrically, $S$ is the product of its factor sets.

Here are my questions.

**1. Why do we need these solution concepts?**

I don't quite get the intuition behind these solution concepts. Moreover, since a solution may not exist in a game, it seems that the concept of equilibrium points can be applied to a broader class of games than these solution concepts. Additionally, it has been mentioned in the paper that if a solution exists, then it is unique. Then it seems that we don't need to differentiate between solutions and strong solutions because there can be only one solution (strong solution if it exists) and those conditions (1) and (2) will be trivially satisfied.

**2. The difference between sub-solution and equilibrium points**

I am a little confused about the difference between sub-solution and equilibrium points. From the definition of sub-solution, it seems that it is exactly the set which contains all the equilibrium points. However, based on Theorem 3, it is the Cartesian product of its factor sets. For example, suppose in a game where the set of all equilibrium points is $S=\{(U,L),(D,R)\}$. Then based on the definition, the sub-solution should also be $S$. But by theorem $3$, the sub-solution should be $\{(U,L),(U,R),(D,L),(D,R)\}$.

I really appreciate it if someone can give me some help on these. Thanks in advance!

**Tags:** game-theory


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## 1 answers

### Answer by username_1 on 2022-05-19. Score: 6 | Accepted answer

The idea of a solvable game is probably inspired by the theory of two-player zero-sum games. Such games are known to be solvable; a simple consequence of the [minimax theorem](https://en.wikipedia.org/wiki/Minimax_theorem).

One way to think about solvability is that using equilibrium points as a solution concept does not tell a player how they should behave—for this they would need to know which equilibrium point is played. If the game is solvable, this issue does not arise.

An undesirable feature of equilibrium points in mixed strategies is that a player playing a mixed best response will always be indifferent among all pure strategies they are randomizing over and, consequently, all mixtures of these pure strategies. Nevertheless, equilibrium points may require a player to randomize in a very special way. In a strong solution, all possible best replies must be included and the mentioned problem does not arise.

Finally, the issue with the definition of a subsolution comes from a typo Nash made, presumably while turning his thesis into a paper. The published paper refers to a condition (1) that is actually the definition of an equilibrium point. [Nash's original thesis](https://library.princeton.edu/special-collections/sites/default/files/Non-Cooperative_Games_Nash.pdf) also refers to (1) in the same place but (1) refers to something different there.

The correct definition of a subsolution is that it is a maximal set $S$ of equilibrium points satisfying the condition:
$$(t;r\_{i})\in S\,\,and\,\,s\in S \implies (s;r\_{i})\in S.$$


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