<p>Your analysis is correct in principle, but your notation is not. A (behavioral) strategy for a player in a perfect information game has to specify a probability distribution over actions for each decision node of that player. Since player 2 has two such nodes, a Left and a Right one, with two possible actions at each node, her strategies can be written as pairs <span class="math-container">$(p_2^L,p_2^R)$</span> of probabilities of choosing <span class="math-container">$A$</span> at the left and the right node, respectively. Strategy profiles are then denoted by <span class="math-container">$(p_1,(p_2^L,p_2^R))$</span>. The three sets of SPNEs are then given by</p>
<ol>
<li><p><span class="math-container">$(1,(1,p_2^R))$</span> with <span class="math-container">$p_2^R&gt;\frac12$</span>,</p>
</li>
<li><p><span class="math-container">$(0,(1,p_2^R))$</span> with <span class="math-container">$p_2^R&lt;\frac12$</span>, and</p>
</li>
<li><p><span class="math-container">$(p_1,(1,\frac12))$</span> with <span class="math-container">$0\le p_1\le 1$</span>.</p>
</li>
</ol>
<p>Your question &quot;<em>why would player 2 ever choose <span class="math-container">$p_2^R&lt;\frac12$</span>?</em>&quot; has already been answered by yourself: Because <em>conditional on player 1 choosing B it is optimal.</em> Note that your claim that &quot;<em>if she instead chooses <span class="math-container">$p_2^R&gt;\frac12$</span> she can be sure that player 1 chooses <span class="math-container">$A$</span></em>&quot; is not true. This would only hold if player 1 could observe player 2's choice <em>before</em> deciding.</p>

CC BY-SA 4.0
2022-04-16T13:29:28.200Z
Thanks! So if I understand it correctly, player 2 cant commit to prob > 0.5 because once player 1 chooses B, there is no point in actually following through, all prob values are optimal. Maybe what I was thinking would work if we keep repeating the game though, to punish player 1? Somewhat like tit-for-tat?
Ishan Kashyap Hazarika
CC BY-SA 4.0
2022-04-16T14:10:38.953Z
@IshanKashyapHazarika Well, player 2 could try to stubbornly play (1,1) until player 1 "learns" to play 1. But the repeated game is a very different thing and I haven't yet thought through it...
VARulle
CC BY-SA 4.0
2022-04-16T12:31:29.633Z
2022-04-16T12:31:29.633Z
VARulle
<p>My question follows from this question: <a href="https://math.stackexchange.com/questions/2132846/game-theory-subgame-perfect-nash-equilibrium-in-a-sequential-game-with-identica">https://math.stackexchange.com/questions/2132846/game-theory-subgame-perfect-nash-equilibrium-in-a-sequential-game-with-identica</a> from Maths stackexchange.</p>
<p>Based on the explanation given in the answer, I tried to find the subgame perfect Nash equilibrium (SPNE) of another game:</p>
<p><a href="https://i.sstatic.net/1khpA.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1khpA.png" alt="enter image description here" /></a></p>
<p>Here, there are 2 players, and both have to choose between A and B. Player 1 moves first, followed by player 2.</p>
<p>Let us define the strategies of the players as <span class="math-container">$(p_1, p_2)$</span> where <span class="math-container">$p_1$</span> and <span class="math-container">$p_2$</span> are the probabilities that player 1 and player 2 choose A respectively.</p>
<p>In the subgame to the lower left, the analysis is simple: player 2 will choose A because it gives a higher payoff.</p>
<p>In the lower right, player 2 faces identical payoffs, so any <span class="math-container">$p_2 \in [0, 1]$</span> is optimal. We divide the possibilities into 3 cases:</p>
<p><strong>1.</strong> <span class="math-container">$p_2 &gt; 0.5$</span>: Here the expected payoff of choosing B for player 1 is <span class="math-container">$2p_2 + (1-p_2)4 &lt; 3$</span>, the payoff of choosing A. Therefore player 1 will choose A. This gives us the SPNE <span class="math-container">$(p_1, p_2) = (1, 1)$</span>.</p>
<p><strong>2.</strong> <span class="math-container">$p_2 &lt; 0.5$</span>: Here the expected payoff of choosing B for player 1 is <span class="math-container">$2p_2 + (1-p_2)4 &gt; 3$</span>, the payoff of choosing A. Therefore player 1 will choose B. This gives us a family of equilibria: <span class="math-container">$(0, p_2), p_2 \in [0, 0.5)$</span>.</p>
<p><strong>3.</strong> <span class="math-container">$p_2 = 0.5$</span>: In this case, the expected payoff of choosing B for player 1 is <span class="math-container">$2p_2 + (1-p_2)4 = 3$</span>, the payoff of choosing A. Thus, any <span class="math-container">$p_1 \in [0, 1]$</span> is optimal. This gives us another family of equilibria: <span class="math-container">$(p_1, 0.5), p_1 \in [0, 1]$</span>.</p>
<p>This brings me to my first question:</p>
<p><strong>Question 1: Is my characterisation of SPNEs correct? That is, are all the equilibria mentioned above actually SPNEs? And also, are there other SPNEs that I have missed?</strong></p>
<p>If the answer to &quot;are they all actually SPNEs&quot; is <em>yes</em>, then I think there is a problem with the family of equilibria in point 2. <em>Conditional on player 1 choosing B</em>, of course all <span class="math-container">$p_2 \in [0, 1]$</span> are optimal including all <span class="math-container">$p_2 \in [0, 0.5)$</span>. But player 2 knows that choosing any such <span class="math-container">$p_2 &lt; 0.5$</span> would mean that player 1 chooses B, and player 2 ends up with payoff 1. But if she instead <em>chooses <span class="math-container">$p_2 &gt; 0.5$</span> she can be sure that player 1 chooses A</em> and then player 2 ends up with payoff 4, which is better than 1. Then why would player 2 ever choose <span class="math-container">$p_2 &lt; 0.5$</span>?</p>
<p>This leads me to my second question:</p>
<p><strong>Question 2: If <span class="math-container">$(0, p_2), p_2 \in [0, 0.5)$</span> are not SPNEs, what went wrong in my analysis? Does backward induction fail in such cases or did I make a mistake? On the other hand, if these are actually SPNEs, then is there any refinement of SPNE, some other solution concept that is immune to my objection? Where can I read about them?</strong></p>

<nash-equilibrium><dynamic-games><extensive-games>
Subgame perfect Nash equilibrium when there is a tie in payoffs seems problematic
Ishan Kashyap Hazarika
# Subgame perfect Nash equilibrium when there is a tie in payoffs seems problematic

(Asked by: username_0 on 2022-04-15)

My question follows from this question: <https://math.stackexchange.com/questions/2132846/game-theory-subgame-perfect-nash-equilibrium-in-a-sequential-game-with-identica> from Maths stackexchange.

Based on the explanation given in the answer, I tried to find the subgame perfect Nash equilibrium (SPNE) of another game:

[![enter image description here](https://i.sstatic.net/1khpA.png)](https://i.sstatic.net/1khpA.png)

Here, there are 2 players, and both have to choose between A and B. Player 1 moves first, followed by player 2.

Let us define the strategies of the players as $(p\_1, p\_2)$ where $p\_1$ and $p\_2$ are the probabilities that player 1 and player 2 choose A respectively.

In the subgame to the lower left, the analysis is simple: player 2 will choose A because it gives a higher payoff.

In the lower right, player 2 faces identical payoffs, so any $p\_2 \in [0, 1]$ is optimal. We divide the possibilities into 3 cases:

**1.** $p\_2 > 0.5$: Here the expected payoff of choosing B for player 1 is $2p\_2 + (1-p\_2)4 < 3$, the payoff of choosing A. Therefore player 1 will choose A. This gives us the SPNE $(p\_1, p\_2) = (1, 1)$.

**2.** $p\_2 < 0.5$: Here the expected payoff of choosing B for player 1 is $2p\_2 + (1-p\_2)4 > 3$, the payoff of choosing A. Therefore player 1 will choose B. This gives us a family of equilibria: $(0, p\_2), p\_2 \in [0, 0.5)$.

**3.** $p\_2 = 0.5$: In this case, the expected payoff of choosing B for player 1 is $2p\_2 + (1-p\_2)4 = 3$, the payoff of choosing A. Thus, any $p\_1 \in [0, 1]$ is optimal. This gives us another family of equilibria: $(p\_1, 0.5), p\_1 \in [0, 1]$.

This brings me to my first question:

**Question 1: Is my characterisation of SPNEs correct? That is, are all the equilibria mentioned above actually SPNEs? And also, are there other SPNEs that I have missed?**

If the answer to "are they all actually SPNEs" is *yes*, then I think there is a problem with the family of equilibria in point 2. *Conditional on player 1 choosing B*, of course all $p\_2 \in [0, 1]$ are optimal including all $p\_2 \in [0, 0.5)$. But player 2 knows that choosing any such $p\_2 < 0.5$ would mean that player 1 chooses B, and player 2 ends up with payoff 1. But if she instead *chooses $p\_2 > 0.5$ she can be sure that player 1 chooses A* and then player 2 ends up with payoff 4, which is better than 1. Then why would player 2 ever choose $p\_2 < 0.5$?

This leads me to my second question:

**Question 2: If $(0, p\_2), p\_2 \in [0, 0.5)$ are not SPNEs, what went wrong in my analysis? Does backward induction fail in such cases or did I make a mistake? On the other hand, if these are actually SPNEs, then is there any refinement of SPNE, some other solution concept that is immune to my objection? Where can I read about them?**

**Tags:** nash-equilibrium, dynamic-games, extensive-games


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## 1 answers

### Answer by username_1 on 2022-04-16. Score: 2 | Accepted answer

Your analysis is correct in principle, but your notation is not. A (behavioral) strategy for a player in a perfect information game has to specify a probability distribution over actions for each decision node of that player. Since player 2 has two such nodes, a Left and a Right one, with two possible actions at each node, her strategies can be written as pairs $(p\_2^L,p\_2^R)$ of probabilities of choosing $A$ at the left and the right node, respectively. Strategy profiles are then denoted by $(p\_1,(p\_2^L,p\_2^R))$. The three sets of SPNEs are then given by

1. $(1,(1,p\_2^R))$ with $p\_2^R>\frac12$,
2. $(0,(1,p\_2^R))$ with $p\_2^R<\frac12$, and
3. $(p\_1,(1,\frac12))$ with $0\le p\_1\le 1$.

Your question "*why would player 2 ever choose $p\_2^R<\frac12$?*" has already been answered by yourself: Because *conditional on player 1 choosing B it is optimal.* Note that your claim that "*if she instead chooses $p\_2^R>\frac12$ she can be sure that player 1 chooses $A$*" is not true. This would only hold if player 1 could observe player 2's choice *before* deciding.

> **Comments:**
> **username_0**: Thanks! So if I understand it correctly, player 2 cant commit to prob > 0.5 because once player 1 chooses B, there is no point in actually following through, all prob values are optimal. Maybe what I was thinking would work if we keep repeating the game though, to punish player 1? Somewhat like tit-for-tat?  
> **username_1**: @username_0 Well, player 2 could try to stubbornly play (1,1) until player 1 "learns" to play 1. But the repeated game is a very different thing and I haven't yet thought through it...  

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