<p><span class="math-container">$$
\begin{array}{c|cccc}
\text{P1} \backslash \text{P2} &amp; \text{Ll} &amp; \text{Lr} &amp; \text{Rl} &amp; \text{Rr} \\
\hline
\text{T} &amp; (3,2) &amp; 0.9\cdot(3,2) + 0.1\cdot(1,1)  &amp; 0.9\cdot(1,1) + 0.1\cdot(3,2) &amp; (1,1) \\
\text{B} &amp; (4,3) &amp; 0.1\cdot(4,3) + 0.9\cdot(2,4) &amp; 0.1\cdot(2,4) + 0.9\cdot(4,3) &amp; (2,4) \\
\end{array}
$$</span></p>
<p>where a strategy Xy of player 2 denotes that they will choose X if they observe T, and Y if they observe B.</p>
<p>Note the coincidence with previous exercise's payoff matrix if we change probability of successful observation from 0.9 to 1.</p>

CC BY-SA 4.0
2024-05-14T05:23:44.923Z
2024-05-14T05:23:44.923Z
Giskard
<p>Consider a game where player 1 must choose T or B, player 2 must choose L or R, and their
payoffs depend on their choices as follows.</p>
<p><span class="math-container">$$
\begin{array}{c|cc}
\text{P1} \backslash \text{P2} &amp; L &amp; R \\
\hline
\text{T} &amp; (3,2) &amp; (1,1) \\
\text{B} &amp; (4,3) &amp; (2,4) \\
\end{array}
$$</span></p>
<p>Suppose that player 1 moves first, and then player 2 makers her choice after observing 1's move.</p>
<p>But also suppose that, whatever 1 chooses, the probability that player 2 will correctly observe 1's action is 0.9, and there is probability 0.1 that player 2 will mistakenly observe the other action (which 1 did not choose).</p>
<p>The payoffs depend on the players' actual choices according to the previous table (so, for example, if 1 chose T but 2 mistakenly observed B and chose R then 2's payoff would be 1).</p>
<p><strong>Show</strong> : The <em>normal representation in strategic form</em> for the extensive-form game.</p>
<hr />
<p><em>Note</em>: This question is a variation of a previous version of a problem, which does not involve player 2 mistakenly observing player 1's move. In this case, the <em>normal representation in strategic form</em> for the extensive-form game is as follows:</p>
<p><span class="math-container">$$
\begin{array}{c|cccc}
1 \ / (2.\text{T}/2.\text{B}) &amp; \text{Ll} &amp; \text{Lr} &amp; \text{Rl} &amp; \text{Rr} \\
\hline
\text{T} &amp; (3,2) &amp; (3,2) &amp; (1,1) &amp; (1,1) \\
\text{B} &amp; (4,3) &amp; (2,4) &amp; (4,3) &amp; (2,4) \\
\end{array}
$$</span></p>

<game-theory>
Finding the normal representation of extensive form game
user9487
# Finding the normal representation of extensive form game

(Asked by: username_0 on 2024-05-13)

Consider a game where player 1 must choose T or B, player 2 must choose L or R, and their
payoffs depend on their choices as follows.

$$
\begin{array}{c|cc}
\text{P1} \backslash \text{P2} & L & R \\
\hline
\text{T} & (3,2) & (1,1) \\
\text{B} & (4,3) & (2,4) \\
\end{array}
$$

Suppose that player 1 moves first, and then player 2 makers her choice after observing 1's move.

But also suppose that, whatever 1 chooses, the probability that player 2 will correctly observe 1's action is 0.9, and there is probability 0.1 that player 2 will mistakenly observe the other action (which 1 did not choose).

The payoffs depend on the players' actual choices according to the previous table (so, for example, if 1 chose T but 2 mistakenly observed B and chose R then 2's payoff would be 1).

**Show** : The *normal representation in strategic form* for the extensive-form game.

---

*Note*: This question is a variation of a previous version of a problem, which does not involve player 2 mistakenly observing player 1's move. In this case, the *normal representation in strategic form* for the extensive-form game is as follows:

$$
\begin{array}{c|cccc}
1 \ / (2.\text{T}/2.\text{B}) & \text{Ll} & \text{Lr} & \text{Rl} & \text{Rr} \\
\hline
\text{T} & (3,2) & (3,2) & (1,1) & (1,1) \\
\text{B} & (4,3) & (2,4) & (4,3) & (2,4) \\
\end{array}
$$

**Tags:** game-theory


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## 1 answers

### Answer by username_1 on 2024-05-14. Score: 1 | Accepted answer

$$
\begin{array}{c|cccc}
\text{P1} \backslash \text{P2} & \text{Ll} & \text{Lr} & \text{Rl} & \text{Rr} \\
\hline
\text{T} & (3,2) & 0.9\cdot(3,2) + 0.1\cdot(1,1) & 0.9\cdot(1,1) + 0.1\cdot(3,2) & (1,1) \\
\text{B} & (4,3) & 0.1\cdot(4,3) + 0.9\cdot(2,4) & 0.1\cdot(2,4) + 0.9\cdot(4,3) & (2,4) \\
\end{array}
$$

where a strategy Xy of player 2 denotes that they will choose X if they observe T, and Y if they observe B.

Note the coincidence with previous exercise's payoff matrix if we change probability of successful observation from 0.9 to 1.


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