 Players’ knowledge is represented as usual in a possible worlds framework. Consider for example the case in which there are two agents, both with one of thresholds 1, 2 or 3. There are nine possible worlds represented by ordered pairs of numbers, representing the first and second player’s individual thresholds respectively: 11, 12, 13,…, 32, 33. If the players do not communicate, each knows her own threshold only. Player 1’s information partition reflects her ignorance about player’s 2 threshold and it consists of the sets \(\{11, 12, 13\},\) \(\{21, 22, 23\},\) \(\{31, 32, 33\};\) whereas, similarly, player 2’s partition consists of the sets \(\{11, 21, 31\},\) \(\{12, 22, 32\},\) \(\{13, 23, 33\}.\) If player 1’s threshold is 1, she revolts no matter what player 2’s threshold is. Hence, player 1 revolts in \(\{11, 12, 13\}\). If player 1’s threshold is 3, she never revolts. Hence, she plays \(s\) in \(\{31, 32, 33\}\). If her threshold is 2, she revolts only if the other player revolts as well. Since in this example we are assuming that there is no communication between the agents, player 1 cannot be sure of player’s 2 action, and chooses the non-risky \(s\) in \(\{21, 22, 23\}\) as well. Similarly, player 2 plays \(r\) in \(\{11, 21, 31\}\) and \(s\) otherwise. Consider now the case in which \(1 \rightarrow 2\) and \(2 \rightarrow 1\). Both players have now the finest information partitions. Thresholds of 1 and 3 yield \(r\) and \(s\), respectively, for both players again. However, in player 1’s cells \(\{21\}\) and \(\{22\}\), she knows that player 2 will revolt, and, having threshold 2, she revolts as well. Similarly for player 2 in his cells \(\{12\}\) and \(\{22\}\). Note, that the case in which both players have threshold 2, yields both the equilibrium in which both players revolt and the equilibrium in which each player stays home. It is assumed that in the case of multiple equilibria, the one which results in the most revolt will obtain.