Two hunters have set off to get food from the forest. They have two possible games that they can hunt: either a stag or rabbits. If both players hunt the stag, then they succeed in bringing it down; this gives them both a payoff of 2 units of food. If one player hunts the stag and the other hunts rabbits, then the stag hunter will fail while the rabbit hunter will succeed; i.e. the stag hunter gets a payoff of O while the rabbit hunter gets a payoff of 1. Finally, if both hunters pursue rabbits, they both receive a payoff of 1. The payoff matrix for this game is shown below. Determine all the Nash equilibria of this game (if there are any). O (Stag, Stag) and (Rabbit, Rabbit) The game has no Nash equilibrium. (Rabbit, Rabbit) O (Stag, Stag) STAG RABBIT STAG (2, 2) (1,0) RABBIT (0, 1) (1, 1)
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- Description: Raghu and Sayan both like to eat (a lot) but since they are also looking after their health, they can only eat a limited amount of calories per day. So when Kuldeep invites them to a party, both Raghu and Sayan decide to play a game. The game is simple, both Raghu and Sayan will eat the dishes served at the party till they are full, and the one who eats maximum number of distinct dishes is the winner. However, both of them can only eat a dishes if they can finish it completely i.e. if Raghu can eat only 50 kCal in a day and has already eaten dishes worth 40 kCal, then he can't eat a dish with calorie value greater than 10 kCal. Given that all the dishes served at the party are infinite in number, (Kuldeep doesn't want any of his friends to miss on any dish) represented by their calorie value(in kCal) and the amount of kCal Raghu and Sayan can eat in a day, your job is to find out who'll win, in case of a tie print "Tie" (quotes for clarity). Input: First line contains…In an astronomy board game, N planets in an imaginary universe do not follow the normal law of gravitation. All the planets are positioned in a row. The planetary system can be in a stable state only if the sum of the mass of all planets at even positions is equal to the sum of the mass of planets at the odd positions. Initially, the system is not stable, but a player can destroy one planet to make it stable. Find the planet that should be destroyed to make the system stable. If no such planet exists, then return -1. If there are multiple such planets, then destroy the planet with the smallest index and return the index of the destroyed planet. Example Let N-5 and planets = [2,4,6,3,4]. Destroying the fourth planet of mass 3 will result in planets = [2,4,6,4], and here, the sum of odd positioned planets is (2+6)=8, and the sum of even positioned planets is (4+4)=8, and both are equal now. Hence, we destroy the fourth planet. 11 MNBASK19922 13 14 15 16 17 18 20 * The function is…This is related to Dungeons and Dragons: Your character has just been reduced to 0 Hit Points (HP) (because of a fireball) and is now dying. They must now make several Death Saving Throws to determine their fate. Death saving throws are rolled every turn until you get three successes or three failures. To make a death saving throw, simply roll a d20. If the roll is 10 or higher, you succeed. Otherwise, you fail. On your third success, you become stable. On your third failure, you die. If you roll a 1, it counts as two failures. If you roll a 20, you regain 1 hit point. In which case, you regain consciousness and can fight once more! Calculate the probability that your character dies.
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