Recall the game “Squad Game”, Suppose that in each round (independent of the past) contestant 1 nominates a uniformly chosen player other than themselves, while all other players always nominate contestant 1.(a) If L = 1, find the probability that contestant 1 is eliminated.(b) Now suppose that n > L = 3. Find the probability that player 1 is eliminated.
Question
Recall the game “Squad Game”, Suppose that in each round (independent of the past) contestant 1 nominates a uniformly chosen player other than themselves, while all other players always nominate contestant 1.(a) If L = 1, find the probability that contestant 1 is eliminated.(b) Now suppose that n > L = 3. Find the probability that player 1 is eliminated.
Solution 1
The game you're referring to seems to be a variation of a game of chance, where the outcome depends on the random selection of players. Let's break down the problem:
(a) If L = 1, this means there is only one other player in the game. Since player 1 can only nominate another player, and all other players always nominate player 1, the only possible outcome is that player 1 nominates player 2, and player 2 nominates player 1. Therefore, the probability that player 1 is eliminated is 1, or 100%.
(b) If n > L = 3, this means there are more than 3 players in the game. Player 1 has a 1/(n-1) chance of nominating each other player, while all other players will always nominate player 1. The probability that player 1 is eliminated is the probability that they are nominated by at least L = 3 other players.
Since all players other than player 1 always nominate player 1, the only way player 1 can avoid elimination is if they nominate themselves, which is not allowed, or if they nominate one of the players who would have otherwise reached the threshold to eliminate player 1.
Therefore, the probability that player 1 is eliminated is 1 - (1/(n-1)), or (n-2)/(n-1). This represents the probability that player 1 does not nominate one of the players who would have otherwise reached the threshold to eliminate them.
Solution 2
The game you're referring to seems to be a variation of a game theory problem. Here's how you can solve it:
(a) If L = 1, that means there is only one other player in the game. Contestant 1 can only nominate this player, and since the other player always nominates contestant 1, the probability that contestant 1 is eliminated is 1.
(b) If n > L = 3, that means there are more than 3 players in the game. Contestant 1 has a 1/(n-1) chance of nominating any other player. However, all other players always nominate contestant 1. Therefore, the probability that contestant 1 is eliminated is the sum of the probabilities that each of the other players is chosen, which is 1/(n-1) + 1/(n-1) + ... + 1/(n-1) = n/(n-1). However, since n > 3, the probability that contestant 1 is eliminated is less than 1.
Similar Questions
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