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A game at a carnival has the following rules: A player rolls a 6-sided die and, if the number is less
than 6, wins twice the number on the die in tokens. On the other hand, the player loses 12 tokens
when a 6 is rolled. What should it cost to play if the game is to be fair?
7 tokens for one game
3 tokens for one game
9 tokens for two games
1 token for two games



Answer :

Using the expected value of a discrete distribution, it is found that the game would be fair with a cost of 3 tokens for 1 game.

What is the mean of a discrete distribution?

The expected value of a discrete distribution is given by the sum of each outcome multiplied by it's respective probability.

Considering the situation described in this problem, that a dice has 6 sides, and x as the price in tokens, the distribution of prizes in this problem is given as follows:

  • P(X = 2 - x) = 1/6. (rolls 1 wins 2).
  • P(X = 4 - x) = 1/6. (rolls 2 wins 4).
  • P(X = 6 - x) = 1/6.
  • P(X = 8 - x) = 1/6.
  • P(X = 10 - x) = 1/6.
  • P(X = -12 - x) = 1/6.

The game is fair when E(x) = 0, hence:

1/6(2 - x + 4 - x + 6 - x + 8 - x + 10 - x) - 1/6(12 + x) = 0.

30 -5x - 12 - x = 0

6x = 18

x = 3.

The cost should be of 3 tokens for 1 game.

More can be learned about the expected value of a discrete distribution at https://brainly.com/question/3316979

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