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Imagine a card game in which you have a 1 out of 10 chance of winning. If you win, you get $10. If you lose, you pay $2. What is the expected value of this game

Respuesta :

jacob193

Answer:

Pay [tex]\$0.8[/tex].

Step-by-step explanation:

Let [tex]X[/tex] denote the observed value of this game. Let positive values denote money received and negative values denote money paid.

There are two possible values for [tex]X[/tex]:

  • [tex]X = 10[/tex] (receive [tex]\$10[/tex],) and
  • [tex]X = -2[/tex] (pay [tex]\$ 2[/tex].)

The probability of each value is:

  • [tex]P(X = 10) = (1 / 10) = 0.1[/tex] as given in the question.
  • [tex]P(X = -2) = 1 - (1/10) = (9/10) = 0.9[/tex].

The expected value of this game, [tex]\mathbb{E}(X)[/tex], is an average of the outcomes [tex]x \in \lbrace 10,\, -2\rbrace[/tex] weighted by the probability [tex]P(X = x)[/tex] of each outcome:

[tex]\begin{aligned}\mathbb{E}(X) &= \sum\limits_{x} x\, P(X = x) \\ &= 10 \times P(X = 10) + (-2) \times P(X = -2) \\ &= 10 \times \frac{1}{10} + (-2) \times \frac{9}{10} \\ &= 1 - 1.8 \\ &= -0.8\end{aligned}[/tex].

The expected value of [tex]X[/tex] is [tex](-0.8)[/tex], meaning that on average, playing this game would require paying [tex]\$0.8[/tex].

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