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Thread: Expected Value: Carnival Game. Probability

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    Expected Value: Carnival Game. Probability

    This is my first post here. I need help with my math homework. I had to create the problem, but I don't know how to answer it.

    At the carnival, there is a board with a spinner. It costs $3 to play this game and you receive $6 if you win. There are 4 green tiles on the board and 6 blue. Each player has 2 spins. In order to win, the player must spin one of each color. They must spin a green, then blue, or spin a blue, then green. If they get two of the same color, they lose.

    - Figure out the expected value for the game (cost of game minus cost of price) to show the profit the carnival should expect to make over the long haul if the game is enticing enough for people to play.

    ... I am assuming I need to find the probability of spinning a green , then blue, or blue, then green.. but I do not know how to do this. Thank you for taking the time to read this and any help showing how to do it is greatly appreciated.
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    Re: Expected Value: Carnival Game. Probability

    Assumption: The result of each spin is independent of the previous ones.
    Probability of spinning green is $\frac4{10}$.
    Probability of spinning blue is $\frac6{10}$.

    Thus \begin{align*}\Pr{(BG)} &= \frac6{10} \cdot \frac4{10} \\ \Pr{(GB)} &= \frac4{10} \cdot \frac6{10} \end{align*}
    And therefore the probability of winning is $\frac{48}{100}$.
    This means that the probability of losing is $\frac{52}{100}$.
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    Re: Expected Value: Carnival Game. Probability

    Quote Originally Posted by needstatshelp View Post
    At the carnival, there is a board with a spinner. It costs \$3 to play this game and you receive \$6 if you win. There are 4 green tiles on the board and 6 blue. Each player has 2 spins. In order to win, the player must spin one of each color. They must spin a green, then blue, or spin a blue, then green. If they get two of the same color, they lose.
    Note that the question asks for winning from the carnival's point of view.
    The carnival wins \$3 if the player loses. The carnival loses \$3 if the player wins.
    So the carnival's expected outcome on play is $(\$3)\cdot\mathscr{P}(\text{ the player loses. })+(-\$3)\cdot\mathscr{P}(\text{ the player wins. })$
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    Re: Expected Value: Carnival Game. Probability

    I see what you are saying about the expected outcome but I can’t figure out the answer
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    Re: Expected Value: Carnival Game. Probability

    Quote Originally Posted by Plato View Post
    So the carnival's expected outcome on play is $(\$3)\cdot\mathscr{P}(\text{ the player loses. })+(-\$3)\cdot\mathscr{P}(\text{ the player wins. })$
    Surely you can do elementary arithmetic, $(\$3)\cdot(0.52)+(-\$3)\cdot(0.48)=~?$
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