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    happyclp's Avatar
    happyclp Posts: 2, Reputation: 1
    New Member
     
    #1

    Oct 21, 2009, 08:35 PM
    Probability of 7 Cards
    I am studying for a test and am struggling with this question. Can anyone help?

    Suppose you have an unusual deck of cards for a game called Seven-Oh! The probabilities
    of drawing each card (1 through 7) in this game are given in the table.

    x prob(x)
    1 0.11
    2 0.08
    3 0.23
    4 0.22
    5 0.05
    6 0.15
    7 0.16

    Suppose someone makes a casino game from these cards. It costs $10 to play, and
    you could win $100 if you draw a 5 or you could win $10 if you draw a 2 or a 7.
    Otherwise, you win $0. Is this a winning game? Determine the probabilities of
    each dollar amount (including $0) and calculate the expected value to answer this
    question.
    Nhatkiem's Avatar
    Nhatkiem Posts: 120, Reputation: 9
    Junior Member
     
    #2

    Oct 21, 2009, 08:51 PM
    Quote Originally Posted by happyclp View Post
    I am studying for a test and am struggling with this question. Can anyone help?

    Suppose you have an unusual deck of cards for a game called Seven-Oh! The probabilities
    of drawing each card (1 through 7) in this game are given in the table.

    x prob(x)
    1 0.11
    2 0.08
    3 0.23
    4 0.22
    5 0.05
    6 0.15
    7 0.16

    Suppose someone makes a casino game from these cards. It costs $10 to play, and
    you could win $100 if you draw a 5 or you could win $10 if you draw a 2 or a 7.
    Otherwise, you win $0. Is this a winning game? Determine the probabilities of
    each dollar amount (including $0) and calculate the expected value to answer this
    question.
    Lets generalize this a bit. Instead of winning 100 dollars 5% of the time, we can make it 10 dollars 50% of the time. (For example if you play 100 games, you should win about 5 times, either way you earn 500 dollars.)

    The probability of winning 10 dollars back then is 0.5+0.16++0.08

    So your probability of winning JUST your money back is 74%, over time, you will end up losing money from not making it back.
    happyclp's Avatar
    happyclp Posts: 2, Reputation: 1
    New Member
     
    #3

    Oct 21, 2009, 08:57 PM

    Okay, but how do you figure it out by using expected value?
    Nhatkiem's Avatar
    Nhatkiem Posts: 120, Reputation: 9
    Junior Member
     
    #4

    Oct 21, 2009, 08:59 PM

    I recall something like this back in high school haha. Something I kind of just thought about.

    A winning game would mean you would make back your money, so I just though about the odds of getting $10 back out of 100 games.
    morgaine300's Avatar
    morgaine300 Posts: 6,561, Reputation: 276
    Uber Member
     
    #5

    Oct 21, 2009, 10:06 PM

    Think of it like this:

    There's .05 probability of getting the 5. That gets you $100

    There's .24 probability of getting the 2 or the 7. (.08 + .16) That gets you $10.

    Now, what's the probability of getting nothing? If you get a 1, 3, 4 or 6. What's the probability of getting any of those?

    That is basically in the end your three possibilities of what will happen. If you add the probabilities together, then instead of having 7 different cards, you have 3 different possibilities of the outcome.

    (You can apply this to all sorts of possibilities, like getting a 5 or higher would be the totals of the 5, 6 & 7, etc.)

    Try putting that together and see if you can go from there.
    Nhatkiem's Avatar
    Nhatkiem Posts: 120, Reputation: 9
    Junior Member
     
    #6

    Oct 21, 2009, 10:55 PM
    Quote Originally Posted by morgaine300 View Post
    Think of it like this:

    There's .05 probability of getting the 5. That gets you $100

    There's .24 probability of getting the 2 or the 7. (.08 + .16) That gets you $10.

    Now, what's the probability of getting nothing? If you get a 1, 3, 4 or 6. What's the probability of getting any of those?

    That is basically in the end your three possibilities of what will happen. If you add the probabilities together, then instead of having 7 different cards, you have 3 different possibilities of the outcome.

    (You can apply this to all sorts of possibilities, like getting a 5 or higher would be the totals of the 5, 6 & 7, etc.)

    Try putting that together and see if you can go from there.
    While this is all correct, the main question is whether this is a winning game or losing game, so finding the probability of drawing the cards you want alone are not enough. Since different cards have different winning values.
    morgaine300's Avatar
    morgaine300 Posts: 6,561, Reputation: 276
    Uber Member
     
    #7

    Oct 23, 2009, 12:04 AM

    I never said it was enough. It was a hint to get the person started. It's the concept of not doing peoples homework for them.

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