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Thread: Probabilty of at least 4 games?

  1. #1
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    Probabilty of at least 4 games?

    Hello, I've encountered a question during my m.a statistics course that I'm having a real time solving. The question goes like this:
    There's the following game: two balls are randomly chosen (without return) from a vase that contains 7 black balls, 3 white and 2 orange. For every black ball chosen, u get a dollar and for ever white ball chosen, u lose a dollar. If u an orange ball was chosen, u neither loose nor gain money.
    a man decides to play this game until he will receive 0 profit twice. what is the probabilty of him playing this game more than 4 games?
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    Re: Probabilty of at least 4 games?

    Quote Originally Posted by ehariel View Post
    The question goes like this: There's the following game: two balls are randomly chosen (without return) from a vase that contains 7 black balls, 3 white and 2 orange. For every black ball chosen, u get a dollar and for every white ball chosen, u lose a dollar. If u an orange ball was chosen, u neither loose nor gain money. a man decides to play this game until he will receive 0 profit twice. what is the probability of him playing this game more than 4 games?
    This wording is very confusing. I realize that it may be a fault of translating. Read this phrase. “a man decides to play this game until he will receive 0 profit twice.” Saying “receive 0 profit” is an active present tense. Thus strictly as written it means that the player stops after receiving zero profit on two turns at play. Note that the player receives zero profit on any play by drawing $\{O,O\}\text{ or }\{B,W\}$

    Note that we are not considering any accumulation of gain or loss. We are only concerned with what happens on any one play. On any play, the player can have an outcome of $\{-2,-1,0,1,2\}$ dollars. Our player stops after s/he gets two zeros.

    Now if that is not what is intended but rather some variation taking into account accumulation of gain or loss. This the problem because a total computational nightmare. I will not spend any more time on this in this case.

    I will be glad to help with the first reading. Please tell us which is correct.
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    Re: Probabilty of at least 4 games?

    If we go with Plato's first interpretation, i.e. we play until we've played two games that have a net winning of 0

    then we have the following

    $P[\text{0 winnings}] = P[(O,O)||(B,W)] = \dfrac{2\cdot 1}{12\cdot 11} + 2\dfrac{7\cdot 3}{12\cdot 11} = \dfrac{44}{132} = \dfrac 1 3$

    so $P[\text{non-zero winnings}]=\dfrac 2 3$

    So now we ask what is the probability that out of 4 games we will have 3 or more that result in non-zero winnings.

    Well the number of games with non-zero winnings in 4 games is distributed with a binomial distribution with parameters $n=4,~p=\dfrac 2 3$

    Thus this probability is simply

    $P[3]+P[4] = \dbinom{4}{3}\left(\dfrac 2 3\right)^3 \left(\dfrac 1 3\right) + \dbinom{4}{4} \left(\dfrac 2 3\right)^4 = \dfrac{16}{27}$

    The second interpretation would involve evaluating probabilities on a markov chain. A significantly more challenging problem.
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    Re: Probabilty of at least 4 games?

    Quote Originally Posted by romsek View Post
    If we go with Plato's first interpretation, i.e. we play until we've played two games that have a net winning of 0
    then we have the following $P[\text{0 winnings}] = P[(O,O)||(B,W)] = \dfrac{2\cdot 1}{12\cdot 11} + 2\dfrac{7\cdot 3}{12\cdot 11} = \dfrac{44}{132} = \dfrac 1 3$ so $P[\text{non-zero winnings}]=\dfrac 2 3$
    So now we ask what is the probability that out of 4 games we will have 3 or more that result in non-zero winnings.
    Well the number of games with non-zero winnings in 4 games is distributed with a binomial distribution with parameters $n=4,~p=\dfrac 2 3$
    Thus this probability is simply $P[3]+P[4] = \dbinom{4}{3}\left(\dfrac 2 3\right)^3 \left(\dfrac 1 3\right) + \dbinom{4}{4} \left(\dfrac 2 3\right)^4 = \dfrac{16}{27}$
    Please note that the OP asked for the probability that the player quits after more than four plays.
    Thus there cannot be two successes (a zero profit) in the first four.
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    Re: Probabilty of at least 4 games?

    Quote Originally Posted by Plato View Post
    Please note that the OP asked for the probability that the player quits after more than four plays.
    Thus there cannot be two successes (a zero profit) in the first four.
    that's why i recast the problem in terms of non-zero profit.

    the 3 and 4 is number of non-zero profit games
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