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    Game Probability Question




    Hi,
    I am playing a game. I win with an 88% chance and lose with 12% probability.

    If I have won 20 games in a row what is the probabilty that I will lose the next game.

    If you can figure out the answer and provide also the calculation logic that would be great.

    Thanks
    B

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    Re: Game Probability Question

    Is each game independent from other games? There's no opportunity for a tie?
    Last edited by Buckeye; 09-18-2016 at 05:08 PM.
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

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    Re: Game Probability Question

    Quote Originally Posted by Buckeye View Post
    Is each game independent from other games? There's no opportunity for a tie?
    No tie is possible its either win or lose.

    Would be interested to see how the numbers come out for 1) games are independent 2) games are not independent. Thanks

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    Re: Game Probability Question

    If we treat the games as independent, we might be able to use a negative binomial distribution where the "success" is thought of as your first loss. http://stattrek.com/probability-dist...-binomial.aspx

    I welcome any other input from other TS members
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

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    Re: Game Probability Question


    Quote Originally Posted by system2016 View Post
    Hi,
    I am playing a game. I win with an 88% chance and lose with 12% probability.

    If I have won 20 games in a row what is the probabilty that I will lose the next game.
    The wording of your question implies that you wish to find P(G_{21}=L|G_1=W , G_2=W , ... , G_{20}=W), the probability of losing the 21st game, conditional on having won all of the first 20. If the probability of winning/losing is the same for each game, and all games are mutually independent, then the probability of losing the next game after having won the previous 20 is still 12%, since independence means that the chance of winning a game is not affected by the chance of winning another game.

    That's a different question though than P(G_1=W , G_2=W , ... , G_{20}=W , G_{21}=L), the probability of the joint event of 20 straight wins followed by one loss. This type of event is described by the geometric distribution (a special case of the negative binomial distribution). Again assuming independence and constant win/loss probability, then

    P(G_1=W, G_2=W , ... , G_{20}=W , G_{21}=L) = P(G=W)^{20}P(G=L) = (0.88^{20})(0.12).

    If there is dependence among the game outcomes, then more information would be needed.
    http://ab-stats.com

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