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    Game Probability




    Hi All - I have not looked at statistics or probability for a number of years but am wondering if someone can provide me with a formula for figuring out a simple game probability.
    The game consists of 25 dots. The dots are in 5 rows and 5 columns.
    You select one of the dots from each column. After you select a dot from each column, the random generator selects a dot from each column and you (the player) get to view the results on how many you had correct.

    It would seem that you have a 1/5 chance on each of the 5 columns to get your selection correct.

    So the odds of getting all 5 picks correct would be 1/5 * 1/5 * 1/5 * 1/5 * 1/5 = 1/3125

    How would I calculate my probability of getting other selections correct, such as 3/5?
    Would it simply be: 4/5 * 4/5 * 1/5 * 1/5 * 1/5 = 16/3125
    where the 4/5 represents the wrong selections and 1/5 represents the right selections.

    thanks
    Andrew

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    You'll want to look into the binomial distribution. But for this particular problem we get

    Code: 
         [,1]    [,2]
    [1,]    0 0.32768
    [2,]    1 0.40960
    [3,]    2 0.20480
    [4,]    3 0.05120
    [5,]    4 0.00640
    [6,]    5 0.00032
    Where the first column is the number of times you get it right and the second column is the probability.

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    I hope you don't mind me correcting some of your statistical grammer. "So the odds..." should actually say "So the chance.." or "So the probability...". Odds are different.

    Going on to your question, 4/5 * 4/5 * 1/5 * 1/5 * 1/5 = the probability of getting only the last three columns correct. If you want the probability of getting exactly three correct, whether it be the 1st,3rd, and 5th columns, or the 2nd, 3rd, and 4th, then the probability is:
    \binom{5}{3}*\left ( \frac{1}{5} \right )^{3}*\left ( \frac{4}{5} \right )^{2}=0.0512

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    hehehe...Dason beat me to the answer.

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    Thank you both!

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