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Math Help - Whats the chances of...

  1. #1
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    Whats the chances of...

    Hi,
    My very first post so please be gentle lol


    Im programming a simple slot machine that has 5 reels and a has 1 "star" on each.

    What i want to know is if the Star can be on, above or below the pay line what is the probabilty of getting ANY 3 stars , Any 4 and any 5 etc


    As for working out the 5 im thinking its a simple as 3 positions on each reel divide by 24 on each =

    3 x 3 x 3 x 3 x 3 = 243
    24 24 24 24 24 = 7962624

    32768 / 243 = 1 in 32768 ??


    but how do i work out the chances of getting ANY 3 or ANY 4 and not just say the first 3 ?

    i did work out 3 as 1 in 512 but i think it must be different to that as you have 5 chances ( reels ) to get ANY 3 etc


    Hope this makes sence to someone
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  2. #2
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    It is quite possible that several of us can help you. But we must be able to understand the problem. Please give a simple description of each ‘reel’ in question.
    How many positions are there on each reel?
    There is exactly one star on each reel: CORRECT?
    The reels operate independently: correct?
    There are only five reels?
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by banditman View Post
    Hi,
    My very first post so please be gentle lol


    Im programming a simple slot machine that has 5 reels and a has 1 "star" on each.

    What i want to know is if the Star can be on, above or below the pay line what is the probabilty of getting ANY 3 stars , Any 4 and any 5 etc


    As for working out the 5 im thinking its a simple as 3 positions on each reel divide by 24 on each =

    3 x 3 x 3 x 3 x 3 = 243
    24 24 24 24 24 = 7962624

    32768 / 243 = 1 in 32768 ??


    but how do i work out the chances of getting ANY 3 or ANY 4 and not just say the first 3 ?

    i did work out 3 as 1 in 512 but i think it must be different to that as you have 5 chances ( reels ) to get ANY 3 etc


    Hope this makes sence to someone
    This is a binomial distribution problem.

    On any reel the prob that the star is in a favourable position is p=3/24=1/8.

    Then the probability of n from N reels show a star in a favourable position is
    a binomial random variable \sim B(N,p) so:

    <br />
p(n;N,p) =\frac{N!}{n!(N-n)!}p^n(1-p)^{N-n}<br />

    RonL
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  4. #4
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    Quote Originally Posted by Plato View Post
    It is quite possible that several of us can help you. But we must be able to understand the problem. Please give a simple description of each ‘reel’ in question.
    How many positions are there on each reel?
    There is exactly one star on each reel: CORRECT?
    The reels operate independently: correct?
    There are only five reels?
    yes here are the answers

    24 positions

    1 star on each reel but can be in any of 3 positions so i suppose it is like haveing 3 ?

    reels are independant and every spin has equal chance as previous spin

    yes only 5 reels


    Hope this is clearer
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    This is a binomial distribution problem.

    On any reel the prob that the star is in a favourable position is p=3/24=1/8.

    Then the probability of n from N reels show a star in a favourable position is
    a binomial random variable \sim B(N,p) so:

    <br />
p(n;N,p) =\frac{N!}{n!(N-n)!}p^n(1-p)^{N-n}<br />

    RonL

    Ok is there any chance i can have the last bit in simpler terms please eg what is the ! all about ??


    Thanks
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  6. #6
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    Hello, Banditman!

    You're working on a Probability problem
    . . and don't know about Combinations and Factorials
    . . or anything about Binomial Probabilities?
    I suggest a Google search . . .


    I'm programming a simple slot machine that has 5 reels.
    Each wheel has 24 positions and has one "star".

    What I want to know is: if the Star can be on, above or below the pay line,
    what is the probabilty of getting: 5 stars, 4 stars, 3 stars, etc.

    The probability of a "win" on a wheel is: . P(\star) \,=\,\frac{3}{24} \,=\,\frac{1}{8}

    The probability of "not win" on a wheel is: . P(\star') \,=\,\frac{7}{8}


    Here are the probabilities:

    P(5\,\star) \;=\;C(5,5)\left(\frac{1}{8}\right)^5\left(\frac{7  }{8}\right)^0 \;=\;\frac{1}{32,768}

    P(4\,\star) \;=\;C(5,4)\left(\frac{1}{8}\right)^4\left(\frac{7  }{8}\right)^1 \;=\;\frac{35}{32,768}

    P(3\,\star) \;=\;C(5,3)\left(\frac{1}{8}\right)^3\left(\frac{7  }{8}\right)^2 \;=\;\frac{490}{32,768}

    P(2\,\star) \;=\;C(5,2)\left(\frac{1}{8}\right)^2\left(\frac{7  }{8}\right)^3 \;=\;\frac{3,430}{32,768}

    P(1\,\star) \;=\;C(5,1)\left(\frac{1}{8}\right)^1\left(\frac{7  }{8}\right)^4 \;=\;\frac{12,005}{32,768}

    P(0\,\star) \;=\;C(5,0)\left(\frac{1}{8}\right)^0\left(\frac{7  }{8}\right)^5 \;=\;\frac{16,807}{32,768}

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  7. #7
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    I do know the basics and i do use spreadsheets to do the work for me it was just the first time i used this kinda win method

    The machine is only going to give you a win for any combination of 3 stars where as all the other wins have to be on the payline and left to right

    Many thanks for taking the time to help me out

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