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Thread: Calculating lottery probabilities

  1. #1
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    Calculating lottery probabilities

    The lottery i am talking about is a 6/49 game. You select 6 numbers from a pool of 49, and in the draw 6 numbers are taken. There is also an additional 'Bonus' number drawn from the remaining 43 balls which is used in some of the prize tiers.

    I have managed to calculate the overall odds of winning in each tier as follows:

    Match Odds
    6 1 in 13,983,816
    5 + 1 BonusBall 1 in 2,330,636
    5 1 in 55,491
    4 + 1 BonusBall 1 in 22,197
    4 1 in 1,083
    3 + 1 BonusBall 1 in 812
    3 1 in 61

    Q1) How do i work out the probability for winning a prize in these divisions from these odds? Ive seen conflicting messages, but i assumed it was 1/odds, so the probability of matching 3 balls would be 1/61= 0.0164

    In the lottery you can play via a system where you choose more than 6 numbers and enter all the combinations of those numbers. So if you did a system entry and chose 1,2,3,4,5,6,7 you would get lines of:
    • 1+2+3+4+5+6
    • 1+2+3+4+5+7
    • 1+2+3+4+6+7
    • 1+2+3+5+6+7
    • 1+2+4+5+6+7
    • 1+3+4+5+6+7
    • 2+3+4+5+6+7


    The lottery offers this all the way up to selecting all 49 numbers, which gives the number of different combinations as so:

    Numbers No. of combinations
    6 1
    7 7
    8 28
    9 84
    10 210
    11 462
    12 924
    13 1,716
    14 3,003
    15 5,005
    16 8,008
    17 12,376
    18 18,564
    19 27,132
    20 38,760
    21 54,264
    22 74,613
    23 100,947
    24 134,596
    25 177,100
    26 230,230
    27 296,010
    28 376,740
    29 475,020
    30 593,775
    31 736,281
    32 906,192
    33 1,107,568
    34 1,344,904
    35 1,623,160
    36 1,947,792
    37 2,324,784
    38 2,760,681
    39 3,262,623
    40 3,838,380
    41 4,496,388
    42 5,245,786
    43 6,096,454
    44 7,059,052
    45 8,145,060
    46 9,366,819
    47 10,737,573
    48 12,271,512
    49 13,983,816

    Q2) How would i work out a) the odds of matching in each prize tier for each the combinations above? so the odds of matching three balls when i have chosen all the way from 6 - 49 balls for my system entry b) get the probability of matching in each prize tier from those odds?

    I realise that at some point with the above combinations there will a point where its guaranteed that you will match the number and thus the probability is 1.0 but i dont know how to get to that point.

    I had tried getting the probability of matching in each tier as 1/odds, then multiplying that value by the number of combinations, but that gave a probability way over 1.0 which i know isnt possible.

    Any help in sending me the right way to calculate this is greatly appreciated!! Id like to learn the steps rather than just being given the answer
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  2. #2
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    Re: Calculating lottery probabilities

    Suppose you choose all combinations of $k$ numbers. Then, to get exactly $r$ balls (no bonus ball), you would have:
    $$\dfrac{\dbinom{k}{r}\dbinom{49-k}{7-r}\dbinom{7-r}{1}}{\dbinom{49}{7}\dbinom{7}{1}}$$

    To get exactly $r$ balls plus the bonus ball:
    $$\dfrac{\dbinom{k}{r+1}\dbinom{r+1}{1}\dbinom{49-k}{6-r}}{\dbinom{49}{7}\dbinom{7}{1}}$$

    In each of these formulas, the denominator is the total number of ways to choose 7 balls from 49, then choose one of them to be the bonus ball. The first numerator is the number of ways to choose the $r$ winning balls from the $k$ numbers you selected times the number of ways to choose the remaining balls from the balls you did not choose times the number of ways to choose a bonus ball among the balls that you did not choose.

    The second numerator is the number of ways to choose the bonus ball from among your k numbers. Then, choose the $r$ winning balls from among your remaining numbers, and finally, choose the remaining winning numbers from the balls you did not choose.
    Last edited by SlipEternal; Jun 28th 2018 at 07:21 AM.
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  3. #3
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    Re: Calculating lottery probabilities

    Quote Originally Posted by SlipEternal View Post
    Suppose you choose all combinations of $k$ numbers. Then, to get exactly $r$ balls (no bonus ball), you would have:
    $$\dfrac{\dbinom{k}{r}\dbinom{49-k}{7-r}\dbinom{7-r}{1}}{\dbinom{49}{7}\dbinom{7}{1}}$$

    To get exactly $r$ balls plus the bonus ball:
    $$\dfrac{\dbinom{k}{1}\dbinom{k-1}{r}\dbinom{49-k}{6-r}}{\dbinom{49}{7}\dbinom{7}{1}}$$

    In each of these formulas, the denominator is the total number of ways to choose 7 balls from 49, then choose one of them to be the bonus ball. The first numerator is the number of ways to choose the $r$ winning balls from the $k$ numbers you selected times the number of ways to choose the remaining balls from the balls you did not choose times the number of ways to choose a bonus ball among the balls that you did not choose.

    The second numerator is the number of ways to choose the bonus ball from among your k numbers. Then, choose the $r$ winning balls from among your remaining numbers, and finally, choose the remaining winning numbers from the balls you did not choose.
    Thanks very much for that, doesnt look too complicated. Would you be able to do one as an example for me? Just so i know im putting the numbers in the right place
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  4. #4
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    Re: Calculating lottery probabilities

    Quote Originally Posted by mikabrown View Post
    Thanks very much for that, doesnt look too complicated. Would you be able to do one as an example for me? Just so i know im putting the numbers in the right place
    Sure. Let's say you want to buy tickets with all combinations of 9 numbers. Your odds of getting exactly 5 balls plus the bonus ball would be:

    $$\dfrac{\dbinom{9}{5+1}\dbinom{5+1}{1}\dbinom{49-9}{6-5}}{\dbinom{49}{7}\dbinom{7}{1}} = \dfrac{120}{3,579,191} \approx \dfrac{1}{29,827}$$

    I fixed it.
    Last edited by SlipEternal; Jun 28th 2018 at 07:23 AM.
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