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Thread: Function to graph probability of expected results for a number of draws of a lottery

  1. #1


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    Function to graph probability of expected results for a number of draws of a lottery

    How was the "Expected Results" table calculated in the picture below? I'm trying graph the probability of different results based on buying N number of scratch tickets. Imagine a game with 4 tiers or prizes.

    Prize Value Number of tickets
    100 4
    10 40
    1 100
    0 846

    What would my expected +/- results be 66% of the time if I bought 10 tickets? How does one figure that out?


    picture-2.jpg

  2. #2


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    OK, I am assuming you are buying tickets randomly and your table indicates the frequency and amount of wins if you buy 990 tickets.

    I assume each ticket cost $1.00

    So, ev would be .909 per ticket. (100X4+10X40+1X100+0X846)/990 tickets = .909

    Therefore ev for your 10 ticket random buy would be $9.09 for every 10 ticket buy. So, you will lose $.91 for every 10 ticket buy.

    Booking a win will require positive variance from the distribution.
    Last edited by Stealth; 11-20-2019 at 10:10 PM.
    Luck is nothing more than probability taken personally!

  3. #3


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    Quote Originally Posted by eihli View Post
    How was the "Expected Results" table calculated in the picture below? I'm trying graph the probability of different results based on buying N number of scratch tickets. Imagine a game with 4 tiers or prizes.

    Prize Value Number of tickets
    100 4
    10 40
    1 100
    0 846

    What would my expected +/- results be 66% of the time if I bought 10 tickets? How does one figure that out?


    picture-2.jpg
    eihli,

    Stealth showed your EV for playing 10 $1 games is -$0.91.

    To answer your question, though, we also need to know the Standard Deviation for your game. I calculated the SD per game to be $0.713. For 10 games, your total SD will be $0.713*sqrt(10), or about $2.25.

    Thus, about 68% (close enough to the 66% you mentioned) of the time, your result will be within one SD of your EV. With the numbers above, we see that for 10 games, 68% of the time you'll end up in the range of

    (-$0.91-$2.25) to (-$0.91+$2.25), so

    -$3.16 to +$1.34.

    Hope this helps!

    Dog Hand

  4. #4


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    Quote Originally Posted by Dog Hand View Post
    eihli,

    Stealth showed your EV for playing 10 $1 games is -$0.91.

    To answer your question, though, we also need to know the Standard Deviation for your game. I calculated the SD per game to be $0.713. For 10 games, your total SD will be $0.713*sqrt(10), or about $2.25.

    Thus, about 68% (close enough to the 66% you mentioned) of the time, your result will be within one SD of your EV. With the numbers above, we see that for 10 games, 68% of the time you'll end up in the range of

    (-$0.91-$2.25) to (-$0.91+$2.25), so

    -$3.16 to +$1.34.

    Hope this helps!

    Dog Hand
    Thanks! How did you calculate the standard deviation? That's the thing I can't figure out.

    The formula I see everywhere doesn't give the answer you got. The answer I get seems unreasonable.

    https://www.itl.nist.gov/div898/soft...2/weightsd.pdf

    In English, I'm taking the sum of (each prize's weight * (each prize's value - expected value a.k.a weighted mean) ^ 2) )

    The weights I get for each ticket are: 0.004, 0.04, 0.1, 0.85 and the weighted mean is -0.09

    So (0.004 * (99 - -0.9) ^ 2 + 0.04 * (9 - -0.09) ^ 2 + 0.1 * (0 - -0.09) ^ 2 + 0.85 * (-1 - -0.09) ^ 2) = 43.72

    Since the weights add up to 1, there's no reason to divide by the sum of the weights like in the formula that I linked. We just take the square root of that and get 6.612 for the standard deviation. Way off from your 0.713. Where did I go wrong?
    Last edited by eihli; 11-21-2019 at 05:34 AM.

  5. #5


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    I'm still trying to figure this out and had another question arise. What function describes this type of graph? I don't think this would be a "normal distribution" (if that's the right term). For example what if the tickets are 9,999 $0 valued tickets and 1 $10,000 valued ticket. In that it doesn't really make sense to think 68% of the time you'll be within +/- $10, 95% +- $10. It's like... 66% of the time you'll be exactly at -$10. Same for 95% of the time. I don't *think* that would be a normal distribution. The graph would have this huge spike on the left-hand side and then a really long tail off to the right.

    I need to take a statistics class :/

  6. #6


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    Quote Originally Posted by eihli View Post
    <snip>We just take the square root of that and get 6.612 for the standard deviation. Way off from your 0.713. Where did I go wrong?
    Oops... you are correct! I had an error in my formula.

    Quote Originally Posted by eihli View Post
    I'm still trying to figure this out and had another question arise. What function describes this type of graph? I don't think this would be a "normal distribution" (if that's the right term). For example what if the tickets are 9,999 $0 valued tickets and 1 $10,000 valued ticket. In that it doesn't really make sense to think 68% of the time you'll be within +/- $10, 95% +- $10. It's like... 66% of the time you'll be exactly at -$10. Same for 95% of the time. I don't *think* that would be a normal distribution. The graph would have this huge spike on the left-hand side and then a really long tail off to the right.

    I need to take a statistics class :/
    Yes... the method we used is for a game (like BJ) whose results approximate a normal distribution. Since you posted a picture containing BJ output, I thought you were actually interested in such "normal distribution" games.

    However, your lottery game's results would be skewed, more like video poker. For such skewed games, simulation is generally used to determine output ranges. Dunbar's Video Poker Analyzer is excellent for calculating video poker.

    Hope this helps!

    Dog Hand

  7. #7


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    As we know from a blackjack game, where the mean is virtually zero and can be ignored, the variance is simply the average squared result of the various payoffs. Then the s.d. is the square root of that. Ignoring the mean for your chart, the square root of the average squared result is 6.67. Close enough.

    Don

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