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Math Help - 4% or 40% that is the question

  1. #1
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    4% or 40% that is the question

    Hi all,
    I am having an argument about the probability of winning a raffle.
    Now I am no math wizard but I would like to think I have a grasp for the simple.
    I may be wrong but we will leave that up to you to decide.


    Thanks for the help.

    The question is what are the odds or the % chance of winning a raffle if;

    There are 10,000 entries
    You have 400 of them and,
    There are 10 entries drawn one at a time.

    What is you chance of winning on any one of the 10.

    So 400 of 10000 and 10 chance to win.
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  2. #2
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    Quote Originally Posted by l6779k@gmail.com View Post
    Hi all,
    I am having an argument about the probability of winning a raffle.
    Now I am no math wizard but I would like to think I have a grasp for the simple.
    I may be wrong but we will leave that up to you to decide.


    Thanks for the help.

    The question is what are the odds or the % chance of winning a raffle if;

    There are 10,000 entries
    You have 400 of them and,
    There are 10 entries drawn one at a time.

    What is you chance of winning on any one of the 10.

    So 400 of 10000 and 10 chance to win.
    The probability that the first ticket drawn is not one of yours is 9600/10000.
    The prob that the 2nd is not one of yours is 9599/9999
    The prob that the 3rd is not one of yours is 9598/9998
    :
    :
    The prob that the 10th is not one of yours is 9591/9991

    So the probability that you don't win is:

    \frac{9600\times 9599 \times ... \times 9591}{10000\times 9999 \times ... \times 9991} \approx 0.6647

    Hence the probability that you do win is: 1-0.6647 =0.3353 or about 33.5\%

    RonL
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  3. #3
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    This is what I came up with...

    So what I gather from that is,
    some one has the same odd of winning with 400 out 10000 entries with 10 chances, as someone with 3300 of 10000 entries with only one drawing?



    This is what I came up with, because each drawing is indipendent of each other.

    Why would this be wrong compared to your answer?

    Starting number of entries Drawing number % of winning with 10

    Chances

    10000 1st 4.000000000%

    9999 2nd 4.000400000%

    9998 3rd 4.000800100%
    9997 4th 4.001200300%

    9996 5th 4.001600640%

    9995 6th 4.002001001%

    9994 7th 4.002401441%

    9993 8th 4.002801961%

    9992 9th 4.003202562%
    9991 10th 4.003603243%
    Sum of all 10 drawings 40.018011248
    Number of drawings / 10

    Sum Divided (/) be the number of chances (10) = 4.0018011248%



    Last edited by l6779k@gmail.com; December 19th 2007 at 12:06 AM. Reason: more info
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by l6779k@gmail.com View Post
    So what I gather from that is,
    some one has the same odd of winning with 400 out 10000 entries with 10 chances, as someone with 3300 of 10000 entries with only one drawing?



    This is what I came up with, because each drawing is indipendent of each other.

    Why would this be wrong compared to your answer?

    Starting number of entries Drawing number % of winning with 10

    Chances

    10000 1st 4.000000000%
    9999 2nd 4.000400000%
    9998 3rd 4.000800100%
    9997 4th 4.001200300%
    9996 5th 4.001600640%
    9995 6th 4.002001001%
    9994 7th 4.002401441%
    9993 8th 4.002801961%
    9992 9th 4.003202562%
    9991 10th 4.003603243%

    Sum of all 10 drawings 40.018011248
    Number of drawings / 10

    Sum Divided (/) be the number of chances (10) = 4.0018011248%


    Rather than ask what is wrong with what you suggest, ask what justification
    do you have for averaging these numbers?

    That it can't be right is clear from the observation that if only one ticket
    was drawn the chance of winning would be 4%, but drawing 10 tickets
    increases the chance of winning substantialy. hence what you have done
    can't be right and we don't have to bother pointing out exactly where it
    goes wrong as:

    1. You have no justification for it
    2. It gives answers that can't be right

    RonL
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  5. #5
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    Why is the answer not right?

    We have spoke with over 30 people, some who are odds makers in a casino, some are engineers, and we are split on which answer is correct... So again the question is, "Why is your answer right over the other one?"
    Please someone explain it to us in plain English.
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  6. #6
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    Quote Originally Posted by l6779k@gmail.com View Post
    We have spoke with over 30 people, some who are odds makers in a casino, some are engineers, and we are split on which answer is correct... So again the question is, "Why is your answer right over the other one?"
    Please someone explain it to us in plain English.
    1. There is no guarantee that my answer is right, at the verry least you
    need to check my arithmetic.

    2. It is easier to compute the probability that you dont win, and as the two
    events : You win, You don't win are mutualy exclusive:

    prob(you win) = 1-prob(you don't win).

    (note: here "you win" means one or more of your tickets are winners)

    3. The tickets are drawn without replacement, so after N non-winning tickets
    are drawn there are 9600-N non-winning tickets left out of 10000-N tickets.

    4. Thus there are 9600x9599x..x9591 (equally likely) ways of picking 10 non
    winning tickets out of a total of 10000x9999x...x9991 (equally likely) ways of
    picking a ticket so the probability that none of the tickets are winners is:

    prob(you don't win)=(9600x9599x..x9591)/(10000x9999x...x9991)~=0.6647

    and so the probability of winning is 1-0.6647=0.3353 or ~33.53%

    RonL
    Last edited by CaptainBlack; December 19th 2007 at 12:56 AM.
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  7. #7
    Senior Member DivideBy0's Avatar
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    I got CaptainBlack's answer too, using the complement.

    1-\left(1-\frac{400}{10000} \right)^{10}= 0.335= 33.5\%

    Since Pr(Win at least one) = Pr(Win NOT 0 times) = 1 - Pr(Win 0 Times)
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  8. #8
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    Likewise, I have the same logic as CaptainBlack.

    P(win at least once) = P(X > 0)
    P(X > 0) = 1 - P(0)

    1 - P(0) = 1 - \frac{9600}{10000}\frac{9599}{9999}\frac{9598}{999  8}...

    P(X > 0) = 1 - \frac{9990!(9600!)}{10000!(9590!)}

    P(X > 0) = 1 - 0.66471

    P(X > 0) = 0.33529

    The entries are drawn one at a time. If you lose on the first drawing, then there are now 9999 total entries left, and you still have 400 of them. To not win, you need any of the remaining 9599 entries. This pattern repeats itself nine more times.

    Divide by zero, your answer yields an actually lower answer, 33.5167%. Luckily, the sampling size is so big, that your answer is correct up to one decimal place.
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  9. #9
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    Quote Originally Posted by colby2152 View Post
    Likewise, I have the same logic as CaptainBlack.

    P(win at least once) = P(X > 0)
    P(X > 0) = 1 - P(0)

    1 - P(0) = 1 - \frac{9600}{10000}\frac{9599}{9999}\frac{9598}{999  8}...

    P(X > 0) = 1 - \frac{9990!(9600!)}{10000!(9590!)}

    P(X > 0) = 1 - 0.66471

    P(X > 0) = 0.33529

    The entries are drawn one at a time. If you lose on the first drawing, then there are now 9999 total entries left, and you still have 400 of them. To not win, you need any of the remaining 9599 entries. This pattern repeats itself nine more times.

    Divide by zero, your answer yields an actually lower answer, 33.5167%. Luckily, the sampling size is so big, that your answer is correct up to one decimal place.
    It's a 2-line Monte-Carlo simulation and 1,000,000 replications gives an
    estimate of 33.51%.

    RonL
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  10. #10
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    OK.. I get it, kind of

    Ok thank you all for your responses,
    I see how the math works, but I still can't believe that some one with 400 and 10 chances has the same odds as some one with 3300 and 1 chance.

    I can see where both answers are correct, seeing the math.
    But on an individual draw each individual draw had an 4% chance + a .0004 increase on each of the fowling chances.

    Right?

    and the guy with the 400 entries won nothing and was pissed be cause he figured he should have won.

    Again thank you for your help
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  11. #11
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    But on an individual draw each individual draw had an 4% chance + a .0004 increase on each of the fowling chances.

    Right?
    sort of. This is right if you assume that he doesn't win any of them. If he does win one then he has less chance to win the following ones because one of his tickets has been removed.
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