Results 1 to 7 of 7

Math Help - Advanced Probability Question

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    4

    Lightbulb Advanced Probability Question

    Hi, I'm completely new to this forum and, in fact, my maths is horrendous to say the least. I do have quite an advanced probability question I would like to work out though, for the purposes of a gambling experiment.

    If someone can help, I would be very very grateful.

    I need to know the probability (%) of winning a game where there are 100 numbers. 100,000 people pick six numbers. However, you can pick the same number twice or even three times and there is also a much higher chance that 10 of the 100 numbers will come up.

    25 numbers are then selected at random. What is your chance of winning the game?

    So to recap, 100,000 people pick 6 numbers each. 25 numbers are picked at random from 100, where each number can be selected twice or even three times, and some of the numbers (say 10 of them) have more weight attached to them so they have more of a chance of being selected.

    If that makes sense to anyone and you might be looking for a good maths equation on a monday, I would be very grateful for your help.

    Cheers,
    Ronan
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jul 2009
    Posts
    593
    Thanks
    4
    What does "However, you can pick the same number twice or even three times and there is also a much higher chance that 10 of the 100 numbers will come up." mean? How many multiples of numbers are there? Are the numbers 1 through 10? Is one number represented more than the others. These questions affect the probability of someone winning.

    From your description, it sounds like 10 of the balls has a higher chance of being selected than the other balls. You would need to know WHAT the probability of selecting those balls are to give an accurate probability of success.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    4

    Question

    Hi ANDS,

    Thanks for getting back to me. The numbers are numbered 1 to 100. Numbers 1 - 10 have a much higher weight attached to them so they are more likely to come up. For every number that's drawn, a number between one and ten would come up 7 times. Does that help or make sense?

    To clarify, a person playing might pick the following 6 numbers:
    1,3,5,5,9,26
    because numbers between one and ten have a much higher chance of being selected AND you can pick the same number twice or three times if you wish - though three times isn't very wise!

    I knew I'd be leaving something important out.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jul 2009
    Posts
    593
    Thanks
    4
    You would still need to know how much more likely these ten numbers are to come up compared to the other 90 numbers: 2 times as likely, three times as likely.

    For example, if I had a "loaded" dice, and the number 5 is 2 times as likely to come up compared to the other numbers then

    P(1, 2, 3, 4, 6) = 5/7
    P(5) = 2/7

    The total probabilitie of all numbers in your game must add up to 1. In a normal game, each would be "equally likely" and have p(being chosen)=1/100; and would be a much simpler game. However, to calculate your odds, we need to know what the weighted probabilities of each of these ten numbers are. Only then can you begin to calculate the combinations necessary to derive some winning chances for you.

    Also, it seems a person is drawing a ball WITH replacement, meaning once they draw a ball that ball goes back into this drawing bin. That would be the only way a person could draw the same number multiple times, although the probability of that happening (even if some of the numbers are weighted) is pretty unlikely. Twice maybe. . .more than twice and we're pushing the bounds of a random draw and where someone starts pointing fingers and screaming "this game is rigged!"
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2009
    Posts
    4
    LOL...


    Okay let's say that the 100 numbers are like so 11111122222233333344444455555566666677777788888899 9999101010101010
    So 6 1's, 6 2's, 6 3's ......6 10's
    Then the other 40 numbers are 11 - 50.

    Does that help? In this instance, the balls are not replaced.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jul 2009
    Posts
    593
    Thanks
    4
    Real quick before we go further - is there any point to you weighting the numbers as such? Because that makes this far more "complex" than it needs to be. If the goal is to "randomize" the draws, you might be better served offering a combination of say 1 letter and 5 numbers which is easier to find the odds for. Essentially, with your current set up, you are asking to find the odds of choosing six corrent numbers, which involves finding the total possible combinations of of these numbers that a person can possibly draw. Since the numbers themselves are not equally likely, we can not use simple counting methods to figure this out. No bueno.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2009
    Posts
    4
    Hi Ands,

    Sorry for delay in getting back to you. Right, without giving too much away... The game involves betting on six horses (akin to the Tote Scoop6 - but different). The issue is that out of 100 horses, favourites are always going to be backed. That's why I have to add significant weight to some of the numbers (horses). Does that make sense?

    If you need any more info, let me know. Appreciate your continued help...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simple probability or more advanced technique?
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 4th 2011, 08:56 PM
  2. Replies: 4
    Last Post: March 28th 2011, 06:50 AM
  3. Advanced Draw Probability issue
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: January 7th 2010, 01:41 PM
  4. Advanced Blackjack Probability
    Posted in the Statistics Forum
    Replies: 1
    Last Post: February 2nd 2009, 03:15 PM
  5. Advanced Probability Help!!!
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 16th 2008, 05:21 PM

Search Tags


/mathhelpforum @mathhelpforum