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Trying to determine probability

  1. Jul 1, 2013 #1
    Hi guys,

    It's been 20 years since I left college so I have forgotten how to calculate probability.

    It is probably very easy so can somebody just explain how I can get the odds/probability of winning.

    If I create prize A with odds = 10/1 (one in ten chance of winning)
    Prize B has odds of winning = 75/1 (one in 75 chance of winning
    Prize C has odds of winning = 90/1 ( one in 90 chance of winning)

    If these odds are layered on top of each other what are my chances of winning a prize.

    can you also explain how its done rather than just posting the odds answer.

    I presume my answer would be less than 10/1 as if there were only one prize (with A being the prize) but I now also have a chance of winning prize B or C also so that would mean it would be fractionally better than one in ten?

    Any help greatly appreciated.
     
  2. jcsd
  3. Jul 1, 2013 #2
    Probably easiest to add the fractions...
     
  4. Jul 1, 2013 #3

    jbunniii

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    It's clear that this won't give you the right answer: suppose the probabilities for A, B, and C were each 1/2. Then adding the fractions would give 3/2, which makes no sense.

    Let's introduce some notation:

    P(A) = probability to win A = 1/10
    P(B) = probability to win B = 1/75
    P(C) = probability to win C = 1/90

    P(no A) = probability to not win A = 1 - P(A)
    P(no B) = probability to not win B = 1 - P(B)
    P(no C) = probability to not win C = 1 - P(C)

    P(win) = probability to win at least one prize
    P(no win) = probability to win nothing

    Note that P(win) = 1 - P(no win)

    Also note that if the prizes are awarded independently, then

    P(no win) = P(no A and no B and no C) = P(no A) P(no B) P(no C) = (1 - P(A))(1 - P(B))(1 - P(C))

    So

    P(win) = 1 - (1 - P(A))(1 - P(B))(1 - P(C))
     
    Last edited: Jul 1, 2013
  5. Jul 1, 2013 #4
    Thanks for the replies guys.So, jbunniii what are my odds (percentage or fractional) of winning a prize. Knowing the answer will help me to understand how you got the answer. I'm still not sure how to derive the final answer from what you have written.

    Thanks,

    Dave.
     
  6. Jul 1, 2013 #5

    jbunniii

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    Just plug the numbers into the final expression:

    P(win) = 1 - (1 - P(A))(1 - P(B))(1 - P(C))

    where

    P(A) = 1/10
    P(B) = 1/75
    P(C) = 1/90

    so

    P(win) = 1 - (9/10)(74/75)(89/90) = 1 - 59274/67500 = 8226/67500 = 0.1218666...
     
  7. Jul 1, 2013 #6
    I know this is going to sound really dumb but its 20 years since I last looked at a maths equation but can you just explain how you got the numbers 59274, 67500 and 8226.
    Once again apologies for my ignorance but I need to fully understand this.

    Thank you again for your time and help.

    Dave.
     
  8. Jul 1, 2013 #7

    jbunniii

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    9 x 74 x 89 = 59274
    10 x 75 x 90 = 67500

    So
    $$\left(\frac{9}{10}\right) \times \left(\frac{74}{75}\right) \times \left(\frac{89}{90}\right) = \frac{59274}{67500}$$

    Now we want 1 minus this result, which is
    $$1 - \frac{59274}{67500} = \frac{67500 - 59274}{67500} = \frac{8226}{67500}$$
     
  9. Jul 1, 2013 #8
    Thank you so much for doing this for me this evening. I really appreciate this.

    Have a great day.
     
  10. Jul 2, 2013 #9
    Just one final comment: "odds of 10/1", usually written 10:1 or 10-1 equates to a one in 11 chance of winning, i.e a probability of 1/11, not 1/10. Consider odds of "evens" or 1:1, this equates to a probability of 0.5, not 1.
     
  11. Jul 2, 2013 #10

    Bacle2

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    Just a short comment: in my experience, odds a/b translate into probability a/(a+b) , not
    a/b ; odds a/b are the ratio of "probability for" vs "probability against" , so that, e.g., odds 1:1
    mean you're just as likely to "win" as to "lose".
     
  12. Jul 2, 2013 #11
    What I am doing at present is designing a smartphone app and the users will have a chance of winning prizes. I just want to ensure that the company developing the app for me understands the probability of winning a prize.

    I will be changing the number of prizes that can be won so for example I may have 3 prizes today
    A= 10:1
    B= 50:1
    C= 100:1

    They will scan a barcode and I am hoping it will randomly pick a prize and assign the correct odds of winning that prize.
    I'm not sure how to have the algorithm written so that when the program accepts the barcode scan and does its random prize picking that when it does choose that it will pick the correct odds of winning that prize.
    The higher the odds the more valuable in a monetary sense the prize will be.
    A may equate to $5 value where the 100:1 prize may be worth $250
    Does anybody know how I can lay out this to work....(I am not sure if its a computer programming issue or a maths equation part that I need to get right to ensure that the correct odds are applied to each prize.....i.e I don't want the program to mess up and if 35 people were to scan I don't want 13-14 of them to win the 100:1 prize.
    This is even hard to verbalise but I am hoping that somebody understands what I am trying to do and offer some advice on what I would need to do from a logic point of view.
    Any answers would be really appreciated.

    Dave.
    B
     
  13. Jul 2, 2013 #12
    1. You need to be sure that your offer complies with the applicable laws in the areas you are making it available. In most places, allocating prizes based on a random outcome of a future event requires a gambling license and is taxed appropriately.
    2. To help avoid this (in the UK at least, I'm not sure that this is relevant in the US and almost certainly not under Islamic Law) such promotions are normally done by pre-allocating prizes to tickets (via a unique identity to each ticket). This has the added benefit that you can allocate exactly 1 x 1 in 100 (i.e. 99:1) prize, 2 x 1 in 50 prizes and 10 x 1 in 10 prizes to each 100 tickets (or 10 x 1%, 20 x 2% and 100 x 10% prizes to each 1000 tickets) so you will know exactly how much you have to pay out. Note that this may only be part of what you need to do to be legal in any particular territory, for instance in the UK you normally have to introduce an element of "skill", hence those ridiculous questions like "Who was the first man on the moon (a) Neil Armstrong (b) Louis Armstrong".
    Note that to avoid confusion you really need to specify your probabilities as 1%, 0.01 or 1/100, if you really mean 100:1 then you need to say 1/101 or 0.009901.
     
    Last edited: Jul 2, 2013
  14. Jul 2, 2013 #13
    Hi Mr.Anchovy,

    The will be no money transaction or any cost to anybody to play so I don't think it would be considered "gambling". The prizes will be offered for free by a store and you have a chance of getting a free "prize" based on a random chance by coming into the store.
    Hope this gives more backround into what I am trying to do with the probability part. I just want to ensure that when it comes to coding the app that the "maths" part is correct and that the developers don't get the "formula" wrong, whereby stores end up giving out more prizes than they had planned to. I am just looking for the correct "formula" so they can code(understand) it correctly so the the correct number of people win based on the odds. ,

    Dave.
     
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