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Items are given a fixed probability of selection per trial

  1. Jul 13, 2015 #1

    benorin

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    Several items are given fixed probabilities of selection per trial, one such item has a probability of 1.5%. What is the probability of selecting this item if 20 trials are permitted?


    This is not homework just a calculation for a game I play, so an explanation would be nice so I can do it myself, but all that is required is an answer, preferably in %.

    Haven't done Probability or Stats in few years now... can't think of which rule I ought to use in this situation.

    Many thanks,

    Ben Orin
     
  2. jcsd
  3. Jul 13, 2015 #2

    RUber

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    Question, are you just looking for selection and if you get it, you stop the trials?
    If so, then you have a sum of binomial probabilities.
    ##\sum_{n=1}^{20} (.015)(.985)^{n-1}##
     
  4. Jul 14, 2015 #3
    The way to look at problems of this type is to think about the probability of NOT selecting the item.

    If the probability of selecting the item in one trial is p, what is the probability of not selecting the item in one trial? So what is the probability of not selecting the item in 2 trials? In n trials? So the probability of selecting the item at least once in n trials is...?
     
  5. Jul 14, 2015 #4

    RUber

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    I second what MrAnchovy wrote. That will get you to the answer much more efficiently than the binomial sum equation I posted.
     
  6. Aug 17, 2015 #5

    benorin

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    Another question regarding the same scenario, specifically in a game I play there is an array of items purchasable for a fixed price each having a fixed probability of be obtained per purchase, now this bit is new: some of the items are called "unique items" of which only one may be obtained while the majority of items may be obtained multiple times, and I'm told that "The uniqueness of items doesn't change the probability distribution in the slightest because if an already obtained unique item is selected in a given trial, then random number generator 're-rolls' it, or simply put, it selects another item until a already obtained unique item is not selected." So is the statement in double quotes beginning with "The uniqueness..." really true? It seemed to me (before I knew it got "re-rolled") that having unique items would mean that the probability distribution would change once they were obtained since there was an item with, say a 1.5% chance of being obtained, that could no longer be selected once obtained initially and this intuitively constituted a change in the distribution though I hadn't considered the possibility of "re-rolling". Can u elaborate on this a bit, whether or not there is a change, and why? My math brain is rusty...
     
  7. Aug 17, 2015 #6

    RUber

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    In your example, if a unique item is no longer available the probability of selecting it goes to zero. That 1.5 percent is clearly going to be distributed to the other items.
    So for simplicity, say 3 items with probability of 50% , 40%, and 10%. Say the 10% item is unique, then your likelihood of drawing any items before the unique item is gone is just as above. Once the unique item is gone, you reroll if it is selected. Essentially making that probability zero. So the new probability for the other two items are 50/90 and 40/90.
     
  8. Aug 17, 2015 #7
    It entirely depends how the game programmer has implemented the selection algorithm. Hint: game programmers are often not very good at probability.

    One way of implementing "unique items" is to assign a probability to the event "wins a unique item", which in this case may be 1.5%, and if this event occurs select at random (usually according to some weighted distribution so that some items are harder to get than others) among the unique items that have not already been won.

    Reverse engineering commercial game algorithms is probably outside the scope of this forum; try the game's unofficial fan site.
     
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