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Silly Lottery Question

  1. Feb 23, 2014 #1
    At least I'm educating myself before playing the lotto, right?

    There's a game in Georgia called "Keno!" and they do an awfully good job at hiding probabilities from you. They'll tell the overall chance to win, but not the probabilities for the specific higher winnings possible. Basically it works like this:

    Pick, let's say "N" numbers from 1-80 but no more than 10 total numbers. 1=<N=<10

    The lotto picks 20 numbers from the set 1-80 randomly.

    What is the probability of having N of your numbers drawn (all of them)? or N-1, or N-2, etc? I tried drawing out a "probability tree" so to speak, but it got out of hand for many reasons.

  2. jcsd
  3. Feb 23, 2014 #2
    I forgot to mention. After the lotto selects one number they won't select it again for that draw.
  4. Feb 23, 2014 #3
    You can assume without loss of generality that you always pick the numbers 1, 2, ..., N.

    The probability of having K of your numbers selected by the keno machine is then equal to the probability that the machine selects K of the numbers 1, 2, ..., N. Out of the [itex]\binom{80}{20}[/itex] ways that it can choose the numbers, there are [itex]\binom{N}{K}\binom{80-N}{20-K}[/itex] ways that have exactly K of them among that set ([itex]\binom{N}{K}[/itex] ways to choose the elements from 1, ..., N and [itex]\binom{80-N}{20-K}[/itex] ways to choose the rest). So the probability that exactly K of your numbers are selected is
    [tex]{\binom{N}{K}\binom{80-N}{20-K} \over \binom{80}{20}}.[/tex]

    This is called a "hypergeometric distribution".

    Here's a table for N=10 and the 80/20 parameters of the problem:

    Code (Text):

    P(K=0)  = 0.0458
    P(K=1)  = 0.1796
    P(K=2)  = 0.2953
    P(K=3)  = 0.2674
    P(K=4)  = 0.1473
    P(K=5)  = 0.05142
    P(K=6)  = 0.01148
    P(K=7)  = 0.001611
    P(K=8)  = 0.0001354
    P(K=9)  = 0.000006121
    P(K=10) = 0.0000001122
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