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Thread: Calculating special lottery odds

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    Calculating special lottery odds




    #s out of a bag!

    Hello, I would appreciate anyone's help in this problem:

    There are 11 balls in a bag numbered 0-10.

    You get 6 picks out of the bag. Pull 1 ball, write down number, and put it back (ie interpreted as with repetition?).

    If person 1 chooses 6 balls out of the bag, IE 123456, what is the probability that person 2 who chooses matches 6 balls, 5 balls, 4 balls, 3 balls, 2 balls, 1 ball and none, considering that order does not matter (ie, combination? 123456 = 6 correct, 654321 still equals 6 correct, whereas 123457 = 5 correct).

    Greatly appreciate assistance with this problem which has me scrambling!!!

    Many thanks!!

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    Re: Calculating special lottery odds

    - If your labels are \{0, 1, 2, \ldots, 11\}, then you have 12 labels.

    - If person 1 is allowed to draw with replacement, and obtain a repeated result say 111234, then how did you define the number of matches with person 2? Or person 1 is drawing without replacement?

    - If you are drawing with replacement, then you are probably working with some multinomial probabilities. But before going on lets clear up the above questions first.

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    Re: Calculating special lottery odds

    Thanks for your reply.

    The labels are {0,1,2,3,4,5,6,7,8,9,10} - 11 labels total.

    Yes, person 1 is allowed to draw with replacement. Person 1 can draw 111111, and person 2 can draw 111222 (resulting in a match of 3 numbers). Another example is person 1 can draw 123455, and person 2 can draw 344455 (resulting in a match of 2 numbers).

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    Re: Calculating special lottery odds

    Sorry missed your reply. The mathematical formulation is as follow:

    Let \mathbf{X} = (X_1, X_2, \ldots, X_k) \sim \text{Multinomial}\left(n; \frac {1} {k}, \frac {1} {k}, \ldots, \frac {1} {k}\right)
    be the counts of each ball picked by person 1, where n is the number of picks and k is the number of balls in the bag. In particular, n = 6 and k = 11 in your question.

    Similarly we can let \mathbf{Y} = (Y_1, Y_2, \ldots, Y_k) be another multinomial vector representing the picks from person 2, which has the identical distribution as \mathbf{X} and they are independent.

    The number of matches, by definition, is

    Z = \sum_{i=1}^k \min\{X_i, Y_i\}

    and you want to find out the distribution of this discrete random variable.

    I have not find a nice method to tackle this problem yet; just use R to simulate 1 million times to obtain a numerical solution first.

    Code: 
    m<-1000000
    k<-11
    n<-6
    pr<-rep(1/k,k)
    x<-rmultinom(m,n,pr)
    y<-rmultinom(m,n,pr)
    count<-rep(0,7)
    for (i in 1:m) {
    	j<-sum(pmin(x[,i],y[,i]))
    	count[j+1]<-count[j+1]+1
    }
    count/m
    
    [1] 0.041412 0.209483 0.371843 0.280376 0.087191 0.009471 0.000224
    where the last line gives you the estimate of probability mass function \Pr\{Z = z\}, z = 0, 1, \ldots, 6 from left to right.

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    Re: Calculating special lottery odds


    Note that when k >> n, the probability of having counts larger than 1 is very small. In such case, Z can be approximated by a hypergeometric distribution (as if considering drawing without replacement), with pmf

    \Pr\{Z = z\} = \frac {\displaystyle \binom {n} {z} \binom {k - n} {n - z}} {\displaystyle \binom {k} {n}}, z = 0, 1, \ldots, n

    Of course it is not very suitable for your question as n, k are pretty close. And you see the resulting pmf deviate from the previous simulation quite a lot.

    Code: 
    dhyper(0:6,6,5,6)
    [1] 0.000000000 0.012987013 0.162337662 0.432900433 0.324675325 0.064935065 0.002164502

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