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I About binomial statistics

  1. Jun 10, 2017 #1


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    Hi all,
    I am solving a practical math problem. There is a sale in one of the shopping mall in my town. The mall gives 10 coupons to a new customer. The customer could use one coupon at a time and when it is used, one could spin a fortune wheel to win more 10 more coupons. If one doesn't win additional coupons, he or she will get a chance next time using another coupon or no more drawing will be. So one customer will have not more than 20 coupons. I want to ask how many coupons on average a customer could make. Let's assume the customer use one coupon per day and the promotion activity going on for long enough for the customer to use all coupons. To setup a model, I assume there is ##p## chance to win the fortune wheel. At the beginning, a customer will get ##N=10## coupons, for the first time he/she uses the coupons, he/she has ##p## chance to win 10 more coupons and get ##q=1-p## chance to lose. If customer win additional 10 coupons, it could be when he/she use the kth coupon. So we need to consider all possible chance that the customer play and win the fortune wheel when using 1st, 2nd, 3rd .... coupon. The average number of coupons is computed as follows (here p is 0.001)

    10 + 10 * (p + qp + q^2p + q^3 p + q^4 p + q^5p + \cdots + q^{10}p)

    the sum in the parenthesis is said to be win at the first time or win at second time or win at third time, etc. I am not quite confident on this but running a computer simulation seems to get a result close to that given by above expression. Now comes to my question. The computation is straightforward when only ONE drawing is allowed. but what happen if there is no limit on number of drawing, i.e. customers will get a chance to win additional 10 coupons every time using a coupon. I find it pretty complicate because every time additional coupons won, it will have more chance to win more, it seems that there will be infinite terms to take care of. Any idea?
  2. jcsd
  3. Jun 10, 2017 #2


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  4. Jun 10, 2017 #3


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    To solve this, get a formula for the expected number of coupons after N steps, then take the limit as N goes to infinity.

    Here's a start:

    N=1 gives 10
    N=2 gives ##10+10p##
    N=3 gives ##10+10p(1+10p)##
    N=3 gives ##10+10p(1+10p(1+10p))##

    Do you see a pattern forming?

    Can you express this in a neat form that allows the limit to be seen as a standard infinite series?

    Hint: the expectation does not exist ('is infinite') if ##10p\geq 1##.
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