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Taken from : answers.yahoo.com/question/index?qid=20100904034737AAfGkYF

A mall has sweep stakes where they give you a coupon for every X dollars you spend (amount is not important). You fill the coupon with your contact details and put it in the box. Every two weeks, they draw 10 coupons, give their owners prizes, and discard the rest of the coupons.

Now, thanks to some very big purchase we made, we have about 300 coupons (going to hurt when filling them). To get the highest probability of winning, should we put all the coupons at once, or split them evenly between the three remaining draws?

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Yes, I posted there earlier today, but realized the people at Yahoo! answers aren't as brilliant as people here. –  imgx64 Sep 4 '10 at 12:04

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up vote 2 down vote accepted

Let's assume that in each box there are n other tickets besides yours. So if you put them all in the first box and 10 tickets are drawn your chance of winning something is

$$q = 1 - \prod_{k=1}^{10} \left( 1 - \frac{300}{n+301-k} \right).$$

Now if you split them, 100 each amongst three boxes (where, again, each box has n other tickets), your chance of winning something is

$$p = 1 - \prod_{k=1}^{10} \left( 1 - \frac{100}{n+101-k} \right)^3 .$$

Now p>q, but only slightly, for large n.

For example, with n=10000, $$p \approx 0.258176 \textrm{ and } q \approx 0.256004$$

So you're slightly better off if you split them up into three lots of 100. However, this assumes that the number of other tickets placed in the box every two weeks remains constant.

Note that the difference between p and q decreases with increasing n.

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