Lotteries, Turan's problem, and minimization of risk Suppose I am a high-volume broker aiming to make some money on a state lottery.  In this lottery, six balls are drawn from a population of (let's say) 50, without replacement.  A ticket is a choice of a size-6 subset of 1,2,..50.
The prize structure of this lottery is such that the jackpot alone doesn't impart much value to the ticket.  But it turns out that lesser prizes are sufficiently large relative to their probability that the ticket has a positive expected value, which is why I'm buying a lot of tickets in the first place.  For instance, I can expect to get a pretty substantial return from tickets which match 4 of the 6 numbers drawn by the lottery.
There is a substantial literature related to the "Turan problem," which asks:  what is the minimum number of tickets I need to purchase in order to guarantee that one of my tickets matches 4 of the 6 numbers in the lottery?
My question is somewhat different.  Let's say I have enough capital to buy a fixed number N of tickets, large in absolute terms but small relative to 50 choose 6.  Then my expected gain is fixed.  But of course as a wise investor I may want to minimize the variance of my winnings.
Thus my question.  
If the random variable X is the number of (4 out of 6) wins among my N tickets, how small can I make Var(X) by judicious choice of ticket purchases?
(Of course, the same question applies for (k out of 6) where k=2,3,5.)
By the way, in case the setup seems unrealistic, let me add that the reason I'm asking this is that the situation described here actually happened, and I'm trying to reverse-engineer what the broker's risk-minimization strategy must have been, and assess whether it was worth it.
 A: This is not a real "answer" but an observation. Each 6-tuple has ${6 \choose 4}$ 4-tuples, so it stands to reason that once $N$ is some smallish multiple $m$ of ${50\choose 4}/{6\choose 4}$ then Bob's your uncle, and you can come reasonably close to equidistribution. This is to be contrasted with (say) the binomial expectation. The number of fours (or more) you expect is ${6\choose 4}m$, and the reasonable size $N=138180$ gives $m=9$ and 135 fours. The binomial distribution gives a variance about of slightly under $135$. I expect that one can essentially ensure about 135, plus or minus a small amount, via some covering selection.
ADDITION: In verities, the binomial model does not well model the random choices, as they have considerably higher variance, 180-185 compared to 134.8. I do not understand the theoretical concepts in toto, but an aspect is that the coverage of fours from the random sixes is already askew.
UPDATE: OK, here's the skinny on covering. I did a rather simple process. Do the following 138180 times. Pick a 4-tuple that so far has not appeared 9 times. Append to it the 2 numbers for which the resulting six minimizes the sum of the current counts of its 15 sub-fours. Accumulate the counts of 4-tuples from this six.
Then apply a bit of post-processing if you want (throw out populous sixes). This gives a set of 138180 6-tuples in which every 4-tuple appears between 7 and 11 times (the average is 135/15 or 9, with random choices of sixes the four-counts will range from 0 to 20 or more). Then simulate the ${50\choose 6}$ lotteries. These give an expectation of 135 fours, the minimum was 120 and the maximum was 149. The variance was a mere 5.8, versus 135 (binomial) or 180 (random). The binomial distro gives less than 120 a 8.9% chance (and more than 149 a 10.7% chance). As added above, the actual random distro is even worse than binomial.
I think this shows that with a (small) bit of work, some quite good variance reduction is possible. You can try to further trim the ends if desired. In the actual example, their edge was about 20-25% when free bets were included (later comments suggest 15-20% over the history). The accounting on page 7 says "12.8%" for just the cash component, but I get 425840/400000 is 6.4%. This analysis also lacks the jackpot, which I guess is equally likely to help/hurt among the big players (it is slightly chancey that only 1 of the 45 jackpots was hit, given there are 2-4 groups each buying up to 1/30 of the pool every time).
alt text http://www.freeimagehosting.net/newuploads/3tygt.jpg
