At the end of this study, we wish to be 95% confident of the interval in which the true winning percentage p lies. We can model the outcome of entering head-to-heads as having a binomial distribution with parameter p (here we assume the contests are independent, which is generally true for head-to-heads). Suppose that we win w games and lose l games. Then, our current estimate of p is r=w/(w+l). We also desire that our error e=|p-r|<0.02. The variance of r is then r(1-r)/n, where n is the number of contests entered. Thus, if a z-score of z is desired, z \sqrt(r(1-r)/n)<e, or n>r(1-r)z^2/e^2. r is approximately p, and we would like z=2 and e=0.02. Thus, n>2400. There are 17 weeks in the NFL season, so we would have to enter around 150 contests per week.
We wish to maintain enough bankroll to play for an entire season. Thus, assuming that we bet x percent of our original bankroll on each week, we would like to have less than a 1% chance of losing our entire bankroll by the end of the season. To perform a worst case analysis, suppose that we lose all of our bets in losing weeks, and break even in winning weeks. Then, we lose all of our bankroll if we have 100/x losing weeks. If we assume, probability of having a losing week is 40%, then we must have x<8 (http://www.wolframalpha.com/input/?i=sum+from+i%3D12+to+17+of+%2817+choose+i%29*0.4^i*0.6^%2817-i%29). Since this is a worst-case scenario, we may set x=10, so we bet 10% of our bankroll each week. This is still a very conservative number though, so we can probably raise this to 15-20% and still be safe.
We plan to play in low-stakes head-to-head ($1, $2, or $5 buy-ins) so we expect our average buy-in to be about $3. Since we play 150 contests per week, we will bet around $500 per week, so our total bankroll will be $5000.
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