The first graph here is wrong. Recall the constraint that only you and people you invite can enter. If you invite n friends, you have n+1 votes. Since each of y

The first graph here is wrong. Recall the constraint that only you and people you invite can enter. If you invite n friends, you have n+1 votes. Since each of your invites also gets a vote, there are (n+1) + n votes total, or 2n + 1. So, for a given n, your chances of winning is (n+1)/(2n+1) . The limit of (n+1)/(2n+1) as n->infinity isn't 1, as the graph implies; it's 1/2. (http://www.wolframalpha.com/input/?i=limit%20%28n%2B1%29%2F%282n%2B1%29%20as%20n%20goes%20to%20infinity http://www.wolframalpha.com/input/?i=limit%20%28n%2B1%29%2F%...). So the more people you invite, the closer your chances of winning get to 50%. If none of your friends accepts your invite, you have one vote, and your friends have no votes. 100% chance of winning. If one of your friends accepts your invite, you have two votes (your original, plus your bonus), and your friend has one vote. 2/3 chance of winning. For two friends, you're at 3/5; for three, 4/7, etc. Note that we're above 50%, so the delta in your chance of winning is actually negative for each friend you invite. The author is also confused whether you get an initial vote or not, but that's less important. The author didn't seem to count the friends' votes anywhere: "If two friends sign up, then you have a 2/3 chance of winning, then 3/4, then 4/5, and so on." If two friends sign up, there are either four or five votes total, neither of which is divisible by three, as that sentence states. But the author's final conclusion is correct -- at least, it's close enough; you want to share with your friends such that they have enough time to enter the contest, but not enough time to share withe their friends. The appropriate time, of course, varies per-friend.