I play a fair amount of poker, both cash games and tournaments, both online and in "brick and mortar" casinos. Thus, of course, the World Series of Poker is of more than passing interest. That interest is shared by my CFO, and he sends me updates on the Main Event via im (internet messenger). He mentioned that, before tonight's dinner break, there were 1431 players left, and then mentioned later that there were 1071 (out of a starting field of 6494). And intuitively, the rate of elimination of players should be proportional to the number of players remaining. Sounds like a first order differential equation to me.
So, let p(t) be the number of players left at time t, then dp/dt=k*p(t) with k a negative constant. Now the time stamp from msm im for 1431 was at 18:54 PDT, and there were 1071 at 20:38 PDT. This is 104 minutes, but the dinner break is 90 minutes. This means that 360 eliminations took place in 14 minutes, absolutely not possible. This must mean that the updates are sporadic. I'll try a different method of finding two times with a known number of players and a known number of playing minutes between them. It looks like Level 13 ended with 945 players, and 90 minutes of play later, Level 14 ended with 810 players. Even this is sketchy, WSOP's own tournament update page has both 810 and 789 players as the current count. Oh well, I'm going with what I have.
Thus: dp(t)/dt-k*p(t)=0, p(0)=945, p(90)=810. This is a very simple problem and the solution is p(t)=945*exp(-0.0017127853*t). So, the tournament is won when p(t)=1, i.e., when one player is left. Now, this model has assumed that p(t) is a continuous function on the real numbers. At p(t)=945, this approximation may be reasonably accurate but when p(t) is, say, 3, it's clearly not. There are not going to be 3.43 or 2.71 players left. Nonetheless, solving the expression for t when p(t) =1 yields t of approximately 4000. That means after about 4000 more minutes of play, we should know a champion. That's about 67 hours and since they play something like 11 more days with 12 hours per day less breaks of about 3 hours, I'd guess there are actually about 100 hours left.
Various strategic considerations lead to certain times (particularly around getting "into the money" and getting to "the final table") when stalling and cautious play takes place.Thus, k, which I've assumed is a constant, is actually a function of p(t). That's a much more difficult problem though, and I'd only be guessing at the values of the relevant parameters. Possibly analyzing p(t) "piecewise" with k constant in each piece would do better. But the error is also partly attributable to the inaccurate reporting of players left at any given time. So predicting 67 hours is gratifyingly close. For a general idea, see the graph below.
This is, as the mathematically inclined will immediately see, a completely trivial problem. But for those who are not, it may be interesting to see the way a backslid mathematician looks at the world. For those whose interest is piqued by this exercise, I'd recommend "Towing Icebergs, Falling Dominoes, and Other Adventures in Applied Mathematics" by Robert B. Banks.