The poker group I play with has the following tournament structure: if 7 or fewer people are in it, the first place player gets 60% of the buy-ins, the second place 40%. If 8 or more people are playing, the third place player gets in on the act, with 60%-30%-10%.
If I’m an average player among mostly better than average players, how much less likely is it that I’ll earn money with 7 or fewer players than with 8 or more? Is there any way to tell?
Some odds junkies might take a crack at finding the real numbers but I’m more of a gut feeling type and I think you might have it backwards.
Adding additional players of a higher skill level will not increase your chances to win money by adding an additional paid spot. My guess is the first one “washes” the paid spot (3rd) and each additional player decreases your chances to cash.
The actual mathematics would be quite complicated, but look at it this way: with 7 or fewer players you go from a guarantied money with 2 players to having to beat 71.4% of the players to money. At 8 players, you have to beat 62.5% of the players. This percentage finally climbs above 71.4% with 11 players. This is a general rule in poker - your odds of making the money is best when it rolls over to a higher number of finalists.
Now the question is: how much worse off are you because a better player entered the game? As I said before, the math is complex, but would it offset the 8.9% advantage you get by adding a new player? I doubt it. So adding one more player will give you at most a 8.9% better chance in moneying.
But you still have chance of beating this person (although less than 50%) thus intuitively adding just one person would increase your chances slightly.
At the very worst, you still need to beat 5 people to money whether there is 7 or 8 people so we can reasonably conclude that adding an eighth person *cannot hurt * your chances of moneying (although I still hold that the chances improve).
-Saint Cad
We can drastically simplify the situation to get a handle on it: suppose that all the other players are in statistically independent competitions with you, i.e. whether or not you win over player A is independent of whether or not you win over player B. (This is, in all likelihood, false, but we forge ahead nonetheless.) Suppose further that your probability of beating another player is some number p. Then the portion of the pot you can expect to win with seven players (including yourself) is
0.6 * p[sup]6[/sup] + 0.4 * 6 * (1-p) * p[sup]5[/sup]
while with eight players, it’s
0.6 * p[sup]7[/sup] + 0.3 * 7 * (1-p) * p[sup]6[/sup] + 0.1 * 42 * (1-p)[sup]2[/sup] * p[sup]5[/sup]
I plotted the difference between these two functions for p between zero and one. It turns out that the eight-player expectation value is better for all p between 0 and 2/3 (i.e. for all poor players), and worse for p between 2/3 and 1. Unfortunately, the difference is very slight: the largest value of the difference is only about 0.007, and that’s for p = 0.53 or so. In other words, adding an eighth player will increase your expected winnings by at most 0.7% of the pot.
You can also repeat this exercise to see what the probability is that you’ll win something — i.e. you don’t care about how much you win, you just want some glory. In that case, you would replace all the 0.6, 0.4, 0.3, and 0.1’s in the above formulas with 1’s. You’ll find that everyone is more likely to get some money in the seven-player game, regardless of how good or bad they are.
Finally, if you take into account the fact that the pot is larger for the eight-player game (assuming that there’s a set buy-in fee for everyone), then you just multiply the eight-player formula by 8/7 to compensate for this. In this case, everyone, regardless of how good they are, can expect to win more in the eight-player game.
I’ve been very sketchy in the above description, so let me know if you want more details.