shockkkk

Granted I haven't played online poker seriously in 3-4 years, but I still like to log onto stars and play a few hours each week. I know I'm late to the discussion, but I feel there is a major flaw in the Bhupan analogy. Essentially, the situation where Bhupan is losing less at the rec-filled high rake table than at the pro-heavy low rake table won't happen because pros will realize they can win more in the rec-heavy high rake games. At equilibrium, both pros and recs can expect to earn the same winrate playing either game, obviously with pros making more than recs. The caveat here is that charging a higher rake merely yields a higher effective rake for both pros and recs. If this winrate for pros drops below zero, pros gradually leave the game (and in reality will be the weakest pros) until the rec:pro ratio becomes large enough for profit to be sustained once again. However, the benefits recs recieve from greater rec:pro ratios are cancelled out by the increases in rake.

Feel free to skip this as the rest is just technical game theory stuff to reinforce my summary in the paragraph above.

Consider the population of poker players, a proportion p of whom are deemed professional. We assume pros win at a rate w off the recs, and zero against each other. Subsequently, recs win zero off other recs. There are two sites running, one of which offers a rake h (the high rake site) and the other a rake of l (the low rake site). For simplicity, I'll assume each site is capable of running the same number of games though the core results do not depend on this. Let p1 and p2 be the proportion of pros on site l and site h respectively. Note that (p1+p2)/2 = p.

Winrate at each site is given by:
(1-p1)w - l (pro at low rake site)
-p1w - l (rec at low rake site)

(1-p2)w - h (pro at high rake site)
-p2w - h (rec at high rake site)

Equilibrium conditions are:
(1-p1)w - l = (1-p2)w - h
-p1w - l = -p2w - h

Using these along with (p1+p2)/2 = p we get:
p1 = p + (h-l)/2w
p2 = p - (h-l)/2w

Winrates at equilibrium are given by:
Pro winrate: (1-p)w - (h+l)/2
Rec winrate: -pw - (h+l)/2

So essentially the outcome is the same as if everyone played on a single site where rake was the average between the two sites. Clearly, if the high rake site chose to offer the lower rake l, both pros and recs would be better off. Implicit is the assumption that the pro winrate exceeds zero, that is (1-p)w - (h+l)/2 >= 0. If it doesn't, the proportion of pros p will fall until the winrate is no longer negative.

*In real life, due to oppurtunity cost, the threshold for the winrate will be some positive number.

Consider the situation where pros can only breakeven. The winrates are given by:

Pro winrate: (1-p)w - (h+l)/2 = 0
Rec winrate: -pw - (h+l)/2

Now consider the high rake site wants to increase rake to drive out professionals. In this case, it may seem like the rec players could benefit from increased rake lowering the amount of pros in the games, but it turns out the benefit to driving away professionals is offset by increases in rake. We can look at how the rec winrate, now denoted by R, changes:

dR/dh = (dp/dh)(-w) - 1/2

Note that changing h results in a new p, and specifically the new p will satisfy:

(1-p)w - (h+l)/2 = 0 ------> p = 1 - (h+l)/2w
and that dp/dh = -1/(2w)

Thus:

dR/dh = (-1/2w)(-w) - 1/2 = 0

And rec players recieve no benefit from this rake increase. Long story short, higher rake screws everyone but Amaya.

I would guess it has something to do with not letting 95/85/65 type stuff from having too profitable of a bluff on the river because they block a large portion of villains calling range. I think this is also the reason behind the plethora of mixed strategies this program recommends as well.