[ QUOTE ]
f(x)= C(1/x)
f(x)= the amount of cash you win/lose ($)
x = the amount of tabels
C = konstant, depends on how fast your brains works [img]/images/graemlins/smile.gif[/img] etc.

^ f($)
|.............
|\................
|..\...............
|....\.............
|.....\............
|......\...........
--------------> x (tabels)

I hope this solved your problems [img]/images/graemlins/wink.gif[/img]

[/ QUOTE ]

I'm not sure that winrate is inversely related to number of tables played. For one thing, the constant C is determined by only one thing, not lots of factors....That is; what is your winrate at one table?

This is because f(1)= C.

Moreover, it must be the case that at some point you will begin losing money by trying to play too many tables....say 50. But the model 1/x approaches 0 as x -> infinity. So this says that if I play 10,000 tables I am essentially breaking even, which is clearly not true.

Also, winrate = 1/x => that by playing 2 tables you half your winrate, and by playing 4 you half that, etc. Which does not happen.

I wasn't sure if this was meant as more of an analogue or a specific model. In the case of the former it's apt enough, for the latter it doesn't work.

I believe a more accurate model is (C+k) - k*x, where C is your winrate at one table, and k is determined by how fast your brain works , etc. (x>= 1 is again # of tables)

Even this is too simplistic though as the loss is definitely not linear, in fact there is virtually no drop off from 1 to 2 tables. A better model (I'm too lazy to make this perfect) would be something that looked like:

2 * exp( -.025 * (x-1)^2 ) - (x-1) / 20

Where exp(*)=e^* and again x>=1. Plot this in matlab or maple and I think you'll see it's a more accurate picture of what happens.