Just a quick question. does one surrender when the chance of winning that hand is less then 25%. Because if you always lose 1/2 your bet every time for that hand thats like winning 1 out of 4 times for that hand...is this right?
In practice however the correct play is determined by calculating the expectation value of each action using combinatorial analysis and or simulations.
An easy way to figure in the pushes: Count each push as 1/2 of a win.
Example: Your chance of winning is 20%, chance of losing 70%, chance of push 10%.
Win ---------- 20
1/2 of push ---- 5
___
Total ------ 25 (the cutoff point at which it doesn't matter whether you surrender or not)
Extreme example: Chance of winning: 0, chance of losing 50%, chance of a push 50%. Nothing gained or lost by surrendering, even though you have ZERO chance of winning!