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[QUOTE="Bombjack, post: 445922, member: 11914"]
[FONT=Verdana][SIZE=2]To put it mathematically, if
X = the amount initially in the pot
p = your chance of winning
S = the larger of your or your opponent's remaining stack

If you both go all-in, your expected return
Ea = p*(X+S) - (1-p)*(S)
    = pX + 2pS - S

If you cause your opponent to fold, your expected return
Ef = 1*X

So getting your opponent to [B]fold [/B]is better for you if [/SIZE][/FONT][FONT=Verdana][SIZE=2]
[/SIZE][/FONT]   [FONT=Verdana][SIZE=2]Ef > Ea
X > pX + 2pS - S
X + S > p(X + 2S)[/SIZE][/FONT][FONT=Verdana][SIZE=2]
[B]p < (X + S)/(X + 2S)[/B]   [/SIZE][/FONT][FONT=Verdana][SIZE=2]

So:
If there’s very little in the pot, or your stacks are very deep compared to the pot, the right hand side tends to 0.5: in other words, you want your opponent to fold whenever your probability of winning is less than half, and call when it’s more than half.[/SIZE][/FONT][FONT=Verdana][SIZE=2]

[/SIZE][/FONT]      [SIZE=2]If the pot is large, or your stacks or short, the r.h.s. tends to 1, so [I]getting your opponent to fold is better for you even when your probability of winning is very high[/I].[/SIZE][FONT=Verdana][SIZE=2]

[/SIZE][/FONT]      [FONT=Verdana][SIZE=2]
In the above example, X = 4.10, S = 17.18[/SIZE][/FONT][FONT=Verdana][SIZE=2]
[/SIZE][/FONT]   [SIZE=2]So getting my opponent to fold is advantageous is my chance of winning is less than (4.1 + 17.18)/(4.1 + 2*17.18) = 55.3%[/SIZE][FONT=Verdana][SIZE=2]
(I was slightly out with the 61% above, but here it is worked out properly.)

[/SIZE][/FONT]    [SIZE=2]A bit tricky to calculate on the fly, but maybe I’ll make a few more examples to practice working out what the best move is.[/SIZE]
[/QUOTE]

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