[CurryLover]

Yes, that is an example. Nice one m8!

After I made my initial post I thought of an example myself. It's not as good as yours and it is a pre-flop example, but here it is:

You are in a 15(!) handed NL Hold'em game with 2 Black Aces. You have the current nuts, but cannot possibly win if all the hands see the river because they are:
A [img]/images/graemlins/spade.gif[/img] A [img]/images/graemlins/club.gif[/img]
A [img]/images/graemlins/heart.gif[/img] K [img]/images/graemlins/diamond.gif[/img]
A [img]/images/graemlins/diamond.gif[/img] K [img]/images/graemlins/heart.gif[/img]
K [img]/images/graemlins/spade.gif[/img] K [img]/images/graemlins/club.gif[/img]
Q [img]/images/graemlins/spade.gif[/img] Q [img]/images/graemlins/club.gif[/img]
J [img]/images/graemlins/spade.gif[/img] J [img]/images/graemlins/club.gif[/img]
T [img]/images/graemlins/spade.gif[/img] T [img]/images/graemlins/club.gif[/img]
9 [img]/images/graemlins/spade.gif[/img] 9 [img]/images/graemlins/club.gif[/img]
8 [img]/images/graemlins/spade.gif[/img] 8 [img]/images/graemlins/club.gif[/img]
7 [img]/images/graemlins/spade.gif[/img] 7 [img]/images/graemlins/club.gif[/img]
6 [img]/images/graemlins/spade.gif[/img] 6 [img]/images/graemlins/club.gif[/img]
5 [img]/images/graemlins/spade.gif[/img] 5 [img]/images/graemlins/club.gif[/img]
4 [img]/images/graemlins/spade.gif[/img] 4 [img]/images/graemlins/club.gif[/img]
3 [img]/images/graemlins/spade.gif[/img] 3 [img]/images/graemlins/club.gif[/img]
2 [img]/images/graemlins/spade.gif[/img] 2 [img]/images/graemlins/club.gif[/img]

Let's say you are UTG. The pot is $0 (okay, there will be blinds in there, but to keep the maths simple we'll call it a pot of zero). You decide to go all-in for $1,000.

How many hands do you want to call your bet? I am not so good at the maths involved, but can work a few things out. For example:

Everyone passes: EV = $0

Only 2 [img]/images/graemlins/spade.gif[/img] 2 [img]/images/graemlins/club.gif[/img] calls: EV = $600
(Rounding the percentages, you will win 4 times out of 5. So 4 times you win £1,000 and one time you lose $1,000. That means your EV is $3,000/5 = $600)

Only 2 [img]/images/graemlins/spade.gif[/img] 2 [img]/images/graemlins/club.gif[/img] and 3 [img]/images/graemlins/spade.gif[/img]3 [img]/images/graemlins/club.gif[/img] calls: EV = $1,100
(You will win about 7 times out of 10. So your EV = ($14,000 - £3,000)/10 = $1,100)

4 [img]/images/graemlins/spade.gif[/img]4 [img]/images/graemlins/club.gif[/img] also calls: EV = $1,200
(You will win about 11 times out of 20. So your EV = ($33,000 - $9,000)/20 = £1,200)

5 [img]/images/graemlins/spade.gif[/img]5 [img]/images/graemlins/club.gif[/img] also calls: EV = $1,250
(You will win about 9 times out of 20. So EV = ($36,000 - $11,000)/20 = $1,250)

6 [img]/images/graemlins/spade.gif[/img]6 [img]/images/graemlins/club.gif[/img] also calls: EV = $1,000
(You will win about 15 times out of 40. So EV = ($75,000 - $35,000)/40 = $1,000)

So, it seems like the optimum number of callers for you in this coup would be 4 since that is the highest EV. Any more than 4 callers and your EV starts to decline. Of course, this is not exact because different combinations of callers might be slightly better or slightly worse for you - but it is a reasonable estimate I think.

For those of you who are good at maths (I am not) can you check my working and my conclusion for me in case I've botched it up? Also, is there a method or formula that would allow me to work out the above in a quicker, more accurate way?

I am not sure what conclusions are to be reached from thinking about this type of theoretical idea. However, I think there might be some understanding to be gleaned somewhere in there about the inner structure of the game. What do others think?