Great to see you're working through this.

You got it right as far as F*P=1.13

After that, you've gone slightly wrong, and to get it right you have to understand the brackets (that's why I said in the other thread it's very difficult to explain it entirely without the brackets).

An example is if I say to you what is 4 x 5 + 3

You could say 4x5=20 + 3 = 23
Or you could say 4 x (5+3) which is 4 x 8 which = 32

So the brackets are important, because they tell you to do the sum inside the bracket _before_ the rest.

So when you do F*P + (1-F) you're doing that wrong - (1-F) on it's own isn't much use, it's just that it's (1-F) of the time that he calls (whenever he doesn't fold), so (1-F) gets multiplied by the EV when he calls.

F*P + (1-F) [E*(P+S) - (1-E)*S]

The square brackets are used just to make it more readable, you could just as correctly write:

F*P + (1-F) (E*(P+S) - (1-E)*S)

To answer your question of "is it 2 separate sums), yeah, you need to evaluate (P+S) first (1.30+9.00), then you multiply that by E. Then you evaluate (1-E) and multiple that by S. Then you take (1-E)*S away from (E*(P+S)). Then you take the result of that, and you multiply it by (1-F).

So,

F*P + (1-F)[E*(P+S) - (1-E)*S]

0.87*1.30 + (1-0.87)[0.42*(1.30+9.00) - (1-0.42)*9.00]

1.13 + 0.13*[0.42*10.30 - 0.58*9.00]
1.13 + 0.13*[4.326 - 5.22]
1.13 + 0.13*-0.894
1.13 - 0.11622 = 1.01378

or rounding that off, you expect to make about $1.01 each time you shove in that scenario.

Look back at the worked example I did above, and you'll see that I work my way out, evaluating things that are in brackets first, then continuing to work my way "out" from the bracketed parts.