I don't play or follow poker, (just thought I'd declare that), but found it interesting from speaking to a few people that do that their way of calculating the Expected Value, $E(x)$, didnt account for the money they had paid up front to play. I asked why and was basically told it doesn't matter. I can't understand why.
Lets say person $A$ and $B$ are playing against each other and it is known player $B$ will fold $35$% of the time, meaning $65$% of the time they won't.
So a hypothetical game is played and each time Player $B$ doesn't fold Player $A$ has really bad luck and looses. So Player $A$ is loosing $65$% of the time and Player $B\ 45$% of the time
$35$% of the time Player $A$, will win $\$20,(\$40$ in the middle - $\$20$ dollars of their own they put in).
$65$% of the time Player $A$ will loose $\$20$ dollars
Therefore: I calculated $E(x)=(0.35*20)-(.65*20)=7-13=-6$.
However I have been told that is wrong
The people I ran this by said when calculating the loss value I should not consider the $\$10$ put in as the initial playing fee. Meaning the calculation should be
$E(x)=(0.35*20)-(.65*10)=7-6.5=0.5$.
To play each player puts in $\$10$ and then at the point where they raise they raise by $\$10$. So each player has put in $\$20$ when it comes time to show their hand (there is now $\$40$ in total sitting in the middle).
This makes no sense to me, despite it looking good by not counting the $10 they are not seeing that their funds are diminishing....
Or have I got this wrong? For the life of me I can't work out why they said not to count the $\$10$ that is put in at the start. One justified it by saying you need to pay the $\$10$ to be in to win the $\$20$ profit. Yes this is true they need to pay the $\$10$ to play and have a chance of making $\$20$, over time it appears they'd loose money??
Is anyone able to clear this up for please ..thanks (or should I be asking in a different forum ?)