In the middle is 10 dollars, someone bets 5. I have to call 5 dollars in order to win 10 dollars.
My Pot Odds - 5:10 = 1:2
I can interpret it as that i can lose 2 games and win 1 and I ll be break-even. Therefore I must win 1/(1+2) of the time (which is 33 and 1/3 %).
And I came with another inrepretationan and I want to ask if it´s correct.
If I have a similar situation with Pot Odds - 2:3, which literally means that I can lose 1 and a half time if I win once, meaning I need to win 1/2.5=40 % of the time
The other way to look at is the amount of money that was given into the pot by myself... which is in case of pot odds 2:3... 2/5=40 % of the pot. So the 40 % of they money in the pot is my money I gave there.
QUESTION 1) In this interpretation, if I have so good cards that I have the chance of winning ALSO 40 % of the time, does that mean that I win on average 40 % of the money in the pot and therefore i am break-even? Because I (on average) win 40 % of the money in the pot and 40 % of the money in the pot is given by me? So it returns back?
QUESTION 2) Does money in the pot * chance of winning mean the amount I win on average (correct me or say if it´s right what I say in the last paragraph)?
formula:
money * probability = 10 dollars * 0,4 = 4 dollars
QUESTION 3) Interpretations of the given formula
FIRST INTERPRETATION: If yes, why does that work? Only because it can be interpreted as that I win 4 times 40 dollars and the average I won during 10 games was 4 dollars per game (10*4 and divede by 10)?
SECOND INTERPRETATION: I think the second interpretation of why this formula works is that it can be seen as weightened average.
(10 dollars * 0,4 + 0 dollars * 0,6) / 1
0 dollars, because in 6 out of 10 you "win/lose, doesnt really matter how u call it" 0 dollars, the other 4 out of 10 times you win 10 dollars.
And therefore we can look at the problem as the weightened average of the amount of 10 dollars which has weight 0.4 and number 0 which has weight 0.6. I am not sure about this one, i just made it up, I am not sure about this one and I´d like to hear if it´s correct and if yes, why. Because i am not 100% sure of the concept of weights in averages.
Your first example is fine. You have made the unstated assumption that your $5$ is the final bet, but under that assumption you need to win $\frac 13$ of the time to break even. You are quoting the odds in the reverse order from Wikipedia, which would call this situation $2:1$. Your second example is correct as well, but the last paragraph may or may not be true and you don't care. It doesn't matter how the money got into the pot. Maybe it all came from the blinds, maybe you bet it on a previous round, maybe some of each. All that matters is how much is there and how much it costs to try to win it. I don't understand your question 1 at all. For question 2, yes, your expected win is money in the pot times the chance to win. Your expected loss is the bet required to play times the chance to lose. If these are equal, the bet is fair and it doesn't matter if you make it or not. If the expected win is greater than the expected loss you should make the bet.