Hello,

I ve read on this site an article about poker odds. There was one thing that I did not understand and I hope someone can explain it to me.

This was written:
I have JT and the flop comes A 8 9 - rainbow.
The pot is current at $8.00, and Player1 bets $2.00. Everyone else folds.
I have a 31.5% chance to hit my hand over the turn and river.
I have to call a $2.00 bet to win what will be a $12.00 pot. Since my bet is only about 17% of the pot, yet I have a 31.5% chance of winning (across the turn & river card!), the "pot odds" justify making the call.

Then a little bit later this was written about the same example only then with implied odds:
Let's say you know Player1 will probably bet another $4 on the turn. So really, you are looking at calling $6.00 against a pot of $16.00. The percentage now is 37.5% (6/16) as opposed to our earlier example where it was 17%. In this case, the call isn't as clear and the current pot odds don't quite justify the call because 37.5% is greater than our 31.5%. However, there are implied odds to consider, and these odds may justify making this call.

The thing that I do not understand is the way they calculate the money with the implied odds. In the first example they are telling me that I have to call a $2.00 bet to win what will be a $12.00 pot ($8 was already in the pot en both players put in 2 dollars, so 12 in total.)
Now in the implied odds part they are telling me that I have to call $6 to win 16$. But why sixteen dollars? Then you are only calculating the extra four dollars put in by player one on the turn. But I also put in an extra four dollars when I call on the turn. So you are looking at a pot of 20 dollars I would say, because that is what they did at the first example (they also took my own bet into the calculation). So I m a bit confused. Now is my question whether you should also take your own bet into the calculation or not?

I hope someone can tell me the answer

Martijn (Netherlands)