Ok, I've been having trouble figuring this up at the table in a few situations, so I want to review it. Easy is if the initial bettor is the only one in the hand: You raise to 3 times his bet plus the original pot.
But it's more complicated if various players have called a bet to you. I want to make a raise here so that the initial bettor is getting 2:1 for my first scenario.
To make this an easy general formula that one can calcuate right at the table, let's call the initial bet b, and the pot, INCLUDING all bets already in, p. My raise is going to be a total of r.
Now, the initial bettor has to call r-b. For him to be getting 2:1 on the call, we have 2*(r-b) = p + r, hence, 2*r - 2*b = p + r iff r = p + 2*b.
So, your raise is simply to total pot (INCLUDING all bets already in) plus twice the initial bet. So, if someone bets $15 into a $25 pot, 2 callers to you, the current pot is 25 + 45 = 70, and you add 30 to that. So, a raise to 100 means that the initial bettor has to call 85 for a shot at a pot of 170. Looks good to me.
I'll put the second case I've been having difficulty with in a separate post in this same thread.