Clarifying bet and raise sizing in relation to the pot once and for all
This question is regularly asked in the forum or irc, so this is an attempt to clarify the issue. If you are more confused after I answered the question than you were before, I will consider becoming a politician.
One of the reasons for betting or raising to a certain amount is to lay pot odds of our choice for our opponent
This is easy to do in unopened pots (when nobody has opened the betting yet on the current street). We simply decide the odds we want to lay and bet the corresponding fraction of the current pot. For example say the pot is $5 and we want to lay 30% pot odds for our opponent. According to the table below, we simply bet 3/4 of the $5 pot ($3.75). Easy game.
Here are a few classic bet sizes, and the corresponding pot odds:
2x pot: 40% (3:2)
1.5x pot: 37.5% (5:3)
1x pot: 33.3% (2:1)
3/4 pot: 30% (7:3)
2/3 pot: 28.6% (5:2)
1/2 pot: 25% (3:1)
1/3 pot: 20% (4:1)
1/4 pot: 16.7% (5:1)
When thinking about raise sizes in relation to the pot size ("pot sized raise", "2/3rd pot sized raise"), we have to twist our mind a bit. The idea is that we want a 2/3rd pot sized raise to lay the same odds as would a 2/3rd pot sized bet in an unopened pot. I will first give you the shortcut to calculate the raise size, and for those who are interested I demonstrate the shortcut at the end of this post.
Let's call pot the actual pot size just before we make our play, including any bet(s) from our opponent(s).
Let's call B the bet size from our opponent. In multiway pots, this is the bet size of the opponent who last bet or raised.
Let's call C the amount we would have to call when it is our turn to play.
Let's define P=pot+C. This amount P is the "pot" in "2/3rd pot sized raise".
To make a 2/3rd pot sized raise, we raise 2/3rd of P extra on top of B. So we raise to B + (2/3) * P.
Here are some examples to make this clear:
Example 1:
pot is $1 (from previous streets)
villain bets $0.8 ($1.8 now in the pot)
-> pot=$1.8, C=$0.8, P=$2.6
-> to make a 2/3rd pot sized raise, we raise to $0.8 + (2/3) * $2.6 = $2.53
Example 2:
pot is $1 (from previous streets)
villain1 bets $0.8 ($1.8 now in the pot)
villain2 raises to $3 ($4.8 now in the pot)
-> pot=$4.8, C=$3, P=$7.8
-> to make a 2/3rd pot sized raise, we raise to $3 + (2/3) * $7.8 = $8.20 (by doing so we lay 28.6% pot odds for villain2 and, incidentally, more than 28.6% for villain1).
Example 3:
pot is $1 (from previous streets)
we bet $0.8 ($1.8 now in the pot)
villain raises to $3 ($4.8 now in the pot)
-> pot=$4.8, C=$3-$0.8=$2.2, P=$7
-> to make a 2/3rd pot sized reraise, we raise to $3 + (2/3) * $7 = $7.67
Note that some poker rooms' interfaces have buttons to help you bet or raise predetermined fractions of the pot. These conveniently perform the above calculations automatically. For poker rooms without these buttons, you can use software such as TableNinja to do the same job.
Now for the mathematically inclined, I will demonstrate all of the above in equations.
Let p be the size of an unopened pot.
Let f be a fraction
When we bet fp into p, fp is the amount our opponent has to call to continue, and p+2fp would be the total size of the pot after he calls. So we are laying fp / (p + 2fp) = f / (1 + 2f) pot odds (or (1+f):f for those who like this notation).
Conversely, to lay X% pot odds we have to bet a fraction f of the pot p such as:
f / (1 + 2f) = X/100
100f = X + 2fX
100f - 2Xf = X
f = X / (100 - 2X)
Now looking at already opened pots:
Let pot be the current amount in the pot just before it is our turn to play including any bet(s) from our opponent(s).
Let B be the bet of our opponent. In multiway pots, this is the bet size of the opponent who last bet or raised.
Let C be the amount we would have to call to continue when it is our turn.
Let R be the total amount that we need to raise to in order to lay the same f / (1 + 2f) pot odds as in the case of the unopened pot. Note that in multiway cases we assume that we want to lay these pot odds for the player who last bet/raised in the hand.
Just after we raise to R into pot, the total money in the pot will be pot+R-(B-C) (B-C being the money that was already in front of us from this round of betting before we raised). Now our opponent would have to call (R-B), and the total pot after he calls would be pot+R-(B-C)+(R-B) = pot+2R-2B+C. So we get this equation which we have to solve to find R:
(R - B) / (pot + 2R - 2B + C) = f / (1 + 2f)
(1 + 2f) * (R - B) = f * (pot + 2R - 2B + C)
R - B + 2fR - 2fB = f*pot + 2fR - 2fB + fC
R = f*pot + fC + B
R = B + f * (pot + C)
If we now define P = pot + C as the "pot" in "f pot sized raise", we find again the equation:
R = B + f * P
er... clear?
