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Bet sizing to break even when stealing from the btn

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  1. #1

    Default Bet sizing to break even when stealing from the btn

    I know that as a general rule of thumb the tighter your two villains in the blinds are, the more widely you should steal from the button. What I'm a bit confused about is how to calculate the sizes of the raises in order to break even when behind facing two tight players.


    For simplicity sake, let's assume both SB & BB fold to steals 80% from their respective position. How do you do the math to figure out the size of your steal that's just high enough to break even? Is there a way to figure this out?


    (Also fyi, my btn steals are usually 2.5 to 3bb, but that's pretty unscientific.)
  2. #2
    spoonitnow's Avatar
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    SB folds x percent, BB folds y percent, our bet size is b in big blinds

    Both players fold: x*y percent of the time (or just xy for short)

    If we're not taking into account anything post-flop and assume we just lose the hand right then if the blinds don't fold:

    EV of Stealing = (1.5)(xy) + (-b)(1-xy)

    We want the EV of stealing to be zero:

    (1.5)(xy) + (-b)(1-xy) = 0
    1.5xy - b + bxy = 0
    -b + bxy = -1.5xy
    b(-1 + xy) = -1.5xy
    b = -1.5xy/(-1 + xy)

    Example: SB folds 75 percent, BB folds 60 percent

    xy is 0.75 * 0.60 = 0.45

    b = -1.5xy/(-1 + xy)
    b = -1.5(0.45)/(-1 + 0.45)
    b = -0.675/(-0.55)
    b = 1.227

    If you want to understand the math of this a little bit better, I wrote an entire series that teaches you how to do EV calculations using only basic addition and multiplication. The first part of the series is here.
  3. #3
    Quote Originally Posted by spoonitnow View Post
    Example: SB folds 75 percent, BB folds 60 percent

    xy is 0.75 * 0.60 = 0.45

    b = -1.5xy/(-1 + xy)
    b = -1.5(0.45)/(-1 + 0.45)
    b = -0.675/(-0.55)
    b = 1.227

    If you want to understand the math of this a little bit better, I wrote an entire series that teaches you how to do EV calculations using only basic addition and multiplication. The first part of the series is here.
    Thanks, spoon. I'll read your article in a minute after I try to digest this. Your example is more realistic, but let me see if I can do the math using my more extreme scenario of two nits in the blinds using your formula. We are solving for b:

    SB & BB each fold 80%, so 80% * 80% = both fold 64%

    b = (-1.5 * .64) / (-1 + .64)
    b = -96 / -.36

    thus...

    b = 2.66.

    I'm guessing that theoretically I'd need the raise to be 2.66 to break even, no? (I understand this isn't necessarily in game thinking, but I'm trying to how the numbers break down.)
  4. #4
    spoonitnow's Avatar
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    That's right. Note that when b is less than 2, we have to make it 2 because we can only minraise.

    The basic idea is that the more someone is folding in any situation, the larger your bet size can be while staying profitable (not that it should necessarily be larger).
  5. #5
    Two follow up questions....

    This is less about the math here, and more about how villains respond to steals (or bets on any street for that matter): Just because we can raise more and technically remain profitable, should we always be raising more? I mean, if we can get away with a smaller raise and still remain profitable vs an opponent(s) -- as in the example above, or let's say on the flop when you can bet in amounts smaller than the pot -- wouldn't we rather risk less to win the same amount of $?

    Also:


    Let's say we look at this from the BB's perspective in dealing with a steal attempt from the Btn with the SB folding. How do we figure out the break even point when we're facing a bet (or a raise, or 3bet, etc.) and considering coming over the top? Assuming we have a ton of hands for villain's fold to resteal %, and assuming we always lose when they call, do we basically simply plug in the numbers to the same formula above?
  6. #6
    spoonitnow's Avatar
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    To answer your first question, one thing you have to think about is the entire range that you're betting or raising with in the given situation and how the change in bet size will affect your total EV.

    On the subject of choosing between multiple bet sizes for bluffs, your bet size will typically change how much the opponent folds. That's a whole subject to itself, but the basic thing to know is that there are elastic and inelastic folding ranges. An elastic folding range will change a lot based on changing your bet size, and an inelastic folding range will not change much based on your bet size. Against an elastic folding range, you can try to figure out the bet size that gives your bluff the highest EV based on the corresponding folding frequency you get. Against an inelastic folding range, you should generally bet the lowest you can get away with in terms of maximizing the profitability of the bluff. I hope that gives you some ideas to think about on your own while answering your question the best I can without writing a novella on the subject.

    And it basically works out kind of the same. Like if the guy raises to 3x from the button and the SB folds, then we 3-bet in the BB to a total of 11x (including our blind), then we're betting 10bb and the pot after we put in our 3-bet is 14.5bb, so we do 10/14.5 = 69 percent, which is how often he has to fold for our bluff to be profitable in a vacuum without taking post-flop play into consideration.

    On this last point, go read the second part of the NLHE Foundations Series since it covers the bluffing situation and figuring out how often the guy has to fold, etc., in a lot of depth.

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