I have a couple questions about an example from Bill Chen and Jerrod Ankenman's book The Mathematics of Poker. Below is a screenshot of the example (Example 4.3).

MOP Example 4.3

My questions are:

  1. If the game is $30-$60, how can the pot be $75? Shouldn't the pot be at least $120 (2x the BB)?

  2. Why should B call if his chance of winning is greater than 30/105? Shouldn't it be 30/135? After A bets $30, if B calls, B is risking $30 to win $105, which implies a 30/135 chance right?

1 Answer 1

  1. They're playing limit hold'em. That pot size means the SB folded, BB checked and the other player. 15$ from the SB, 30$ from the BB and 30$ from the other player. Limit hold'em or fixed-limit hold'em has a fixed amount that is the max bet and raise. That max changes depending on the street. In the case of this 30$-60$ game, the limit of betting is 30$ pre-flop and on the flop. It'll be 60$ on the turn and river. Important to also note as it's very important in limit hold'em is there is a cap on betting of 1 bet and 3 raises. So at the 30$ stage the max any play will be able to make their bet would be 120$, unless there is only two players in the hand. I understand the confusion as limit hold'em is not popular these days, but you do still get it spread in Vegas a decent bit.

  2. It is correct that it is 105 and not 135. The player's money isn't in the pot yet, therefore they can add that to the calculation as an amount to win. An example further on this from the next street, if we call and then the other player again bets on the turn the calculation would be - $105 + $30 from the flop we called, then + 60$ of the bet on the turn. We'd have to call 60 to win 195 - 60 / 195. Basically you can't count the money you need to put in as part of the pot you'll win, it's not in the pot yet.

  • For question #2, I am not counting the money you need to put in as part of the pot you'll win. The pot that you'll win is still 105 (75 plus the 30 that the other player bet). However, the probability this implies (when you have to risk 30) is 30/135 still though right? Since p(win)*(amt won) + p(lose)*(amt lost) = (30/135)*(105) + (105/135)*(-30) = 0, so indeed when (amt won) is $105, the implied p(win)=30/135. Is this incorrect?
    – Tony Bai
    Commented Aug 26, 2021 at 20:55

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