This book at once thrills and infuriates me. It thrills me in that it is the book I wish I had written myself. It infuriates me in that the little niggling errors—typographic and logical—drive me to distraction as I read it. This thread is to focus on the logical errors; if you have found any that I have not, feel free to add them. It would be nice to have everything in a single place to point people at. (If I am mistaken in any of my corrigenda, please correct me too.) Because this would be massively long if I posted all the corrigenda at once, I've split them up into separate replies in this thread.

1. Ch. 11, Ex. 11.3, [0,1] Game #2, p. 118, para. 1

MoP: "...Y could unilaterally improve his equity by checking hands from x_1 [sic] to y_1 and instead bluffing an equal amount of hands near 1. By doing this, he improves the equity of those bluffs because X will more often fold the winning hand against those bluffs..."

While it is true that this strategy change improves Y's equity, the reason is wrong: his bluffs are still worth the same in equity, i.e., $0$. What improves is the equity of his checked hands; the hands near $1$ lose to a larger range of X's hands than the hands just above $x_1^*$.

2. Ch. 12, Ex. 12.3, Jam-or-Fold No-limit Holdem, p. 133, paras. 2-8

MoP: "...we can see that jamming with A5s is actually better than jamming with ATs when the opponent will call with both AA and AKs...We add hands marginally now as they perform best – so A5s goes to 100% as well:..."

I would not have caught this flaw if I had not done up my own solution of the jam-or-fold game (spreadsheet linked to in another thread). My values for the thresholds are slightly different from those in the book, and I'm not sure whose values are correct; but the book's description of what happens at the second threshold is wrong.

The fallacy the book commits is to look at the value of ATs vs. A5s against the indifferent hand alone rather than the entire calling range. Although it is true that A5s performs better against AKs alone, ATs does better against the range, which at this point is 100% AA with an almost-vanishingly-small sprinkling of AKs. Because the book is considering what happens at 833 big blinds, just below the threshold, I'll do the same. Let's start with Caller calling 1% of AKs and move it down from there.

With 100% AA and 1% AKs in the calling range, ATs and A5s both are underdogs against the entire range and will fold. So Caller has to lower the AKs percentage. As it decreases, the first critical value is at 0.1969%; this makes Jammer indifferent with ATs. But even when Jammer jams with 100% of ATs's, Caller still doesn't get enough equity to call with AKs, so he reduces its call percentage even further. At 0.05839%, Jammer is indifferent with A5s, and now she adds enough A5s hands (85.508%) to make Caller indifferent with AKs. This is the Nash equilibrium.

It is true that at a certain stack size, there will be enough AKs in Caller's range so that A5s becomes the equity favourite against the range as a whole, and ATs becomes the mixed hand. Again by my calculations, this occurs at 819.20 big blinds, when Caller calls AKs 3.1690% of the time. It is also true that there are thresholds where some new hand shoots up to 100% while the marginal hand stays marginal; this happens at the very next threshold in fact, 807.52 big blinds, when AKs switches from pure fold to pure jam while ATs remains Jammer's marginal hand with a reduced percentage.

3. Same example, p. 135

At exactly 3 big blinds, my calculations give a mixed strategy. (I agree with the given strategy at 2 big blinds.) I have T2 is a mixed (34.4%) jam, and 63 is a mixed (46.2%) call. MoP gives these as a pure jam and pure fold respectively; the rest of our range is in agreement. I do not have MoP's exact jamming and folding ranges at any stack size, although two nearby pure strategies are close: Just above between 3.005–3.013 big blinds, the call range is the same but the jam range is missing 85; and just below between 2.935–2.985 big blinds, the jam range is the same but the call range adds 63.

4. Ch. 15, Ex. 15.2, AKQ Game #5, p. 169, table

The third-to-last line of the table should go:
Q K b1—fold 4 4/75 4/75
(the three numbers are too small by a factor of 4)

5. Ch. 16, Ex. 16.1, [0,1] Game #4, p. 179, para. 1

MoP: "Y gains value when he bets a hand after X checks. Otherwise, Y's strategy is forced. So to prevent Y from exploiting him, X can simply bet all hands. If he does this, he guarantees that the game will have equal equity for both players (one bet will go in)."

Again, correct conclusion, but wrong reasoning. Y does not gain value when he bets his good hands—he gains value when he can check his bad ones. X's optimal strategy, then, is to deprive Y of checking space! And yes, she does this by betting all her hands, since Y can only check behind if X has checked in front.

6. Ch. 18, pp. 222–3

At the bottom of p. 222, the first three equations are incorrect; the isolated 'g' term should be a gx_1^2. (Somehow, though, the fourth and following equations are correct.) Also, in the third equation there is a 1 that should be subscripted but isn't.

The equations at the top of p. 223 have a more serious error: In the second equation, the values from Game #4 are used, when Game #6 values should be used instead. The equation should read:

\< Y \> = (r/(1+2r))^2 + (r/2)(1/(1+2r))^2

This gives a game value of approximately 0.11327 instead of 0.11428.

7. Ch. 18, pp. 228–9, table et seq.

After the first two lines, the ordering of hands and the "Number of Ways", plain and cumulative, are in error. They should be:
JJ,99,88,33,22 – 15 – 47
J9,J8,J3 – 27 – 74
worse 2 pair – 63 – 137
AA–QQ – 18 – 155
AJ–QJ – 36 – 191
JT,J7–J4 – 60 – 251
TT – 6 – 257
9x – 96 – 353
8x – 96 – 449
Other hands – 632 – 1081

The first paragraph of p. 229 should be adjusted to reflect these values.

8. Ch. 19, Fig. 18.7, The [0,1]×[0,1] Game, p. 231

The regions are on the wrong side of the diagonal.

9. Ch. 19, p. 236, para. 12

Not necessarily an erratum, but definitely a question. When the bet size changes from 1 unit on the flop to 2 units on the turn, could check-bet-? now be a valid strategy for Y? It is clearly dominated when the bets are identical, but does some weird AKQ-Game-#5-like scenario arise where we underbet our strong hands to extract more equity from our weaker ones? Whether this happens or not, clarification at this point might be in order.

10. Ch. 19, Ex. 19.6, pp. 246–7

Everything inside the Sigma is confusing and it took me three full days' work to convince myself that the book's solution was, in fact, the correct one. No explanation or derivation is given for the second formula, and in fact I found a small error that doesn't change the result (the $ΔsΔx$ should be $2ΔsΔx$). Where did it come from? I have no idea, but here's my derivation:

<Y, bets s with a losing hand> = –sx(s) – \int_0^s (1+t)x'(t) dt + k_x

The first term is Y's loss when X bets $s$ or more; the integral evaluates the value of X's bets on $(0,s)$ that Y wins; and $k_x$ is the proportion of hands that X bets exactly $0$ with. For indifference between $s$ and $s+Δs$ on $(0,S)$, we have (with $Δx$ having its usual meaning rather than reversed as in the book):

ΔEV = 0 = –(s+Δs)x(s+Δs) + sx(s) – \int_s^{s+Δs} (1+t)x'(t) dt
0 = –(s+Δs)x(s+Δs) + sx(s) – x(t)|_s^{s+Δs} – tx(t)|_s^{s+Δs} + \int_s^{s+Δs} x(t) dt
0 \approxeq –2sΔx – 2Δs x(s+Δs) + 2Δs x(s) – 2Δs x(s) – Δx + Δs x(s)
0 \approxeq –(1 + 2s + 2Δs)Δx – Δs x(s)
Δs x(s) \approxeq (1 + 2s + 2Δs)(–Δx)
As Δs→0:
x(s) ds = (1 + 2s)(–dx)
and now the argument follows the book from the 7th to the 10th equation
x = k/\sqrt{1+2s}

When $s$ decreases to exactly $0$, Y loses the $k_x$ term since his losing hand will lose the showdown. This is the reason that X chooses $k = 1$—otherwise, Y would never bet exactly $0$ but would instead bet a small positive value to pick up an extra $k_x$ in equity. (The book simply presumes continuity at $0$, which is not necessary, and in fact $y(s)$ is usually discontinuous at $0$.) If $x(s) = 1/(1+2s)$ has a negative game value, then X will instead always bet $0$ and simply give up the pot. This is the case if $y > 1/\sqrt{1+2S}$ as we will now show.

Mutatis mutandis, the same argument as above shows that $y(s) = k/\sqrt{1+2s}$ as well. Y chooses $k = y\sqrt{1+2S}$ so that the limit at $S$ equals $y$, to make X indifferent to betting $S$ or $S – Δs$. Y will bet $S$ with all winning hands and only winning hands; if he bet $S$ with any other hands then X would always bet $S$ to gain the equity of beating those hands.

If $y < 1/\sqrt{1+2S}$ then $y(s) < 1 – \epsilon$ for all positive $s$ and Y will bet $0$ with a nonzero proportion of hands. (This does not spoil the indifference for X, since she will beat those hands in a showdown even if she bets $0$.) These hands, that Y essentially gives up on, represent X's value in the game; so the value of the game to Y is $\lim_{s→0^+} y(s) = y\sqrt{1+2S}$. If $y > 1/\sqrt{1+2S}$, then X would lose value by betting her $k = 1$ strategy, so instead she will always bet $0$ and essentially refuse to play.

As a follow-up—and this comment applies to other games in the book as well—the c.d.f. itself is of no practical use "at the table"; to actually use the strategy, the player really wants the inverse function and a [0,1] random number generator. Using the inverse function s(x) = 1/(2x^2) – 1/2, X can now play perfectly—and in this particular game, this function applies universally for X, regardless of stack sizes.

11. Ch. 20, Ex. 20.3, pp. 251–3.

This example needs an edit to keep the numbers consistent. At the start on p. 251, the book says, "For the sake of simplicity, assume that the flush hits 20% of the time and the weak draw hits 4% of the time (these events are independent ...", and then on p. 252 para. 7, it says, "We'll assume that the set is made with a flush draw 1/4 of the time." This contradicts the events being independent! (This decision appears to have been made partway through calculating the problem in order to make the denominators 100 instead of 125.) It would be clearer if it said, "For the sake of simplicity, assume that the flush hits 20% of the time, the weak draw hits 4% of the time, and the weak draw and flush both hit 1% of the time."

Moving on to p. 253, the sixth equation has an error; the right-hand side should be $93/100$ instead of $61/100$. Using this value gives us $x_f = 227/1169$ which is substantially smaller than $3/7$. In the following paragraph, then, "Y is able to force X to call on the river three times as often" is incorrect; it is more like one-third more often.

12. Ch. 20, Ex. 20.4, pp. 255–6.

The values in this example appear to have been calculated assuming that there is a dead queen somewhere. There should be 6×QQ and 8×AQ in Y's range, giving him 44 hands instead of 39. Also there is a bizarre processing error throughout this example and the rest of the chapter: A function that I believe is supposed to be $y^–$ instead appears almost everywhere as "yø".

Also, it would be extremely helpful at this point to provide a graph displaying the relationship between $y$ and $y^–$. Once I drew one up for myself, a whole lot more things made sense.

13. Ch. 20, Ex. 20.5, p. 262

The second $\alpha =$ equation has a bad value: $(P+2)^\dagger – P^\dagger$, if you work it out, is equal to $2 + 2/((P+3)(P+1)) > 2$. Using $P = 5$ gives us $49/24$, not the $15/64$ that appears in the equation.

This results in a cascade of bad values and bad logic. The correct value for $\alpha$ is $127/867$ or approximately 0.14648, substantially less than the given 0.19561—and much more importantly, less than $1/6$! X actually folds fewer hands against Y in this situation, not more. The following bullet point then says: "...it is correct for X to fold almost 1/5 of his hands (instead of 1/6). This is because X is at a disadvantage not only on this, but on future streets. When he folds, he doesn't have to pay off the river. Hence, he can fold more often optimally." (If Y is playing optimally, then X is indifferent to calling or folding the turn, hence the mixed strategy! X is at no more of a disadvantage when she calls as when she folds.)

In fact, if one goes back to the formula for $\alpha$ (the second equation on p. 260)—which is correct but for a small typo—a little algebra shows that it is less than $1/(P+1)$ as long as the denominator stays positive. Why does X's fold percentage goes down, rather than up? Because X's aim is to make Y's worst betting hand indifferent, here and now on the turn. Y's worst betting hand is –EV against X's hand in a showdown, so X has to give it enough fold equity to make it indifferent. But because it is not a pure bluff and has some showdown equity, it needs less fold equity to compensate.

14. Ch. 21, p. 272, para. 5

"...and so his opponent can reasonably fold a large portion of his hands (more than $\alpha$) because he need not fear being exploited."

See corrigendum 13.

Here is the list posted by the publisher's website (last updated in 2007).

Why should it be zero and the same?
have you made calculations?
for me in first case his bluff region equity - negative. In second - positive.

I take it back; the equity is not exactly 0 in the situation given. I stand by "the same", though.

If both players play optimally, the value of a bluff is the same as the value of the worst check-fold hand. In a half-street game such as the one given, this is small but positive.

While it may not be zero, it is pretty obvious that any hand below opponent's worst calling hand, whatever the hand's actual value, has the same value when you bet it. The only way such a hand will win is if opponent folds, and at that point it doesn't matter if you are a single hand rank below the worst hand in their call range, or if you have the complete pits that beats nothing else at all.

Thank you for that. It appears that the errors they caught and the ones I caught are mutually exclusive.

so if X bets 50%, we just call 50% and check back the other 50% because betting doesnt give us any value and we are allowed to check bad hands, right?

I'm not sure you understand that game correctly; folding is not allowed.

If X bets 50%, Y is forced to call those bets, regardless what he holds. If X is checking her worst 50%, then Y gets to check behind with his worst 25% (the worst half of X's checking range) and bet the other 75%.

If you compare this to the Nash equilibrium, the hands that Y bets perform exactly the same. The 25% of hands that Y gets to check lose less—not less often but less value. Specifically, they lose 1 unit 3/32 of the time and win 1 unit 1/32 of the time where they win or lose 2 units in the Nash equilibrium, for a net gain of 1/16 unit for Y.

Oh, I get it, you were trying to be clever and snarky. Of course Y would lose value by checking his good hands; that isn't my point. Compared to the Nash equilibrium, Y doesn't (and can't) gain value by betting more, because there is no way to bet more. The only way Y can gain value compared to the Nash equilibrium is to check poor hands (not all hands!), and this is what X is preventing by betting everything out front.

Thanks for writing this up.

Thanks for writing all of this up. I've often felt similar about the book, and am just now finishing the game theory section. As for this specific issue:

It turns out that it does matter, but they set up the section in a way that it isn't clear why. It took me a very long time to understand what they were doing, but essentially they introduce this concept of ex-showdown value without fully defining it in terms of anything useful, and then they proceed to implicitly assume that raising the ex-showdown value raises it's actual value, which turns out to be true.

The reason that the nut low "wins more" than the second nut low when the opponent folds the third nut low is that the ex-showdown value of betting the nut low and having them fold better is higher than when you bet the second nut low and have them fold.

The authors set the EV of checking down to essentially be zero without telling anybody, and then use that. They are fully justified in doing this, but they don't explain it at all. To see what they are doing, just express the value of a strategy as the sum of it's expectations over every possible situations, and then group the situations where betting happens into one summand, and the situations where no betting happens into another.

What you end up with is that the EV ends up being commensurate with this term that they call "ex-showdown" value (this expresses the value of a strategy in terms of the difference between the value obtained by having bets go in the pot and the value of checking down).

Sorry to necro this thread, I'm going through the MOP and working every problem. I read the stated correction from the link that was posted in this thread but I think it is still wrong. On page 51, the example state the pot is 75, player A bets 30; then it says player B has immediate pot odds of 30/105. Should it be he has odds of 30:105? And doesn't that convert to a probability of (2/9)?

this is the copy/paste of the "correction" off the publisher's website:

Page 51: where it says "So from an immediate pot odds standpoint, B needs to be getting
more than $75 to $30 (or 40%) to win the pot." it should be "So from an immediate pot
odds standpoint, B should call if his chance of winning the pot is greater than 30/105, or
28.5%." (12/6/06)

Please correct me if im wrong, but in order to make AKs indifferent to calling, we only calculate the attacker's range vs AKs. For example, in Chapter 13 page 143, where the author is figuring out how to make X indifferent to calling with Kings, his calculations only include A,Q vs K.

Also, when I tried plugging in the ATs and A5s separately versus the entire range {AA, AKs} into pokerstove/equilab, A5s has a higher win percentage that ATs.