I originally posted this in Beginners' Questions, but I was advised to post here:

Hi guys, new poker player here, I'm currently reading Play Optimal Poker to get a grasp on the fundamentals, and I have a couple of questions. I hope someone here can help me.


First question:

In the chapter Reciprocal Ranges, Half-Street Game, Question 5, the answer says that betting 1/2 pot is +EV for Ivan, but a bet size of 1 pot makes the game 0 EV. So, not betting at all is 0, betting pot is 0 too, and in between there's a +EV bet size. It doesn't say so in the book, but that clearly means that there has to be an optimal bet size for maximum EV.

I worked out that given a value hand percentage v (in the example v = 1/2, meaning half of the hands in Opal's range are value hands she will always call with, the rest are bluff catchers that are indifferent), the bet size (as a fraction of the pot) at which the game reaches 0 EV is 1/v - 1, and the optimal bet size for maximum EV is 1/sqrt(v) - 1.

For instance, in the book example with 1/2 of Opal's range being +EV calls, that means that betting pot makes the game 0 EV, and betting about 0.41 pot is optimal. If 1/3 of Opal's range are +EV calls, a bet size of 2x pot makes the game 0 EV, and betting about 0.73 pot is optimal. If 2/3 of Opal's range are +EV calls, a bet size of 1/2 pot makes the game 0 EV, and betting about 0.22 pot is optimal.

So, long story short, this makes sense to me in the toy game, but what I'd like to know is whether and how this concept of optimal bet sizes is applicable to actual poker, what are the necessary adaptions, and what's the best book to read up on this?


Second question:

In the chapter Reciprocal Ranges, Full Street Game, Question 3, the answer says that "not coincidentally, the frequency with which Ivan calls with a K is exactly the same as the frequency with which he bluffs a Q". The way I understand this, though, it actually is total coincidence.

Crunching the numbers, I'm coming up with (-bv) / (b - bv - v + 1) for Ivan's frequency to call with a bluff catcher (b being the bet size as a fraction of the pot, and v being the percentage of +EV calls in his range). For Ivan's frequency to bluff, I'm coming with (bv + v - 1) / (b - bv - v + 1). Those two are equal only if v = 1 / (2b + 1), or formulated differently, v = 1 - 2PO (PO being the pot odds the caller gets). Another way to say this would be this: For the frequencies to be equal, the bet size has to be exactly half of the bet size that makes the game 0 EV (see previous question; this is not the optimal bet size).

So, this clearly works out in the book example with v and b both being 1/2. As soon as the bet size changes, though, particularly when choosing the optimal bet size, it seems those frequencies aren't equal anymore.

My question is, am I totally off here or is this an error in the book? Not sure what to make of this, please tell me where I'm wrong.

Thanks, these are excellent questions/observations!

On the first point, you're correct that there's an optimal bet size for the toy game. Real poker is messier, especially before the river, because ranges aren't perfectly polarized, different hands in the opponent's range will have difference incentives, etc. They key takeaway, I think, is not finding the exact perfect bet size but rather understanding what makes the bet size optimal in the first place. It's the size that applies maximum pressure to the target hand (a K, in this case).

Bigger bets actually provide less pressure because they don't force Villain to defend as often with his bluff-catchers. That's an important lesson for real poker games, where we do tend to see bigger bet sizes when betting into capped ranges than when betting into uncapped ranges.

The sequel to his book, POP2, talks more about bet sizing on earlier streets. You might also like Chen and Ankenman's Mathematics of Poke, which covers more complex toy games, including one where (if memory serves) they derive a formula for the optimal bet size for any hand in an infinite, continuous range on the river.

As for the second point, you are correct. Given that Opal is indifferent, it's not a coincidence that Ivan's bluffing and bluff-catching frequencies are the same, but at other bet sizes Ivan doesn't try to make her indifferent with her As.

Thanks again! Please do post here if you have other questions/comments along the way.

-Andrew

Thanks for the reply, Andrew, and for writing the book in the first place. I almost didn't buy it because of a couple of Amazon reviews that said it wasn't about "real poker", but of all the books I've started reading since I got interested in poker three months ago, this is the one I feel is helping me most at figuring out how to think about the game.

Thanks for the book recommendations, too, I've already bought POP2, and I'm looking forward to reading it once I've finished POP1 (it's a slow read if you try to work through the questions yourself before reading the answers). I just read the introduction to The Mathematics of Poker, looks like that's right up my alley, too.


As for my second question, I'm not sure I get it. I'll try to go through my assumptions and conclusions step by step, and I'd be grateful if you could tell me at which step I'm wrong:

1. You're saying that at equilibrium Ivan's frequency of bluffing with a Q is the same as his frequency of bluff-catching with a K.
   
   
2. "At equilibrium" means that the frequency at which Ivan bluffs with a Q makes Opal indifferent to calling with a K, and the frequency at which Ivan bluff-catches with a K makes Opal indifferent to bluffing with a Q.
   
   
3. Both of those frequencies depend on which bet size is chosen for this toy game, which is an arbitrary choice. We could choose, say, 1/3 pot instead of 1/2 pot as the only possible bet size for this toy game, which would still result in an equilibrium, but at different frequencies.
   
   
4. If the only allowed bet size is 1/3 pot, Ivan has to have 1/5 bluffs in his betting range to make Opal indifferent to calling with a K. Having 1/5 bluffs in his betting range means that he will bluff with 1/4 of his Qs.
   
   
5. If the only allowed bet size is 1/3 pot, Ivan has to call with 3/4 of his range to make Opal indifferent to bluffing with a Q. Calling with 3/4 of his range means that he will call with 1/2 of his Ks.
   
   
6. Ivan bluffs with 1/4 of his Qs and calls with 1/2 of his Ks at equilibrium. Those frequencies are not the same.

Addendum to my last post: The formulas I came up with have sign errors and are switched around, which doesn't really change anything, but I'd like to correct that—I can't edit the post though, does anyone know why? Am I too new a member to be granted those rights?

Sorry, I should have been more clear on the second point. You're right that the equilibrium will always involve indifference to calling with a K and bluffing with a Q. It will not always involve indifference to checking an A, even though with the half pot bet size it does. The reason Opal is indifferent with her As is that Ivan's bluffing and bluff-catching frequencies are the same for a half-pot bet.

I believe you can only edit posts within a few minutes of making them. A moderator might be able to edit for you, though.

Thanks, I get it now. I'm just going to write down what got me confused in the first place and restate what you said in my own words, so I can order my thoughts on this and hopefully get called out if there's still some misunderstanding on my part (and maybe it can help someone else who might be stumbling on this):


I was confused by this section: Not coincidentally, the frequency with which Ivan calls with a K is exactly the same as the frequency with which he bluffs a Q: 1/3. If it weren't, then Opal would be incentivized always to play her As in whatever way would benefit from Ivan's imbalance.

The reason I was confused was because I interpreted the first sentence "Not coincidentally, ..." to mean that since it's not a coincidence, it has to be an intrinsic property of this sort of toy game, which it isn't. What it's actually saying is that given that checking and betting As has the same EV it's not a coincidence that the frequencies are the same. This seems quite obvious and sort of the wrong way round, since the equal frequencies are the very reason for the equal EVs.

So, why are the frequencies equal in the first place? Because the fixed bet size we chose for this toy game happens to be the single exact bet size where this happens to be the case.

So, is it a coincidence that we chose this exact bet size? I'd argue yes, it's a total coincidence, because the bet size doesn't represent any meaningful equilibrium: If we allowed both players to choose their own bet size, they would immediately deviate from it towards a smaller bet size in order to increase their EV, and Opal would start betting all her As. I suppose this smaller bet size of about 0.41 pot could be called the "equilibrium bet size", because neither player has any incentive to deviate from it when given the chance.

The second sentence "[if the frequencies of calling with Ks and betting with Qs weren't the same], then Opal would be incentivized always to play her As in whatever way would benefit from Ivan's imbalance" confused me because I interpeted it to mean that whenever those frequencies are not the same, Ivan is somehow playing an imbalanced strategy that can be exploited by Opal, which is not the case. At almost all bet sizes, and particularly at the equilibrium bet size, Opal isn't indifferent with her As, but that doesn't mean she's exploiting Ivan—she's just assuming Ivan is playing perfectly, and she's making max EV decisions based on that assumption.


Summary: My reading of this section was that in the general setting of reciprocal ranges, it never matters whether OOP bets or checks their value hands, as long as IP is playing equilibrium. That was wrong, it totally matters and depends on whether IP's bet size is expected to be larger or smaller than (1-v)/2v (v being the amount of value hands as a fraction of the total range that could call a bluff). In the example in the book, IP's bet size is exactly (1-v)/2v, because the rules say so, but if IP is allowed to choose a bet size that maximizes their EV, betting value hands becomes strictly better than checking value hands for OOP.

Any chance to see table of content of Play Optimal Poker 2?

Chapter 1: Leverage
Chapter 2: Protection And Semi-Bluffing
Chapter 3: Range Construction
Chapter 4: Using Leverage
Chapter 5: Shallow Stacks
Chapter 6: Continuation Betting Without Range Advantage
Chapter 7: Continuation Betting With Range Advantage
Chapter 8: Barreling The Turn
Chapter 9: Attacking a Missed Continuation Bet
Chapter 10: Continuation Betting From Out of Position
Chapter 11: Adapting to Tournament Play

Edit: Or just look it up on Amazon for a more detailed TOC.

Thanks for your interest! If you look at the paperback listing on Amazon, you can use the "Look Inside" feature to see the ToC (which includes hyperlinks that will enable you to preview a few sections of text as well).

Thank you!

Hoping I can tag onto your thread here with a question of my own about this book. I'm just getting into the concept of equilibrium and I am a bit hung up on the Clairvoyance Game. The mechanics of the game are (quote):

The out-of-position player, Opal, is always dealt a K. The in-lisition player, Ivan, is dealt an A 50% of the time and a Q 50% of the time

...

Each player antes $1. Opal acts first and may either bet $1 or check. Facing a check, Ivan may bet $1 or check. Facing a bet, either player may only call or fold; there is no raising permitted. If there is not bet or if a bet is called, then there is a showdown and the high card wins the pot.

In the explanations, the author (Brokos) describes that Opal should check 100% of the time. My question is:

Should Opal instead bet 100% of the time?

I calculate Opal's EV as follows:

EV = %Fold*$Fold -%Call*$Call

Since Ivan will call 100% of the time when he has an A and fold 100% of the time when he has a Q, Opal's EV should be:

(0.5)*($2)-(0.5)*($1) = $1

In other words, she breaks even at $1 on her $1 bet and guarantees that there will be no more betting since Ivan cannot raise.

There is inherent disadvantage in allowing Ivan to employ an equilibrium betting strategy, so the best Opal can do is turn the game into a coin flip.

Am I correct or incorrect in my assessment above? I am not asking because I want to point out a mistake in the book - I want to make sure I'm not misunderstanding the concept at play.

(0.5)*($2)-(0.5)*($1) = 1-0.5 = 0.5

Haha of course. Thanks!

Hi Jrichmo! Thanks for picking up the book, and for posting here. Glad you got sorted out on this, though don't be shy about pointing it out if you do spot (or think you spot) a mistake in the book. I hope there aren't any (more), but if there are, I want to know about it!

I find the intuitive explanation here helpful as well. Yes, checking can lead to a difficult decision, but betting guarantees you lose the maximum when behind and win the minimum went ahead. No matter how badly you play after checking, you couldn't possibly lose more than you would if you bet.

This is a key point: sometimes it's OK to end up in tough spots where you might make some mistakes. That's a part of the game. Sometimes the things you'd have to do to avoid those spots are so costly that you're better off just making the best of it when the tough spots arise rather than trying to avoid them at all cost.

This comes up with, for instance, defending your big blind with a weak hand in a tournament. Or calling a 3-bet with a marginal hand pre-flop.

Thanks so much for the quick and detailed response. I will definitely come back with more questions and comments. Hopefully no worries to OP (atomicashes) if I piggy pack in his/her thread

In the Reciprocal Ranges, Full Street game, Question 1, it says that Ivan has a positive EV, but lower than in the half street game.

But, why is it lower? If Opal is indifferent to checking/betting with aces, she can check with all her aces, and then we revert to the half street game. So, Ivan gets the same EV of 1/18 (~0.06).

The book also states that Opal betting all her As and 1/3 of her Qs is one of many equilibriums. Let's calculate the EV.

Bet scenarios
Aces: Opal always bets with aces
She wins the pot always (+$2). 1/2 of the time, Ivan has a King, and his calling freq with King is 1/3, so Opal wins an addition 1/6 (total of 13/6).

Queens:
1/3 of the time, Opal bluffs with Qs. Since she is indifferent to this, the EV is 0 (same as if she folds).

Check scenarios
Kings:
Once we check with Kings, Ivan will bet all his aces and 1/3 of his queens. We are indifferent when Ivan bets, so we only win when he doesnt check with Queens. This is 1/2 * 2/3 of the time, or 1/3 of the time. So we win the $2 1/3 of the time, for an EV of 2/3.

Queens:
2/3 of the time, we check our Queens. We never win, so EV is 0.

Total:
Aces is 13/6, Kings 2/3, Queens 0. Average of 17/18. Buyin is 1, so our overall EV is -1/18. So, Ivan's ev is 1/18 (same as half street game).

I was wondering if I'm missing something, because I have been confused for a wile. Did I make a mistake anywhere? plz lmk