The subject is often discussed but I haven't found this question answered clearly anywhere. Possibly because the terminology used in discussions is confusing to me. It seems that a lot of times when someone talks about "losing money", they really mean "not winning as much money as we could".

It is pretty clear (and I understand) that there are many situations in which shoving a nash range is not nearly the best approach to take, and other options will net more $ in the long run. As a trivial example, if we are heads up and opponent is sitting out, just minraising any two will make us much more money than shoving nash (obviously).

But my question is simply: Can an opponent that knows we are shoving nash ever adapt in any way such that we actually lose chips in the long run?

As an extension to this question: Can there ever be any reason to fold a hand that is inside the nash range for the given spot? Besides ICM considerations in tournaments, the only situation I can think off is a situation where we can make villain think that we shove way tighter than nash when in fact we are not shoving as tight as villain thinks we do. Thus villain reacts by tightening up their calling range, but too much vs our actual range. The EV gained from this could theoretically be higher than just shoving nash. Is this actually correct? Are there other reasons for shoving tighter than nash? So let's say I know villain is very good and can't be fooled, but there are 8 other fish at the table. Is shoving nashin situations where this good villain has to be considered just a save way to go?

No, you can't lose money, unless rake is involved

If you're asking if you can play better than nash given your opponent has exploitable tendencies like calling or folding too much then obviously the answer is yes.

Can you lose playing nash? No. You can give your opponent the range you are shoving, let him calculate his equity against that range and make a perfect decision and you will both be breakeven. If he deviates from the equilibrium he will lose money.

Yes, you can def lose money.

In 3+ player situations, Nash play doesn't come with any of the nice guarantees about not losing money.

In heads up, you can be losing money in one position, but you should make it up in the other position. (However, if by "shoving Nash" you mean using the shove/fold charts, keep in mind that those aren't the Nash equilibrium strategies for the full game, so you can definitely lose money in both positions using those.)

If we're talking heads up, it is impossible to lose money when you shove a nash range. By losing I mean not losing as much as the (small) blind is when you fold so on the SB a nash range will assure your EV to be > -0,5bb.

If you are shoving a nash range heads up it is literally not possible for your opponent to make a calling range that gets him any profit and thus assures you don't lose. Of course this requires you to shove the whole nash range because that is what the model is based on.

Try it yourself; look up a nash range for example for 10BB and try to come up with a calling range that have a positive expectation when shoved on. There isn't one.

This is correct. Unfortunately, losing 0.5bb/hand (aka 50BB/100hands) is tremendously bad . Certainly, it's losing money.

In deep stacks, you're losing money playing shove/fold from the SB due to all the open-folding you have to do. You can make that up when in the BB if villain is also trying to play shove-fold in his SB, but you likely can't if he's playing better taking advantage of all his options.

its not so much that you should just blindly shove nash , you should only do this if you think your opponent is playing gto, if he's a monkey you can construct a range of hands to shove vs him exactly

I don't get to play much NL cash, only NL tournaments. Do cash players really treat SB, BB confrontations as shove / fold decisions?

Not unless they've decided to play 6bb deep cash.

As a tournament player, it's useful for you to know about, but certainly don't start play shove/fold above 10bb, and probably not before 5 or 6

Edit: In reply to the original question.

Heads up, you are guaranteed to at least break even before rake using the nash push/fold chart if your opponent is also playing push/fold. You are not guaranteed to break even in the SB alone playing push/fold, and you can't play push/fold in the BB if your opponent doesn't want to.

Edit 2:
Also, "lose money" is relative term. If your opponent folds 100% to minraises and you decide to open 60% of your hands you are losing money even though your expectation is above 0. Making a small profit when a huge one is available IS losing money.

You are only at a Nash equilibrium when a change in your strategy will be unprofitable. If your opponent plays strategy A, you can approach Nash equilibrium by playing strategy N(A). But if your opponent changes to strategy B, YOU CANNOT BE AT A NASH EQUILIBRIUM, unless you change your strategy!!!

If you are at Nash equilibrium with your opponent, neither of you can make a profitable strategy change, and you will both go broke from the rake.

There are so many conditionals with the "Nash = unbeatable" idea that it should really just be erased from people's memories. Nash simply isn't unbeatable unless you're specifying a specific type of poker where it is.

What do you mean by "changing your strategy" to be at nash equilibrium?

The whole point of nash is that you can play a fixed strategy regardless of what your opponent is doing. There is no "changing strategy", it doesn't exist in nash.

Agreed. The game you're playing in won't be at an equilibrium, but that doesn't really matter.

Too tired and on my phone so from memory

1. Using nash is profitable. If opponent plays optimal result is 0 ( or loss of rake if ...)

2. In tournaments huge number of situations where applying said plan can be -ev.

Why shove at huge stack(bb) when xx and yy are about to be blindet out in pay out jump spot.

Gibberish.

A Nash equilibrium exists when NO PLAYER can make a profitable strategy change. If ANY of the players can change his strategy to decrease his losses, you are NOT AT A NASH EQUILIBRIUM.

IF you are not at a Nash equilibrium, it may be possible to change your own strategy to increase your EV. In other words, playing the strategy you would play at Nash equilibrium may be suboptimal if you are NOT at Nash equilibrium.

I.e. the Nash equilibrium strategy is a defensive strategy; it is optimal when your opponents play optimally. That doesn't mean it's optimal when they don't.

IMO OP pertains to one player playing a Nash strategy, not that both opponents are playing a compatible pair of Nash strategies which form a Nash equilibrium.

Let's say it's heads up, and your opponent calls with any 2 cards from the bb vs your sb shove at 10bb deep.

54 suited is a nash shove at 10bb deep.

However, if your opponent is calling any 2 cards, then 54 suited will lose money.

Granted your opponent will lose more in the long run by calling any 2 cards, I understand that.

But 54 suited in this situation vs such an opponent will lose money, no doubt.


So my question is, aren't all hands supposed to be +EV shoves in a GTO shove range?

OR

Does it mean that the range as a whole just needs to be +EV, even if some hands actually are minus EV?

Because 54 suited, even though it's in the nash shove, it will lose money vs a guy who calls 100% of the time @ 10bb vs your shove

Again,

And to answer your question, yes, each decision made using a Nash equilibrium strategy will be optimal [but only when you are at Nash equilibrium!!]

Simple example. You're playing the Monty Hall game.

Before you pick a door, there isn't anything Monty can do to affect your EV, and there isn't anything you can do to improve your EV over picking at random. At that point in time, as long as you pick one of the doors at random, you and Monty are at Nash equilibrium.

But after you pick a door, if Monty intentionally shows you a goat behind one of the other two doors, Monty has deviated from a Nash equilibrium strategy, and you should always exploit his mistake, by changing your strategy away from random selection.

Monty's error did not change the EV of your Nash equilibrium strategy, but sticking to your original strategy, after Monty deviates, would be suboptimal.

What you mean is correct but "optimal" has come to be used to describe a strategy that may be used at a Nash equilibrium regardless of whether you actually are at that equilibrium. This makes the usage of "optimal" different here than the colloquial usage which simply means "best". So generally "game theory optimal" is used to specify that we're using the game theory usage not the colloquial usage.

"Game theory optimal" or "GTO" is completely interchangeable with "Nash equilibrium strategy". So saying "Nash equilibrium strategy is always optimal", is like saying "GTO is always GTO" which is obviously correct. But of course, GTO is not always best.

Why would you choose one of the most commonly misunderstood logic problems in history as a "simple example"?

Haha that is an amazing point.

Except that traditionally, in game theory, an "optimal strategy" IS DEFINED as better than any other possible strategy, i.e. best.

The example is simple because there are only two decision points (the minimum required) and it completely illustrates the entire problem.

There is a sticky at the top of the threads in this forum that describes the meaning of "game theory optimal" and "Nash strategy" as used in the Poker Community.

Are there other reasons for shoving tighter than nash?

If we have a pretty noticeable edge over the rest of the table then we can adjust our jamming range and take out certain hands. The EV we expect to gain from the rest of the table vs our relative edge should out weigh the lost EV that the player gets from us tightening our jamming range and the EV of the Jam itself.