I think MDF is a difficult concept for a lot of people, so I have attempted to formulate a small challenge about it. I am not sure of the level of difficulty here, but I hope it interests somebody and isn't trivially easy.

Construct a situation with the following conditions, or prove the impossibility of such a situation:

1. Two players are HU
2. The OOP player is facing a raise after betting on the current street
3. The OOP player's betting strategy is part of a Nash Equilibrium
4. The OOP player folds more than MDF as part of a Nash Equilibrium

Any number of streets or stack sizes may remain, and any FL, NL or PL betting structure is allowed. You may either use a solver to find a real poker example or construct a toy game to serve as your example. After someone posts a solution I may add further stipulations for an additional challenge. I'll provide any clarifications if needed.

I've done a simulation in pio once. It went something like this:

- 2bb in the pot.
- 100.000bb deep.
- IP player is 100% polarized (with as many bluffs as he wants).
- OOP player 100% capped.
- The IP player can bet POT on the flop and any size turn/river.

The solution for this situation was that the OOP player folded 99,9% of the time vs the flop pot sized bet.

The OOP player never donks the flop of course here. It's just an interesting solution.

If a player has a betting range on the flop however (or any other street), he has to defend vs a raise at least MDF. Otherwise the other player would just bluff 100% there and the first player's original bet would be -EV for all the folding hands. Therefore he would never do it in the Nash Equilibrium.

So, the answer to your question is "No, its not possible for such situation to happen".

That's interesting. Toy games suggest that OOP should fold exactly at MDF of 50% against the flop bet if IP's range consists of only 100% or 0% equity hands, and IP has enough bluffs in his range to balance his value range on future streets. Even if IP's bluffs have equity, why shouldn't OOP defend at a frequency which makes IP's bluffs indifferent between betting and checking, or does such a high folding frequency accomplish this somehow?

This is the reply I expected, but there's more to it. I don't want to give away any nuances yet so I'm going to wait for a few replies before I share my own thoughts.

I don't think that there's more to it. If the bettor folds more than the MDF vs a raise, his opponent has no reason to ever fold a hand vs the bet (raising has a higher EV than folding). This would mean that the bettor should have no bluffs in his range either (bluffs are always -EV here), which obviously means that he can't have any value range either. Therefore he would never be betting in a spot like this.

Perhaps you're thinking that this would change if the raiser doesn't have enough bluffs in his range, but that is not true. The raiser can always turn the hands that would normally be folds or break even calls into bluffs.

There are circumstances where OOP should bet despite knowing IP will never fold, such as when IP has a draw and is getting at least direct odds to call. In NL this may mean the bet should be bigger, but in FL only one sizing is allowed

This shifts the indifference equations to compare raising/calling rather than raising/folding. I am imagining a situation where OOP has made hands only, IP has a range of immortal nuts and draws.

Yeah, but if the bettor is folding more than the MDF, the raise will still always have a higher EV for the draws than the call. Therefore the draws would raise 100% and the bettor would never want to bet/fold anymore.

Is there a logical reason this must be true? The math seems complicated to me. What if the draws have worse equity against the b/c range than the overall betting range?

If the draws have the same equity vs both ranges, raising will 100% be more profitable in that spot. It's pretty simple math.

If the equity changes however, it gets a bit more complicated:

If the equity of the draws was theoretically reduced from ~40% to 0% when called, then yes, the call would be a bit more profitable than a raise. So I guess that a situation like that could theoretically exist after all. I'm just not sure if such spot could actually exist in NLH.

Here's a situation that could happen in NLH:

Flop A37o
check/check

IP always has Ax or KQ/KJ/KT here.

Turn 3o
OOP bets 50% of the pot. His range is only: Ax and bluffs
And he calls a raise with only Ax.

When IP raises here with something like KJo, his equity is reduced from ~30% to 0%.
He has the pot odds to call the 50% bet size, so GTO would let OOP fold a few % more than the MDF vs a raise.

Just a clarification: In stipulation 4 above, shouldn't the folding tendency beyond MDF always by definition equal the value of position at NE? Or is this tendency already subsumed into your definition of NE in this scenario? Thanks!

I’m not sure what you’re asking. A Nash Equilibrium is a strategy pair such that neither player can increase their EV by changing any part of their strategy. MDF is defined by:
MDF = P/(b+P) where b is the opponent’s bet size and P is the size of the pot prior to his bet.

Right, so is it correct under NE that we can define the value of Position mathematically as the difference between the amount we fold over and above MDF and still be playing within NE?

Not sure position and MDF are correlated in any way.

As browni said MDF is just some calculation that helps give someone the idea of how much they need to defend vs a bet. It tells you nothing about position, future betting, etc. It only says here is the betsize, here is the potsize, and this is how much you have to defend to make villain's bluffs (that have 0 equity) 0 EV from stealing the pot.