Why does poker only have 1 Nash equilibrium.

Who said poker only has one equilibrium?

There was an interesting discussion on this topic in r/Poker_Theory the other day: https://www.reddit.com/r/Poker_Theory/co...

There are trivial cases where there are clearly multiple equilbria, (e.g. royal flush on board).

But I guess your question is more like, why do most nodes seem to converge to one strategy? That's a far more interesting question.

Lets define "strategy buckets" as a class of hands combined with some strategy profile such that when following this profile one can exchange frequencies among the given bucket without losing any EV and exchanging frequencies with any other bucket strictly loses EV. Then by a unique equilibrium I just means that there exist only one strategy bucket which maximizes our EV. It seems tome that if we take a 2-player,zero sum spot that poker has a unique strategy bucket equilibrium. Perhaps this definition is equivelant to saying that only one set of payoffs is possible as a Nash Equilibrium.

I mean if you view finding the equilibrium in poker as the convergence of a simulation then I assume you will always be left with at least perfect if not proper equilibria. It seems to me that non-perfect equilibria should not exist in poker but I couldn't ever reason why. I am quite knew to learning game theory formally so this is still all very abstract to me. Perhaps the best way to think about it would be that the convex hull of Nash equilibria should only contain one element. Therefore all nash equilibrium payoffs are equivelant and furthermore proper

There are likely many equilibriums—specially if we allow the exploitability to be >0%, but still small (or basically ~0%). So, for practical purposes there are likely many equilibrium if we allow the exploitability for a node to be between 0 to 0.1% of pot basically.

I meant Nash equilibria with exactly 0% exploutability. That being said I think I understand what you are saying and how your question is more interesting and relevant from a practical perspective. Like Can there exist .1% exploitable strategies which aren't "close" to the theoretical equilibria. Or even more generally is exploitability increasing with distance from the theoretical equilibria where closeness and distance are defined in terms of mixing frequencies

Arriving at perfect equilibrium is absolutely impossible, right? Just like it's impossible to find the last decimal of pi (or any other irrational number).

Like if we play the polarized river game and I shove for pi times the pot, how often are you supposed to defend?

1/(1+pi) = 24.14....%

But if you only call 24.14%, I will always shove because you are under-defending. If you round up to 24.15%, I never bluff because you are over-defending.

So you move on to the next decimal, and the next one... and so on for all eternity.

But then again, I'm never gonna be able to bet exactly pi times the pot to begin with anyway...

I imagine this is the case for instances in which we allow bet sizes to continue for infinite decimal places. The equilibria would asymptotically approach 0 exploitability, but not ever *technically* reach it.

Very wonderful statement, Bob. I'm a little disconcerted I didn't think of stating this.

While your statement is certainly true, I’m not sure it constitutes an example of multiple Nash equilibria. Consider a GTO rock, paper, scissors strategy. On any given play, we can choose any of the three options so long as we choose each with probability of exactly 1/3. We don’t regard this as three NE strategies for RPS, but rather as a single NE strategy. In like fashion, in your example, suppose we have N combos in our folding range and our optimal bluff frequency is p. Is it not valid to say that a single equilibrium strategy of bluff raising any combo in our folding range with probability p/N would be a single NE strategy for this spot?

I'm not sure, but I would have considered it a different equilibrium. Your argument is valid, though.

seems legit at a glance. are there other spots where applying probability to the appropriate range can be equilibrium? river bluffcatching call probability for example could manifest in what appears to be infinite equilibria, but really its just one.

I just built the following version of the nuts/air vs bluffcatcher toy game. Pot is 100, Stack sizes are 100, OOP has the nuts/air range and he can only go B100 or check. Board is 22223.

OOP's range is AsAh and all combos of 76o. IP's range is AsKh, AsKs and all combos of 99.

For OOP, the solver is definitely betting at the right 2 combos of value for 1 bluff, 100% for AsAh and ~4.17% for each of the 12 combos of 76o. We can definitely rearrange which combo of 76o bluff for what frequency, in infinite ways, as long as we reach 2 combos of value for 1 bluff.

IP may be a bit more interesting. The solver always calls AsKh and AsKs, which should be obvious (they block the only combo of AA), and 1/3 of each combo of 99, reaching 50% calls, to make OOP bluffs indifferent. The combos of 99 can be rearranged in infinite ways, and the reason is analogous to the reason why we could rearrange 76o.

Now, can we make AK fold, let's say the AKss now always fold, and every single combo of 99 call 50% of the time, reaching the exact same overall frequency?

The answer is no. OOP doesn't change his strategy at all, but he gained EV, and IP lost EV.

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In most realistic river scenarios of the game, there must be blocker/unblocker effects going on everywhere, so the hands you choose to bluff raise, even if in the end you still reach the correct frequency, should matter for the overall EV, correct?

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I took issue with the following statement, "it doesn't matter which hands you choose to bluff raise in equilibrium"

nice reply.

it was so long ago that i forget where i learned it. im pretty sure it was in some 2p2 literature, but it could have been an extrapolation of properties of a (0,1) game without card removal. thus it may be dated info.

to the example provided

ak should be a +ev call because oop is bluff heavy, and thus swapping it for a 0ev call with 99 cant be equilibrium. make it kq instead of ak and i think only the frequency will matter.

your reply brings up a good point, that +ev calls(whether +ev because of removal or because you beat part of the opponents value range) count on the call side of indifference calculations at 100% frequency. if these +ev calls are present in our river range, we dont need to call as many pure bluffcatchers that can only beat a bluff, hence the 1/3 call probability with 99.

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i think +ev bluff raises have a similar effect on limiting the combos in our 0ev river bluff raise range, but those hands between the very bottom of our range and the top of our our folding range without removal benefits may be interchanged infinitely without ev loss.

edit, i remember now that bluff raising with the top of your folding range was shown to perform better vs humans because humans are prone to curiosity calls.

Thank you for the reply Bob148. I just would like to mention that KQ and 99, for that particular toygame, have the same relative hand strength 😀

Other than that, I got your point now.

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Think I found an interesting way to manipulate the toy game, that might have real applications somewhere in real life:

If I make IP range be only one combo of 99, 9s9h, so that it's impossible to achieve 50% calls without folding AK, will the solver fold AK sometimes to make 76o indifferent?

No, because AK is a +ev call. It will always call AK, 99 will never call, the overall call% will be 66.666...% and OOP will now never bluff because 76o is not indifferent anymore. Also, by accident, this toy game has only one equilibrium.

Cheers

I don't think this discussion is really related to the question. What I'm trying to show is that the convex hull of all Nash Equilibria is unique. That's basically to say that " mixing which combos of 99" to call isn't a meaningful strategic change. I think the way you would prove it is to show that if you correctly split hands into "hand buckets" by the definition I had above that then the optimal response function always has a unique maximum. I actually think that that's a pretty hard condition to impose and so it's very likely that there could exist strategically different equilibria although unlikely on the river. It would need to be on an earlier street.

In some simple hands there are more than 1. If your range is only boats on the river against a range of only flushes you can bet (and they always fold), or you can always check, or any mix of the two.

In complex hands it's hard for 2 completely different strategies to both be equilibriums because the strategy is driven by underlying principles of poker and by changing one equilibrium strategy for that player they become more exploitable

i asked grok some questions about this. heres what grok said:

seems theres no proof for poker having 1 and only 1 equilibrium.