In a hypothetical and very unrealistic scenario, you're playing some heads-up NLHE and you know your read is 100% accurate that your opponent has either AA or AK. Maybe you're a telepathic mutant from the future or something.

Anyway, facing a bet, you have a borderline decision to make. Exact pot size, board cards, etc. aren't important but you're ahead of AK, behind AA and calling/folding is break-even EV. However, you know your opponent likes to talk during hands and you think you can get some helpful information from him. You decide to say "I know you have at least one Ace." [pause] "Do you have the Ace of spades?"

His answer allows you to easily make the correct fold or call. Explain why.

I'm also interested if anyone can think of a more plausible situation using similar logic, maybe something that could actually happen IRL.

Does he always answer with the truth?

Yeah, sorry I should have made that clearer. In fact, let's say he shows you the Ace of spades after you ask (and you somehow know that this action isn't dependant on the hand he has).

If he has the ace of spades, then there are 4 ways for him to have pocket aces, and 4 to have AK.

If he doesn't, then there are 3 combos of AA and 12 combos of AK.

The decision of whether or not to call is entirely dependent on the amount to call and the amount in the pot. Not sure how you can say otherwise.

Fold if he answers yes and call if he answers no. Knowing he has the As doubles the number of combos he has of AA relative to AK.

Yes, this was the answer I was going for.
Unspecified ace: 16 combos AK vs. 6 combos AA.
Ace of spades: 4 combos AK vs. 3 combos AA. Fold.

However, I've been going back and forth about whether it works like that. I think it does as set out in the OP, but it depends how exactly you get the information. For example, what if:

1. You know V has at least 1 Ace: Prob(AA) = 6/22 = 3/11.
2. V shows you a specific ace. Now Prob(AA) has gone up to 3/7.

This is where I was struggling a bit. Obviously, if 1 is true then 2 is always possible. So why does the probability change? After thinking it over I realised that it doesn't. I wasn't modelling the "shared" combinations of AA where V has the option to show one of two specific Aces correctly. Therefore, V is twice as likely to show, for example, the Ace of spades when he has AsK than when he has AsA. So the probability for 2 is actually 3/(8+3) = 3/11, the same as 1.

In summary, I think the example in the OP is fine. If you ask him to show specifically the Ace of spades (or h/d/c) then the probability changes and you can gain an advantage. However if you ask him to show any Ace, or if he voluntarily shows you the Ace of spades (or h/d/c) then you don't learn anything new. The wording is important.

Apologies if all of that was boring or elementary. I found it interesting to think about and it took a little while to completely get my head around it all, so hopefully someone else will get some usefulness out of it.

The decision point of whether to fold or call has to do with your pot odds. Just because your equity has now gone down because of the ratio of AA combos to AK combos has increased doesn't mean it goes from a call to a fold.

You raise to 100 with QQ. He reraises to 300. You reraise to 700 and he shoves for 1100. You need to pay 400 to win 1800. That's 4.5:1.

Literally even if he turns over AA, you still should call.

He specified in the OP that it's a neutral EV decision before asking the question. So with any additional information that influences the decision one way or the other we can say this is a call or fold.

Yeah, I know about pot odds. This part of the OP was supposed to cover that.

With the unspecified ace, it's 16 combos of AK and 6 combos of AA.


With the ace of spades, it's 4 combos of AK and 3 combos of AA.

In both of those cases, there are more cases of AK than AA.

What is the justification that the former scenario, where the ratio is 8:3, is a call, but the latter scenario where the ratio is 4:3 is a fold?

And how do you know where to draw the line?

I'm not sure I follow. Surely you can contrive any number of scenarios where this is the case? I left it deliberately vague in the OP because EV calculations weren't the focus of the puzzle, but okay...

Hero has red 22 in BB, Villain has any AK or AA in SB and shoves all in for 6.85 BB total. With 2 BB in the pot and 5.85 BB to call, Hero is 42.7% to win and has neutral EV: 0.427*7.85 - 0.573*5.85 = 0. So it doesn't matter if Hero calls or folds.

Now, if we know Villain has the As then Hero is only 37.5% to win because, as you say, there is a higher ratio of AA:AK combinations. Calling is now clearly negative EV: 0.375*7.85 - 0.625*5.85 = -0.71. Hero can make the easy decision to fold.

This all seems rather dull and trivial, and wasn't the reason why I started the thread. The situation just seemed counter-intuitive, I was surprised that knowing he has a specific Ace can lead to different decisions than knowing he has an arbitrary Ace (ignoring obvious flush/flush draw scenarios).

8:3 is breakeven because OP said so. If villain answers that he doesn't have the As then it's 12 combos of AK and 3 of AA, 4:1, which improves his odds from breakeven. If villain answers that he does have the As, it's 4:3 odds which is worse than breakeven.

For some reason this didn't sink into my thick skull when I read OP.

Ty