It seems like this game would be easy to figure out the correct strategy to, but all of my numbers keep coming up as gibberish. I've put together some example questions, could someone please show the math behind the answers.

2-8 players. Each player gets 2 cards. Each player declares simultaneously if they are calling, or folding. Best 2 card high hand wins the pot. All players who entered the pot and lost must match the pot up to the capped amount.

A)
$1 in the pot. 2 players in the hand. $1 cap. What percentage of hands do we call with?

B)
$2 in the pot. 2 players in the hand. $1 cap. What percentage of hands do we call with?

C)
$1 in the pot. 4 players in the hand. $1 cap. What percentage of hands do we call with?

D)
$5 in the pot. 8 players in the hand. $1 cap. What percentage of hands do we call with?

What have you done so far?

Typically, finding the optimal GTO solution requires writing down a condition (or multiple conditions) that must be satisfied at the solution. Then solving the equation(s).

Start with the simplest Question A. What condition(s) must hold at the GTO in this case?

Toy game approximation (continuous hand spectrum, no card removal)

A) $1 pot, two players, $1 cap

- In:
- Opponent is in, we win -> win 1
- Opponent is in, we lose -> lose 1/2
- Opponent is out -> win 3/2
- Out:
- Opponent is in -> 0
- Opponent is out -> win 1/2

Having to match the pot effectively costs only half, because we then get to play for it again. Similarly, if the opponent folds, we also win not having to ante.
By symmetry, our opponent plays the same strategy, so if we're both in it's 50-50 to win. Also, they're in as often as we are. This means we only have to solve for one variable: The chance they (and we) are in, ie. the percentage of hands we call with. Call it x.

Is this what you meant by weird numbers? I'm going to interpret this as 'always stay in', but I preferred my original answer, when I made an error and got 4/7

(Followup posts deleted as they all had the same mistakes)

I got the payouts wrong; the 'pot-plus-next-ante' big payoff is if we both stay and we win, not if the opponent folds.

But more importantly my assumption that we simply win half the hands doesn't make sense at all: with our 'equilibrium hand', the worst we stay with, we never win a showdown. Got **** to do but I'll get back to this..


EDIT: Scenario A revisited: cutoff hand, which we never win a showdown with:

IN:
- Opponent in: lose 1/2
- Opponent out: win 1

OUT:
- Opponent in: 0
- Opponent out: win 1/2

B

IN:
- Opponent in: lose 1/4 (we pay half the pot, then get to play for it)
- Opponent out: win 1

OUT:
- Opponent out: win 1/2 (play for this pot again)

EDIT: Play top 2/3 - the cap makes it cheaper to 'bluff'

Ok taking a crack at the multiplayer one..


C (4 players, $1 pot, $1 cap). Assuming no ties

IN:
- Three in: lose 1/4 (pay 1, but get 1/4 equity in a $3 pot)
- Two in: lose 1/2 (pay 1, 1/4 * 2 back)
- One in: lose 3/4
- All out: win 1

OUT:
- Three in: win 1/2 (two losers, $2 pot)
- Two in: win 1/4
- One in: 0
- All out: win 1/4

Probabilities:
- Three in: x^3
- Two in: 3 x^2 (1 - x)
- One in: 3 x (1 - x)^2
- All out: (1 - x)^3

OK this is starting to look a bit tedious.. I'm throwing it in Wolfram Alpha:
(x^3)(1/4 + 1/2) + (3 x^2 (1 - x))(1/2 + 1/4) + (3 x (1 - x)^2)(3/4) + (1 - x)^3 (-1 + 1/4) = 0


(I subtracted the 'in' scenario from the 'out' scenario for each coefficient).

This gives me x = 1 - (cube root of 1/2), or about 0.206; ie. play a bit over the top 20%