Worst LC/NC discussion ever? I haven't read one post on this topic.

They have to agree before hand which subset corresponds to which card, but for example they could say that {011, 100} means the first card, {001, 110} means the second card, and {010, 101} means the third card. If the dealer keeps all the cards face down (000) and then chooses the first card, player 1 would then flip the first card up to get state 100. if the dealer chooses the second card player 1 would flip the third card to get state 001.

Thanks, that explains it a lot. Quite a huge problem space, but sounds like it would work.

So a mapping the players agree upon beforehand could be:

Card 1: 100, 011
Card 2: 010, 101
Card 3: 001, 110

And then any of the following original dealer states can be converted to one of the above depending on which card you want to indicate:

000
001
010
011
100
101
110
111

Neat. I find it interesting that 000 and 111 would not map to anything, not sure why, but OK.

Popetonite(tm)

Not really threadworthy but I have a question about not 3-betting HU in position, but always calling and then raising the turn. I like this concept and I've been doing it. But wouldn't an exception to this be when we have a big draw?

Say we have an OESFD or bottom pair+FD. If we call and then brick the turn, we'll just be calling the turn instead of raising - and then we missed out on putting more money in on the flop when we had an equity advantage.

Thoughts?

sometimes you get in >1 bet on the turn with 100% equity after you call flop

Confirmed. But I don't understand the original question at all.

LC-Just cooked a bunch of Brussels Sprouts.

nom

yuk

It's official-the sprouts were epic

Hey TSB,

Having still not had any chance to really sit down with a piece of paper 'cause turkey, I'm still stuck at the same spot in my head where I can't find a way to prove to myself that there won't be any unsolvable setups using your method (same with permutation ordering in base-n) for any integer n. I'd like to prove it logically as opposed to procedurally or brute forcing it or whatever, but I just can't see it yet.

I guess I'm still unclear whether you see if/how it can/can't work for all scenarios, but in the case where you do, you should prob show me 'cause that would be awesome.

Yes, but how?

Did you steam them?

Did you toss them in olive oil and salt and roast them on a cookie sheet?

Did you cut them in half and sautée them with a minced clove of garlic?

Or what?

Most importantly, did HammerinHank approve? He's our resident expert in the kitchen.

HU Rock Paper Scissors, you vs me.

I have to do a hidden die roll before each throw. 40% of the time it will say I must throw Rock. The rest I can do whatever I want. All this is common knowledge. What's your best strat?

I haven't done the analysis to prove it, but I am thinking that the right strategy within the constraints is for you to go 40% rock, 20% paper, 20% scissors, and for your opponent to go 20% rock, 40% paper, 20% scissors.

100% paper for opponent.

He'd just throw 100% scissors when he has a choice and beat you.

Misread it, just saw die roll and assumed he randomized other 60%

These don't add up to 100%. You probably meant 30 wherever there is a 20, in which case I think I agree.

The problem with going R:0.2, P:0.6, S:0.2 is that the guy w/ random dice can exploit this strategy by going R:0, P:0, S:1 when die role indicates random selection, and the guy who would be in a fixed strategy 40% of the time would actually win this match unless the pure mixed strategy player adjusts.

I decided to write a quick simulation (because I'm way too lazy to derive an analytic solution) and had the players play 150k games. The pure mixed strategy player played the following for the last 100k hands:

51.171% paper
30.540% rock
18.289% scissors

It doesn't make sense for unrestricted player to ever choose scissors since it starts off losing 40 percent of the time so the optimal strategy for him will be some combination of rock and paper.

The unrestricted player should go paper 2/3 of the time and rock 1/3 of the time. Any counterstrategy by the restricted player will be no more effective than allowing unrestricted player to win 6.6667% more often than restricted player. Any other combo of rock and paper allows a more effective counterstrategy.

The problem with this is that it assumes a nash equilibrium, and I don't see this being NE. NE is defined as neither player having a profitable deviation from the current game state. Yes, player 1's optimal state of [1,0,0] in Game state 1 and [0, 1/3, 2/3] in Game state 2 is the best adjustment given Player 2's chosen game state, but that doesn't mean Player 2 is in an optimal state himself.

First let's discuss how player 2 is playing, which is [1/3, 2/3, 0].

In Game state 1, player 2 will win 2/3 and tie 1/3, an EV of 2/3

In Game state 2, player 2 will tie 2/9, win 2/9, and lose 5/9, an EV of -1/3.

2/3 * 2/5 - 1/3 * 3/5 = 4/15 - 3/15 = 1/15, as you calculated.

But, player 2 is smart and can figure out that player 1 is playing [1,0,0] 40% of the time, and [0, 1/3, 2/3] 60% of the time. Or:

[2/5, 0, 0] + [0, 1/5, 2/5] = [2/5, 1/5, 2/5].

Can't we maximize by playing:

[1, 0, 0]?

Now we tie w/ his rock 40%, lose to his paper 20%, and beat his scissors 40%. Our EV has jumped to 0.2! So clearly, we aren't in a NE state.



Essentially, the trick for this problem is that the mixed strategy player is trying to exploit the forced strategy of the restricted player, while the restricted player is trying to exploit the adjustment the mixed strategy player is making. The analytical formulation is (I think):

R1 + P1 + S1 = 1
R2 + P2 + S2 = 1
P2 = R1
(R1 - 0.4) / 0.6 = S2
P1 / 0.6 = R2
S1 / 0.6 = P2

I'm not saying we can't adjust to exploit his counterstrategy but as soon as we do it opens us to getting exploited by a new counterstrategy (the 40/60 JL suggested would be the correct new counterstrategy vs. the 1/0/0 you mentioned). The solution has to be the best strategy where he can't adjust to improve his odds.

We are looking for a GTO solution if we are assuming our opponent will always adjust optimally. Keep in mind GTO will not be as profitable as an exploitative line vs. someone who does not adjust correctly.