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I was considering an interesting situation. Imagine you are playing poker against someone who is always raising in a tournament game (both start with a stack of chips, and win when they have the other persons chips). We are talking pot limit omaha. This means that people get dealt 4 cards, and they are supposed to make a best match with 5 cards on the board. Person with the highest cards according to poker (pair, two pair, straight etc), wins.

What makes this question particularly interesting is that blinds (the amount you have to pay to stay in the game) increase over time (so the maximum pot size automatically increases over time). Every 10 rounds, the big blind increases in size (first +10, later +20, then +40 etc)

Contrary to texas holdem, omaha has a lot more spread since people are dealt 4 cards instead of 2.

Basically, this means you can not just sit and wait for getting the best possible cards, especially since it cannot be too clear at the start (a lot can happen: e.g. in texas holdem the best starting cards AA has something like 78% chance to win, while in omaha the best starting cards would only have like 65% chance to win).

I am trying to find a way to find the optimal strategy when you know someone would always raise.

An example:

You start with a big blind of 10. You know that in the subsequent turns, the pot size would triple for each turn (disregarding the fact that you can raise as well). Since the opponent will never fold, this means that it will be quite expensive to see the cards after all the rounds.

A certain way to play would be to only call when you believe you have higher than 50% of a chance to win, but this is certainly not optimal.

What would be the way to solve this problem analytically? I was trying to consider some simulation study where I try to be more picky initialially.

In fact, I came up with 3 criteria:

-How deep is your stack (how many times can you call the opponent and 
 lose a round before you lose the whole game)
 --> You can be more picky when you have "more to lose". Then again, desperate
    times could call for desperate measures (why slowly bleed to death?). Does
    this even have an influence at all?
-What is the chance to win given your current cards, and given the current board, 
compared to "random" cards (yields a probability to win on each turn in a round)
 -->You could for example quit halfway in a round when your probability 
    to win dropped dramatically
-What is the ratio of your stack divided by the size of the stack of the opponent?
 -->Also some kind of measure that can either be used to be "picky" or "desperate" 
    or neither.

I am pretty sure the problem can be solved with these 3 criteria (together with the increasing-over-time blind), I just do not know how I can come up with a design that would test this adequately. Perhaps some people even know if some of these criteria do not apply?

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  • $\begingroup$ If someone always raises, then they are giving up the advantage of bluffing. How many people are assumed to be in the game? $\endgroup$ – DWin Sep 22 '14 at 19:55
  • $\begingroup$ @DWin It's a two player game, just you and a person (villain) who always raises. $\endgroup$ – PascalVKooten Sep 22 '14 at 20:39
  • $\begingroup$ Then I'm not sure why you think that only raising when your odds of winning are better than even is suboptimal. $\endgroup$ – DWin Sep 22 '14 at 20:54
  • $\begingroup$ I used some simplified conditions and simulated with it, where being "picky" gave something like 58% chance to win, while only calling above 50% really just only gave 50% chance to win. Don't forget: if at any stage of assessing the probability to win would drop below 50%, you would fold. $\endgroup$ – PascalVKooten Sep 22 '14 at 20:58
  • $\begingroup$ If you are folding to to 50% then you are not playing pot odds. If you are getting 3:1 on your money then you only need to be 3:1 to be even. $\endgroup$ – paparazzo Feb 10 '16 at 2:31
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Let's simplify this a little:

  • Each hand, you start with $D$ in chips, and your enemy starts with $E$;
  • The ante is $A$;
  • If you call your enemy's pre-flop raise and fold on the flop, you'll have to put in $P$ total;
  • If you call your enemy's raise on the flop and fold on the turn, you'll have to put in $F$ total;
  • If you call your enemy's raise on the turn and fold on the river, you'll have to put in $T$ total;
  • If you call your enemy's raise on the river, you'll have put in $R$ (and you'll have a chance to win);
  • At any point (and apologies if this is off; I'm not too solid on the rules of pot-limit poker) you can re-raise, in which case your enemy will re-re-raise until one of you is all-in.

What makes this problem tractable is that at each step (pre-flop, post-flop, post-turn, post-river) you know exactly how much you have in chips, and you know the probability that your hand + the community cards will win against any randomly selected hand.

For example, before the flop, you see that your cards have probability $p_0$ of winning against a randomly selected hand. Then your decisions are:

  • Fold pre-flop and lose $A$ with probability 1,
  • Pay $P$ to see the flop,
  • Re-raise until you're both all-in, which you will win with probability $p_0$.

If you pay to see the flop, then your probability of winning changes from $p_0$ to $p_1$, and your options become:

  • Fold on the flop and lose $P$ with probability 1,
  • Pay $F$ to see the turn,
  • Re-raise until you're both all-in, which you will win with probability $p_1$.

etc.

The final challenge of this problem is determining your probability of victory when there is nothing in the pot, and you have $D$, and your enemy has $E$. While there's probably some non-linearity when $D >> E$ or $E >> D$, you can at least set a lower bound on the probability of you winning with $\frac{D}{D+E}$.

To compute the value exactly, you can use Backward Induction. This is a rather large game because of the number of potential hands, but it has a small number of moves, so it's much smaller than chess or Go.

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  • $\begingroup$ Note that there can be more moves though: you could quit at any time. Going to see the next stage but then discovering your probability dropped could be very costly. From the little simulation I've done, it seems like you don't have to be so picky pre-flop, but most importantly on the flop you need to be picky, only call with p>0.8 or p>0.9. I varied the probabilities for each possible moment (pre-flop, flop, turn, river), but it's hard to optimize and I doubt if it is efficient. $\endgroup$ – PascalVKooten Sep 23 '14 at 6:55
  • $\begingroup$ Also, do you think the optimal strategy shifts as the stack sizes change change during a game? $\endgroup$ – PascalVKooten Sep 23 '14 at 6:58
  • $\begingroup$ Good point--I updated my answer. $\endgroup$ – Michael K Sep 23 '14 at 16:09
  • $\begingroup$ You seem to really nicely sum up the situation. It seems awesome, though I'm sad to say I'm still not sure how to continue. Is there a way to use sampling to then approximate the probability values under which one should fold at each time period? $\endgroup$ – PascalVKooten Sep 23 '14 at 17:28
  • $\begingroup$ Note that I created some example data here: pastebin.com/VwshxDUk, where h1-h4 are the cards in the hand, and the simulated probabilities against a random hand are shown when considering the board cards known at each moment. $\endgroup$ – PascalVKooten Sep 23 '14 at 17:30
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This has been up for a while

Bind size makes no difference. Yes it put you at a higher risk if getting stacked but the ratios are the same.

Lets say have 150 bb
You can live 100 orbits before blinding out (ignore blinds go up)

Let say you only play if you are 50% or better and the average is 55%

So you play exactly 1/2 the hands. We can ignore the blinds as you lose and many as you win.

You only check if you are in

The average ending pot is 73 bb
(that is kind of dangerous as you only start with 150 bb)

So for every two orbits the EV for two orbits is -.45(54) + .55(54) = 5.4 bb
Lets just ignore you will win more pre-flop action

Notice that is in terms of bb so bb going up is not a factor

So lets look at you go cold and loose every hand
Every 2 round
+72/2 + .75 (blind) = 37

If you start stone cold you can live 8 orbits
You would need to lose a .45 4 times .45 * .45 * .45 * .45 = 4.1%

On average you are going win 2.7 bb so about the only way to go out is go stone cold early.

You have the option of folding the flop, turn, or river if cards don't hit, and can raise if they hit then things go up.

If you have re-buys then you cannot lose.

If it is a tournament with starting stack of 150bb then you will win 95% of the time.

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