The Shortstack Game (Part 4)

First of all....I hate Blogger. I had a lengthier post written that was for some reason deleted by Blogger when I tried to change the font color. Thanks Blogger.

Anyway, it seems like nobody's really interested in this Shortstack Game. But having come this far, I should give some sketch of a solution for equilibrium in the game. If you are enterprising and want to find it (or at least one of them):

1) Show that there is no equilibrium when player 2 shoves less than 1/3 of the time

2) Show that there is no equilibrium when player 2 shoves more than 1/3 of the time

3) Find an equilibrium where player 2 shoves exactly 1/3 of the time.

In the equilibrium I found, player 1 raise/folds 27% of the time, raise/calls 29% of the time, and open folds 44% of the time. Player 1's expected payoff in equilibrium is .42, while player 2's is 1.08, showing the disadvantage of having to act first.

I can also solve for an n-player game where two of the players have to post blinds, and the other players sequentially must raise or fold if no-one has yet raised, shove or fold if a player has raised already, and fold or call if a player has shoved already.

If you are interested, post a comment or email spritpot.

-BRUECHIPS

The Shortstack Game (Part 3)

Just in case anybody is still reading this...having set up our game and solved for player 2's best response function, we now need to figure out player 1's best reponse function. I ended the last post with a couple of questions to get you started considering player 1's best response function:



What's player 1's best response if player 2 folds every time?



Pretty obvious that if player 2 always folds, player 1 should always raise.



What's player 1's best response if player 2 shoves every time?



It's pretty obvious that if player 2 shoves every time, player 1 should never raise/fold. So depending on his hand, he should either raise/call or fold. If he folds, he gets zero, whereas if he raise/calls, he gets 21.5e - 20*(1-e). Set this expression equal to zero and solve for e, you get 20/41.5. So he should raise/call anytime his hand has more than (20/41.5) equity against a random hand. We already determined in the last post that these hands are 22+,A2+,K2+,Q3o+,Q2s+,J7o+,J3s+,T8o+,T6s+, and 97s+, for 54.8% of all hands. Player 1's average payoff is then:



.452*0 + .548*(.572*21.5 - .428*20) = 2.05



Whereas player 2 has an average payoff of -0.55. Not coincidentally, this is the exact inverse of the payoffs when player 1 raise/calls every time and player 2 best responds.



What if player 2 shoves 22+,A9o+,A5s+,KQo,KJs+ (14.6% of all hands) and folds everything else?



We will have to solve this by "backward induction", considering the last decision first and then using that to figure out the payoffs involved in making earlier decisions. So let's assume player 1 has raised and player 2 has shoved. Which hands should player 1 be calling with? He would be calling 17 to win 24.5, so he needs 24.5*e - 17*(1-e) > 0, or e greater than 40.9%. These hands are: 33+, ATo+, A9s+,KQs (9.7% of all hands). All other hands, had player 1 raised them, would be folded. Player 1's calling range then has 53.8% equity against player 2's shoving range. Now we can consider player 1's decision to raise or fold as his first move. Say he has the most marginal hand with exactly 40.9% equity. Should he raise? His expected payoff to raising is:



.854*1.5 + .146*(.409*21.5 - .591*20) = .84



This is clearly better than 0, the payoff from folding. Also note something else. If player 1 has a hand that he will NOT call a raise with, his payoff to raising is:



.854*1.5 + .146*(-3) = .84



This is also better than folding. That the payoff to raise/calling and raise/folding with the 40.9% equity hand is no coincidence. The 40.9% mark is exactly the point where calling player 2's shove and folding to it are equal in expected value.



Player 1's expected payoff to this strategy is now:



.854*1.5 + .146*(.097*(21.5*.538 - 20*.462) - .903*(1.5)) = 1.1



What is player 1's best response to any given strategy for player 2?



Let f be the % of the time that player 2 folds when player 1 raises. Then player 1's expected payoff of raise/calling is:



1.5*f +(1-f)(-3) = 4.5f - 3



This obviously equals 0 when f equals 2/3. So if player 2 is folding two-thirds of the time or greater, player 1 should raise every time. If player 2 shoves, then player 1 should call if he has greater than 40.9% equity vs. player 2's shoving range.



If player 2 folds less than 2/3 of the time (and therefore shoves more than 1/3 of the time), player 1 should never raise/fold, and should raise/call if:



1.5*f + (1-f)*(21.5e - 20*(1-e)) > 0



Where e is the equity of player 1's hand against player 2's shoving range.



-BRUECHIPS

The Shortstack Game (Part 2)

WARNING: This post is long and has a lot of math in it. If you don't understand something or just want to discuss, I encourage you to post in comments. I will try to respond and clarify.

In an earlier post I described a game representing the interaction between a cutoff raiser, player 1, and a shortstack on the button, player 2. Towards the end of the post, I asked two questions - what is player 2's best response if player 1 plays RAISE/FOLD every hand? What if he plays RAISE/CALL every hand?

If player 1 plays RAISE/FOLD every hand, it's pretty clear that player 2 should be shoving every hand. That is, SHOVE is player 1's best response to RAISE/FOLD. It doesn't matter what player 2's cards are, since player 1 will fold every time.


What if player 1 always plays RAISE/CALL? Player 2 has two options, FOLD and SHOVE. FOLD gives a payoff of zero, whereas SHOVE gives a payoff of:


(1) f*(4.5) + (1-f)*(21.5*e - 20*(1-e))


where f is the percentage of the time player 1 folds and e is the equity that player 2 has when player 1 calls. This e is obviously a function of the cards that player 2 holds, and the range of cards that player 1 is calling with. If player 1 plays RAISE/CALL every time, then f = 0, and e for a given hand for player 2 will just be the equity that hand has against a random hand (in Poker Stove, just click the 'RD' button to give a player a random range). Setting f = 0, we reduce (1) to:


(2) 21.5*e - 20*(1-e)


For SHOVE to be the correct play, it has to offer a higher payoff than the alternative, which is FOLD. Set (2) equal to the payoff for FOLD, which is zero, and solve for e :


(3) 21.5*e - 20*(1-e) = 0


=> e = 20/41.5


That is, if player 2's hand has equity of more than 20/41.5 against a random hand, he should shove. These hands are 22+,A2+,K2+,Q3o+,Q2s+,J7o+,J3s+,T8o+,T6s+, and 97s+. This is 54.8% of all hands. So how is each player doing under these strategies?


Player 1 raises every time. 45.2% of the time, player 2 folds and player 1 gets 1.5. The other 54.8% of the time, they get all in and player 1 has an average of 43% equity:

equity win tie pots won pots tied
Hand 0: 57.233% 55.68% 01.55% 847980578496 23586063110.00 { 22+, A2s+, K2s+, Q2s+, J3s+, T6s+, 97s+, 87s, A2o+, K2o+, Q3o+, J7o+, T8o+ }
Hand 1: 42.767% 41.22% 01.55% 627684857684 23586063110.00 { random }


So his total payoff is: .452*1.5 + .548*(.428*21.5 - .572*20) = -.55


Player 2 folds 45.2% of the time for a payoff of zero. The rest of the time, he has 57.2% equity for a total payoff of: .452*0 + .548*(.572*21.5 - .428*20) = 2.04 (Note that the two average payoffs add to 1.5. This must be the case under any strategies the players play, since no matter what happens, the two players' payoffs add to 1.5 - go back to the earlier post if you want to check)


Again, since we have figured out player 2's BEST response for player 1's strategy, this must be the highest payoff player 2 can get. Obviously this is not the case for player 1. If he just folded every time and got zero, he'd be better off than he would be playing the strategy of RAISE/CALL for every hand. So this is not an equilibrium (remember, an equilibrium is where both players are best responding to each other. Here, player 2 is best-responding to player but not vice versa).


Now that we've figured out player 2's best responses for some very simple player 1 strategies, let's figure it out for a strategy where player 1 varies his strategy based on his hand. Say, for instance, Player 1 plays RAISE/CALL with 22+, A8o+,A2s+,KT+, RAISE/FOLD with 45s-QJs, 68s-QTs, 89o-QJo, and T8o-QTo, and FOLD with all other hands. (1) still holds for determining player 2's payoff. To get the answer exactly right, we would have to calculate a different f for every hand player 2 has, due to card removal effects (e.g., if player 2 holds A2, it is less likely that player 1 holds AA, etc.). But we will sacrifice some accuracy in exchange for ease of calculation and calculate f independent of player 2's holdings. Notice that f is just the probability player 1 plays RAISE/FOLD divided by the sum of the probabilities player 1 plays RAISE/FOLD plus the probability he plays RAISE/CALL. That is, it's the probability he folds given that he has raised. 22+,A8o+,A2s+,KT+ is 18.6% of all hands. 45s-QJs, 68s-QTs, 89o-QJo,T8o-QTo is 10.2% of all hands, so


f = 10.2/(18.6 + 10.2) = 35.4


Plug this into (1), set equal to 0, and solve for e as earlier. you get e = .413. That is, any hand with at least 41.3% equity vs. player 1's calling range of 22+,A8o+,A2s+,KT+, player 2 should shove. Here are all such hands: 22+,A9o+,A5s+,KQo,KJs+. So player 2 is shoving 14.6% of the time. When player 1 calls, he has an average equity of 59.7%.


Now player 1's payoff is: .712*0 + .102*.146*(-3) + .186*.146*(.597*21.5 - .403*20) + .288*(1-.146)*1.5 = .45


So at least player 1 has improved upon his RAISE/CALL strategy and has now found a strategy that's better than folding every time.


Player 2's payoff is: .712*1.5 + .288*(.01*0 + .99*(.646*4.5 + .354*(.36*21.5 - .64*20)) = 1.05


So now we have figured out how to determine player 2's best response to a given strategy for player 1. Step 1: Calculate f from player 1's strategy. Step 2: Set equation (1) equal to zero and solve for e. Step 3: Go to Poker Stove and find all the hands that have at least e equity vs. player 1's calling range. This mapping from player 1's strategy to player 2's best response is called player 2's "best response function". In the next post on The Shortstack Game, we will figure out player 1's best response function. Then we will find a point where they intersect, which will then be an equilibrium. If you want to get a head start, try and find player 1's best response to player 2 playing FOLD every time, playing SHOVE every time, and playing SHOVE with the 22+,A9o+,A5s+,KQo,KJs+.


-BRUECHIPS

The Shortstack Game, Part 1

A recent post by gnome got me thinking, in part because I think he made an error in his post, which I describe in my comment there. As you might know if you follow the blog closely, I am a grad student in Economics. One of the things I study is game theory. When game theorists "solve" a game, what they look for (or at least the first thing they look for) is a "Nash equilibrium". When two players's strategies are in Nash equilibrium, each one knows the other's strategy, and even with that knowledge, neither one would choose a different strategy. Here's another way of saying it. Take some game with two players. Fix a strategy for player 2. Call player 1's "best response" to that strategy as the strategy available to him that maximally exploits player 2's strategy. Two strategies strategy1 (player 1's strategy) and strategy2 (player 2's strategy) are in Nash equilibrium if and only if strategy1 is player 1's best response to strategy2, and strategy2 is player 2's best response to strategy1.

As a whole, poker is far too complicated a game to find equilibrium strategies. There is not any guarantee that there exists a unique equilibrium. But sometimes we can take toy games that mimic at least some situations in poker, with some simplifications, and find an equilibrium. You'll find many such games in The Mathematics of Poker by Ankenmann and Chen, which I highly recommend. In any case, here we will solve 'The Shortstack Game', which works as follows:

Player 1 and Player 2 receive two cards. Player 1 can either fold or raise. If player 1 folds, he gets a payoff of zero and player 2 gets a payoff of 1.5. If player 1 raises, player 2 can either fold or shove. If player 2 folds, player 1 gets a payoff of 1.5 and player 2 gets a payoff of zero. If player 2 shoves, player 1 can either call or fold. If player 1 folds, he gets a payoff of -3 and player 2 gets a payoff of 4.5. If player 1 calls, a Hold 'Em board is dealt out and the player with the best hand wins. The winner gets a payoff of 21.5, whereas the loser gets a payoff of -20.

This game preserves the important features of a Hold 'Em situation where a player raises on the cutoff with a shorty on the button. I have made some simplifying assumptions, some of them important, some not. For instance, I say that when player 1 folds, player 2 also gets a payoff of 1.5. In reality, of course, player 2 won't always win the blinds when player 1 folds, but insteads enters into a new game with the players in the blinds. But that is inconsequential to our analysis here (although I plan to come back to it later), as all the decisions we're interested in are 1) whether player 1 decides to raise, and 2) what happens after player 1 raises. Neither of these depend at all on what player 2's payoff is when player 1 folds (convince yourself this is true or ask for clarification in comments if it is unclear).

Another assumption I've made is that the blinds fold every time (if player 1 raises and player 2 folds, player 1 wins 1.5, the blinds, automatically, and if player 2 shoves, the action is immediately back on player 1). This is obviously consequential as the potential for blinds calling or re-raising affects the payoff of player 1 when he raises and player 2 folds, and therefore how good raising is relative to folding for player 1. I could eliminate this need for simplification by making player 1 the small blind and player 2 the big blind, but: 1) If I recall correctly, there is already analysis of some very similar game in Ankenmann and Chen, and 2) Often as a good deep-stacked player you're going to want to be raising in position vs. the blinds so you can take a flop in position. I'd venture that 1.5 (i.e., winning the blinds for sure) is actually much lower than player 1's true EV of player 2 folding on the button to his raise.

Anyway, this post is plenty long enough already, so I think I'll stop there for now and allow for digestion and questions of the game setup before proceeding to solving the game. But if you want to do some work on it before next post, note that a strategy for player 1 consists of choosing either FOLD, RAISE/CALL, or RAISE/FOLD for every hand possible. A strategy for player 2 consists of choosing FOLD or SHOVE for every hand possible. What is player 2's best response to player 1 choosing RAISE/FOLD for every hand? What is player 2's best response to player 1 choosing RAISE/CALL for every hand?

-BRUECHIPS