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