R$ is a constant negative expectation to the caller.The caller calls much more narrowly because of this. All the caller EV calcs in this thread take that same unchanging amount into account.

This is the last post that I make that does not account for every chip on the table. What do you think the total chips on the table, including the rake drop box, must add up to?

Or, you could take as long as you want, check it yourself, and post what you find.

Let's look at two different sets of strategies at 20% rake:
Pusher 100%, caller 0%: Sum of expected payoffs for both players equals +10$.
Pusher 100%, caller 100%: Sum of expected payoffs for both players equals -32$.

Since +10$ and -32$ are not equal, we don't have a constant sum game. Do you disagree?

This is why poker is located in game theory textbooks (older cool ones) under the chapter ‘Zero-Sum non-cooperative games’

This confusion arises because of the way game theorists came up with the term ‘zero-sum’. They had to call it zero-difference or zero-sum. They went with zero-sum. Today, we typically call this ‘indifference’, instead of using the older term ‘zero-sum’



When comparing two non-cooperative players, Player1 and PLayer2, we assign them strategies based upon the actions they can implement when competing against one another.

Here, Player1 is the shover, Player2 is the caller.

Player1 may bet or fold. He has bet gains and losses (including rake) and never folds.

Player2 may call or fold. She has calling gains and losses(including rake), and folding losses.



According to the Minimax Theorem:

For every two-person, zero-sum game with finite strategies, there exists a value V and a mixed strategy for each player, such that (a) Given player 2's strategy, the best payoff possible for player 1 is V, and (b) Given player 1's strategy, the best payoff possible for player 2 is −V.



This means that Player1 gains = Player2 losses

So...

Player1 gains – Player2 losses = 0

And...

[Player1 gains – Player1 losses] - [Player2 gains – Player2 losses] = 0



^^^This is the modern method of indifference, which has half as many equations in half as many variables

Now, this below will demonstrate that it was always about the ten bucks in the middle, and that both players started the hand with a net worth of 105, not 100. When I said somewhere that they start with 100, that was my mistake, that I corrected in my post earlier today.

Given that players start with net worth of 105...

20 percent rake scenarios:

Scenario A:

Player1 always bets and Player2 always folds...

Players start at 105 each.

Neither player pays anything in rake..

Player1 steals the 5 dollars ante that belongs to Player2...

This means Player2 has a 5 dollar loss. This is a positive loss due to strategy. <---------IMPORTANT

[Player1 gains – Player1 losses] – [Player2 gains - Player2 losses] = 0

[+5 – 0] - [0 -(- 5)] = +5 – 0 – 0 –5 = 0 -------> zero-sum



Scenario B:

Player1 always bets and Player2 always calls... (roughly a 50/50 flip at high rake)

Both players flip for stacks and both players pay half the rake...

[Player1 gains – Player1 losses] – [Player2 gains - Player2 losses] = 0

[(105(.5) - 105(.5)) - 21dollars rake ] - [(105(.5) - 105(.5)) - 21 dollars rake] = 0 <-----rake is a mandatory fee to BOTH PLAYERS and is not a strategy loss

[52.5 - 52.5 - 21] - [52.5 -52.5 - 21] = 0

[-21] - [-21] = 0

-21 + 21 = 0 ------>zero-sum



Every possible scenario, since the game is finite, can be enumerated this way. Poker chips are not created or destroyed. They just move around.

There is 210$ on the table at the beginning of the hand and depending on the actions that either stays the same or is reduced by some amount of rake.

Poker is located in the zero sum category of textbooks because the books usually look at unraked 2-player games. In unraked games the sum of chips indeed remains the same. In our game the house literally takes chips off the table.

You can model this game as:
2-players, variable sum
or
3-players, constant sum (we consider the "house" as a player in that variant)

In your calculations you are just arbitrarily assuming that the average payoffs over all players in any given state is the game's supposed "fair outcome" and you then measure payoffs relative to that fair outcome. (Two values have equal distance to the average of the two values? Amazing!) You could apply that same process to "Game of Chicken" or "Prisoners Dilemma" and wrongly conclude that they are zero sum games.)

Well, it would seem that now I am requested to do the homework support.

So, to be clear, this is what you bekieve:

Because 2 players HEADS-UP pay an equal amount of rake in a single hand of poker, they can somehow have competing non-cooperative strategies where they both SIMULTANIOUSLY HAVE A POSITIVE NET WORTH.

Read the OP. There is an added 10$ in the pot that doesn't come out of players' stacks. And the rake taken out of the pot depends on the action, so it's not constant. So yes, it's possible (and indeed expected) that both players end up +EV.

It wouldn't make sense for one of the players to have a negative "net worth" since he can always just fold 100% of hands and be at 0 EV.

They should always both be making money here. The higher the rake is, the higher the EV of the pusher will be. At 100% rake the EV of the caller would be 0 and the EV of the pusher would be 10. At 0% rake, the caller would be making more money.

Using your math, for your shover, at your equilibrium, at your 20.61 percent rake:



The net results of your long term maximin + EV winning strategy are thus:



[starting stack] - [rake paid] + [ante wins] + [showdown wins] - [showdown losses] = [final stack]



What numbers go in these blanks, and what is the final stack of your shover?



Hint: the house charges each player (stack x rake) for placing a bet that is called, and also deducts (antes x rake) whenever a bet is placed and called.







I suggest you double check before posting.

I'm not interested in arguing for another 50 posts about whether or not this is a zero sum game, it obviously isn't and if you disagree then please pick up any intro to game theory. The stuff we've been going over for the last 50 posts is super basic and if you aren't trolling then you can figure this out by yourself in a few hours. gl & hf

So, you do not even know what the shover ends up with?

Just not interested in taking another detour before you understand the definition of a zero sum game. We are not going anywhere by hopping around like that. You've posted hundreds of times in Nash/GTO threads and still get the basic definitions wrong, if you are trying to learn something here then that was obviously not a very efficient use of anyone's time.

Players A & B are at the Turn and the pot is $60.
Player A has top pair.
Player B has a flush draw.

Player A goes all-in for $10.
If Player B calls, what is each player's EV?

With respect, we are dealing with preflop range vs range calculations. I will be glad to discuss other post-flop scenarios another time.

Thank you heehaww.

The concept is the same: there is already a pot, which is why both of their EV's can be positive.

heehaww,

I can demonstrate that preflop poker is zero-sum, even when raked.

The caveat required, is that both players are at equilibrium, and that neither player can improve by varying strategy. Both players equally contribute to the rake, if there is a bet that is called. House collects rake on the antes, if there is a bet that is called.

Both players start with equal net worth, both players finish with equal net worth.

Do you think I can do this?

We got 5 different payoff scenarios for this game:
1) Pusher folds: P: 0, C: +10
2) Pusher pushes, caller folds: P: +10, C: 0
3) Pusher pushes, caller calls; pusher wins: P: 210*(1-rake) - 100, C: -100
4) Pusher pushes, caller calls; caller wins: P: -100, C: 210*(1-rake) - 100
5) Pusher pushes, caller calls, tied: P: 105*(1-rake) -100, C: 105*(1-rake) -100

Whether or not the game is zero sum is not dependent on any specific strategies by the players. You only need to check if the sum of EVs of P + C is zero for every possible scenario. (Or at least constant, as constant sum games can trivially be transformed to zero sum).

In our game, the sum of payoffs is not constant unless rake=0%.

@Robert, I was only contesting your apparent view that the players can't both be +EV. Also, if there's always a preexisting pot preflop, from a source external to the players, then it can be a positive-sum game for the players (eg if there's no rake).

Without a preexisting pot, a raked game is negative sum because either both players are losing money, or the winning player's profits are less than the losing player's losses.

However, with rake AND a preexisting pot, the game's sum depends on the parameters, and I would think hypothetically there can be parameters that cancel each other out to make the game zero-sum. I have not attempted the problem(s) posed ITT because it sounds like the case is already closed.

That's not what zero-sum means.

Fair enough.

Starting net worth of both players added together will be a number (within a few pennies)

Ending net worth of both players added together will be the same number (within a few pennies)

This will also show that both players equally own the free antes at the start.

All of this, without the handy aid of Minimax Theorem.

A game is classified as zero sum if the sum of payoffs is always zero, that's for every possible strategy profile. Finding one specific strategy profile where the sum of EVs is zero is not sufficient.

Since I am no longer using minimax theorem, the interpretation of the term zero-sum will not matter.

This will be maximin, each player maximizing their own minimum gains.

At your 20.61 percent rake, your players are not optimal. Shover can do better by selecting hands to fold, and claim more than his fair share of the antes.

When he does this, the ending net worth of both players added together will be all of the chips in play within a few pennies ~210.

Are we finally in agreement that this is not a zero sum game?

For non zero sum games, Nash Equilibrium and maximin are different things. Showing that you can find a higher guaranteed minimum does not mean that the Nash Equilibrium solution is wrong.

Player1 gains will equal Player2 losses.

Rather than banter back and forth if that is the definition of zero-sum, I will do the above since this will make the matter clear.

You keep using theoretic results from zero sum games in a game that isn't zero sum. Let's get this one topic squared away before we go on. It's really simple, you just need to look at the payoff definitions. Idk what's so difficult here.

Friends,

This is absolutely crucial. This is a foundational building block of economics. Forget about my magic calling range (not magic, really works), if anyone does not get this question below right, they are in serious jeopardy of making HUGE mistakes both on and off a poker table.

This is a basic exercise in game theory. This is usually in the first few pages of guess which chapter.



Below, I will use the 20.61 rake and the ranges from others in the thread, probabilities are only slightly adjusted for tiny occurrences of chops (every possible outcome must add to 100 percent).

First, we are the bettor. We will sit in the bettor’s seat and watch his betting line. When his chips cross the line we will count them, and when they come back we will count them. (We don’t trust the dealer).


Betting strategy is all-in always:

With probability 100 percent we push 100 into the middle, so BEHIND THE BETTING LINE there is zero, and across the line there is now 100 of our chips plus the free 10 in antes.

With probability (1326-188)/1326 the dealer gives us back the whole 100 plus the 10 so 1138/1326(110) = 94.40422

So, 94.40422 is back in front of us, as we sit in the bettor’s seat.

With probability 188/1326 there is a call and the dealer collects exactly 43.281 from the pot and keeps it.

This leaves precisely 166.719 in the middle of the table.

With probability (.3298) we win that pot:

(188/1326)(166.719)(.3298) = 7.79560945

And...

With probability (.0176) there is a tie, and we get back 83.3595

(188/1326)(83.3595)(.0176) = 0.20801

We get back zero otherwise.

We have watched every chip from the bettor’s chair, and now let’s count everything we won with our betting strategy:

94.40422 + 7.79560945 + 0.20801 = 102.40784

Neato! We won a net increase of 2.40784!



Next, same for caller. We count every chip.

We fold with probability (1326-188)/1326 and we keep all 100.

100*(1326-188)/1326 = 85.82202 kept behind the betting line.

With probability 188/1326 we call and push our 100 in the middle.

The dealer collects 43.281 and keeps it.

Pot is now 166.719...

With probability (188/1326)(.6526) we win and collect 166.719

(188/1326)(.6526)(166.719) = 15.4257572


With probability .0176 there is a tie:

(188/1326)(83.3595)(.0176) = 0.20801

We collect zero otherwise:

We have watched every chip from the caller’s chair, and now let’s count everything we kept or won:

85.82202 + 15.4257572 + 0.20801 = 101.4557872



Sweet! We won money too!



We started with 210 chips in play, but since this is raked poker, it is supposedly ok that the totals we finish with are Shover + Caller = 101.4557872 + 102.40784 = 203.8636272 = 204 Bucks.

And, if you set rake at .00001 percent, suddenly you get back a total of 210 bucks.

I must admit, that is a pretty slick trick. I can understand why there would be books, videos, software etc. That would be marketed as if all of the above is true.



SO, WHY IS IT THAT:



If we keep this same sequence of actions, shover always shoves, caller calls sometimes, why is it that when there are no antes at all our players only get back a total of 194.49 chips? (the calculation is below).

You can assign ANY ranges you want, shover has aces, caller has ATC, both players have ATC, whatever. You can NEVER FIND ALL the missing chips.


The house is only charging the same percentage price on the 200 pot as it did on the 210 pot.

Somehow, money has disappeared??!?

So. Where did the money go?


If no one on the worlds greatest poker forum will post the solution to the missing chips, then I will post the answer in 24 hours.

Thank you for reading.





Bettor:

Caller folds with probability (1138/1326) and we get back (1138/1326)(100) = 85.82202

The dealer collects EXACTLY 41.22 and keeps it.

Pot is now precisely 158.78.

(188/1326)(.3446) we win and collect 158.78

(188/1326)(.3446)*158.78 = 7.75756

With probability .0176 there is a tie:

(188/1326)(79.39)(.0176) = 0.19810

Add up every chip:

85.82202 + 7.75756 + 0.19810 = 93.77768





Now the caller:

100*(1326-188)/1326 = 85.82202

(188/1326)(.6526)(158.78) = 14.69120

(188/1326)(79.39)(.0176) = 0.19810

Adde ‘em up:

85.82202 + 14.69120 + 0.19810 = 100.71132



Total money back to players from 200 in play: 93.77768 + 100.71132 = 194.489 = missing 5 bucks!