This is a new bar top computerized holdem game called “Shove-A-Dub"



Here are the rules:

The Human deposits a minimum of 105 dollars.

The computer and the human are ‘dealt’ two Hold'em cards each, face own.

The computer always plays ‘all-in’ and always plays with any two cards.

The human can check his hand and decide to play or fold.

If the human folds, he loses 5 dollars.

If the human plays, community cards of a flop, turn, and river are dealt as in normal Hold'em.

Wins and ties are as normal Hold'em hand rankings.

If the human plays and wins he gains 85 dollars.

If the human plays and loses he loses 105 dollars.

In the rare event of a tie, the human loses 12.50 bucks.





Can this game be beaten?

If so, what range(s) can beat the game?

If not, why not?

Here's my take on this:

Clearly, over the long run, if the human always played, he has a 50/50 win/loss chance. Thus, his expected value for always playing is 0.5*85 -0.5*105 = -10.

But, the human can fold and lose 5. So, there will be folding range where folding is better than calling.

Since human can fold every hand, he wants to choose a strategy that has EV > -5.

If we let R be the folding range, then we want to choose it so the following EV is maximized:

EV(R) = P(R)*(-5) +(1 - P(R))*(eq(R)*(85)- (1-EQ(R))*105),

where

P(R) is the probability of being dealt a hand in range R and therefore means a fold

eq(R) is the probability that a hand not in R wins when played

Let’s take the extreme case where R is all hands except AA. Human folds all hands but AA. This hand has about an 85% chance to beat any other hand (except AA of course and some others) but is a fairly constant value. If played the EV is about 56.5 but AA only occurs 0.5% of the time. So the overall EV is

0.995(-5) + 0.005*(56.5) = -4.7

Whatever the best player folding range is, I think it almost 100% certain that this is a -EV situation for the human for the obvious reason that it would not be offered if that was not the case.

BTW - where have you been all this time?

Shouldn't be tough to calculate using just Equilab.

We're risking $100 to win $90, so we need 100/190 = 52.63% equity to call. According to Equilab, the following range has sufficient equity:

{33+,A2s+,K2s+,Q5s+,J8s+,T9s,A2o+,K5o+,Q8o+,J9o+}

This range represents 520/1326 = 20/51 of all hands.

Equilab breaks down the equity like this:

win: 58.54%
tie: 1.37% I think this is the portion of equity which is accounted for by ties, so the actual tie % would be double, 2.74%. Otherwise the sum of the numbers are short of 100% by 1.37%.

EV = 31/51*-5+20/51*(.5854*85+.0274*-12.5+.3872*-85) = $0.3963 per game

It's possible that since ties are handled weird (you're actually paying extra rake since you'd expect to only lose $10 to a tie) a different range might be very slightly more +EV.

Double check my math, please!

+EV casino games are pretty unheard of so I wonder if there's more to it or if I did something wrong, but this is a really small player edge given you're risking $105 in the hands you call for a measly 40 cents per game.

Edit: I just checked the lowest equity hand, Q5s, and the EV of calling is +$0.16, still positive, accounting for the extra rake on ties, so this range should be accurate if I did everything else right. Just based on equity the EV of calling would have been +$0.263.

You should get the equity of each holding, from the strongest (AA) to the weakest (32o??) vs a 100% range (ATC). With that, you can calculate the EV for each one assuming you´re always calling.

You can also determine the breakeven point of folding vs calling for each combination of %call/%fold (for example, if it´s 50/50, then obviously you should only call with hands possesing 55.26% or better equity).

I´m ignoring ties as it complicates and obv makes things worse, but even here it´s very easy to see this is unprofitabe, as your 50% best holding is far from the required equity. (T7o if I´m not mistaken, and it has 47.91% equity vs any 2 cards)

Your worst hand with at least 55.26% to win is JTo, and it´s in the top 20% only

I´d use equilab here, and would do it to you guys if it wasn´t too time consuming. Anyway, intuitively, it seems you can´t find a combination of %call/%fold and find enough hands to make it breakeven, so my educated guess is this game is very unprofitable for the player, and highly profitable for the house.

I think it's $100 to win $85 but anyway, the real problem is that you're paying a $5 rake on every hand, so every time you fold you lose $5. So the profitability of a single hand isn't the only consideration - like, AA has a massive advantage but if you only play AA you'll lose big.

I'm pretty sure I've seen this game before, maybe even posted in this forum, although possibly with different dollar amounts and maybe slightly different rules

I thought of it as posting a $5 blind and facing an all-in for $100 more. The blind counts towards what we can win since it's a sunk cost.

The only purpose of this was to get the required equity to call, and we can verify that 52.63% is correct, so it was fine to model it that way. You could also set up an EV equation and solve for equity.

.5263*85+.4737*-105 = -5, the same price of folding.

I calculated the range of hands we can profitably play.

I think it's probably OK to think of it as a live blind, but you still have to consider that you lose it whenever you fold.

If you don't agree then consider: is playing only AA a profitable strategy? How much will you profit over time?

You need to consider not only the profitability of each individual hand in your +ev range, but what percentage of the time you'll be dealt that, and how much you'll lose waiting for a hand in your range.

I accounted for that in the EV calculation.

I did all of that.

Do you have a problem with any part of this? I think my first post made it clear enough where all the numbers come from, but if not I'll explain better.

Edit: I have just noticed that I had a typo in post #3 and I highlighted the change. However the calculation used the correct numbers so the result doesn't need to be fixed AFAIK.

Nicely done browni3141. (Bolded includes your typo correction)

By using free publicly available software you have quickly arrived at a profitable range to call with and beat the machine.


I’m going to rewrite your equation so that maybe we can all keep track more clearly what is being calculated:

pCall*pWin*85 - pCall*pLose*105 - pFold*5 - pCall*pTie*12.5

(520/1326)(.5854)(85) - (520/1326)(.3872)(105) - (806/1326)(5) - (806/1326)(.0274)(12.5) = $0.32

The gambler starts with $105 and has an expectation of ending with $105.32.

Whomever has designed this game will have to iron out that kink for sure.




Now, it turns out that sometimes when 'shove-a-dubbers' go busto they just walk away from the machine and leave small amounts of free money.

Say for instance, a busto gambler has left 5 dollars on the Shove-A-Dub and we were next in line.

Now there is 5 dollars free plus our 100 we deposit. Now what range do we use?

Small correction needed. You used the pFold when you should have used pCall here.

I don't see why the strategy would change. Isn't it just free money that you could immediately cash out or is it different from a regular slot? I would assume someone who leaves money in a machine is coming back, btw.

Agreed. If there was any way to finish the hand with more EV than $105.40 then you would have included that in your original range. I guess I owe the previous player 5 bucks, but I will keep the $0.40 in EV.


browni3141,

The marketing team has convened a focus group of potential 'shove-a-dubbers'.

When surveyed for what hands they would play if they played the machine, Q5s (and similar hands) were thrown in the virtual muck. When asked why, different shove-a-dubbers would have different complicated mathematical systems that say to fold it. Some of it was fanciful and bizarre, but the more thinking types could be quite convincing.

Usually something about "That hand does not win often enough versus a range of ATC". I checked, and the four combos of Q5s only win with probability pWin=.5063 when run against a range of any two cards.

Would you mind showing your system or the math that justified calling with Q5s and that folding it would be losing money? Then that info can be used in the focus group.

According to Equilab, vs. ATC Q5s:

wins .5071
ties .0406
loses .4517
EVcall = pWin*85+pTie*-12.5+pLoss*-105
EVcall = .5071*85+.0406*-12.5+.4517*-105
EVcall = -$4.8325
EVfold = -$5, so calling is $0.17 better (I got $0.16 last time but it's probably just rounding differences.)

Thanks browni3141! I will take your math to the focus group. Words are pretty easy to disagree with, but math is math. I will report back after the focus group and let you know how it goes.

Browni3141,

OOF! There were three more meetings of the focus group. It did not go well. Here are my notes from day 1:

There were three different types of potential Shove-A-Dub customers. There are the general gamblers types that like to play electronic casino games. Then there are the sports betting types that generally bet on the outcome of live events. Then there are the poker player types that usually play poker live with cards or online with software.

The general gamblers types were not very enthusiastic about the game, and don’t like to do a bunch of math anyway. However, they played the game pretty darn well! They called with anything they considered ‘pretty’ and folded anything that looked ‘ugly’.

The sports betting types were very enthusiastic about the game. They seem to be familiar with odds and math, but I did not get to talk to them very much. They are a close-knit group. However, I did overhear them talking about combined bankrolls and planning to play the game in shifts 24/7!?!

The poker player types are the target demographic of the Shove-A-Dub machine. That is why the prototype is located near the poker tables. However, they disagreed about the payout of the machine. Some who thought it could be beat were trying to beat it but losing money. Some thought it could not be beat at all and would not play. I showed them all your range and your math, including your model that translated the machine into a regular poker hand.

The disagreement was about the $EV of folding being equal to -$5.

I mean, it is obvious that when you press the fold button on the machine it deducts $5 from your balance. But the thinking type poker players kept saying “ev of folding is always zero.” It seemed like this was a rule in their mathematical models of poker. To be fair, your poker model is what you used to derive your range, and then you used that range to beat the machine. So, the model needs to be absolutely sure to be correct before poker players will agree.

Now, in my opinion, your model is correct and is identical to the machine. The action is the same, the risk is the same, and the payout is the same. I did tell them as much.

That was the end of that day of the focus group.

I was planning to come back here to the forum with all of my notes and get more help for the shove-a-dubbers, but the focus group was abruptly cancelled during the third meeting (more on that later). So, I was only able to refer them to this forum in hope that they could come here and get some advice. To my surprise, several of the shove-a-dubbers were already aware of the forum. I guess the very fine reputation of this forum is known among poker players. It is only fair to assume that there are shove-a-dubbers reading this thread now or in the future.

What do you think browni3141? In my opinion your model is correct and your range obviously works. How can we convince the poker playing types from the focus group to play the machine correctly?

Why do the poker player types say “ev of folding is always zero” when the fold button costs 5 dollars?

Is there any difference between $EV (capitalized with a dollar sign) and +/-ev (lower case with a plus/minus sign)?

I actually used two different models in my posts here and it may be confusing. I just used whichever model I felt made the math the easiest.

EV of folding is only zero when you model it that way. It makes sense to do so in traditional poker because when you fold you're putting no additional money into the pot, and the money you already contributed is a sunk cost and irrelevant to the decision. IMO it makes less sense to model it that way here because the game charges you $5 when you fold.

BTW even in traditional poker folding isn't always modeled as 0EV. Consider ICM in tournaments.

This implicitly treats the EV of folding as zero, since the cost of folding is modeled as a live blind. The rest of my post(s) considers the EV of folding -$5. It really doesn't matter as long as you use the correct payouts for your model. If you model the $5 cost as a live blind all the math works out to the same results. I am not going to do the math though because you need to change the payouts in the EV calculation to account for the blind and I think that will just lead to more confusion.

Yes, I agree that is confusing.

It seems to me like the pot is always divided between the players according to the range of cards the players hold on that street (plus possibly other considerations). Then the pot is divided equitably among the players and counts towards their net worth in $EV (capitalized with a dollar sign). And, as you have shown, it does not matter if the money is free (from other players pockets) or comes from our pocket.



Onward to day two of the focus group notes:

This time in addition to the marketing team there were also representatives from the legal team. Apparently, the exact nuances of gaming laws vary in different jurisdictions. The Shove-A-Dub team wants to be able to offer the game in as many places as possible. Some jurisdictions have laws about ‘gambling machines’ that go back to the 1950’s!

In some places, the machine has to pay out a minimum percentage versus ‘rational play’ but in other places the game has to pay out a minimum percentage versus ‘random play’. Finally, in some places the machine has to pay out a minimum percentage period versus ‘worst play’.



Statmanhal has already quoted the price for worst play which is to always call and lose ten bucks on average.

We already know that always folding loses 5 bucks exactly.

What does the machine pay out if the human randomly tosses a coin? Heads means call, and tails means fold.

Worst play would be to take the opposite action of optimal play. I reference the range before to find that these are all the hands not in it: {22,Q4s-Q2s,J7s-J2s,T8s-T2s,92s+,82s+,72s+,62s+,52s+,42s+,32s,K4o-K2o,Q7o-Q2o,J8o-J2o,T2o+,92o+,82o+,72o+,62o+,52o+,42o+,32o}, this range is 31/51 of all combos.

Vs. ATC, this range:
Wins: .4115
Ties: .0492
Loses: .5393

EV = 20/51*-5+31/51*(.4115*85+.0492*-12.5+.5393*-105)
EV = -$15.49 per game

For random play, the EV is just the average between folding and playing ATC (when you do play it's still a random hand). Statman calculated the EV of playing ATC as -$10 and I don't feel I need to doublecheck that. The EV of folding is -$5, so EV of random play is -$7.5

Sorry, I forgot to come back to this. I didn't look closely at your formula at the time and was just going by your description, which I thought was leading you to make a mistake. Let me take a minute to look at your calculation and see if I agree with it. Also I'll probably do a simulation and see how that goes.

I just saw this thread and since I have all of the HU vs Random equities on my computer I thought I would do a simple "brute force" analysis. Caveat: my equities subsume ties so I cannot separate out the tie payment. I am not sure if that will affect the final answer.

First I sorted the 169 hand combos by equity vs a random hand. Then I calculated the "expected" profit for playing the top N hand combos (N ranges from 1 to 169), and, of course, select the N with the highest "expected" profit.

"Expected" profit is simply the hand equity times the amount if won minus (1-hand equity) times the amount if loss minus the amount you must pay for folding, summing over all 1326 possible hands. Again, as mentioned at the outset, I "ignore" ties.

My program finds that you should play all hands from AA down to Q5s. This is 72 hand combos (out of the 169 total) and 520 hands (out of the 1326 total). Total expected profit (over all 1326 hands you could be dealt) is $565.94.

I have not checked to see if this result agrees/disagrees with anything posted above.

$565.94/1326 = $0.4268 profit per hand. I think the small difference between your result and mine is because I accounted for ties. The quirky tie payout makes a small enough difference that it doesn't affect the optimal calling range, but it changes the EV by a few cents. I also found that Q5s is the lowest call.

Chaos on day three of the focus group!

An entirely different company has bought out Shove-A-Dub and are totally changing the machine.

Now the machine matches your deposit and plays against you!

The machine is now called Tourney Ballers.

It has a basketball theme. I don’t really get the comparison of poker to athletics, but maybe they are hoping to cater towards the sports betting types.

At this point the sports betting types said “ugh” and left the room, except for one guy. I think his name was Kelly, but his nickname was Half Kelly so maybe he has a twin brother or something.

The general gambler types just got bored and left, taking their vouchers for the free buffet with them.

That left only the thinking poker players types, the original target demographic of the original Shove-A-Dub machine (now known as the Tourney Ballers machine).



Here are the rules:



The human deposits a minimum of 100 dollars into the machine.

This puts 100 points on the scoreboard.

The computer starts with the same score, 100 points.

The players are dealt two cards each, face down.

The computer always plays and always with any two cards.

The human can check his cards and decide to “shoot” or “pass”.

If the human passes, the score of the computer is increased by 1, and new cards are dealt.

If the human shoots, then community cards of flop turn and river are dealt as in normal holdem.

Wins and ties are as in normal holdem hand rankings.

If the human wins, the shot goes in the hoop and the human score increases by 5.75 points, while the computer score is decreased by 10 points.

If the human loses, the shot clanks off the rim and the human score is decreased by 10 points, while the computer score is increased by 5.75 points.

If there is a tie, the ref calls a foul and scores do not change.

The game continues like this until the human has zero points or 200 points. If the human gets to 200, the machine pays out 200 bucks. If the human goes to zero, the game ends. No payouts of any other amounts are allowed.



Can you help the shove-a-dubbers beat the Tourney Ballers machine?



If so, what range(s) beat the machine?

If not, why not?

Sorry I read the instructions wrong. The game also ends if the computer gets to 200 points.

Well, the designers seem to have trouble vetting their machines. However it sure can be fun to use poker mathematics to ‘beat the machine’.

Is this game broken (+ev for the human)?

(Hint) Mr. Kelly “aka Half Kelly” and all the gamblers from the sportsbook are back and are playing the game in shifts 24/7!





It is called ‘Tourney Ballers' and has a basketball theme.



Here are the rules:


The human deposits 100 dollars into the machine.

This puts 100 points on the scoreboard.

The computer starts with the same score, 100 points.

The players are dealt two cards each, face down.

The computer always shoots (bets all-in) and always with any two cards.

The human can check his cards and decide to “shoot” or “pass”.

If the human passes, the score of the computer is increased by 1, and new cards are dealt.

If the human shoots, then community cards of flop turn and river are dealt as in normal Hold’em.

Wins and ties are as in normal Hold’em hand rankings.

If the human wins, the shot goes in the hoop and the human score increases by 5.75 points, while the computer score is decreased by 10 points.

If the human loses, the shot clanks off the rim and the human score is decreased by 10 points, while the computer score is increased by 5.75 points.

If there is a tie, the ref calls a foul and scores do not change.

The game continues like this until either the machine or the human has zero points or 200 points. If the human gets to 200 or the machine gets to zero, the machine pays out 200 bucks. If the human goes to zero or the machine gets to 200, the game ends with no payout. No payouts of any other amounts are allowed.




EVhuman = pCall(pWin*5.75 - pLose*10)



EVmachine = pCall(pLose*5.75 - pWin*10) + pFold* 1



In both equations:

pCall = how often the human calls

PFold = how often the human folds

PWin = how often the human wins at showdown

PLose = how often the human loses at showdown


What range or ranges of hands can the human play to beat the machine? (If Any)





There are multiple ways to check and see if the human can find a winning strategy. These include:

Brute force exhaustive analysis
Trial and error
Theory of Games

i'll take a crack at this basketball one.

I think you can treat this like an even money bet for 15.75 points since the computer's losses are our gain and vice versa. We want to call with hands that have greater than -1 points EV.

(PWin - Plose)*15.75 > -1
Minimum equity > 0.468

So any hand that has at least 46.8% equity can call in a vacuum. That gives us the following range:

[22+, A2s+, K2s+, Q2s+, J2s+, T5s+, 96s+, 87s, A2o+, K2o+, Q2o+, J5o+, T7o+, 98o]

call (combos) 798/1326
win% 54.76%
lose% 42.03%


EV human = -0.635 points
EV Machine = -1.44 points


Realistically this is a race to the bottom. Yes we will lose points, but the machine will lose points faster, giving us the edge.

I don't know if is the optimal solution, but it seems to be a winning strategy anyway. It doesn't use the Kelly Criterion so I'm assuming that I've missed something here.