perhaps u guys have played, or have another name for it. this is a gambling game and seems way to easy to crush souls because ppl are not good at it.

rules: fresh deck, one shoe (multiple would be badass but have never played with more than one, making it even easier) everyone antes whatever amount u want. two cards are flipped up and it is the player thats in turn (left of the dealer and continues in clockwise position like normal poker) to wager an amount within the pot, his objective is to have the next card (3rd card in deck in this case) flipped up and be "in-between" the first two cards. if it is, he wins the amount from the pot he wagered, if not he pays it. if there is money in the pot, play continues in a clockwise order. pretty simple.

some variations or common themes we play:

- if the "in between" card matches one of the "range" cards u pay 2x ur wager.

- cant bet the pot the first orbit or two to give the chance to grow(i see this rule as irrelevant mostly but it's pry good vs the donks)

- if ur range cards are runners like 9 10 - u pay the ante to the pot, if u want to wager on top of that so be it lol

thats about it, do ur own variations but my point of the post was the math behind it. pretty simple to, no? if there is $100 in the pot on a fresh deck and we have a queen and 4 as our range cards, we have 28 of 50 cards to hit "in between," we calculate the pot-odds - we win 28/50 = 56% so we bet 56% of the pot, $56 right?? that simple, solved game, right? middle of the deck and later on the good card counters will excel to.

The above is not factoring in the "2x" wager rule where if I were to hit a Q of 4 as my inbetween card, I would pay 2x my wager, $112 in this instance. I didnt run this math but may try it unless some smarty pants beats me to it

brag: shipped $56 bones in a 6 way dealer calls game - most ppl bought in for like $10 initially, i would have guessed there to be maybe another $8 winner max. ppl passing on j3 and stuff, betting $5 on like 85% hands when there is $50 in the middle. its fun to play friends and family

I don't quite get this. You say, that if you win, you get even money from the pot, not the whole pot? ("he wins the amount from the pot he wagered")

If that is correct, the EV for a bet of $X in your example (using the 2x rule) is:

(28x - 12x - 16x)/50 = 0. This means, that no matter what amount amount you bet, your EV is always zero.

If the two cards are more favourable, lets say 3 and K, the EV of a bet of $x is:

(36x - 12x - 8x)/50 = 8x/25. In this case the more you bet, the bigger your EV, so you should always bet the pot.

Conversely, if your odds are worse, lets say with 5 and J, the EV of a bet of $x is:

(20x - 12x - 24x)/50 = -8x/25. In this case the more you bet, the bigger your negative EV is, so you should always bet the minimum.

Or did I completely misunderstand the rules?

well, shoot, did i spew that hard? i am kind of looking at it more now like "well if i only have 50% chance to win and i am not forced to bet, why bet at all?" ha. i think i have to be quite a bit off then, makes sense when i think about it more simply.

what are ur 12x/16x/8x/24x variables representing?

also, with 52 cards, the first round there will be "6 double" cards. ex: Q and 7 are up, 8,9,10,J are our outs, 6 cards (Qs and 7s) result in us doubling our wager amount.

correct, if u win you take ur wager amount from the pot, if u lose u put it in the pot.

Let's say the two cards are 4 and Q. Your bet is x.

6 out of 50 cards will result in you losing twice your bet:
6/50*2*-x = -12x/50

28 out of 50 cards will result in you winning the amount of your bet:
28/50*x = 28x/50

16 out of 50 cards will result in you losing your bet:
16/50*-x = -16x/50

putting everything together:
(28x - 12x - 16x)/50

etc.

This is a game that has been around a long time, and is generally called Acey-Deucey. Get any group of older poker players who played a lot of dealers choice games and you will find people who have played this game, and also who have amazing stories about how they started out with a pot of $2 and ended up with someone losing $300 (or whatever) because they went pot with A2 or A3 and hit the ace. This can be a very wild game.

Also, it is one where memory is very important - because as you go through the deck you do get to see the cards as they are played. With a really good memory and good, quick math skills, you can have an edge when it comes to the marginal plays. With the easy decisions (A3, K3, etc.) it really doesn't matter - you make your bet and take your chances.

If you estimate your EV 56%, you should not bet 56% of the pot if your goal is maximize expectation. You should bet the whole pot.

There is one other complexity. You get some value from money left in the pot. The larger the table the smaller this value, but you should still factor it in.

With a fresh deck and no "double on matches" rule, with Ace high only, by my quick calculation you bet 18% of the time and have 2/3 chance of winning when you bet. That means if you are the bettor, your equity is 6% of the amount in the pot.

Suppose you are second to act, 82% of the time you get the pot intact, and 6% of the time you get it doubled. So the average pot you get is 94% of the original pot (of course that's equal to the original pot minus the 6% equity for the first player), your equity is 6% of that. This ignores the card subtraction effect from the first deal.

You get additional equity for the chance that the pot comes around back to you. So if you are first to act at a table of 6 your equity is 0.06 + 0.82*0.94^5*.06 + 0.82*0.94^10*0.06 + . . . which comes out to 15% of the pot.

So if you have a 56% chance and bet, your EV is 12% of the pot. But if you don't bet, your EV is 9% from the chance of the pot coming back around to you, possibly much bigger. You still bet, but even two removed cards would destroy your edge. The bet would still win more than 50% of the time, but you would gain more by waiting.

That calculation assumes everyone plays perfectly and the deck is reshuffled each time. After a round or two, the computations change a lot due to card subtraction effects.

But the real kicker, the reason the game causes so much trouble, is the pot equity of the player is so low, and probably negative for average players. The game is designed to be unstable and to result in the extreme events you mention.

I posted too quickly. While it's true that you have equity if you leave money in the pot, you also have equity if you bet and lose.

But my main conclusion is the game is a way for people who know a little math and keep their heads to extract money from people who don't.

thanks for the feedback guys. So, is there a nutshell formula to decide how much to bet? Does it make sense that if a player has a greater than 50% chance of winning, pt should be the default bet to maximize expectation? i will re-read some of the posts, some of them are over my head

Yes. Count the number of in-between cards versus losing cards and bet the pot if the first is greater, bet nothing if the first is lesser. In a tie, don't bet.

If you have to pay double on matches, you have to figure that probability in as well.

Bankroll matters a lot in this game. If you don't have the bankroll to keep betting the pot after several doubles, you give up significant EV.

the 2x rule, which we always play, will cut down on our betting immensely eh? is there a quick formula on how to weight this into whether to bet or pass?

The EV of a given bet is pretty easy to determine. Let's start with a simple case: the first two cards are dealt and they are J and 4. There are 50 unknown cards. The cards in between are 5-T, which is 6 * 4 = 24. The cards outside are A-3 and Q-K, which is 5 * 4 = 20 cards. The remaining 6 cards are *double* cards, 4 & J.

So if we bet 1 unit we have the following possible outcomes:
win 1 unit with a probability of 24/50
lose 1 unit with a probability of 20/50
lose 2 units with a probability of 6/50

1 * 24/50 + -1 * 20/50 + -2 * 6/50 = -0.16 units. Making a bet of 1 unit has a negative expectation. Of course betting more just has a larger expectation (i.e., betting the pot has an expectation of -0.16 pots). Usually players are forced to win their ante back (at least once) before they can bet the pot though (and the OP has confirmed this rule as well).

The above formula can be used for any bet, so long as you know how many winners, losers, and ties are available. So at face value, the bet in this example is -EV. However, given that it might give you the first +EV pot bet opportunity (should you win it) it could be +EV overall to bet here. The answer probably depends on how large the pot is (i.e., how many people are in the game).

Regardless, when it comes to deciding whether to bet the pot or not, the above method should be used. If it is +EV to bet 1 unit, then it is +EV to bet the pot.

word, thanks Sherman. S'more(hersheys)in between a year later, had to reflect

I ran one calculation. Can it be confirmed that a 7 gapper on a fresh deck is break even.

So King and 5 dealt: we have 28 winners, 16 losers, and 6 x*2 losers.

i plugged that equation into wolfram alpha(Shermans) and got exact result of 0, confirmed?

If so, basically, ur always betting anything better than a 7 gapper. No matter where you are in the deck(assuming no card counting) the odds of your outs, wins, and doubles all have had the same chance of being laid up until whatever point you are at in the deck. thus, anytime by nature if your two showing cards are 8 gaps or more you bet them, 7 you can bet them for fun to show your big time and break even , 6 we pass. Card counting is the biggest factor in this game that can swing your ev, when u start remembering and analyzing the deck, taking heavy note on your chances to pay 2x the pot. how many 2x cards are there, i think thats the x factor. Thats when u can start betting the 7 gappers and perhaps even down to the 6 gapper if u have no 2x cards to come. Do you guys agree?

Damn, I butchered my OP math/train of thought hard though