New poker game

Yes I agree with your total of the combinations, but the EV is -1 not -.5 : i.e. you stand to lose your initial Ante if you fold (i.e. can accept a loss of 1 unit) or for the same expected return raise and either Win 2 or Lose 2 (75% Lose, 25% Win).I get that you lose 17489.5 units per 270725 hands (and personally always quote against the initial bet rather than total bets staked, in this case an average of 2.134 Units).

I don't really understand your columns. In fact, I don't really understand any of it.

We could be calculating two very differnt things.

My current way of thinking is like this:

1. You make the ante
2. After receiving your cards, you calculate your odds of winning.
a. If your odds of winning are less than 33%, fold.
b. If your odds of winning are 33% to 40%, raise x1
c. If your odds of winning are 40% to 42.85%, raise x2 (if that is allowed)
d. If your odds of winning are 42.85% to 44.44%, raise x3 (and anything above if that is the max
e-z. Continue all the way down as accordance to this chart:

Raise	Payout	Odds
1	2	33.33%
2	3	40.00%
3	4	42.86%
4	5	44.44%
5	6	45.45%
6	7	46.15%
7	8	46.67%
8	9	47.06%
9	10	47.37%
10	11	47.62%
20	21	48.78%
50	51	49.50%
100	101	49.75%
1000	1001	49.98%

Your odds of winning are calculated using the different styles of hands and card values.
Styles of hands include:
4 suits (1,1,1,1)
3 suits (2,1,1,0)
2 suits (2,2,0,0)
2 suits (3,1,0,0)
1 suit (4,0,0,0)

I've already discovered that 1 suit occurs around 1% of the time and 4 suits is 11% of the time. I performed calculations on the other combinations, however I'm positive I screwed up, as the values don't add to 1.

My plan is to be able to make an optimal strategy based on the cards received. As I said, with four suits, 33% corresponds to a combined value (points of cards) of 24. 50% corresponds to 32. The other percentages of importance would obviously be 42% and 42.86%.

In the end, you'll end up with a strategy similiar to video poker.

The math is easy for the 4 suit scenario and the 1 suit scenario. Sadly, that's only 12% of the hands.

EDIT: Just realized that your rows correspond to the suit scenario. Can I ask you what your math on that is?

What I do is as follows - for simplicity take the four different suit option.

List every combination starting AAAA, AAAK, etc. Obviously AAAA only has one way, AAAK has four etc.
Then count how many cards the dealer can get that (a) you win (b) you lose. In this case it is A=12, K=11 etc for each suit.
Remember you have four cards, so there are 48 possible cards the dealer can have, some of these are winners for you and others losers. Thus these will be two numbers that add up to 48; so AAAA = 48:0, AAAK = 47:1 ... 2222 = 0:48.

You have three choices (i) "fold" (ii) raise x1 (iii) raise x3 - see which means you win the most or lose the least (jn some cases two options are equal).

(i) "fold" - you will therefore lose all 48 possibilities of the dealer cards and hence lose 48 units.
(ii) "Raise 1" - you will win 2 units for every time (a) you win - you will lose 2 units for every time (b) you lose. Simply put if this means you lose less than 48 units, then it's better to raise than fold even if it means you often lose 2 units.
(iii) "Raise 3" - you will win 4 units for every time (a) you win - you will lose 4 units for every time (b) you lose. Simply put if this means you're better off than "Raise 1", then do it!

Now add them up and see for all the combinations what you won and what you lost.

Do the same for the other suit combinations, except remember where you have two in a suit then Ax=11, Kx=10 etc.

The columns are how many times you (A) "Raise x3" (AB) equal (B) "Raise x1" (BC) equal (C) "Fold".

Personally I don't agree with your maths - here's a few examples.
Best 48:0 "Fold" = -48 "Raise x1" = {+96:-0} "Raise x3" = {+192:-0} Expected payback = 192/48 = +4 (i.e. you're bound to win 4)
Good 36:12 "Raise x1" = {+72:-24} "Raise x2" = {+144:-48} Expected payback Raise x1 (48/48) = +1 Raise x3 (96/48) = +2.
Close 24:24 "Fold"=-48 "Raise x1" = {+48:-48} "Raise x3" = {+96:-96} Expected payback = 0/48 for both Raise options.
Raise 20:28 "Fold"=-48 "Raise x1" = {+40:-56} "Raise x3" = {+80:-112} Expected payback = -16/48 (so "Raise x1" not fold)
33% 16:32 "Fold"=-48 "Raise x1" = {+32:-64} Expected payback = -32/48 (still "Raise x1" not fold, as on average you lose less than 1 unit).
25% 12:36 "Fold"=-48 "Raise x1" = {+24:-72} Expected payback = -48/48 So either "Raise x1" or "Fold"

Quote: charliepatrick

Yes I agree with your total of the combinations, but the EV is -1 not -.5 : i.e. you stand to lose your initial Ante if you fold (i.e. can accept a loss of 1 unit) or for the same expected return raise and either Win 2 or Lose 2 (75% Lose, 25% Win).

Raise x3 Raise Raise x1 Even    Fold
13546    1469	11726    455    1365
32604    7800	96096   5304   16380
    0       0   25740   2574    8190
    0       0   25740   3432   15444
    0       0       0      0    2860
				
46150    9269  159302  11765   44239

I get that you lose 17489.5 units per 270725 hands (and personally always quote against the initial bet rather than total bets staked, in this case an average of 2.134 Units).

I get the same except that you would lose 103402 per hand. Average bet is 2.134 if you raise 1x on 1x or 3x options and fold on 1x or fold options. I like to bet more in those situations so I get 2.246 average bet.

Alright. I found out basic strategy. Pick out the highest ranking card in each suit. Take it's value, and multiply it by 2.083%. The values are based on the table below:

Card	Value
A	12
K	11
Q	10
J	9
10	8
9	7
8	6
7	5
6	4
5	3
4	2
3	1
2	0

For the lower rank cards, subtract 2% for each one.
>33%, fold
33% to 42.86%, raise x1
42.86%+, raise x3.

Adjust if casino allows for raising of different denominations.

Quote: ChesterDog

I get a player EV of -0.381944778.

Sorry I had a typo in my spreadsheet when I copied the total columns from the 1111 spreadsheet to the others, so I hadn't counted enough negatives at the bottom. I now get the same figure - so it's a lousy game percentage wise.

Quote: Triplell

Quote: Carl1946

Quote: MrCasinoGames

Here is a Similar Game: Block

The Block bet is based on the player's first two cards and the dealer's up card. It can be found in Africa, South Africa, Egypt, Latvia, Estonia, Ireland, Morocco and Eastdend-EU. The bet wins if the dealer's up card matches the suit of one of the player's cards, and the player's card is higher. There are higher pays if the player's cards are a pair, suited, or both. Here is how the various winning hands are defined.Read More

Does the player bet an ante to play and a side bet on the odds of a pair, suited, etc. ?

Block is a blackjack sidebet.

Yes, Block is a blackjack side-bet.
It has been out there in casinos, since 2003.

Other Variations of this game.

High Card Pool

A standard 52 card pack is used with cards ranking from ace (high) down to two (low). There can be from three to about eight players; each in turn will be the dealer - the first dealer is chosen at random.

Before the deal each player puts an equal stake (ante) into the pot. The dealer then deals five cards to each player, and the players look at their hands.

Starting with the player to dealer's left, going around the table clockwise and ending with the dealer, each player has a chance to bet. At your turn you choose how much to bet - you must bet at least the amount of the ante, and may bet anything up to the entire pot - and you place your stake next to the pot. When you have bet, the dealer burns one card (i.e. takes it from the top of the deck, shows it to the players, and adds it face up to the bottom of the deck) and then deals one card face up in front of you.

If you have a card in your hand which is the same suit as the dealt card and higher in rank, you show this card, take back your stake, and the dealer gives you an equal amount out of the pot.
If you have no card which is the same suit as and higher than the dealt card, you must show your whole hand, and the dealer sweeps your stake into the pot.
Your cards and the dealt card are now set aside face down and it is the next player's turn to bet against the pot, whatever it now contains. If the pot is empty, or contains less than the minimum bet, each player immediately puts in the initial stake again.
After everyone has bet, the turn to deal passes. Whatever was in the pot stays there for the next hand, and each player adds another ante to it (but if the players agree that the pot has become too large they may split it between them and ante to a new pot).

Variations

Some play that only four cards are dealt to each player. This enables a larger number of players to take part without the cards running out. Some play with only three cards dealt to each player.

Red Dog can be played with a bank put up by the dealer, as in Shoot or Slippery Sam; in this case the players do not ante, and if the pot becomes empty the deal immediately passes to the next player.
------------------------------------------

Shoot

The game is similar to High Card Pool, but there are the following differences.

At the start, the dealer alone puts up a stake, which can be any amount between an agreed minimum and maximum.

The dealer deals just three cards to each other player. Players may not look at their cards until just before their turn to bet (when the previous player's turn is over).

At your turn, you may bet anything between an agreed minimum and the amount currently in the pot (obviously the agreed minimum for a player's stake must be less than the minimum that the dealer has to put into the pool). The dealer then turns up a card and you win if you can show a card from your hand of the same suit and higher in rank; otherwise you lose.

If the pot becomes empty, anyone who has not yet bet in that deal does not have a chance to do so. The deal immediately passes to the next player to the left.

If money remains in the pot at the end of a deal, the same dealer deals again. When the same person has dealt three times in succession they can choose whether to pass the deal on to the next player, keeping whatever is in the pot, or to deal a fourth time and pass the deal on after that, keeping whatever remains in the pot after the fourth hand.
-------------------------------

Slippery Sam

This is also called Six-Spot Red Dog. The betting mechanism is the same as in Shoot, but the players bet on the basis of the dealer's turned up card, without having seen the cards in their hands.

The dealer deals just three cards to each other player, but they must not look at their cards. Then the dealer continues by dealing cards face up in the centre of the table until a six or lower appears. Each player bets on having a higher card in the same suit as the face up card. After the player has decided how much to bet, the player's whole hand is then exposed and the player wins the amount of the stake from the pot if it contains a higher card of the same suit as the dealer's card; if not, the player's bet is added to the pot.

If the pot becomes empty in the middle of a hand, the deal passes to the next player, who creates a new pot. At the end of a hand, the dealer retains anything that is left in the pot, and the deal passes to the next player.

Variations

In his Encyclopedia of Games (1973), John Scarne says that the following variation is popular in Fort Wayne, Indiana. The dealer deals three cards to each player and one face up in the centre. The player can either bet against this card or ask the dealer to deal another card - for which the player has to pay into the pot one fifth of what it contains (or some other agreed amount). If the player does not wish to bet against the second card, a third card can be called for at the same price. After paying for the third card the player may either bet against it or pass without betting, and the turn passes to the next player.
----------------------------------------

Polish Red Dog

This is similar to Slippery Sam, except that the players bet without seeing either their own cards or the dealer's card. It is described in Ostrow's Complete Card Player (1945), and the description is reprinted in various later books. Alternative names are Stitch and Polski Pachuck.

The banker's initial stake is a fixed amount, and a player's maximum bet is half of what is in the pot at the time. When the player has bet, the dealer burns one card (faces it and puts it on the bottom of the pack), and turns up the next card, and the player's three cards are exposed. If the player has a higher card of the same suit as the dealer's card, the player is paid from the pot twice the amount of the bet; otherwise the bet is added to the pot.

If the pot is busted (runs out of money), the deal immediately passes to the next player. Otherwise the same player continues dealing until at the end of a hand the pot is at least three times its initial size. At this point the dealer declares a stitch round - a final deal after which if the pot is still not busted the dealer collects whatever is in it and the turn to deal passes to the left.

Quote: Triplell

Alright. I found out basic strategy. Pick out the highest ranking card in each suit. Take it's value, and multiply it by 2.083%. The values are based on the table below:

Card	Value
A	12
K	11
Q	10
J	9
10	8
9	7
8	6
7	5
6	4
5	3
4	2
3	1
2	0

For the lower rank cards, subtract 2% for each one.
>33%, fold
33% to 42.86%, raise x1
42.86%+, raise x3.

Adjust if casino allows for raising of different denominations.

I would modify your basic strategy to read "if P(win)<25%, fold; if 25%<P(win)<50%, raise x1; if 50%<P(win), raise x3". Also, if P(win)=25%, either fold or raise x1, and if P(win)=50%, either raise x1 or x3.

Also, to make the calculations easier, I would just add your card values for the highest card in each suit and subtract 1 for each non-highest card in each suit and compare the result to the numbers 12 and 24 (which are 25% and 50% of 48, respectively). If the sum is less than 12, I would fold; if greater than 12 but less than 24, I would raise x1; and if greater than 24, I would raise x3.

You're right. I was trying to correlate the act of adding more money would be like adding more to the pass-line bet after the come out, but because folding is 100% to lose (and not adding money to the pass-line still has a probability of winning), they are different concepts.

I am still back to my basic question. Can't anyone figure what the basic house odds are on just the Ante bet of 1 unit and Play bet of 1 unit?? I think once this is determined I can firure out the betting skeem that will reduce the house odds. I am not a mathematician but it looks like the house has almost a 10% edge. So I think the probability is that the house pays out 90%. Can anyone help ??

Please forgive me but unless you're very lucky (i.e. whoever invented roulette) but maths is needed if designing a game. Especially if there's more than one betting option for the player, as the process needs the strategy for the player (or run all permutations and let the computer do it) to calculate the house edge.

Also this thread has a figure that has been obtained by two of us and shows the current game is a very lousy bet and non-starter in its present form. I suspect you have some ideas on improvements, but you will need to be able to perform the same calculations/process for the various combination of ideas to see which ones work.

Quote: Carl1946

...Can't anyone figure what the basic house odds are on just the Ante bet of 1 unit and Play bet of 1 unit?...it looks like the house has almost a 10% edge...

I assume that if the player wins he gets paid 1:1 on his 1 unit ante bet and 1:1 on his 1 unit play bet. If the player always plays regardless of his cards, I get a house edge of 49.1092%. If the player folds and loses his 1 unit ante bet when he has less than a 25% probability of winning, I get a house edge of 44.5474%. These house edges are based on the original 1 unit ante, so the player would lose an average of 0.491092 unit or 0.445474 unit, respectively.