Hard Luck

Quote: MrCasinoGames

I don't think a new dice game can make it to the Casino.

How about Hard Luck?

Two players: dealer (designated dice thrower) and banker, two dice each. Object: throw a higher sum than opponent. Win pays even money.

Tie-sums (146 in all) happen three ways:

(1) hard-hard (all four dice same, a quadruple) = pay even money to both Dealer and Banker bets (6 ways)

(2) easy-easy, same combo (two-pair X&Y) = take half from both (60 ways)

(3) easy-easy, diff. combo (one-pair X&Y=S, one pair W&Z=S) = higher-order combo wins (80 ways in all, 40 behooving each).

Definition: higher order = having smaller difference between numbers on each die. For example: 3&3 beats 2&4 and 1&5, 2&4 beats 1&5. Thus, hardways would be a special case included in the rule. Very simple to use, actually.

House edge on the above description = 1.85%

Side bets

(1) Spread - difference between dice sums - pays on a scale as follows(for HA = 11.73% / for HA = 2.78%):

10-Spread (maximum, 2 possible ways) 80 to 1 / 84 to 1
9-Spread (8 ways) 20 to 1 / 21 to 1
8-Spread (20 ways) 8 to 1 / 42 to 5
7-Spread (40 ways) 4 to 1 / 21 to 5
6-Spread (70 ways) 2 to 1 / 12 to 5
5-Spread (minimum, 112 ways) 1 to 1 / 3 to 2

(2) Chance - four-die outcome, regardless of which dice are whose

Any Triple (X&X&X&Y, 120 ways) - pays 2 to 1
Any Two-Pair (X&X&Y&Y, 90 ways) - pays 3 to 1
Any Straight (1&2&3&4, 2&3&4&5, or 3&4&5&6, 72 ways in all) - pays 4 to 1
Any Quadruple (X&X&X&X, 6 ways, obviously) - pays 23 to 1 (HA = 5.56%) or 29 to 1 (HA = 2.78%)

(3) Above- or Below-14: four-die total of greater than or less than 14, each pays even money (HA = 11.27%) or 6 to 5 (HA = 2.39%)

Dealer chooses which dice to be his own and throws both his own and the banker's dice (to keep the appearance of absolute honesty) and keeps dice as long as the Dealer bets continue to win (Like-combo tie with banker counts as a loss).

Hi NowTheSerpent,

Hard Luck, Sounds like a good game.

I thank MathExtremist have a game like that and he was looking for a name for the game and he was offering it for free to casinos.
(Player and banker, two dice each. Object: throw a higher sum than opponent. Win pays even money.)

Quote: NowTheSerpent

Definition: higher order = having smaller difference between numbers on each die. For example: 3&3 beats 2&4 and 1&5, 2&4 beats 1&5. Thus, hardways would be a special case included in the rule. Very simple to use, actually.

For whatever reason, the thought of subtraction at a dice table doesn't sit well with me.

But that's OK. You'll get the same ordinal sequence if ties are decided by multiplying the two numbers.

---

FYI: There's no need to start two threads about this...

Quote: DJTeddyBear

For whatever reason, the thought of subtraction at a dice table doesn't sit well with me.

But that's OK. You'll get the same ordinal sequence if ties are decided by multiplying the two numbers.

---

FYI: There's no need to start two threads about this

The duplication of the thread was accidental.

The method of subtraction or of multiplication would just be to minimize true ties. I didn't like the idea of taking half a wager, but I didn't want to have to take full wagers as mutual losers all that often, either. One could have a re-roll in the event of tied sums until one side won within a specified number of rolls (for HA's sake), thus removing both disflavors. I also, then, recommend pushing (0 to 1) the Spread bets on ties, so they can carry over automatically through the tie-break rolls. Perhaps one could adjust the paytable as follows (top first): 70-to-1, 35-to-2 (or just 17-to-1 for $1 bets), 7-to-1, 7-to-2 (3-to-1), 2-to-1, 1-to-1, and 0-to-1 for ties. Total ways to win or push: 252 + 146 = 398. from 1,296 gives 898 ways left to lose. The total return would be:

HA = [(70)(2) + (17.5)(8) + (7)(20) + (3.5)(40) + (2)(70) + (1)(112) + (0)(146) - 898]/(1,296) = (812 - 898)/(1,296) = (-86)/(1,296) = -6.6358%

assuming a $2 Spread bet. A $1 bet would carry an HA of 8.4877%.

Quote: MrCasinoGames

Hi NowTheSerpent,

Hard Luck, Sounds like a good game.

I think MathExtremist has a game like that and he was looking for a name for the game and he was offering it for free to casinos.

I just wanted something simple and catchy. If he thinks it'll work, he can go ahead and use it, since the concept is already out there.

Quote: NowTheSerpent

I just wanted something simple and catchy. If he thinks it'll work, he can go ahead and use it, since the concept is already out there.

It is a Three Dice Each, Total 6-Dice game.
Below is the link to the game:
Dice Game

Quote: MrCasinoGames

Hi NowTheSerpent,

Hard Luck, Sounds like a good game.

I thank MathExtremist have a game like that and he was looking for a name for the game and he was offering it for free to casinos.
(Player and banker, two dice each. Object: throw a higher sum than opponent. Win pays even money.)

It was actually 3 dice each, not thrown but shaken in Pai Gow dice cups. The point was to create a game playable at a standard BJ table with existing casino equipment (pai gow dice and shakers), not something that requires a dice table and multiple dealers to play. It would work equally well if you just removed one die from both cups, or you could use three in one and two in the other and make it an uneven chance.

It's all public domain but I'm not really "offering" it per se -- I'm not spending any effort promoting this. I just think it'd be fun to see in a carnival game pit because I love dice games.

Edit: I understand the pun behind "Hard Luck", but the phrase itself conjures bad mojo for gamblers. I'd pick a different name.

Quote: MathExtremist

Edit: I understand the pun behind "Hard Luck", but the phrase itself conjures bad mojo for gamblers. I'd pick a different name.

That's a good point. How about Skirmish or Shootout?