Table top dice game idea

Thoughts or suggestions appreciated
The game is played on a blackjack table using a chuck a luck roller
Player needs to first qualify to play by rolling a 6 7 8
Player makes an ante wager and the dice are rolled
If player qualifies the dice are rolled again and player wins on 6 7 8
If player missies they can surrender or make another wager equal to the first.
If player missies again they can surrender both wagers or make a third bet
If player missies the 3rd try they loose half the original wager and the game starts over
House edge is .56 on roll once qualified

Quote: BigJ

Thoughts or suggestions appreciated
The game is played on a blackjack table using a chuck a luck roller
Player needs to first qualify to play by rolling a 6 7 8
Player makes an ante wager and the dice are rolled
If player qualifies the dice are rolled again and player wins on 6 7 8
If player missies they can surrender or make another wager equal to the first.
If player missies again they can surrender both wagers or make a third bet
If player missies the 3rd try they loose half the original wager and the game starts over
House edge is .56 on roll once qualified

Is the first part out of order? I assume the player makes an ante wager and then rolls to try to qualify, otherwise the player is just pointlessly rolling dice. Does this look like what you're going for?:

Rules(?)

1.) Player makes an Ante wager.

2.) The player qualifies by rolling a 6, 7 or 8, otherwise the player loses immediately.

3.) Having qualified, the player must roll another 6, 7 or 8.

4.) If the player rolls an additional 6, 7 or 8 on the first try, the player wins and is paid the amount of original bet.

5.) If the player rolls anything but a 6, 7, or 8 after qualifying, then the player may surrender his original bet or make a wager equal to the first.

6.) If the player makes a second wager and rolls a 6, 7, or 8, then the player wins and is paid on both(?).

7.) If the player fails to roll a 6, 7, or 8 again, then the player may either surrender both wagers or make a third wager equal to the first bet.

8.) If the player makes a second additional wager and wins, then the player is paid on all wagers.

9.) If the player rolls anything but a 6, 7, or 8, then the player loses half of the original wager?

Okay, just to keep this simple, I'm going to make the original wager two units.

The player bets two units and can lose immediately by rolling anything other than a 6, 7 or 8:

(20/36) * (-2) = -1.11111111111

Probability: 0.55555555555

(16/36)---Game continues

(16/36) * (16/36) * 2 = 0.39506172839 (Player wins two units on the roll after qualifying)

Probability: 0.19753086419

(16/36) * (20/36)---Game Continues. The player would never want to surrender.

(16/36) * (20/36) * (16/36) * 4 = 0.43895747599 (Qualify-Miss-Hit---Player wins four units)

Probability: 0.10973936899

(16/36) * (20/36) * (20/36)----Game Continues. The player would never want to surrender.

(16/36) * (20/36) * (20/36) * (16/36) * 6 = 0.36579789666 (Qualify-Miss-Miss-Hit---Player wins six units)

Probability: 0.06096631611

(16/36) * (20/36) * (20/36) * (20/36) * (-1) = -0.07620789513 (Player loses half of original two unit wager)

Probability: 0.07620789513

Sum of Probabilities: 0.55555555555 + 0.19753086419 + 0.10973936899 + 0.06096631611 + 0.07620789513 = 1

Expected Return: 0.39506172839 + 0.43895747599 + 0.36579789666 - 0.07620789513 - 1.11111111111 = 0.0124980948 (Two Units Bet)

PLAYER Advantage: 0.0124980948/2 = 0.0062490474

I'm getting a Player Advantage of .625%, so that makes me think I am misunderstanding something about the rules. This player advantage (the way I'm interpreting the rules) could be resolved by the player just losing the original bet on series: Qualify-Miss-Miss-Miss and the player, obviously, would still never surrender.

The Expected Return would then look like:

Expected Return: 0.39506172839 + 0.43895747599 + 0.36579789666 - (0.07620789513*2) - 1.11111111111 = -0.06370980033

Overall House Edge: 0.06370980033/2 = 0.03185490016 or 3.1855%

In order to determine the House Edge per roll, we must come up with an average number of rolls that assumes the player will play optimally by never surrendering.

Game Over One Roll: 0.55555555555 * 1

Game Over Two Rolls: 0.19753086419 * 2

Game Over Three Rolls: 0.10973936899 * 3

Game Over Four Rolls: (0.06096631611 + 0.07620789513) * 4

Average Number of Rolls: (0.55555555555 * 1) + (0.19753086419 * 2) + (0.10973936899 * 3) + ((0.06096631611 + 0.07620789513) * 4) = 1.82853223586

House Edge Per Roll: 0.03185490016/1.82853223586 = 0.01742102191 or 1.742%

The Element of Risk (EoR) for this would be based on average amount bet, which would be:

AVERAGE TOTAL BET:

(0.55555555555 + 0.19753086419) * 2 = 1.50617283948

(0.10973936899 * 4) = 0.43895747596

(0.06096631611 + 0.07620789513) * 6 = 0.82304526744

0.82304526744 + 0.43895747596 + 1.50617283948 = 2.76817558288

Element of Risk: 0.06370980033/2.76817558288 = 0.02301508644 or 2.30151%

Conclusion

I assume that I am misunderstanding something about the rules. I'm trying to figure out exactly what that might be, but I think it is one of two things:

A.) The player does not lose immediately if the player fails to qualify.

---If this is true, why bother even having the player need to qualify?

B.) The player loses HALF of the original wager AND any additional wagers that were made if the player misses three consecutive.

OR:

C.) Both.

Anyway, we are coming up with something different for House Edge, so please clarify my understanding of the rules, if you would be kind enough.

THOUGHTS ON THE GAME

Whatever the House Edge, I must respectfully submit that I do not think this game would work for a number of reasons:

1.) In general, there appears to be no need for a dice game that might take the place of Craps. Other efforts in this vein have not gone very far.

2.) The Blackjack Table doesn't leave a heck of a lot of room for side bets, which you are certainly going to want to have for this game to have any chance.

3.) Unless there is a separate area for the rolling of the dice, one concern is that the dice might knock over chip stacks. Granted, that can also happen at the Craps Table.

4.) If the player does not immediately lose having failed to qualify, then I don't understand what the point of the qualifying process is.

ADDENDUM

I think I might have figured this out. The player loses additional wagers AND half of the original wager if Qualify + Miss-Miss-Miss. This assumes the player loses immediately if he fails to qualify.

(16/36) * (20/36) * (20/36) * (20/36) * (-5) = -0.38103947569 (Player Loses Extra Four Units Bet AND half of original wager)

0.39506172839 + 0.43895747599 + 0.36579789666 - 0.38103947569 - 1.11111111111 = -0.29233348576 (Original Bet of Two Units)

AVERAGE TOTAL BET:

(0.55555555555 + 0.19753086419) * 2 = 1.50617283948

(0.10973936899 * 4) = 0.43895747596

(0.06096631611 + 0.07620789513) * 6 = 0.82304526744

0.82304526744 + 0.43895747596 + 1.50617283948 = 2.76817558288

House Edge (Initial Wager): -0.29233348576/2 = -0.14616674288 or 14.61667%

Element of Risk (Based on Average Wager): -0.29233348576/2.76817558288 = -0.10560510957 or 10.5605%

Average Number of Rolls: 1.82853223586

Element of Risk (Per Roll) 0.10560510957/1.82853223586 = 0.05775403216 or 5.7754%

Conclusion

I'm now coming up with a House Edge way higher than yours. Can you please explain the rules of the game step-by-step and preferably with line breaks?

A hard 6 or 8 pays 3 to 2 A 6 7 8 pays 6 to 5

Player needs a 6 7 8 to qualify Player has 3 tries to qualify before losing

Player makes an ante wager

1st try 6 7 8 hits the player qualifies and the dice are rolled again. Player wins on 6 7 8 losses all other numbers

1st try 6 7 8 missies player may surrender or add a second wager equal to the first to try again

2nd try missies player may surrender both wagers or make a third bet equal to the first and try again

3rd try missies player losses half of the original wager. The other two wagers are a push

Dice are contained in a cage

The qualifying round is a way to build.

Quote: BigJ

A hard 6 or 8 pays 3 to 2 A 6 7 8 pays 6 to 5

Player needs a 6 7 8 to qualify and has 3 tries to qualify before losing

Player makes an ante wager

1st try 6 7 8 hits the player qualifies and the dice are rolled again. Player wins on 6 7 8 losses all other numbers

1st try 6 7 8 missies player may surrender or add a second wager equal to the first to try again

2nd try missies player may surrender both wagers or make a third bet equal to the first and try again

3rd try missies player losses half of the original wager. The other two wagers are a push

Dice are contained in a cage

The qualifying round is a way to build.

Okay, so there are different payouts now; that is definitely helpful information.

Thank you for the more detailed description of the rules. I will try this again in the post to follow. Maybe a half hour, or so?

Step 1

Player makes a $5 (simplicity) ante wager.

At this point, the player either rolls a 6, 7, or 8 and has the opportunity to do it again, or the player fails to do that.

A.) The probability that the player rolls one of these numbers is .444444 and the probability that the player does not is .555556.

B.) If the player fails to roll a 6, 7 or 8, then the player may surrender the initial wager, but the player would never want to do that.

C.) If the player successfully rolls a 6, 7 or 8, then the player can do it again and get paid 6:5 for a soft 6, 8 or Any Seven and get paid 3:2 for a Hard Six or Eight.

D.) The player loses the $5 initial bet in the event that the player qualifies, but then fails to roll, a 6, 7 or 8:

(16/36) * (2/36) * 7.5 = 0.18518518518-----Probability: 0.02469135802

(16/36) * (14/36) * 6 = 1.03703703704-----Probability: 0.17283950617

(16/36) * (20/36) * (-5) = -1.23456790123----Probability: 0.24691358024

Step 2

In order to be on this step, the player now has a total of $10 in bets on the table. The player has failed to qualify (20/36) and has decided to play again. The rules are otherwise the same as Step 1.

(20/36) * (16/36) * (2/36) * 15 = 0.20576131687-----Probability: 0.01371742112

(20/36) * (16/36) * (14/36) * 12 = 1.15226337449-----Probability: 0.09602194787

(20/36) * (16/36) * (20/36) * (-10) = -1.37174211248-----Probability: 0.13717421124

Another possibility is that the player fails to qualify for a second time, which would result in having to do it again.

Step 3

In order to be on this step, the player now has a total of $15 in bets on the table. If the player fails to qualify for a third consecutive time, then the additional bets will be pushed back to the player AND the player will lose only half of the original $5 bet. The rules are otherwise the same as Step 1:

(20/36) * (20/36) * (20/36) * (-2.5) = -0.42866941015-----Probability: 0.17146776406

(20/36) * (20/36) * (16/36) * (2/36) * 22.5 = 0.17146776406-----Probability: 0.00762078951

(20/36) * (20/36) * (16/36) * (14/36) * 18 = 0.96021947873----Probability: 0.05334552659

(20/36) * (20/36) * (16/36) * (20/36) * (-15) = -1.14311842707-----Probability: 0.07620789513

RESULTS

Sum of Probabilities

0.02469135802+0.17283950617+0.24691358024+0.01371742112+0.09602194787+0.13717421124+0.17146776406+0.00762078951+0.05334552659+0.07620789513 = 1

Sum of Expected Values ($5 Base Bet)

(0.18518518518 + 1.03703703704 + 0.20576131687 + 1.15226337449 + 0.17146776406 + 0.96021947873) - (1.14311842707+0.42866941015+1.37174211248+1.23456790123) = -0.46616369456 ($5 Base Bet)

House Edge Based on Initial Bet

-0.46616369456/5 = -0.09323273891 or 9.3233%

Expected Number of Rolls

Two Rolls (Resolves on Step 1): (0.02469135802 + 0.17283950617 + 0.24691358024) * 2 = 0.88888888886

Three Rolls (Resolves on Step 2): (0.01371742112+0.09602194787+0.13717421124) * 3 = 0.74074074069

Three Rolls (Resolves on Step 3): 0.17146776406 * 3 = 0.51440329218

Four Rolls (Resolves on Step 3): (0.00762078951 + 0.05334552659 + 0.07620789513) * 4 = 0.54869684492

Expected Number of Rolls: 0.88888888886+0.74074074069+0.51440329218+0.54869684492 = 2.69272976665

House Edge Per Roll ($5 Base Bet)

0.09323273891/2.69272976665 = 0.03462387502 or 3.4623875%

Average Total Amount Bet

Resolves on Step 1: (0.02469135802 + 0.17283950617 + 0.24691358024) * 5 = 2.22222222215

Resolves on Step 2: (0.01371742112+0.09602194787+0.13717421124) * 10 = 2.4691358023

Resolves on Step 3: 0.17146776406 * 15 = 2.5720164609

Resolves on Step 3: (0.00762078951 + 0.05334552659 + 0.07620789513) * 15 = 2.05761316845

Average Total Amount Bet: 2.05761316845+2.5720164609+2.4691358023+2.22222222215 = 9.3209876538

Element of Risk (Expected Loss/Average Total Amount Bet)

-0.46616369456/9.3209876538 = -0.05001226392 or 5.0012264%

Element of Risk Per Roll (Element of Risk/Expected Number of Rolls)

0.05001226392/2.69272976665 = 0.01857307203 or 1.85731%

CONCLUSION

I'm still coming up with a higher house edge than you are, unless I am missing something else. Does a Hard Six or a Hard Eight on the first roll cause the player to win the 3:2 immediately without qualifying?

MY OBSERVATIONS:

1.) I like this version better than what I thought you were saying before, but I still don't think there's any need to replace Craps with anything. I think Scossa might have had the most success in that regard, and that was something like two tables in one or two casinos.

2.) I have been made to understand that this would not be patentable without there being some sort of electronic component to it.

3.) The layout not having much room for side bets, as before.

4.) The thing about the dice knocking over chips, unless there is a walled-off rolling area.

Thank you

Might make a good one roll side bet in craps but as a game it's a no go house edge is too high

Quote: BigJ

Thank you

Might make a good one roll side bet in craps but as a game it's a no go house edge is too high

You're welcome!

By the way, since I was looking at it anyway, here is something that occurred to me:

The game would still have:

House Edge: 0.00749885688 or 0.75%

Element of Risk: 0.00402256561 or 0.4023%

If instead of losing half of initial bet on failing to qualify three times in a row, the rules instead made failing to qualify three times in a row a total push.

With the modified rules, that would probably make for a really fun home game that can be played without a House Edge (assumes no cheating) as long as players acted as, "The House," on a rotation and got to do so an equal number of times...that also assumes that all players bet equal dollar amounts.

I think the modified rules would also work as a fast-paced online game. The House Edge and EoR are both quite low with the modified rule that three non-qualifying rolls in a row is a total push, but the game would play pretty quickly, so that would make up for the extremely low House Edge.

As far as a side bet, the very rules of the bet prohibit it from being a one roll bet. It cannot resolve in less than two rolls. Technically, it can, but that would require the player to roll a non 6, 7 or 8 (20/36) and then surrender, which would be a terrible decision. Craps also already has a lot going on, so it would be tough to incorporate this on a physical Craps game.

I still think this is a great effort, overall, in coming up with a game. I think it's a theoretically viable online game and would make a great home game.

That's a great idea

Quote: BigJ

Well I like that idea so if I'm understand correctly a single roll player wins or losses is a .56 house edge

But by making three qualifying rounds that push this increases the house edge to .75

(Quote above prior to OP edit)

Oh, you meant a single roll by which the player gets:

(14/36) * 6 = 2.33333333333

(2/36) * 7.5 = 0.41666666666

(20/36) * (-5) = -2.77777777778

2.33333333333 + 0.41666666666 - 2.77777777778 = -0.02777777779/5 = -0.00555555555 or 0.5555555556% House Edge

Now I see where that number is coming from! I couldn't figure out why my House Edge was nowhere near yours no matter what I did, but you were talking about two different bets---one single roll and one multi-roll.

Anyway, yes, I think your single-roll idea might work as a Craps side bet. It's kind of like a reverse Field Bet, in a way. I could maybe see where some players would play this and the Field simultaneously where:

2, 6, 7, 8 or 12 = Profit

3, 4, 9, 10, or 11 = Push (One wins, one loses)

5 = Loss

So, I think that's an excellent idea for a one roll side bet.

Multi-Roll

Yes, doing the same exact thing in my long post, but simply changing the $2.50 loss on three consecutive non-qualifying rolls to a Push would reduce the House Edge on the initial $5 bet to 0.75%. The Element-of-Risk is even lower because it is based on the average total amount bet.

Quote: BigJ

That's a great idea

To add on that what if after three fails the player then makes a second bet to continue or losses the original bet

(Quote above is before an Edit by OP)

The player already has three bets out there, at that point. I think it's best just to stay with what is above rather than have the bet go to potentially five rolls.

thanks for the help

Quote: BigJ

How many qualifying rolls would be needed to raise the house edge to Baccarat

(Quoted Before Edit by OP)

I'm not sure, but you can add additional rounds by doing exactly what I did in the long post.

I'll do the fourth round for you.

Instead of being a loss:

(20/36) * (20/36) * (20/36) = 0.17146776406

Becomes the probability that a fourth qualifying roll is needed, so if you wanted to stop at four, you would get:

(20/36)^ 4 = 0.09525986892 (Push after four rolls, or probability of Step 5, if you wanted to keep going)

(20/36) * (20/36) * (20/36) * (16/36) * (2/36) * 30 = 0.12701315856----Probability: 0.00423377195

(20/36) * (20/36) * (20/36) * (16/36) * (14/36) * 24 = 0.71127368795-----Probability: 0.02963640366

(20/36) * (20/36) * (20/36) * (16/36) * (20/36) * (-20) = -0.84675439042----Probability: 0.04233771952

Added Expected Loss

(0.12701315856+0.71127368795) - 0.84675439042 = -0.00846754391

Original Expected Loss (Except Three Non-Qualifiers No Longer Loses)

(0.18518518518 + 1.03703703704 + 0.20576131687 + 1.15226337449 + 0.17146776406 + 0.96021947873) - (1.14311842707+1.37174211248+1.23456790123) = -0.03749428441

New Expected Loss on Four Round Proposition

-0.03749428441 - 0.00846754391 = -0.04596182832

New House Edge ($5 Initial Bet)

-0.04596182832/5= -0.00919236566 or 0.91923657%

More Rounds

1.) You'll need at least one more round to get to over 1% House Edge.

2.) This would require an additional spot on the layout for a potential fourth bet, as would all other bets.

3.) As you continue, you will find that less and less gets added to the House Edge, because all added results become increasingly less likely.

Quote: BigJ

thanks for the help

You're welcome.

Sorry but reading your initial post you lost the 30 second rule, that is the game (or at least the idea) has to be described and gain interest by someone walking by within 30 seconds. I think the confusion was using the term "qualify", so I was looking for a two-part game where you get past the first stage to play a second game.

Later in the thread I could see a one-roll side-bet that pays if you roll a 6 7 or 8, else loses. The odds payable are 6/5 or 6/4 if you roll a hard 6 or 8. This has a house edge of 0.56%.

While we are talking dice... Are there approved devices for shaking dice on a blackjack style table? I have seen the Pai Gow cups used to determine order of the deal, are those able to be used for any game? I am assuming so since I have seen games like "Yo" discussed from the Cutting Edge conference but wanted to check with those who know more than I do.

Dice Down
Football theme dice game

Offence
Pay table 6 7 8 pays 6 to 5
Hard 6 8 pay 6 to 4

Game uses two sets of dice player green dealer red

Player makes 4 wager equal amount

Player has four downs to score by rolling a 6 7 8 all other numbers are a incomplete pass with wagers stacking on top of each other. After four incomplete passes it's a push player now on defense

If player rolls a 6 7 8 and dealer rolls a 2 3 4 5 9 10 11 12 it's a interception player losses wager

Defense
Pay table 2 12 pay 1 to 1

Player makes 3 wager equal amount

Dealer scores on a 2 or 12 all other numbers are incomplete pass with wagers stacking on top of each other after three incomplete passes it's a push player back on offense.

Player intercepts with a 7

If dealer rolls 2 or 12 and player rolls a 7 player wins 1 to 1.

If dealer rolls 2 or 12 and player rolls anything but seven player losses wager

Dice are in a chuck a luck roller with a camera in it . Only dealer uses
Screen displays numbers graphics if touchdown or interception