## Poll

13 votes (50%) | |||

10 votes (38.46%) | |||

5 votes (19.23%) | |||

2 votes (7.69%) | |||

8 votes (30.76%) | |||

3 votes (11.53%) | |||

5 votes (19.23%) | |||

4 votes (15.38%) | |||

10 votes (38.46%) | |||

7 votes (26.92%) |

**26 members have voted**

Quote:DogHandgordonm888,

That's a unique way to demonstrate your question ;-)

Dog Hand

LOL, I had no idea it triple-posted. Maybe we can all catch up to EvenBob if the site remains buggy.

Quote:Ace256% is not a feasible payout. Okay, it would be on $100 but it needs to be an easy ratio. Let’s say the min/max bet is $10 / $100Quote:unJonQuote:ThatDonGuyI'm working on an easier solution, but I think I have one for the pass - well, sort of; it only works with bets that are even numbers

A missed point of 6, 8, or 10 pushes

A 6-1 on the comeout pushes

A missed point of 5 loses half

A missed point of 9 where the comeout was 6-3 (instead of 5-4) loses half

That’s complicated. I found a simple but unsatisfying one:Making a point with a hard 6 or hard 8 pays an extra 56%

If any payout ratio is allowed, you could just pay a winning bet at 251 to 244 without changing any rules.

I don't think a solution is possible (at least, not on the pass side) without any sort of fractional bet payout.

Here is a table of probabilities of results broken down by comeout roll and resolution roll; the numbers are the probabilities multiplied by (36 x 495); the losing results for pass are negative numbers.

For example, the probability that the comeout is 3,1 and the result is a 7 is 660/(36 x 495), and the probability that the comeout is 4,4 and the point is also made the hard way is 45/(36 x 495).

"N/A" means that the result is not applicable as it was resolved on the comeout.

The sum is -252, so in order to make the house advantage zero, one or more of the results needs to be modified to increase their values by a total of 252. This is not easy to do when you notice that each number is a multiple of 55, 90, or 198. If you allow half-losses, these become 55, 45, and 99.

For Don't Pass, change all of the signs and replace the -495 in the last row with zero; the sum becomes -243. (To make it easier, leave the values as they are, except for the last row; the sum now needs to be reduced by 243.)

Comeout | Point 1 | Point 2 | Point 3 | 7 | ||||
---|---|---|---|---|---|---|---|---|

1,1 | N/A | -495 | ||||||

2,1 | N/A | -990 | ||||||

3,1 | 3,1 | 220 | 2,2 | 110 | 7 | -660 | ||

2,2 | 3,1 | 110 | 2,2 | 55 | 7 | -330 | ||

4,1 | 4,1 | 198 | 3,2 | 198 | 7 | -594 | ||

3,2 | 4,1 | 198 | 3,2 | 198 | 7 | -594 | ||

5,1 | 5,1 | 180 | 4,2 | 180 | 3,3 | 90 | 7 | -540 |

4,2 | 5,1 | 180 | 4,2 | 180 | 3,3 | 90 | 7 | -540 |

3,3 | 5,1 | 90 | 4,2 | 90 | 3,3 | 45 | 7 | -270 |

6,1 | N/A | 990 | ||||||

5,2 | N/A | 990 | ||||||

4,3 | N/A | 990 | ||||||

6,2 | 6,2 | 180 | 5,3 | 180 | 4,4 | 90 | 7 | -540 |

5,3 | 6,2 | 180 | 5,3 | 180 | 4,4 | 90 | 7 | -540 |

4,4 | 6,2 | 90 | 5,3 | 90 | 4,4 | 45 | 7 | -270 |

6,3 | 6,3 | 198 | 5,4 | 198 | 7 | -594 | ||

5,4 | 6,3 | 198 | 5,4 | 198 | 7 | -594 | ||

6,4 | 6,4 | 220 | 5,5 | 110 | 7 | -660 | ||

5,5 | 6,4 | 110 | 5,5 | 55 | 7 | -330 | ||

6,5 | N/A | 990 | ||||||

6,6 | N/A | -495 |

If the shooter wins with a point of 5 (must be with a 1,4 combination), then he rolls for a bonus as follows:.

If he rolls a 4, then he gets paid a bonus of seven times his original wager and the round is over. If he rolls a 10, then he rolls again. If he rolls any other number then he gets no bonus and the round is over.

1/9 chance of establishing a point of 5 times 1/5 of winning it (with 1,4) times 1/11 chance of winning the bonus is 1/495. Paid at 7x will eliminate the edge of 7/495

Adding a bonus round isn’t ideal but...it only lasts an average of 1.09 rolls, it only occurs once every 45 resolutions and the original game/payout isn’t changed at all

After running through numbers in my head I tend to agree with Don that there probably isn’t a way to eliminate the edge exactly using just the original game and “reasonable” fractional payout adjustments like 1/2. In a real world scenario I think my original proposal of the craps-2 losing only half is probably the best solution. Easy to understand, no material effect on variance and and edge of 2.5 basis points is essentially zero...that’s less than the edge difference between pass and DP

Quote:Ace2Here’s an idea

If the shooter wins with a point of 5 (must be with a 1,4 combination), then he rolls for a bonus as follows:.

If he rolls a 4, then he gets paid a bonus of seven times his original wager and the round is over. If he rolls a 10, then he rolls again. If he rolls any other number then he gets no bonus and the round is over.

1/9 chance of establishing a point of 5 times 1/5 of winning it (with 1,4) times 1/11 chance of winning the bonus is 1/495. Paid at 7x will eliminate the edge of 7/495

Adding a bonus round isn’t ideal but...it only lasts an average of 1.09 rolls, it only occurs once every 45 resolutions and the original game/payout isn’t changed at all

After running through numbers in my head I tend to agree with Don that there probably isn’t a way to eliminate the edge exactly using just the original game and “reasonable” fractional payout adjustments like 1/2. In a real world scenario I think my original proposal of the craps-2 losing only half is probably the best solution. Easy to understand, no material effect on variance and and edge of 2.5 basis points is effectively zero...less than the edge difference between pass and DP

Ran out of time but you can cut off the 11 and a 36 by saying the come out is a hard 8 and it gets made with a hard 8. If that paid 5.6 times bonus we would be edgeless without a bonus roll. Though it’s not an ideal round bonus. If you can work in the 5 to that same roll (so it’s 1/5 as likely) then the bonus could pay a round 28 times.

Quote:ThatDonGuy

I don't think a solution is possible (at least, not on the pass side) without any sort of fractional bet payout.

I stand corrected - I found one:

Comeout 12 is a push for both sides (+495)

Point of hardway 8 made hardway is a push (-45)

Point of 6,3 made with 6,3 is a push (-198)

Total adjustment to house edge: +252

Since Pass is now house edge zero, making the rules the same for Don't Pass makes that bet zero HE as well.

It should be fairly easy to use two rolls to create such a game, e.g. Odd followed by not (2,7,12) Even followed by not (2,3,7). Anything else replay the game.

1) Point is established with easy 6

2) 4 is rolled before winning point and before rolling a 2

3) Point is won with hard 6

You're a contestant on "Let's Make A Deal" and confront the famous Monty Hall Problem. You're shown 3 doors and are told that behind one of the doors is a (giant novelty) check worth $10,000. Behind each of the other 2 doors is a goat.

You select door #1.

Monty (who knows which door hides the check and which doors contain zonks) opens door #2 to reveal a goat. He then asks if you want to stay with door #1 or switch to door #3. Should you switch?

Fortunately, you're familiar with game theory and know what to do but, before you can respond, a bell rings. DING! DING! DING! Monty tells you this is now a special zonk redemption round. If you lose the game, you get to play second round except that the check will then be written for $20,000!

Eh? Should you now try to lose round 1 to go for twenty grand in round 2?

As this thought crosses your mind Monty casually adds that if you also lose the second round, you'll play a third round for a check worth $30,000!

You weren't expecting this. Probabilities begin shooting through your head...

What are your best decisions to maximize your potential winnings?

If I was on stage, and had to make a decision right then and there, I would NOT switch initially, and hope I did not pick the right door, hoping to win in Round 2 or Round 3.Quote:GialmereWhat are your best decisions to maximize your potential winnings?

And given the opportunity, assuming if I did indeed not pick the right door in Round 1, and thus now advanced to the special zonk Second Round, at that point I think I would switch doors.

If I did not guess right here either, and thus advanced to the 3rd round, I'd definitely switch of course, hoping for the best shot at that final prize of $30,000.

I suspect it's right not to switch in Round 1 and of course you always switch in Round 3. The one round I'm least confident in is the Second Round, but yea, let's switch here too, for the best chance at the $20,000.