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gordonm888
gordonm888
Joined: Feb 18, 2015
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Thanks for this post from:
DogHand
May 11th, 2021 at 8:39:52 PM permalink
Quote: DogHand

gordonm888,

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.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
ThatDonGuy 
Joined: Jun 22, 2011
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May 12th, 2021 at 8:06:40 AM permalink
Quote: Ace2

Quote: unJon

Quote: ThatDonGuy

I'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



Thatís complicated. I found a simple but unsatisfying one:

Making a point with a hard 6 or hard 8 pays an extra 56%

56% is not a feasible payout. Okay, it would be on $100 but it needs to be an easy ratio. Letís say the min/max bet is $10 / $100

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.)

ComeoutPoint 1 Point 2 Point 3 7
1,1 N/A-495
2,1 N/A-990
3,13,12202,2110 7-660
2,23,11102,255 7-330
4,14,11983,2198 7-594
3,24,11983,2198 7-594
5,15,11804,21803,3907-540
4,25,11804,21803,3907-540
3,35,1904,2903,3457-270
6,1 N/A990
5,2 N/A990
4,3 N/A990
6,26,21805,31804,4907-540
5,36,21805,31804,4907-540
4,46,2905,3904,4457-270
6,36,31985,4198 7-594
5,46,31985,4198 7-594
6,46,42205,5110 7-660
5,56,41105,555 7-330
6,5 N/A990
6,6 N/A-495
Ace2
Ace2
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May 12th, 2021 at 8:53:37 AM permalink
Hereís 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 isnít ideal but...it only lasts an average of 1.09 rolls, it only occurs once every 45 resolutions and the original game/payout isnít changed at all

After running through numbers in my head I tend to agree with Don that there probably isnít a way to eliminate the edge exactly using just the original game and ďreasonableĒ 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...thatís less than the edge difference between pass and DP
Last edited by: Ace2 on May 12, 2021
Itís all about making that GTA
unJon
unJon 
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May 12th, 2021 at 9:26:24 AM permalink
Quote: Ace2

Hereís 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 isnít ideal but...it only lasts an average of 1.09 rolls, it only occurs once every 45 resolutions and the original game/payout isnít changed at all

After running through numbers in my head I tend to agree with Don that there probably isnít a way to eliminate the edge exactly using just the original game and ďreasonableĒ 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 itís not an ideal round bonus. If you can work in the 5 to that same roll (so itís 1/5 as likely) then the bonus could pay a round 28 times.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
ThatDonGuy 
Joined: Jun 22, 2011
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Thanks for this post from:
unJon
May 12th, 2021 at 9:30:05 AM permalink
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.
charliepatrick
charliepatrick
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May 12th, 2021 at 9:54:30 AM permalink
Assuming you're only considering the Pass Line Bet. I think one way to do it is when you throw what would be a standoff for Don't Come (say 1-1) then rather than losing you enter a new game to resolve the bet. As per above this happens 495 times and you want to ensure the player only loses 243 units rather than 495. This means that you need a game with one of 55 outcomes, 28 of them the player doesn't lose, 27 of them the player does lose.

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.
Ace2
Ace2
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May 12th, 2021 at 8:42:29 PM permalink
Passline win is paid at 8:1 instead of 1:1 if:

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
Itís all about making that GTA
Gialmere
Gialmere
Joined: Nov 26, 2018
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May 17th, 2021 at 7:09:15 AM permalink
It's easy Monday. Let's make a deal...



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?
Have you tried 22 tonight? I said 22.
EdCollins
EdCollins
Joined: Oct 21, 2011
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Thanks for this post from:
Gialmere
May 17th, 2021 at 11:47:54 AM permalink
Quote: Gialmere

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.

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.
Ace2
Ace2
Joined: Oct 2, 2017
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Thanks for this post from:
Gialmere
May 17th, 2021 at 12:08:04 PM permalink
Round 1: donít switch
Round 2: either
Round 3: switch
Itís all about making that GTA

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