david862
david862
  • Threads: 1
  • Posts: 2
Joined: Sep 8, 2019
September 8th, 2019 at 11:30:40 PM permalink
Given that craps has a house edge... if two friends wanted to play an even game where neither the thrower or the bettor would have an advantage... what is the easiest way to do that?

I figured basing a win on rolling a 5, 6, 8 or 9 would be a 50/50 chance. Is this true?
ChumpChange
ChumpChange
  • Threads: 118
  • Posts: 4917
Joined: Jun 15, 2018
September 8th, 2019 at 11:46:34 PM permalink
I'd stick with just a side bet where the 5, 6, 8, or 9 is a single bet against all the other numbers as a single bet. I've never played street craps so I don't know how players fade each others bets.
kubikulann
kubikulann
  • Threads: 27
  • Posts: 905
Joined: Jun 28, 2011
September 9th, 2019 at 2:38:28 AM permalink
Quote: david862

Given that craps has a house edge... if two friends wanted to play an even game where neither the thrower or the bettor would have an advantage... what is the easiest way to do that?

I figured basing a win on rolling a 5, 6, 8 or 9 would be a 50/50 chance. Is this true?

Or play Heads-or-Tails.
Or throw 3coins and win if more heads than tails. Or 1001 coins, for that matter.

Where’s the fun?

Craps or 421 attract because they introduce some sequence and choices by the player.
Reperiet qui quaesiverit
unJon
unJon
  • Threads: 14
  • Posts: 4654
Joined: Jul 1, 2018
September 9th, 2019 at 3:08:21 AM permalink
Quote: david862

Given that craps has a house edge... if two friends wanted to play an even game where neither the thrower or the bettor would have an advantage... what is the easiest way to do that?

I figured basing a win on rolling a 5, 6, 8 or 9 would be a 50/50 chance. Is this true?



If you want to keep the flavor of craps you can do the following:

1) Shooter can make only a pass line bet. Non-shooter books that action.

2) On the come out roll make the 11 a loser instead of a winner. Keep all else the same.

3) If a point is established, convert the original PL bet to an odds bet that pays based on what point is established just like an odds bet in craps.

Voila. No house edge craps.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
kubikulann
kubikulann
  • Threads: 27
  • Posts: 905
Joined: Jun 28, 2011
September 9th, 2019 at 3:15:20 AM permalink
Voilà is the right spelling.

(Voila is the preterit form of the verb ‘’voiler’’, to veil.)
Reperiet qui quaesiverit
odiousgambit
odiousgambit
  • Threads: 327
  • Posts: 9618
Joined: Nov 9, 2009
September 9th, 2019 at 3:54:17 AM permalink
Kubikulann is the real Hercule Poirot? Voila instead of Voilà is the sort of thing that would bother Poirot too

Everybody wants to reinvent the wheel. The way Craps is played fairly is to pass the dice when the shooter sevens out
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
unJon
unJon
  • Threads: 14
  • Posts: 4654
Joined: Jul 1, 2018
September 9th, 2019 at 4:08:12 AM permalink
Quote: odiousgambit

Kubikulann is the real Hercule Poirot? Voila instead of Voilà is the sort of thing that would bother Poirot too

Everybody wants to reinvent the wheel. The way Craps is played fairly is to pass the dice when the shooter sevens out



That wouldn’t be “fair.” Fair would be passing the dice every time a pass line bet resolves no matter how it resolves. So that each player makes and books the same number of pass line wagers.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 2967
Joined: Jun 17, 2011
September 9th, 2019 at 4:37:23 AM permalink
Quote: david862

Given that craps has a house edge... if two friends wanted to play an even game where neither the thrower or the bettor would have an advantage... what is the easiest way to do that? ....


Charlie's Craps - as follows

Come out roll as normal (wins on 7 11 lose on 2 3 12)

Any other number is then considered either High (8-12) or Low (2-6), and you win if you throw the same group again before a 7 or the other group.

As an example if you rolled a 8 9 or 10 then you need to roll a High number (8 9 10 11 or 12) before you roll a 7 or Low number (2 3 4 5 6).

Also after coming out, there's no waiting around as the bet will be resolved on the second roll.

Edit: Thanks BBB, changed typo
Last edited by: charliepatrick on Sep 9, 2019
beachbumbabs
beachbumbabs
  • Threads: 100
  • Posts: 14267
Joined: May 21, 2013
September 9th, 2019 at 6:36:46 AM permalink
Quote: charliepatrick

Charlie's Craps - as follows

Come out roll as normal (wins on 7 11 lose on 2 3 12)

Any other number is then considered either High (8-12) or Low (2-6), and you win if you throw the same group again before a 7 or the other group.

As an example if you rolled a 8 9 or 10 then you need to roll a High number (8 9 10 11 or 12) before you roll a 7 or Low number (2 3 4 5 6 7).

Also after coming out, there's no waiting around as the bet will be resolved on the second roll.



That's pretty good. Just a typo (sort of) in the instructions, repeating the 7 in the set of low numbers.

Curious why you don't include the 11 as a come-out loser in this variation? 8 winners, 4 losers, 24 point rolls this way. Why not 6 winners, 6 losers?
If the House lost every hand, they wouldn't deal the game.
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 2967
Joined: Jun 17, 2011
September 9th, 2019 at 7:25:14 AM permalink
Quote: beachbumbabs

Why you don't include the 11 as a come-out loser in this variation? 8 winners, 4 losers, 24 point rolls this way. Why not 6 winners, 6 losers?

Thanks - I didn't really think that much about it, except I wanted to leave the come out asis (c.f. crapless craps where people might be unhappy not winning on an 11) and played around on a spreadsheet needing to get back some House Edge on the points (i.e. they have to be less that 50% to win).

(i) If 7 and 11 win, that's 8 chances
(ii) if 2, 3 and 12 lose, that's 4 chances.
(iii) So from 24 other outcomes we need 10 winners and 14 losers - i.e. Pr(winning with a point) needs to be 5/12.
(iv) {2 3 4 5 6} = 15 ways, so Pr(Winning) = 15/36. Yippee!

btw ignoring 2 and 12, gives a fairly small House Edge, ignoring 2 3 11 and 12 gives a larger House Edge.

As to the original idea, another way of getting even is that 7 wins, 2 3 11 12 loses, but all points only lose on their opposite (e.g. 6 loses on an 8). Similar numbers are (i) 6 winners (ii) 6 losers (iii) half win, half lose.
david862
david862
  • Threads: 1
  • Posts: 2
Joined: Sep 8, 2019
September 9th, 2019 at 10:41:50 AM permalink
Thanks to everyone for answering.

Quote: kubikulann

Or play Heads-or-Tails.
Or throw 3coins and win if more heads than tails. Or 1001 coins, for that matter.

Where’s the fun?

Craps or 421 attract because they introduce some sequence and choices by the player.



Well... the idea isn't to replace craps... more like I am hanging out with friends and we don't have a coin... just 2 dice. How could we do a fair single-roll game that is easy enough to explain/prove.

I do like the ideas above though that keep the flavor of craps.
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
September 9th, 2019 at 1:12:53 PM permalink
When a 2 is rolled on the comeout roll, passline bettors lose half their bet and don’t bettors are paid at 1 to 2.

This brings the passline house advantage down to 2 basis points, which is effectively zero.

For comparison purposes, the normal passline HA is 5 basis points higher than the don’t, yet 99% of players choose pass over don’t. Even I play the pass 90% of the time just because I generally enjoy betting dogs over favorites. I generally don’t like laying odds on anything, which is just a preference
It’s all about making that GTA
ChesterDog
ChesterDog
  • Threads: 8
  • Posts: 1565
Joined: Jul 26, 2010
September 9th, 2019 at 3:06:15 PM permalink
Quote: david862

Thanks to everyone for answering.



Well... the idea isn't to replace craps... more like I am hanging out with friends and we don't have a coin... just 2 dice. How could we do a fair single-roll game that is easy enough to explain/prove.

I do like the ideas above though that keep the flavor of craps.



Here's one without craps-flavor: You pick any two numbers out of 1, 2, 3, 4, 5, or 6. Suppose you choose 3 and 5, for example. Then you roll the dice once. If one of the dice is 3 and/or if one of the dice is 5, you win; but if both are 3 or both are 5, you lose. And if neither is a 3 or a 5, you lose.

Voilà!
ThatDonGuy
ThatDonGuy
  • Threads: 118
  • Posts: 6405
Joined: Jun 22, 2011
September 9th, 2019 at 4:20:51 PM permalink
Quote: david862

Thanks to everyone for answering.



Well... the idea isn't to replace craps... more like I am hanging out with friends and we don't have a coin... just 2 dice. How could we do a fair single-roll game that is easy enough to explain/prove.

I do like the ideas above though that keep the flavor of craps.


The easiest one I can think of: bet on odd or even.

Easy proof: of the 36 rolls, one 2 + three 4s + five 6s + five 8s + three 10s + one 12 = 18 evens, and two 3s + four 5s + six 7s + four 9s + two 11s = 18 odds.

Slightly harder proof: roll the dice one at a time. The first die has a 50/50 chance of being even or odd.
If the first die is even, there is a 50/50 chance of the second die being even (so the total is even) or odd (so the total is odd).
If the first die is odd, there is a 50/50 chance of the second die being odd (so the total is even) or even (so the total is odd).
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 2967
Joined: Jun 17, 2011
September 9th, 2019 at 11:48:40 PM permalink
Quote: ThatDonGuy

...odd or even....Easy proof...Slightly harder proof...

Rolling the dice one at a time is an easier way to prove it. When you roll the second die it's 50/50 whether that die is odd or even, so when you add that to the first roll it still will be 50/50 whether the total is odd or even.

Quote: ChesterDog

...pick any two numbers out of 1, 2, 3, 4, 5, or 6....If one of the dice is 3 and/or if one of the dice is 5, you win; but if both are 3 or both are 5, you lose. And if neither is a 3 or a 5, you lose...

Nice idea, would work very well if three people were playing! Also if 33 55 or 53 were a standoff - i.e. none correct lose (16), one correct win (16), two correct standoff (4).
kubikulann
kubikulann
  • Threads: 27
  • Posts: 905
Joined: Jun 28, 2011
September 10th, 2019 at 12:14:35 PM permalink
Suppose you don’t have dice or coins, just some random generator of which the P is not 50/50.
Throwing a thumbtack for instance. It can fall on its Axis or on its Back.

Throw it twice. The probability of observing AB is equal to that of observing BA, whatever the unknown P of A and B.
So one player bets AB, the other BA, and if AA or BB throw again.
Reperiet qui quaesiverit
unJon
unJon
  • Threads: 14
  • Posts: 4654
Joined: Jul 1, 2018
September 10th, 2019 at 12:54:15 PM permalink
Quote: kubikulann

Suppose you don’t have dice or coins, just some random generator of which the P is not 50/50.
Throwing a thumbtack for instance. It can fall on its Axis or on its Back.

Throw it twice. The probability of observing AB is equal to that of observing BA, whatever the unknown P of A and B.
So one player bets AB, the other BA, and if AA or BB throw again.



Nice one. And reminds me of my favorite method to approximate Pi: Buffon’s Needle.

https://en.m.wikipedia.org/wiki/Buffon's_needle_problem
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
  • Jump to: