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.Quote:david862Given 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 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.

Quote:david862Given 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.

(Voila is the preterit form of the verb ‘’voiler’’, to veil.)

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

Quote:odiousgambitKubikulann 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.

Quote:david862Given 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

Quote:charliepatrickCharlie'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?

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).Quote:beachbumbabsWhy 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?

(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.