Fair Two Dice Game
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.
Quote: kubikulannOr 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.
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
Quote: david862Thanks 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!
Quote: david862Thanks 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).
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: ThatDonGuy...odd or even....Easy proof...Slightly harder proof...
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).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...
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.
Quote: kubikulannSuppose 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
