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