The debate is over likelihood of outcome.

Some have stated that there are MORE ways to roll EITHER a 6 or 8 than a 7.

To me this makes sense. There are 6 ways to roll a 7. There are only 5 ways to roll a 6 or 8. BUT, if wanting a 6 OR an 8, there would be more ways for that to come up then a 7. i.e. 10 ways for a 6 OR 8, but only 6 ways for a 7.

Why would this be wrong?

Thank you.

Chance of 7 before 6 or 8 = 6/16.

Doesn’t make it a good bet to place the 6 and 8 for a roll, of course, because of how much you win 10/16 of the time versus how much you lose 6/16 of the time.

That one is somewhat counterintuitive since you’re more likely to roll the 6 and 8 first

Quote:Ace2Are you sure you’re not referring to the probability of rolling a 6 AND an 8 before two 7s?

That one is somewhat counterintuitive since you’re more likely to roll the 6 and 8 first

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That's what I was going to say. If you roll two dice, keeping a count of totals of 6, 7, and 8, you're more likely to see both a 6 and 8 before you see two 7's.

if the casino offered a bet that *either* 6 or 8 would come before 7, then the fair odds would be 3:5 [6:10] . They would probably have you lay 2 units to earn 1, to have a house edge.Quote:FlyNdadWe have been having a lively debate about the odds of a particular Craps dice odds situation over in a large Facebook group.

The debate is over likelihood of outcome.

Some have stated that there are MORE ways to roll EITHER a 6 or 8 than a 7.

To me this makes sense. There are 6 ways to roll a 7. There are only 5 ways to roll a 6 or 8. BUT, if wanting a 6 OR an 8, there would be more ways for that to come up then a 7. i.e. 10 ways for a 6 OR 8, but only 6 ways for a 7.

Why would this be wrong?

Thank you.

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But they don't have that bet. You can bet 6 to come before 7, or 8 to come before 7, bets that are independent. Some of your brainstorming friends may imagine that betting both on 6 and 8 at the same time creates a 6:10 situation that pays 7 to 6. This is a fallacy.

So this is my differing view of what might be going on here in all this discussion in the OD's group.

The argument by some is that there is a 27.78 % chance a 6 OR an 8 would be the result of a single dice roll. While only a 16.67% a 7 would be rolled.

The other side of the argument basically states that you can only have one result, not two. Therefore, your odds don't change.

My take is that while the 2nd argument is true, YOUR MONEY is saying that it will be one OR the other, therefore your odds of winning ONE of your individual bets is 27.78%.

10 chances for that, 6 chances for a 7 rolled, is all that matters. If something else is rolled it's the same as if it was never rolled.Quote:FlyNdadPerhaps I should clarify a bit.

The argument by some is that there is a 27.78 % chance

So not realizing the above gets you this place that has no meaningQuote:.... a 6 OR an 8 would be the result of a single dice roll. While only a 16.67% a 7 would be rolled.

I guess you can say it's trying to apply single roll betting probability to a multi-roll bet. This does not compute.Quote:The other side of the argument basically states that you can only have one result, not two. Therefore, your odds don't change.

My take is that while the 2nd argument is true, YOUR MONEY is saying that it will be one OR the other, therefore your odds of winning ONE of your individual bets is 27.78%.

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You should be buying the 2 and 12 for $39 each. Commission on win onlyQuote:ChumpChangeSo should I be betting the PB 6 or the PB 6 & 8? I could bet $48 on the PB 6, or $24 on the PB 6 & 8 each. Somebody else is shooting the dice so I won't be dice influencing my wins, and I won't have a PL bet. If I want to half-press my bet on a win though, I'll be looking for 2+ hits on the same number in either scenario.

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If laying is allowed, you could lay them for $234 each. Same edge as above, same absolute variance but far lower relative variance