place betting 6 & 8 together

I know that place betting either 6 or 8 gives the casino a 1.52% advantage (losing $1 out of $66 bet using a $6 bet and calculating on a normal 36 count distribution) but it seems to me that if I place both 6 & 8 together they only have a 1.04% advantage (losing $2 out of $192 bet using a $6 bet on each number and calculating on a normal 36 count distribution). Is this correct and a good strategy? I usually place both the 6 & 8 when I have a pass line bet that has a 4,5,9 or 10 as a point.

Quote: ek

I know that place betting either 6 or 8 gives the casino a 1.52% advantage (losing $1 out of $66 bet using a $6 bet and calculating on a normal 36 count distribution) but it seems to me that if I place both 6 & 8 together they only have a 1.04% advantage (losing $2 out of $192 bet using a $6 bet on each number and calculating on a normal 36 count distribution). Is this correct and a good strategy?

I do not think it is correct. They both do not resolve at the same time except on a loss.
One adds ev and variance for multiple wagers.
Here is my table

bet	wager	win	ways to win	total	ways to lose	wager	total	net (ev)	action	edge	variance	sd
6	$6	$7	5	$35	6	-$6	-$36	($1)	$66	-0.015152	41.90082644628100	6.47308477051560
8	$6	$7	5	$35	6	-$6	-$36	($1)	$66	-0.015152	41.90082644628100	6.47308477051560
							total	($2)	$132	-0.015152	83.80165289256200	9.15432427285389

"a good strategy?"

depends on what you mean by "good"

Is it *better*(good) than placing the 4&10 for $5. By HE and long term play, yes.
Than placing the 5&9 for $5 for long term play, I agree.

Long term play placing the 6&8. good?
The 1/66 edge can not be improved upon no matter the amount bet.
It could be worse with breakage on an improper bet.($25 pays $29)

Is it better than making 2 come bets with odds. The math (ev/sd) says no.

But almost everyone bets the pass with odds and places the 6&8.

I understand that I will lose $1 every 36 rolls placing either the 6 & 8 and that the calculation I have seen refer to bets resolved. That is betting either 6 or 8 means that there have been 11 bets resolved in 36 rolls and I have bet $66 and won only $65 for a 1/66 house advantage. Why doesn't that work for betting both 6 & 8 and having 16 bets resolved out of 36 rolls and losing $2 but having now wagered $192 ($12 x 16) for a house advantage of 2/192 = 1.04% ?

Still not as good as 0.6% come bets, which I continue to do if the shooter is hot, but have I really improved my place bet odds? I think my change of strategy occurred when I did 2 come bets after the point and if I had only outside numbers, the shooter could roll for 10 minutes without me getting any payout, whereas the 6 & 8 obviously hit more often and "seemed" to give me more return over the session and I still had the outside pass line bet working. Since I could only take 2x odds the additional house advantage to 1.04% seemed acceptable but would have to rethink if it is really 1.5%.

Quote: ek

Why doesn't that work for betting both 6 & 8 and having 16 bets resolved out of 36 rolls and losing $2 but having now wagered $192 ($12 x 16) for a house advantage of 2/192 = 1.04% ?

It is not about the money wagered or risked, it is about resolved bets.
Action or the handle is the important variable.
A simple computer simulation in WinCraps show this to be true.

Yes, but I have also seen the 6&8 NOT hit when the pass line and come bets are hitting. All betting the 6&8 when the 7out came were just pissed off.
That is craps. each distribution of X number of rolls is different, never text book every time.

Vegas time!

Quote: ek

I understand that I will lose $1 every 36 rolls placing either the 6 & 8 and that the calculation I have seen refer to bets resolved. That is betting either 6 or 8 means that there have been 11 bets resolved in 36 rolls and I have bet $66 and won only $65 for a 1/66 house advantage. Why doesn't that work for betting both 6 & 8 and having 16 bets resolved out of 36 rolls and losing $2 but having now wagered $192 ($12 x 16) for a house advantage of 2/192 = 1.04% ?

Still not as good as 0.6% come bets, which I continue to do if the shooter is hot, but have I really improved my place bet odds? I think my change of strategy occurred when I did 2 come bets after the point and if I had only outside numbers, the shooter could roll for 10 minutes without me getting any payout, whereas the 6 & 8 obviously hit more often and "seemed" to give me more return over the session and I still had the outside pass line bet working. Since I could only take 2x odds the additional house advantage to 1.04% seemed acceptable but would have to rethink if it is really 1.5%.

maybe I'm lost but........If you are only betting the 6 OR 8...with 11 bets resolved betting $6 you would have 5 wins totaling $35 and 6 losses totaling $36......how did you come up with winnings of $65 over 11 bets resolved betting the 6 OR 8? or am I missing something?

The OP is making the mistake of including bets at risk per wager resolved, not just resolved wagers.

This is a common error when calculating many multi-roll bets.

One also has to consider is the results all for resolved bets - that would be one value.
One can also calculate the HE for every roll. That would be another value.
can not mix these up.

The Wizard's craps page at WoO shows both values per roll and per resolved bet

Quote: ek

but it seems to me that if I place both 6 & 8 together they only have a 1.04% advantage (losing $2 out of $192 bet using a $6 bet on each number and calculating on a normal 36 count distribution). Is this correct

The Wizard's answer from
ask-the-wizard #141
4th question down

"Your mistake is that both bets are not resolved all of the time.
When you win either the 6 or 8 you are taking the other bet down, which brings down the expected loss because you are betting less.
So your math is right but you are comparing apples to oranges."