New to the forum, first post so here goes:
Would like help running some scenarios or getting thoughts:
let assume 3,4,5x odds, with a $5 table, all odds working on the come out roll
players try to optimize the CB/DC odds, putting them on the best numbers, so the do player on the 6 &8 and the don't player with the 4 & 10 concentrations
can odds optimization: do side side focus 6 & 8 , and dont side focus 4 & 10 help? increase wins? reduce variance? both?
3 point molly from the DO side,
1 unit odds, when 4 or 10, then 2 units of odds 5 or 9 and starts with 3 units on the 6 or 8, going to 5x odds if 2 hits on 6 or 8, never increasing on the 5,9,4,10 until the 6 & 8 are maxed, then branching out first to the 5 & 9s, then the 4 & 10s, if the 6 or 8 ever falls and a new come bet travels there, it gets athe extra odds money from the 4,5,9,10 first above the starting units (so these odds reduced to fund the 6 & 8 odds)
also optimizing the line odds, the same way, depending on what the point is
3 point polly from the Dont side:
just the reverse
1 unit win on the 6 & 8, so laying &7, win 2 units on the 5 & 9 so laying the 3 units, focusing on the 4 & 10, starting with laying 6 units to get 3, working the 4 & 10 up to max of laying 10 unites, before going any higher above the base starting lays for the 5,6,8,9
same odds strategy on the DP line odds
RS, thanks for responding
I definitely have no illusions, that a negative expectation game can be beaten in the long run, just enjoy playing craps, with lose-able, discretionary funds, when and if I do go gamble, which is a rarity
but I still want to be educated and as hard to beat as possible (if that can happen in a neg edge game)
i understand that the odds affect the variance, win more or lose more
what i was wondering, does increasing the odds on higher frequency base come bet numbers, say 6 & 8 from the do side, help, vice versa 4 & 10 from the donts, then minimizing odds on the less rolled numbers
by help meaning: lessen drawdowns in a session, increasing the chance of being there for a longer roll, with your bankroll
or just a wash overall, no matter what is tried, since a neg expectation game?
so if I decide that my bankroll only warrants 15% of my table session money at one time
ex 300, so .15% is 45
line bet 5, with 10 odds
and then 2 come bets with 10 odds each
is it better to follow that blindly or to optimize
example, point is 4 or 10, so might not want to max line odds,
where as on a point of 6 or 8, should go to the 25 right away, then use the other 15, (45 minus the 30), and do 2 come bets, with the other 5 odds on the better of the 2 come bet numbers
just want to know if it helps overall, dampen session variance to focus line and come bet odds on the 6 & 8 when possible, and minimize on the 4 & 10 as examples
I have been reading the forums and I am still a bit confused on the house edge per roll vs. house edge per bet resolved. I am trying to determine hourly expected loss based on a few different methods of playing. Any help would be greatly appreciated!
I typically play $10 tables and I will play the pass line with single or double odds followed by placing the six or eight.
In calculating the hourly expected loss using this strategy, I am assuming 100 rolls/hour (is this average?)
--Pass Line: ($10 x .42%) x 100 = $4.2 hourly expected loss
--Placing Six: ($12 x .46%) x 100 = $5.52 hourly expected loss
--Placing Eight: ($12 x .46%) x 100 = $5.52 hourly expected loss
Total Hourly Expected Loss = $15.24
Is this the correct calculation?
I am trying to compare this expected loss scenario with playing a three point molly with single odds and double odds. I am a bit confused as to how the house edge is reduced when utilizing odds. My understanding is that odds are a free bet and thus have no house edge but the pass line bet has a house edge of 1.41% no matter what, i.e., if I play a $10 pass line bet with $20 in odds, my expected loss is still 1.41% or .42% per roll.
I am confused when I see that the house edge is .83% when utilizing single odds and .61% when utilizing double odds. Does this percentage apply to the combined wager, i.e., when playing double odds at a $10 table, .61% house edge on the total $30 wagered? If so, how would I calculate the house edge per roll as I am interested in trying to determine expected hourly loss?
The reduction of expected loss does not change, so if your calcs are correct for 100 rolls, for 100X odds the expected loss stays the same. Someone started to include the free odds bet and combined it with the line bet to make it appear you were losing less. Buying half The Brooklyn Bridge for $100 has the same expected loss as buying it all for $100!