July 30th, 2013 at 2:46:52 PM
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How to calculate the Pass line house edge when betting single Odds? The Wizard gives the answer 0.00848. I am getting: 0.00789
July 30th, 2013 at 3:45:27 PM
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formula for figuring the combined HA is:
(-28/55) / (36 + 24x)
where x is the odds multiple.
so (-28/55) / 60 = combined HE with 1X odds
https://wizardofvegas.com/forum/questions-and-answers/math/1579-free-single-odds-bets
the Wizard's formula is also in that thread (yours)
https://wizardofodds.com/games/craps/appendix/1/
I get -28/3300 using the perfect 1980 table
added:
in order to use that combined house edge value to find out the EV of the combined wager (line plus odds)
one must know the average bet
what would the average bet be making $5 pass line bets and taking 1x odds?
maybe $5 or $10 or something between the two?
Good Luck
(-28/55) / (36 + 24x)
where x is the odds multiple.
so (-28/55) / 60 = combined HE with 1X odds
https://wizardofvegas.com/forum/questions-and-answers/math/1579-free-single-odds-bets
the Wizard's formula is also in that thread (yours)
https://wizardofodds.com/games/craps/appendix/1/
I get -28/3300 using the perfect 1980 table
added:
in order to use that combined house edge value to find out the EV of the combined wager (line plus odds)
one must know the average bet
what would the average bet be making $5 pass line bets and taking 1x odds?
maybe $5 or $10 or something between the two?
long way
1/3*$Bet + 2/3* total$bet
or
(3*$bet) + (2* $Odds bet) / 3
1/3*$Bet + 2/3* total$bet
or
(3*$bet) + (2* $Odds bet) / 3
Good Luck
winsome johnny (not Win some johnny)
July 30th, 2013 at 6:38:16 PM
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I get what Wizard gets - here's how:
Assume you are betting $10 pass plus $10 odds.
Look at all 36 come-out rolls:
There is one way to roll 2; you bet a total of $10, and lose $10.
There are two ways to roll 3; you bet $10 and lose $10 on each, for a total bet of $20 and loss of $20.
There are three ways to roll 4; each has $20 bet ($10 plus $10 for the odds). 1/3 of the time, you win $30; the other 2/3, you lose $20. That's a total of $60 bet ($20 for each way to roll 4), and a total loss of $10 ((3 x 1/3 x 30) + (3 x 2/3 x (-20)).
There are four ways to roll 5; each has $20 bet. 2/5 of the time, you win $25; the other 3/5, you lose $20. That's $80 bet, and a total loss of $8 ((4 x 2/5 x 25) + (4 x 3/5 x (-20)).
There are five ways to roll 6; each has $20 bet. 5/11 of the time, you win $22; the other 6/11, you lose $20. That's $100 bet, and a total loss of $(50/11) ((5 x 5/11 x 22) + (5 x 6/11 x (-20)).
There are six ways to roll 7; you bet a total of $60 and win $60.
8, 9, and 10 are equal to 6, 5, and 4 respectively.
There are two ways to roll 11; you bet a total of $20 and win $20.
There is one way to roll 12; you bet $10 and lose $10.
Add up the total amounts bet: $600.
Add up the total wins and losses: you lose $56/11.
In other words, for every $600 you bet, you "should" lose $56/11.
The house edge is (56/11) / 600 = 0.00848485.
Assume you are betting $10 pass plus $10 odds.
Look at all 36 come-out rolls:
There is one way to roll 2; you bet a total of $10, and lose $10.
There are two ways to roll 3; you bet $10 and lose $10 on each, for a total bet of $20 and loss of $20.
There are three ways to roll 4; each has $20 bet ($10 plus $10 for the odds). 1/3 of the time, you win $30; the other 2/3, you lose $20. That's a total of $60 bet ($20 for each way to roll 4), and a total loss of $10 ((3 x 1/3 x 30) + (3 x 2/3 x (-20)).
There are four ways to roll 5; each has $20 bet. 2/5 of the time, you win $25; the other 3/5, you lose $20. That's $80 bet, and a total loss of $8 ((4 x 2/5 x 25) + (4 x 3/5 x (-20)).
There are five ways to roll 6; each has $20 bet. 5/11 of the time, you win $22; the other 6/11, you lose $20. That's $100 bet, and a total loss of $(50/11) ((5 x 5/11 x 22) + (5 x 6/11 x (-20)).
There are six ways to roll 7; you bet a total of $60 and win $60.
8, 9, and 10 are equal to 6, 5, and 4 respectively.
There are two ways to roll 11; you bet a total of $20 and win $20.
There is one way to roll 12; you bet $10 and lose $10.
Add up the total amounts bet: $600.
Add up the total wins and losses: you lose $56/11.
In other words, for every $600 you bet, you "should" lose $56/11.
The house edge is (56/11) / 600 = 0.00848485.
July 31st, 2013 at 5:35:12 AM
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Thank you.
August 3rd, 2013 at 2:14:44 AM
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Quote: 7crapsformula for figuring the combined HA is: (-28/55) / (36 + 24x) where x is the odds multiple.
Wins to Losses are 244:251 in pass line bet. So the house edge is (244-251) / (244+251) / (mean bet)
mean bet = 1 if you take no odds
mean bet = 1+24/36*X if you take X odds
The reason is that 12/36 of the time you finish before hitting your point, but 24/36 of the time you hit your point and place an odds bet. The odds bet has no house edge
Normal House Edge is (244-251) / (244+251) = -7/495 = -1.414141...%
With 1X odds the house edge is -7/495/(1+24/36*1) = -0.848484...%
With 10X odds the house edge is -7/495/(1+24/36*10)= -0.184453228%
It is the same formula, but with some explanation.
June 19th, 2020 at 6:11:45 AM
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Quote: pacomartinWins to Losses are 244:251 in pass line bet. So the house edge is (244-251) / (244+251) / (mean bet)
mean bet = 1 if you take no odds
mean bet = 1+24/36*X if you take X odds
The reason is that 12/36 of the time you finish before hitting your point, but 24/36 of the time you hit your point and place an odds bet. The odds bet has no house edge
Normal House Edge is (244-251) / (244+251) = -7/495 = -1.414141...%
With 1X odds the house edge is -7/495/(1+24/36*1) = -0.848484...%
With 10X odds the house edge is -7/495/(1+24/36*10)= -0.184453228%
It is the same formula, but with some explanation.
So if I have 3X odds on the PL, the HA would be:
-7/495/(1+24/36*3) = -0.0141414 / 1+ (0.6667 x 3) = ( -0.0141414 / 3) x 100 = -0.47138%
If I have 6X odds on the PL, the HA would be:
-7/495/(1+24/36*6) = -0.0141414 / 1 + (0.6667 x 6) = ( -0.0141414 / 5) x 100 = -0.282828%
If I have 9X odds on the PL, the HA would be:
-7/495/(1+24/36*9) = -0.0141414 / 1 + (0.6667 x 9) = ( -0.0141414 / 7) x 100 = -0.202020%