Which figure is right
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Given that there is a 18/36 chance of hitting one of the inside numbers and a 6/36 chance of sevening out, I can expect that my odds of hitting one of the inside numbers before sevening out is (18/36) / (18/36+6/36) = 75.75% Yet if I had it so that I'd win $70 on each of these inside numbers ($50 on 5,9 and $60 on 6,8) My expected win would be $70*3.03 followed by a loss of $220 thus leaving me down $8.9 on average for every roll. Yet -$8.9/$210= -.0423 This is a lot worse than the 2.5% h.e. we'd initially expect. Where exactly is my math wrong here?
I'm not the best with place bets. My math expertise has revolved mostly around DP/DC bets/odds... After a quick look on WOO I believe these are the correct numbers:
Assuming $50 is your unit... $60 would be worth 1.2 units. For one generic unit on any number we see the odds:
1 unit on 5... HE = (4/36)(7/5)(1) + (6/36)(-1) = .11111(7/5) + -.16667 = .15554 + -.16667 = -.01113 ...or -1.11% as the Wizard also has.
1 unit on 6... HE = (5/36)(7/6)(1) + (6/36)(-1) = .13889(7/6) + -.16667 = .16204 + -.16667 = -.00463 ...or -.46% as the Wizard also has.
Don't forget, you're not betting 1 unit on the 6 and 8... You're betting 1.2 units:
1.2 units on 6... HE = (5/36)(7/6)(1.2) + (6/36)(-1.2) = .16204(1.2) + .16667(-1.2) = .19445 + -.20000 = -.00555 ...or -.56%
The same goes for their corresponding numbers.
Thus, if you're going to bet 5, 6, 8, 9, the HE would be:
(4/36)(7/5)(1) + (5/36)(7/6)(1.2) + (5/36)(7/6)(1.2) + (4/36)(7/5)(1) + (6/36)(-4.4)
= .15554 + .19445 + .19445 + .15554 + -.73333 = .69998 + -.73333 = -.03335 ...or -3.34%.
If you check this that should be the sum of the individual EV's: (-1.11*2) + (-.56*2), which = -2.22 + -1.12 = -3.34%
Quote: ThundershockHow can -3.34% be the answer if 6 and 8 and pop up more frequently than the 5 and 9 and the h.e. is 1.52% on 6,8 and 4% on 5, 9. Logically that doesn't make sense.
They pop up 1/36 more frequently... and you also lose more on 6/8 than your 5/9 when a 7 is rolled; 1.2 units compared to 1 unit.
Quote: ThundershockAlso, not to nitpick, but the amount of units you bet shouldn't determine the house edge.
I wasn't trying defining the house edge, although at a generic "1" unit the EV will mimic the HE as shown on the Wizards page I linked because you're multiplying by (1). With the house edge set the more you bet the more you can expect to lose... more in terms of EV.
Here is what I get per bet resolvedQuote: ThundershockI was wondering though if you might help settle a nagging question for me. On a craps table, the house edge for a place bet of 6&8 is 1.52% and the h.e. for a 5,9 is 4% Given this, it would seem that the combined house edge for all 4 numbers is about 2.5%
also matches a simulation
6&8
ev:-$1 action: $66
5&9
ev:-$2 action: $50
total ev = -$6
total action = $232
ev/action = he
-0.025862
(this is close to just (10/18 * 1/66) + (8/18 * 1/25) not the proper way to do this btw)
now,
per roll edges will be different
and
per roll edges not counting the come out roll will be different too (and more the way these bets are played)
do not even attempt to mix and match the math for the 3 different methods unless you want to create a new life form
Sally
Quote: ThundershockI was wondering though if you might help settle a nagging question for me. On a craps table, the house edge for a place bet of 6&8 is 1.52% and the h.e. for a 5,9 is 4% Given this, it would seem that the combined house edge for all 4 numbers is about 2.5%
Given that there is a 18/36 chance of hitting one of the inside numbers and a 6/36 chance of sevening out, I can expect that my odds of hitting one of the inside numbers before sevening out is (18/36) / (18/36+6/36) = 75.75% Yet if I had it so that I'd win $70 on each of these inside numbers ($50 on 5,9 and $60 on 6,8) My expected win would be $70*3.03 followed by a loss of $220 thus leaving me down $8.9 on average for every roll. Yet -$8.9/$210= -.0423 This is a lot worse than the 2.5% h.e. we'd initially expect. Where exactly is my math wrong here?
Well, you made a few mistakes in arithmetic, and one in probability theory. But I figured enough people would point that out.
I'd like to give you an analogy to show you the error of your ways. And how you shouldn't trust your intuition on this. :)
Suppose you decided to use the same reasoning, but instead of your chosen bets, you chose 7 and 5.
Okay, let's average the house edge on those: (16.7%+4%)/2 = 10.35%.
So you figure, hey, I can live with that (I really hope you can't). So I'll bet it. But I'll bet a lot more on the 7. So I'll take $100 and put $10 on the 5 and $90 on the 7.
Now, would you presume that you still have a combined house edge of 10.35% under those circumstances?
10/36 * 70 + 8/36 * 70 - 6/36 x 220 = -$1.667 per roll.
House edge = -$1.667 / 220 = 0.758% per roll.
The bet is resolved on 24 of 36 rolls. So, the total house edge is 1.136%.
Reason that is is lower than say a six or eight is because of the rate the bet resolves itself.
HE per roll on a 6 or 8 is 5/36 * 7/6 - 6/36 = 0.463% per roll
.463% * 36 / 11 = 1.515% or 1.52%
Thank you and good night.
meaninglessQuote: boymimboCombined HE is <snip>
in other words
useless
imo
why do youQuote: boymimboThank you and good night.
calculate
(and all the other guys too)
the edge per roll for place bets
working on the come out roll
when the rules of the game states they are off on the come out roll?
I mean, does the majority of craps players that make the place bets always have them working on the come out roll?
answer:
Is it just because?
if 165/557 is the probability of the roll being a come out roll
I would rather know the 1 minus value so a correct per roll not working on the come out roll value can be calculated
It is only fair
girls rock!
Sally
Quote: mustangsallymeaningless
in other words
useless
imo
why do you
calculate
(and all the other guys too)
the edge per roll for place bets
working on the come out roll
when the rules of the game states they are off on the come out roll?
I guess that means you don't consider me a guy. :)
