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the game, is betting the field and the 7 a good bet?
There are 36 possible outcomes and the field numbers
and the 7 cover 22 of them.
If 3,4,9,10,11 hit and you bet $15 on the field and
$5 on the 7, you win $10. If 2 hits you win $25, and
if 12 hits you win $40. A 7 wins $5. Does this bet at
least come close to breaking even in the long run?
EV = $40 x 1/36 + 25 x 1/36 + 10 x (14/36) + 5 x 6/36 - $20 x (14/36) = -$1.25 = 6.25%
Of course, if you can think you can beat roulette, then you probably think you can beat craps ;)
You can see your expected loss is $1.25 per roll. Below the big table are the edges taken from Wizard of Odds - Craps. You can see the EV matches up as expected.
EDIT: If you were to bet evenly on both then:
Field $10 * 2.78% = $0.28 EV
Seven $10 * 16.67% = $1.67 EV
Combined EV: $1.95
Combined Edge: 9.73%
between -3% and -4%?
The any seven is a lousy bet. It should pay 11/2 if you ask me. Which you didn't... :)
Quote: odiousgambitBob, just figure any combination of negative expectation bets is just going to suck, in any game.
Some are worse than others.
Quote: EvenBobSome are worse than others.
Any seven being a poster child for that.
$3 hop seven pays $13 6/36 of the time and loses 30/36 of the time (-3)
-3 x 30/36 + 13 x 6/36 = -12/36 / 3 = 11.11%.
There really isn't a bet that would let you combine the 7 with the field and has have a net win (ie a hedge) with a 7 and the field that would yield you 4%.
The field bet with triple on a 12 only has a house edge similar to single zero roulette.
7, the 7 would be nullified. Whats the HE?
Quote: EvenBobIf you bet $20 on the field and $5 on the
7, the 7 would be nullified. Whats the HE?
If you're asking where the house edge comes from, it's those 14 times in 36 that the 5, 6, or 8 comes up. You lose $25 each of those times.
Quote: DocIf you're asking where the house edge comes from, it's those 14 times in 36 that the 5, 6, or 8 comes up. You lose $25 each of those times.
Is the HE better than the -6.25 of the other bet?
Let x be the field bet that pays 3:1 on the 12, 2:1 on the 2, and even on the 3, 4, 9, 10, 11.
Let y be the any seven bet, that pays 4:1
EV = -(x+y)*14/36+(4y-x)*6/36 + (3x-y)*1/36 + (2x-y)*1/36 + (x-y)*14/36
= (-14x - 14y + 24y - 6x +3x - y +2x - y + 14x - 14y)/36
= (-x - 6y)/36
Total Bet = x + y
HA = EV / Total Bet = (-x - 6y)/36(x+y)
So, Bob:
$20 on Field and $5 on 7 yields:
EV = (-20 - 6*5)/36 = -50 / 36 = -$1.38889
HA = -$1.388889 / (20 + 5) = 5.556%.
Quote: boymimbo$20 on Field and $5 on 7 yields:
EV = (-20 - 6*5)/36 = -50 / 36 = -$1.38889
HA = -$1.388889 / (20 + 5) = 5.556%.
Thanks a lot.
6 and 8: Place bet always better.
5 and 9: Place bet always better, unless commission on Buy is on win only.
4 and 10: Buy bet always better.
A field bet is only slightly worse than single zero roulette. So if you want to make eve money bets, then play the field in craps over red/black or odd/even in roulette. If you want a jackpot, then play roulette over boxcars or snake eyes in craps.
If you want a lot of action and variance, while keeping a low expected loss, just make lots of pass line and come bets and load up on free odds.
Quote: boymimboHopping a 7 is betting than big red.
$3 hop seven pays $13 6/36 of the time and loses 30/36 of the time (-3)
-3 x 30/36 + 13 x 6/36 = -12/36 / 3 = 11.11%.
There really isn't a bet that would let you combine the 7 with the field and has have a net win (ie a hedge) with a 7 and the field that would yield you 4%.
The field bet with triple on a 12 only has a house edge similar to single zero roulette.
Thanks, I thought the answer should be -11.11%, but kept getting -33.333%, but that's because I wasn't dividing by the total bet. Duh.
---- So hopping the 7 (a one roll bet on a 7 any of the three ways it can come up, pays 15:1 on each combination for a 7) for $3 and $17 on the field would give an edge of -4%, where a 7 results in $4 loss, a 5/6/8 an $20 loss, 3,4,9,10,11 a $14 win, 2 $31 win and 12 $48 win.