VideoBJ
Joined: Jul 21, 2010
• Posts: 6
April 13th, 2017 at 8:12:44 AM permalink
Quote: SanchoPanza

To figure the house advantage on doey-don't, multiply 1.41 X 2.

The house edge for pass / don't pass bettor (assuming equal bet amounts) cannot be 1.41 x 2 -- that makes no sense.

A quick approximation is 1 bet out of 72 bets since you lose 1 bet, on average, in 36 rolls but since you are making 2 opposing bets at the same time, the edge is 1 in 72 bets as a guesstimate.

35 times (or 70 bets) that are non-box cars are a wash or net to zero. One bet wins, the other bet loses. It's the box car where one bet loses and one bet pushes, thus losing one bet in 72 bets.
kmumf
Joined: Jul 5, 2011
• Posts: 182
April 13th, 2017 at 9:44:37 AM permalink
I had a guy at my table doing this the other night. He was laying the 4 and 10 for \$100+ each he did have a min bet on the don't pass with 0 odds. He went up over 1k at first when the table was point 7 out a multiple times in a row. Then proceeded to get crushed to \$0 after a nice lady went for a nice long roll and a few points.
slackyhacky
Joined: Jan 18, 2012
• Posts: 359
April 27th, 2017 at 10:27:12 PM permalink
Just lay the 4 OR the 10.

Laying both - you have the same chance of a 4 or a 10 hitting as you do a 7.

Chance of a 4 hitting = 3/36.
Chance of a 10 hitting = 3/36

Chance of either of them hitting = 6/36.

Chance of a 7 rolling = 6/36

Chance of a 7 rolling before a 4 OR a 10 is twice as great.
slackyhacky
Joined: Jan 18, 2012
• Posts: 359
April 27th, 2017 at 10:38:04 PM permalink
I worked on a strategy that I ended up not liking -

But basically, you lay 4 and 10, then one hits, you triple the bet (you have to triple in order to cover the loss since it pays only 50%), but you split the increase between the 4 and 10.

Example, \$100 on each 4 and 10 lay. a 4 rolls. You triple it it to \$300, but since \$100 is still on the 10, total is \$400 - you split that and so now you have \$200 on 4/10 lay. Then another 4 rolls, you triple it (\$600), but spread out the total between the two so now you have \$400 each. You have now lost \$300. If a seven hits, you will win \$400. It's kind of a fun martingale variant that spreads the increase and lessons the impact of an exponential increase with each doubling - but maintains the value of martingale system because assuming you still have head room (table limits) and bankroll, you always stay ahead.

Assuming the Lay limit is \$5000, you would need to roll six 4/10's before a 7 rolls before running out of room. I guess that is a 2% chance of happening (on the next series) and 99% chance after 1000 rolls.
Last edited by: slackyhacky on Apr 27, 2017
betwthelines
Joined: Jan 2, 2015
• Posts: 171
April 27th, 2017 at 11:38:14 PM permalink
Quote: VideoBJ

The house edge for pass / don't pass bettor (assuming equal bet amounts) cannot be 1.41 x 2 -- that makes no sense.

not only does it make sense, but it is more or less correct...1.41 x 2 = 2.82 but the correct calculation is actually 1.41 + 1.36 since the HE for the dp is 1.36 or 2.78 (with rounding), iow the poster you were responding to had it pretty much correct as both results are of minuscule practical difference...

Quote: VideoBJ

A quick approximation is 1 bet out of 72 bets since you lose 1 bet, on average, in 36 rolls but since you are making 2 opposing bets at the same time, the edge is 1 in 72 bets as a guesstimate.

35 times (or 70 bets) that are non-box cars are a wash or net to zero. One bet wins, the other bet loses. It's the box car where one bet loses and one bet pushes, thus losing one bet in 72 bets.

but 12 loses on average 1/36 not 1/72 ...1/36 = ta dah! 2.78 (percent)

as others have also pointed out, the doey-don't is two bets, not one thus increasing the overall EV.

tom p
"You can't EXPECT to win. But you CAN play Tough"...tom p, 1974