Question about craps
Darth
Quote: KB1no it is more house edge than just betting one or the other.
That's not true. The HE is about double.
Quote: Beethoven9thThat's not true. The HE is about double.
That is what I meant
Quote: KB1That is what I meant
Ah, my apologies. :)
Quote: BJMacay1Will a casino allow you to bet the minimum on the pass and the don't pass and take or lay odds. Giving the house no advantage.
Advantage - House
You can try your strategy using the Wizard's Craps game by clicking the "Games" tab above.
You might also invest a few dollars in WinCraps Pro which allows you to code any strategy and test it over a few or millions of rolls.
The graph will look something like steps leading to a basement.
Quote: Mission146If I were inclined to Doey-Don't, which I'm not, I'd Take Odds on 4&10, 5&9 and Lay Odds on 6&8, got to get me some of that free 2:1 and 3:2.
Would hate to roll 6 or 8 then seven out...
As the venerable Goatcabin once said, the thing to remember is that the EV of each bet cannot be erased by another bet. Thus, there is no point in looking at combinations of negative EV bets, perhaps seduced for the moment by what appears to be some sort of magical combination.
I really need to find that remark by Goatcabin and bookmark it.
Quote: Beethoven9thThat's not true. The HE is about double.
The edge is actually:
(-1 [lose pass line] * 1/36 [12 on comeout roll]) / 2 [total units bet] = 1.39%
Quote: mdsHow about 31.00 Doey/Don't with max odd on the Don't, 1.00 12? At least for 31 come out rolls.
At $31 you are going to lose $1 every come out roll. Whether you lose the bet on the 12, or a 12 rolls. A 12 pays 30-1, so that will replace $30 of your $31 lost on the pass line.
That gives
-1 / 63 = 1.587%
Make it 30 doey/don't so you lose $1 every non-12 and break even on a 12, and you get
(-1 * 35 / 36) / 61 [total bet] = 1.594%
Moral of the story - adding a higher edge bet to hedge is only going to result in a higher edge overall. (funny that hedge = higher edge)
Quote: BJMacay1Will a casino allow you to bet the minimum on the pass and the don't pass and take or lay odds. Giving the house no advantage.
Right. The doey-don't system means that you lose 1 unit when the 12 comes out on the comeout. So, on the comeout, you have a 1/36 chance of losing one unit with a HE of 1.38889% (1/72).
Quote: wudgedAt $31 you are going to lose $1 every come out roll. Whether you lose the bet on the 12, or a 12 rolls. A 12 pays 30-1, so that will replace $30 of your $31 lost on the pass line.
That gives
-1 / 63 = 1.587%
Make it 30 doey/don't so you lose $1 every non-12 and break even on a 12, and you get
(-1 * 35 / 36) / 61 [total bet] = 1.594%
Moral of the story - adding a higher edge bet to hedge is only going to result in a higher edge overall. (funny that hedge = higher edge)
Some casinos in vegas pay 31/1.
Quote: mdsQuote: wudgedAt $31 you are going to lose $1 every come out roll. Whether you lose the bet on the 12, or a 12 rolls. A 12 pays 30-1, so that will replace $30 of your $31 lost on the pass line.
That gives
-1 / 63 = 1.587%
Make it 30 doey/don't so you lose $1 every non-12 and break even on a 12, and you get
(-1 * 35 / 36) / 61 [total bet] = 1.594%
Moral of the story - adding a higher edge bet to hedge is only going to result in a higher edge overall. (funny that hedge = higher edge)
Some casinos in vegas pay 31/1.
Even then would make it
(-1 * 35 / 36) / 63 [total bet] = 1.543%
Quote: wudgedQuote: mdsQuote: wudgedAt $31 you are going to lose $1 every come out roll. Whether you lose the bet on the 12, or a 12 rolls. A 12 pays 30-1, so that will replace $30 of your $31 lost on the pass line.
That gives
-1 / 63 = 1.587%
Make it 30 doey/don't so you lose $1 every non-12 and break even on a 12, and you get
(-1 * 35 / 36) / 61 [total bet] = 1.594%
Moral of the story - adding a higher edge bet to hedge is only going to result in a higher edge overall. (funny that hedge = higher edge)
Some casinos in vegas pay 31/1.
Even then would make it
(-1 * 35 / 36) / 63 [total bet] = 1.543%
I realize 1.543% is a losing proposition. But it is only 1.543?
In life, if something is only a 1.54% edge I'm willing to take that gamble. With what I can afford to lose of course. Yes, you could say the same about the "right" side as well. But, how many times have you had a shooter roll 4-8 numbers and 7 out? Seems like more times than not. So, im in! What is the math on that one?
Quote: mdsIn life, if something is only a 1.54% edge I'm willing to take that gamble. With what I can afford to lose of course...What is the math on that one?
to say "I'm willing to pay for entertainment" is hard to make into a mathematical question
Quote: wudgedThe edge is actually...
Nope. If you bet $1 on PL and $1 on the Don't, you will lose $1 every 36 rolls. The HE is 2.78%.
Many other sources do not count ties in the house edge calculation, especially for the Don’t Pass bet in craps and the banker and player bets in baccarat. The rationale is that if a bet isn’t resolved then it should be ignored. I personally opt to include ties although I respect the other definition.
Quote: DeMangoExactly however that $1 is divided by $72!
Give Beethoven his due.
The house edge is defined as the ratio of the average loss to the initial bet. The house edge is not the ratio of money lost to total money wagered.
Read it here:WizardOfOdds
+1
