craps house edge question

[GrassHopperAA]

If you buy the ten for 25 dollars, what's the edge?

Buying the ten for 25 dollars, when you win you pay 1 dollar commission.

I calculate 1.32 percent, but I'm not sure if I'm correct.

[FearIt]

You bought the odds. They pay at true odds now so the edge is zero. But keep in mind you lost the money you bought the odds with.

At least from what I understand the definition of buying to be. (Craps isn't my game)

[SheetWise]

Normally the breakage goes to the house -- but if you can buy the 4/10 for $25, and only pay $1 vig when you win, then your expectation for the $25 is $25. You will pay $1 once for every time you lose twice.

You should expect to win 3 times and lose 6 times every 36 rolls -- so you will lose $3 every 36 rolls.

So -- you could look at it using only the 9 rolls that make a decision. Your total handle in those 9 rolls will be $225.

Winners (3 * [50-1])
less losers (6 * 25)

= (3 * [50-1]) - (6 * 25) = (147 - 150) = -3

The 9 decisions should take 36 rolls, so you can count it any way you want.

You've lost 3 units in 9 decisions which took 36 trials. Your handle is $225, so your loss rate is 3/225.

3/225 = .013333

[Jimbo]

You are correct, most joints will let you buy the 4 or 10 for up to $35 charging only $1, if you can afford the additional variance always buy for $35.

Jimbo

[mhcmarty]

[ QUOTE ]
only pay $1 vig when you win.


[/ QUOTE ]

How difficult is it to find a house on the strip in LV that allows you to pay the vig when you win. I know the Boyd properties downtown do this. Some places make you pay the $1 to book the 4/10.

[SheetWise]

It's a variation in a competitive craps environment like LV -- I have no idea how common it is, as I don't make buy bets.

I'm not aware of clubs that don't charge a vig on a partial wager -- I would charge a $2 vig on a $25 buy 4/10. If that can be pushed to $35, that's news to me -- but, then again, I don't make buy bets. If that's a common variation, and it means you can lay $70 for $35 on a 4/10 for $1 vig only when you win -- that's a nice change-up play.

[mingorama]

Vegas is unique in that you pay for the "buy" once you hit. Most other places you have to pay up front. From my recollection, you can buy for $1 up to $35 on the 4/10, and you might as well max it out if you can.

[SheetWise]

[ QUOTE ]
... you can buy for $1 up to $35 on the 4/10 ...

[/ QUOTE ]
I find that really interesting. As long as they're paying in dollar units, I wonder why it's not $39?

Does anyone know if this applies to minimum wagers only? Can you, on higher minimum tables for example, wager seven quarters and pay a $5 vig? Or nine hundred and pay a quarter? Is it based on units, table minimums -- or is it just the $15 break?

[jh12547]

[ QUOTE ]
[ QUOTE ]
... you can buy for $1 up to $35 on the 4/10 ...

[/ QUOTE ]
I find that really interesting. As long as they're paying in dollar units, I wonder why it's not $39?

Does anyone know if this applies to minimum wagers only? Can you, on higher minimum tables for example, wager seven quarters and pay a $5 vig? Or nine hundred and pay a quarter? Is it based on units, table minimums -- or is it just the $15 break?

[/ QUOTE ]

At Foxwoods its 1$ for every 20 you bet no matter how high you go. Thats why there are alot more 35$ bets then there are 40 as the 40 will cost 2$. In your question seven quarters will cost 8$ and 900 will cost 45$. The only breaks you may get are say you buy the 10 one roll for 35 and then wait a few rolls and buy the 4 for the same amount, 90% of the time you will pay 1 $ for each bet. In reality you should be charged 2 for the second bet because youre total on the 2 bets is 70 and the rule is 1$ for every 20.Depends on either how much of a douche the player is or even the floor for that matter.

[BigBluffer]

[ QUOTE ]
Normally the breakage goes to the house -- but if you can buy the 4/10 for $25, and only pay $1 vig when you win, then your expectation for the $25 is $25. You will pay $1 once for every time you lose twice.

You should expect to win 3 times and lose 6 times every 36 rolls -- so you will lose $3 every 36 rolls.

So -- you could look at it using only the 9 rolls that make a decision. Your total handle in those 9 rolls will be $225.

Winners (3 * [50-1])
less losers (6 * 25)

= (3 * [50-1]) - (6 * 25) = (147 - 150) = -3

The 9 decisions should take 36 rolls, so you can count it any way you want.

You've lost 3 units in 9 decisions which took 36 trials. Your handle is $225, so your loss rate is 3/225.

3/225 = .013333

[/ QUOTE ]

So if the house allows a $35 buy, taking only $1 juice when I win, would the house edge be only 0.952%?

Win = 3*(70-1) = 207
Lose = 6*35 = 210
Net = -3
Total Bet = 9*35 = 315
House Edge = (-3)/315 = 0.952%

Better than a pass line bet (ignoring so-called "free odds").