Can A Math Wizard Please Help ???

I play craps at an indian casino that uses cards. There are two sets of one through six. Each set has a different colored back. The so-called shooter picks a number from one to six. The boxman who deals the game then burns that number of cards. Now one card of each color is turned up and this becomes the shooters roll. Can someone tell me the house edge and or if it would vary depending on the number the shooter chooses for the dealer to burn.Thank you for your time and effort.

Burns that number from each pile? (It must be one less if it's one through six.) It shouldn't matter - if it's a good shuffle, any card should have an equal chance of being in any position, just like a cut. All the math should be the same as well, since just like two fair dice, each pile has an equal and independent chance of turning up any number one through six.

The Wizard has a section of his variants page on what sounds like just this, as well as another on various other ways to play craps with cards. (Note that the "Card Craps" on another page is something else, where the "rolls" aren't quite fair by the standards of dice, but it sounds like they are in the game you're describing.)

There used to be a web page with a Java app that went into details on how to exploit the don't on this.

The edges are skewed, but not due to the player selecting how many cards to burn.

I think you're best off if you know how the skew on the edge works to always draw one card, and count the cards, and know how it affects the chances of getting a repeater.

The repeat numbers don't repeat as much with cards as they do with dice because it's not random. It's shuffled from a shoe with an equal number of cards for each value.

I haven't looked at searches on this in a couple of years, but it looks like there are more links on this that have popped up in the last year or so.

Here's one:

http://discountgambling.net/category/card-craps/

Yeah, I think this is the same guy .. it was just unavailable for a while. But definitely some interesting stuff if you like to exploit systems, this one is exploitable AFAIK.

The number is burned from all the cards which are shuffled in a machine before each so-called roll.

Quote: Ahigh

There used to be a web page with a Java app that went into details on how to exploit the don't on this.

The edges are skewed, but not due to the player selecting how many cards to burn.

I think you're best off if you know how the skew on the edge works to always draw one card, and count the cards, and know how it affects the chances of getting a repeater.

The repeat numbers don't repeat as much with cards as they do with dice because it's not random. It's shuffled from a shoe with an equal number of cards for each value.

Don't get his hopes up. What you're describing is only if the cards are dealt from the same deck, meaning they're not quite independent, plus often a few rounds are dealt before it's loaded into the shuffler, meaning it's countable. Since they have two different backs (even though they're shuffled together), they're independent in this case, and since they're freshly shuffled for every roll, it's not countable, either.

" It's shuffled from a shoe with an equal number of cards for each value. " where as dice have 6 different values on each. "