Calculating California Card Craps

Let’s say we’re playing California card craps, and the table has six cards spread out, face up corresponding to the following die roll:

Die Roll Card Value
1 = 5
2 = 3
3 = 2
4 = 6
5 = 1
6 = 4

So if a 2-4 rolls (ostensibly a six), the actual number in play is a 9 (3+6). A hard 6 roll (3-3) would now be a hard 4 (2-2). The WOO website says this particular California rules craps carries the same HE of each bet as regular, bone rollin’ craps. Based on my hypothetical layout above, would someone be willing to go through the math for each of the 36 different roll combinations? As well, is there any value in quickly scanning the card layout, figuring out which rolls produce which card combinations and altering betting strategy with respect to place bets or how much to put in odds behind?

Again, no hurry, and please note I am not doubting in any way those who say the probabilities are the same, I’d just like to see the math step by step. Thanks in advance.

Gene

The math is the same whether you use the dice values or the card values as chosen by a roll of the dice. You are still picking two random numbers 1-6 independently of each other giving 36 possible outcomes.

The craps probability and outcomes don't change whether you use two six-sided dice, randomly draw from two decks of six cards, have a computer generate two random numbers 1-6, or even if you use dice to choose a two random cards from a deck of six cards. The probability distribution of the 36 possible outcomes is the same for all methods.

Maybe a thought experiment will save the trouble of someone laying out the math for something that I think can be intuitively grasped.

Take a pair of dice and imagine the faces are stickers with pips on them. Imagine pulling off a sticker and moving it to a different face.

Die Roll Card Value
1 = 5
2 = 3
3 = 2
4 = 6
5 = 1
6 = 4

1 = 5 Find the 5 sticker. Pull it off and move it to the face that has the 1 sticker.

5 = 1Take the 1 sticker and move it to the face where the 5 was.

2 = 3 Find the 3sticker. Pull it off and move it to the face that has the 2 sticker.

3 = 2 Take the 2 sticker and move it to the face where the 3 was.

4 = 6 Find the 6 sticker. Pull it off and move it to the face that has the 4 sticker.

6 = 4 Take the 4 sticker and move it to the face where the 6 was.


Done. You have two modified die. But they each still have faces 1 through 6 pips, just in different places. But they will still roll 1-6 each 1/6 of the time. So you will still have exactly the same distribution of rolls 1-12 when you throw two dice.

Quote: GenoDRPh

Again, no hurry, and please note I am not doubting in any way those who say the probabilities are the same, I’d just like to see the math step by step. Thanks in advance.

Gene
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This is pretty much just confirming the other replies, but as long as all six numbers have a different result, the probability of any particular number from 1 to 6 being generated from a die's roll is 1/6, just as it is in regular craps. The reason for the cards is, California's state Constitution doesn't allow the tribal casinos to run any table games that don't use cards in some way. (The state bends this slightly by treating Pai Gow tiles as cards.)

Well with the HE being the same, I think I would still place the 9-4 and 8