Swingline
Swingline
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November 16th, 2015 at 5:57:18 AM permalink
I went to a California Casino and saw a version of craps where dice are used but the outcome of the game is determined by cards. For example, the dice roll 2-3... the cards in spots two and three are then selected and the combination of those two cards is the actual outcome. My question is how would mucking the cards and resetting them after each roll effect the game's hold? I am thinking that on a real craps game if the player rolls a 2-3, he or she must re roll that same total to either win or lose. On this California game, a 2-3 rolled might equal a five or ten or six and so on...... I am sorry if this is not crystal clear... I am not very good a explaining myself in writing.
MathExtremist
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November 16th, 2015 at 10:48:42 AM permalink
Two differently-colored dice, yes? If the cards are red 1..6 and blue 1..6 to correspond to red and blue dice, mucking the cards has no effect compared to leaving them where they are.

Technically, the cards could be placed in order, in which case the game would look and feel exactly like craps because you wouldn't need to do any outcome mapping. But I think the commission would frown upon that.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
Ayecarumba
Ayecarumba
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November 16th, 2015 at 12:54:10 PM permalink
I recall that the six cards corresponding to each die were reshuffled after each roll, but even if they weren't , there isn't an advantage since the mapping is still one die face to one card. It's just odd that a roll of 3&4 could be a hard 4.
Simplicity is the ultimate sophistication - Leonardo da Vinci
Swingline
Swingline
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November 16th, 2015 at 11:13:42 PM permalink
In this game the die are the same color... I am just looking to find out if there is an equation that could show the risk/benefit of mucking after each roll.
Doc
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November 17th, 2015 at 6:04:03 AM permalink
Quote: Ayecarumba

It's just odd that a roll of 3&4 could be a hard 4.

I think this thread has confused two different versions of craps that are played in California tribal casinos and use both dice and cards to establish the outcome of the roll. The OP is describing the version in which six cards (Ace through 6) are shuffled and then dealt to six spaces numbered 1 through 6 in front of the box man. My recollection is that, in this version of the game, the cards are not reshuffled until the dice progress to a new shooter.

Upon the roll of a standard pair of dice, each of the resulting die faces is matched to one of the six numbers printed on the felt in front of the box man. That leads to two cards that specify the result, with the box man sliding those two cards forward to identify the result. That is, unless the two physical dice show matching faces, in which case the physical hardway gives a result that is also a hardway, with only a single card being slid forward. So no, a 3&4 roll cannot provide a result that is a hard 4. But a roll of 2&2 could indeed be a result of a hard 4, if the card shuffle had placed the 4 card in space 2.

I would have to re-read the original post, but I don't think the math is any different from standard craps. It just makes things a little more complex for the dice setters -- they have to think in terms of the die faces being rearranged according to the order in which the cards are dealt.
Ayecarumba
Ayecarumba
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November 17th, 2015 at 9:13:23 AM permalink
Quote: Doc

I think this thread has confused two different versions of craps that are played in California tribal casinos and use both dice and cards to establish the outcome of the roll. The OP is describing the version in which six cards (Ace through 6) are shuffled and then dealt to six spaces numbered 1 through 6 in front of the box man. My recollection is that, in this version of the game, the cards are not reshuffled until the dice progress to a new shooter.

Upon the roll of a standard pair of dice, each of the resulting die faces is matched to one of the six numbers printed on the felt in front of the box man. That leads to two cards that specify the result, with the box man sliding those two cards forward to identify the result. That is, unless the two physical dice show matching faces, in which case the physical hardway gives a result that is also a hardway, with only a single card being slid forward. So no, a 3&4 roll cannot provide a result that is a hard 4. But a roll of 2&2 could indeed be a result of a hard 4, if the card shuffle had placed the 4 card in space 2.

I would have to re-read the original post, but I don't think the math is any different from standard craps. It just makes things a little more complex for the dice setters -- they have to think in terms of the die faces being rearranged according to the order in which the cards are dealt.



Here's an older thread on the topic.

I was recently at San Manuel, and they don't even use dice. Two cards (A-6) come out of a shuffler to make a "roll". They still use a craps table, but the "stickman" is completely misnamed.
Simplicity is the ultimate sophistication - Leonardo da Vinci
Swingline
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November 23rd, 2015 at 5:19:11 AM permalink
Thanks for all the help
RS
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November 23rd, 2015 at 9:34:09 AM permalink
The odds are exactly the same as in regular (Las Vegas) craps.
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