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I have a question on how some of the southern Cal Casino play Craps by flipping Cards after tossing dice. The question I have is, is the probability of getting a 7 (~.167) is the same when they flapping two cards summing the outcome? My opinion is the odds is not the same. I Don't claim to be a math wizard but this is how I figure the number:
The Carps game at one of the casino is:
-spreading six cards face up from A-6 into 6 open slots, one card per-slot.
-rolling two dice and take the dice value and select the card value as the outcome of the roll.
My math is:
It is 1/6 to select the first card (.167) and 1/5 to select the second card (.2), it does not matter what you selected for the first card, the second selection is 20% chance to select a card to 7 out. 20% is higher than ~16.7% for a 7 outcome in the regular craps.
Any thoughts?
Quote: DocThat is the OP's error that gave him an answer of 20% chance of a 7 rather than 16.667% chance. I was responding not to his question but to yours about what happens if both dice show the same number.
Yes.
Odds are exactly the same.
Imagine if they took the cards, got some tape, folded them, and made them into cubes. Each cube has 6 unique sides with 6 different numbers.
Now I *believe* that shuffling the cards after every roll shouldn't impact the odds at all. But that hasn't been my experience. I find that it's extremely difficult to catch a hot roll at Barona now and other regulars agree. The craps tables are significantly less busy now. Personally my loss rate went up by about a factor of 5x after they changed procedures. I can't explain it. Yet I've also seen one of the best rolls ever there a few months ago. Seems like the shuffling adds volatility but I can't explain the math behind it.
YMMV
Barona is the only casino that shuffles after every roll. Sycuan and Viejas only shuffle after a 7 out. I haven't played at Harrah's but I understand they use a completely different method involving two different colored dice and two sets of 6 cards based on the color of the dice.
If its not the hard way, then the selection is from 2 cards (i.e 1/6 odds to select the first and 1/5 odds to select the second). There are some sequences of the cards layout that some number outcome are not possible. For Example, if you are looking for a natural 9 outcome, and if the dice tossed with anything that has a 1 or 2 for a die, the outcome cannot be transposed to a 7 for the following layout. Position 1: "5" position 2: "2" Position 3: "3" Position 4: "1" position 5: "6" and position 6: "4". Thats 7 out on any natural 9 and 3.
If the first card selected was a 1 or 2, as long as the second card selection does not provide 3, then it will not be equal 7 for 7 out.
Quote: iedIf its hard way then only one card flipped.
If its not the hard way, then the selection is from 2 cards (i.e 1/6 odds to select the first and 1/5 odds to select the second). There are some sequences of the cards layout that some number outcome are not possible. For Example, if you are looking for a natural 9 outcome, and if the dice tossed with anything that has a 1 or 2 for a die, the outcome cannot be transposed to a 7 for the following layout. Position 1: "5" position 2: "2" Position 3: "3" Position 4: "1" position 5: "6" and position 6: "4". Thats 7 out on any natural 9 and 3.
If the first card selected was a 1 or 2, as long as the second card selection does not provide 3, then it will not be equal 7 for 7 out.
Here's a simple way that you can prove your assumption was wrong: what are the odds when the cards are laid out in sequence 1 through 6? Then you just look at the dice to determine the outcome just like in Vegas.
I must know you from one of these casino. Because what Barona did, they make me really thinking about the math. For a random shooter, the odds should be very close (appoxm .000x off) from the true odds for each number. Since there is just one set of cards, there are sequences of a layout that some particular number is not possible as you holding one die fixed.
For the reason of Barona, I feel the selection of the 2nd card is 20% given is not the hard way. We are not playing dice anymore, we are play cards. It is a different game of Craps. The payout is the same kind of payout as regular craps, but casino edge is very high.
Quote: PBguyHere's a simple way that you can prove your assumption was wrong: what are the odds when the cards are laid out in sequence 1 through 6? Then you just look at the dice to determine the outcome just like in Vegas.
It feel like 3 different rules are at play, the hard way rule, the natural 7 rule, ( its the worst layout), the any other combination of the layout.
Quote: RSIn Las Vegas, if the second die that settles lands on a number that is not the same as the other die (in other words, a hard way nor 2 nor 12 landed), then there's a 20% chance of a 7 occuring. Same as with the cards, eh?
No because the odds are set. 36 possible outcomes for tossing two dice (16.7% a 7 outcome). If you change the sequence after every roller or every time the dice will be toss, I feels like it added the coefficient of the game by flipping card.
Quote: iedMy math is:
It is 1/6 to select the first card (.167) and 1/5 to select the second card (.2), it does not matter what you selected for the first card, the second selection is 20% chance to select a card to 7 out. 20% is higher than ~16.7% for a 7 outcome in the regular craps.
Mine is this:
Assume one die is red and the other is blue.
Each number on the red die has a 1/6 chance of coming up, so each card has a 1/6 chance of being the red die's number.
If the red die's card is 1, then you have a 1/6 chance of the blue die's number being the 6 card.
If the red die's card is 2, then you have a 1/6 chance of the blue die's number being the 5 card.
If the red die's card is 3, then you have a 1/6 chance of the blue die's number being the 4 card.
If the red die's card is 4, then you have a 1/6 chance of the blue die's number being the 3 card.
If the red die's card is 5, then you have a 1/6 chance of the blue die's number being the 2 card.
If the red die's card is 6, then you have a 1/6 chance of the blue die's number being the 1 card.
The probability of rolling a 7 is (1/6 x 1/6) + (1/6 x 1/6) + (1/6 x 1/6) + (1/6 x 1/6) + (1/6 x 1/6) + (1/6 x 1/6) = 1/6.
Your math would apply only if there was a rule saying that the same card could not be selected twice - in other words, all rolls of doubles would be ignored.
Side note: any casino with this kind of craps needs to think twice before asking for any expansion or anything else that would cause the compact with the California government to be renegotiated; the last tribe that did this ran afoul of the new "no roulette and no games that use actual dice" policy that has affected the newer casinos. Of course, they could always switch to a version of craps that used only cards, or even one with "electronic dice" - just not one with physical dice. (The same applies to Sic Bo.)
Quote: iedNo because the odds are set. 36 possible outcomes for tossing two dice (16.7% a 7 outcome). If you change the sequence after every roller or every time the dice will be toss, I feels like it added the coefficient of the game by flipping card.
You feel? Or it does? Big difference. If you feel it changes, OK -- I'm not going to say that you're wrong (you feel what you feel). But that does not make it so. The odds do not change.
Quote: DocThat is the OP's error that gave him an answer of 20% chance of a 7 rather than 16.667% chance. I was responding not to his question but to yours about what happens if both dice show the same number.
Yes, thank you. I was attempting, in a round about way, to help him realize his mistake himself by thinking about the answers to my questions.