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Thread: Playing with Probability in Craps

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    Default Playing with Probability in Craps

    I'm sure that most of the players in here at one time or another in their lifetime have played the game of craps in the casino. As we know the game is played with 2 dice of equal dimensions whereas there are 36 different possible dice combinations. There are 6 variations of 7's in the game so that makes for 6:36 = 1:6 ratio. So that basically says that in a random roll a seven should roll 6 times out of 36 times or 1 in every 6 throws. Of course a seven isn't always going to be thrown every 1 in 6 rolls, which I would call variance. As you can see there are more combinations of 7's than any other dice combination.

    Any shooter who has a Seven to rolls ratio above or below 1:6 could possibly be controlling the dice. It's obvious that the house has a built in monetary edge to the game as some bets are not paid off at the true odds. Besides that there are many bets that can be made in the game in which the house edge over the player varies depending on what bet or bets the player is making.

    Now I'll explain a scenario: A random shooter establishes a point of 10, then proceeds to roll a 6,6,9, 11, 4 (which was thrown the easy way 1:3 or 3:1), 5, and is waiting for the dice to get passed to him via the stickman. Then a different player on the opposite end of the table lays odds on the 4 (betting wrong aka as the darkside) at $60 while simultaneously hopping the easy 4's for $4 ( a hedge on the lay bet) which pay at 15:1. That creates a scenario where if a 7 is thrown on the next roll, the player will win the lay bet minus a 5% commission known as the vig but will come out of the situation ahead. If the easy 4 gets thrown the lay bet will be taken down, but the easy 4's will return the $60 where the player could place his lay odds bet back up on the 4 if he wished, a tie. If the hard way 4 gets thrown the player will lose both bets but... it will be a 1:35 probability of that dice combination showing. However if the rolls are independent from each other the Hardway 4 could show on the next roll.

    6 combinations of 7's creates a positive result (the bigger lay bet wins, which is the purpose), 2 result in a tie (easy 4 gets thrown, with the option of going right back up on the lay odds bet), and 1 way to lose the bigger bet in which the Hardway 4 gets thrown. The theory is, is to get the bigger lay bet to get acted upon in a favorable way while reducing the houses chances of picking off the lay bet. The hedge bets protecting the lay bet can grind on a player if a 7 is not thrown, it isn't fool proof and timing or using dice control is critical. If the player wanted to hedge the hardway 4 he can do so by hopping the bet at 30:1. I know by laying odds on the 4 or 10 it can get expensive because they will only return half of the bet minus the vig and minus everything that was used to hedge the bigger bet. Perhaps a play on the 5 or 9 would be better.

    There are other variations of playing this while implementing dice control/random rollers and placing of the inside numbers 5,6,8,9 (take note that those 4 numbers alone make up 50% of the 36 possible dice combinations).
    Last edited by Blitzkrieg; July 1st, 2014 at 09:49 PM.

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