Sanderson M. Smith

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RED DI

(Casino game constructed by Cate students Virginia, Ethan, and Alexa)

Here is the game description as written by the inventors:

Our game is called Red Di. All players start with a numerical value of 13. Seems unlucky but it isn't necessarily. Each player then rolls one red die and adds the number on the die to 13. Each player then rolls the die once more with the goal of reaching 20. Players who go over 20 lose. The players under 20, but over the dealer's number win. The dealer starts with an equal value of 13. Dealers roll the di as well. Dealers must stay when they reach a value of 18. A dealer then takes takes all the money from all of the losers and matches the bets of the winners. Out of the winners the one closest to 20 is matched 2:1. And the others are matched 1:1. If more than one player gets the highest number each player is matched 1:1. If the players tie with the house, no players win. If the dealer goes over 20 all players which do not go over as well, will end up getting their money matched. The player closest to 20 will get a 2:1 matching value. Unfortunately the game will not make as much for the house as some other games, but this could end up making our game more popular amongst the players.

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This game is definitely interesting. However, from my perspective, there are too many possible outcomes from this game to come up with precise mathematical expectations. One could get results through simulation. Below are some probabilities associated with the game of Red Di.

As I understand the rules, a player must roll the die twice after starting with a total of 13. This is equivalent to rolling two dice and adding the sum to 13. A total of 7 or less will result in a final total of no more than 20. Using the chart at the right, a player's probability of getting a total that does not exceed 20 is 21/36, or 58.33%

The rules seem to indicate that the dealer does not have to roll a die twice. If dealer rolls a 5 or 6 on first roll, he "stays" with his total of 18 or 19. If dealer rolls less than 5 on first roll, he must roll again. The only way the dealer can exceed 20 is to

  • Roll 4 (for total = 17), then roll 4, 5, or 6. Probability = (1/6)(1/2) = 1/12 = 8.33%.
  • Roll 3 (for total = 16), then roll 5 or 6. Probability = (1/6)(1/3) = 1/18 = 5.56%.
  • Roll 2 (for total = 15), then roll 6. Probability = (1/6)(1/6) = 1/36 = 2.78%.

The probability that the dealer will exceed 20 is 1/12 + 1/18 + 1/36 = 16.67%. Hence, the probability that the dealer will get 20 or less is 100% - 16.67% = 83.33%.

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I don't know if the probabilities above would be useful in coming up with expectations for Red Di. Any comments or suggestions from readers would be appreciated.

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