The owner of a casino fakes two dices so that in dice $$A$$ we can never get a $$6$$ (and get twice as many ones), and in dice $$B$$ we never get a $$5$$ (and twice as many twos).

- Fill in the following table of probabilities for every dice:

result dice A | probability |

$$1$$ | ? |

$$2$$ | ? |

$$3$$ | $$1/6$$ |

$$4$$ | ? |

$$5$$ | ? |

$$6$$ | 0 |

result dice B | probability |

$$1$$ | ? |

$$2$$ | ? |

$$3$$ | $$1/6$$ |

$$4$$ | ? |

$$5$$ | ? |

$$6$$ | ? |

See development and solution

### Development:

- The impossible events have zero probability $$(A=6, B=5)$$. As we are been told, there is twice the probability of observing events $$A=1$$ and $$B=2$$ (probability $$2/6$$):

result dice A | probability |

$$1$$ | $$2/6$$ |

$$2$$ | $$1/6$$ |

$$3$$ | $$1/6$$ |

$$4$$ | $$1/6$$ |

$$5$$ | $$1/6$$ |

$$6$$ | $$0$$ |

result dice B | probability |

$$1$$ | $$1/6$$ |

$$2$$ | $$2/6$$ |

$$3$$ | $$1/6$$ |

$$4$$ | $$1/6$$ |

$$5$$ | $$0$$ |

$$6$$ | $$1/6$$ |

### Solution:

result dice A | probability |

$$1$$ | $$2/6$$ |

$$2$$ | $$1/6$$ |

$$3$$ | $$1/6$$ |

$$4$$ | $$1/6$$ |

$$5$$ | $$1/6$$ |

$$6$$ | $$0$$ |

result dice B | probability |

$$1$$ | $$1/6$$ |

$$2$$ | $$2/6$$ |

$$3$$ | $$1/6$$ |

$$4$$ | $$1/6$$ |

$$5$$ | $$0$$ |

$$6$$ | $$1/6$$ |