holst | hehe matlab is the best calculator you have |

Menkaure | i found one. TTCalc. its quite easy to handle and enough for my calculations. that other ones are much to hard for me. i do just a few simple calculations. :) 1-(48/50)*(47/49)*(46/48)*(45/47)*(44/46)=0.191836734693877551020408163 so thats my calculation. but looks like ~20% |

zoomzoom | hay |

Menkaure | my poker programm told me ~12%. mmh is this calculation correct? |

Kasadkad | For what? |

Menkaure | http://pastebin.ca/429083 this calculation of possibilty for a texas hold em poker game this was my first calculation then someone told me to do it that way 1-(48/50)*(47/49)*(46/48)*(45/47)*(44/46)=0.191836734693877551020408163 |

Kasadkad | I don't really get your question Oh If you have a pair already, what's the probability you get a card to make it three of a kind ? |

Menkaure | yes |

Chewie[] | So the probability of one of those 5 cards being an ace. If you have 2 in your hand. |

Menkaure | i have a programm wich calculate every option. but i am not sure if it is working correct and want to know how to calculate that yes |

Chewie[] | These aren't independent events, so I don't think you can just add. (each card of the 5 being drawn that is.. Those are not independent.) |

Kasadkad | Why is it not just 2/50? |

Chewie[] | Well it migh tbe. I'm not sure. |

Kasadkad | I don't know how the game works |

Chewie[] | It's gonna be the number of ways to have an ace divided by the number of ways you can draw 5 cards out of the 50. |

Kasadkad | You have two cards, and then you get another card? I mean |

Chewie[] | Say you have 2 aces in your hand. You take 5 cards randomly from the deck and lay them on the table. So the deck will have 45 left, 5 will be on the table, 2 in your hand. |

Menkaure | yes |

Chewie[] | Menkaure: You want the probability of having either 3 or 4 aces out, right? Or *only* 3 and not 4? :-P |

Kasadkad | Oh |

Menkaure | 3 is interesting |

Kasadkad | *sigh8 |

Menkaure | 4 is about ~1% |