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Category: | Probability |

Submitted By: | Zag24 |

Fun: | (1.91) |

Difficulty: | (2.97) |

In the Massachusetts Megabucks lottery, six different numbers from 1 to 42 (inclusive) are selected. When you buy a ticket, you can ask for a "quick pick" in which the computer chooses the numbers for you, and you can purchase up to five games on a single ticket. We'll assume that the computer's random number generator is fair, giving each possible combination an equal probability of being chosen.

1. If I "quick pick" for two games, what are the chances that the two games have the same combination of numbers?

2. If I "quick pick" for five games (one five-game ticket), what are the chances that there are two games on that ticket with the same combination?

3 (The toughie). How many five-game quick-pick tickets would I have to buy in order to have a greater than 50% chance of having at least one ticket with two games on it that match exactly?

Show Hint | Show Answer |

Posted by Paladin | 04/23/09 |

Probabilities always make me stop and think. I haven't had any probability courses, but I can usually muddle through with the math I do know and get a decent shot at it. This was fun but very challenging. Well done! | |

Posted by Shadows | 04/23/09 |

Well . . . I think I'll just look on the bright side. I got the first one! :lol: I see you put quite some work into this. It's a nice teaser. When I saw it in proofreading, I rated it fun and hard. Good job! | |

Posted by bradon182001 | 04/30/09 |

Waaaaay over my head.Excellent teaser. Thanks for posting. :o | |

Posted by gooberbaby1 | 06/11/10 |

I would have just put down an answer,NOT an answer & a HUGE explanation :oops: :oops: | |

Posted by racoonieboy | 07/25/10 |

Actually, the explanation... well, explained a lot and I like it! | |

Posted by Zag24 | 08/04/10 |

I'm betting, goober, that you wouldn't have put the answer at all, since I bet you couldn't have come up with it. | |

Posted by Jaypuzzle12 | 07/21/11 |

THe answer was well explained. Good job! :D | |

Posted by ali_p | 09/13/12 |

I don't think this is right...what accounts for the fact that you're more likely to match 2 games when you have 5 games per ticket instead of just having to match one game to another? I think you're using the wrong "P"...otherwise, what would account for it being less likely to match 3 out of 5 versus just 2 out of 5?? | |

Posted by Zag24 | 02/15/13 |

all_p, sorry I didn't see your question until just now. In the explanation, step 2, there are four fractions we are multiplying: (5245785 * 5245784 * 5245783 * 5245782) / 5245786 ^ 4 That is, we are confirming that all 5 combos on the ticket are different. If we had only, say, three combos on one ticket, that step would only be (5245785 * 5245784) / 5245786 ^ 2 and if there were only 2 combos on one ticket, it would be the same as the first step, 5245785 / 5245786 Remember that these are the chances that all the combos are different. The chances that there are two the same is 1 minus this. |

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