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 Dice Proposition (Posted on 2003-02-22)
If a pair of dice are thrown repeatedly until a 7 is thrown, what is the probability that all the other possible numbers from 2 through 12 will have been thrown by the time the 7 has been thrown? As a preliminary, what is the probability that 2, 11 and 12 will all have been thrown before the first 7?

 Submitted by Charlie Rating: 4.2000 (5 votes) Solution: (Hide) In the preliminary, P(2 and 11 and 12) = P(2) + P(11) + P(12) – P(2 or 11) – P(2 or 12) – P(11 or 12) + P(2 or 11 or 12). You can verify the formula by seeing how many times each possible truth value is added in or subtracted out. All except the desired probability are cancelled by equal adding and subtracting. The individual probabilities are based on the number of ways of the sought combinations divided by that plus the 6 ways of getting a 7: Thus 1/7 + 2/8 + 1/7 – 3/9 – 2/8 – 3/9 + 4/10 = 2/105 = 0.0190476… The main problem has a similar formula, involving 1023 terms, and has the result 0.005257704… By the way this corresponds to 189.197 to 1 odds against.

 Subject Author Date Solution Michael 2005-10-12 19:00:42 A try armando 2005-05-01 23:14:00 Hint Charlie 2003-02-23 04:46:10 one more mehtod....... Ravi Raja 2003-02-23 04:03:52 I think.......... Ravi Raja 2003-02-23 04:00:58

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