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Dice Proposition (Posted on 2003-02-22) Difficulty: 5 of 5
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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionMichael2005-10-12 19:00:42
A tryarmando2005-05-01 23:14:00
Hints/TipsHintCharlie2003-02-23 04:46:10
Hints/Tipsone more mehtod.......Ravi Raja2003-02-23 04:03:52
Hints/TipsI think..........Ravi Raja2003-02-23 04:00:58
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