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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?

 See The Solution Submitted by Charlie Rating: 4.2000 (5 votes)

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 Solution Comment 5 of 5 |
This is easy, really.  The chance that a 2 comes before 7 is the number of ways to get 2 divided by the number of ways to get either 2 or 7, that is, 1/7, since 1 time there will be 2 and 6 times there will be 7.  Similarly, for 11, it's 2/8, since there'll be 2 11's and 6 7's.  All these are separate probabilities.  So:
For 2, 11, and 12, the probability is 1/7*2/8*1/7 = 1/196 and I didn't even need a calculator.
For 2-6, 8-12 we have 1/7*2/8*3/9*4/10*5/11, squared to get the other 5; this multiplication comes out to 1/7*1/4*1/3*2/5*5/11 = 1/(7*2*3*11) = 1/462, but as I said this must be squared:
00462
00462
00924
02772
18480
22176

The chance of all other numbers appearing before a 7 is 1/22176.

 Posted by Michael on 2005-10-12 19:00:42

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