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?
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 26, 812 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 20051012 19:00:42 