Ashley, Bon, Carol, and Doug are rescued from a desert island by a pirate who forces them to play a game. Each of the four, in alphabetical order by first

To find the probability that Doug survives, I'm going to find the probability that he walks the plank and subtract that number from 1.

 

For Doug to die in the first round, the other three must live and he must roll either an 8 or a 9.

 

First, what's the probability that anyone rolls either an 8 or a 9?

An eight can be rolled in 5 ways; either 6&2, 5&3, 4&4, 3&5, or 2&6.

A nine can be rolled in 4 ways; either 6&3, 5&4, 4&5, 3&6.

So, there are 9 different ways that anyone can lose out of 36 possible combinations: 9/36  =  1/4.

 

This also means that here is a 3/4 probability that the person wins.

 

For Doug to die on the first round, the other three must live and he must die:  

     (3/4)(3/4)(3/4)(1/4)  =  (3/4)³(1/4)  =  27/256

 

For Doug to die on the second round, all must live on the first round ( (3/4)⁴ ), the other three must live on the second round and he must die on the second round:  (3/4)⁴ · (3/4)³(1/4)  =  (3/4)⁷(1/4)

 

For Doug to die on the third round, all must live on the first two rounds ( (3/4)⁸ ), the other three must live on the third round and he must die on the third round:  (3/4)⁸ · (3/4)³(1/4)  =  (3/4)¹¹(1/4)

 

This can go on forever; what we must do is to add all these fractions together:

     (3/4)³(1/4)  +  (3/4)⁷(1/4)  +   (3/4)¹¹(1/4)  +  ...

 

This is an infinite geometric sequence whose first terms is  (3/4)³(1/4)  and whose common ratio is  (3/4)⁴.

 

The formula for the sum of an ininite geometric sequence with a ratio in the range:  -1 < r < 1 is:

     Sum  =  a / ( 1 - r)          where a is the first term and r is the common ratio.

     Sum  =  [ (3/4)³(1/4) ] / [ 1 -  (3/4)⁴ ]  =  27/175

 

But, this is the probability that he loses; the probability that he lives is  1 - 27/175  =  148/175.