Question 1094875
<pre>
I'm thinking that's wrong and that you should make a probability 
tree diagram of all possible outcomes, with the probability of 
going along each branch (line) written on the branch:

{{{drawing(800,300,-1,15,-1,10,

green(line(6,-.75,10,-.75)),green(line(6,-.75+1.25,10,-.75+1.25)),

locate(10.8,8.9,sum=2), locate(10.8,7.9,sum=3),locate(10.8,6.8,sum=highlight(4)),locate(10.8,5.9,sum=highlight(5)),locate(10.8,4.9,sum=3),locate(10.8,3.9,sum=highlight(4)),locate(10.8,2.8,sum=highlight(5)),locate(10.8,1.7,sum=highlight(6)),locate(10.8,.9,sum=3),locate(10.8,-.35,sum=highlight(4)), 

line(0,4,5,6.5),line(0,4,5,3),line(0,4,5,.5),line(0,4,5,-.5),
line(6,7,10,8.5),line(6,7,10,7.5),line(6,7,10,6.5), line(6,7,10,5.5),
line(6,3,10,4.5),line(6,3,10,3.5),line(6,3,10,2.5),line(6,3,10,1.5),

locate(5.3,7.1,1), locate(5.3,3.3,2), locate(5.3,.8,3),locate(5.3,.8-1.2,4),
locate(10.2,8.9,1), locate(10.2,7.9,2),locate(10.2,6.8,3),locate(10.2,5.9,4),
locate(10.2,4.9,1), locate(10.2,3.9,2), locate(10.2,2.8,3), locate(10.2,1.7,4),

locate(2.3,6,"1/4"),locate(2.6,4,"1/4"),
locate(2.8,2.5,"1/4"),locate(3.6,1.3,"1/4"),
locate(8.3,8.6,"1/4"),locate(8.3,7.9,"1/4"),
locate(8.3,7.3,"1/4"),locate(8.3,6.6,"1/4"),
locate(-.7,4.3,START), 

locate(8.3,8.6-4,"1/4"),locate(8.3,7.9-4,"1/4"),
locate(8.3,7.3-4,"1/4"),locate(8.3,6.6-4,"1/4") 


 )}}}

The successful outcomes are boxed in red.

The probability of getting the highest highlighted 4 on the 
right is to go along the branch from START to 1 (roll 1 first),
with probability 1/4, and then go along the branch from 1 to 
the 4 (roll 4 second), with another probability of 1/4.  So
the probability of getting the 4 by rolling a 1 first and then
a 4, is gotten by multiplying the 1/4 to go from START to 1, times
the 1/4 to go from the 1 to the 4. It is similar for the other 
outcomes on the far right

So 

the probability of the upper highlighted 4 is (1/4)(1/4) = 1/16.
the probability of the upper highlighted 5 is (1/4)(1/4) = 1/16.
the probability of the other highlighted 4 is (1/4)(1/4) = 1/16.
the probability of the other highlighted 5 is (1/4)(1/4) = 1/16.
the probability of the highlighted 6 is (1/4)(1/4) = 1/16.

plus

the probability of the highlighted 4 on the bottom, the case where
a 4 was rolled first and then you stopped is 1/4, so the desired
probability is the sum of all those probabilities:

1/16 + 1/16 + 1/16 + 1/6 + 1/16 + 1/4 = 9/16.

Edwin</pre>