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From, this you can get the average of 5.294.

From, this you can get the average of 5.294 as the sum of \$n p(n)\$.

for \$n>1\$.

$$ p(n) = q_n \prod_{i=0}^{n-1} (1-q_i) $$ where \$0 \le q_i \le 1\$ is the probability of success on the i'th roll. And the convention \$q_0=0\$ (you can't have succeeded before your first roll)

        q_0=0
       /   \
     q_1  (1-q_1)                    # first roll
            /  \
           q_2 (1-q_2)               # second roll
                /   \ 
               q_3  (1-q_3)          # third roll
                      ...

for \$n>1\$, and \$p(n)\$ indicates the probability that the success occurs on the n'th roll.

$$ p(n) = q_n \prod_{i=0}^{n-1} (1-q_i) $$ where \$0 \le q_i \le 1\$ is the probability of success on the i'th roll, with the convention \$q_0=0\$ (you can't have succeeded before your first roll)

        (1-q_0)=1
       /   \
     q_1  (1-q_1)                    # first roll
            /  \
           q_2 (1-q_2)               # second roll
                /   \ 
               q_3  (1-q_3)          # third roll
                      ...