On average, how many times must a 6-sided die be rolled until the sequence 65 appears (i.e., a 6 followed by a 5)?
i tried it using this way. let say there are E rolls until we get 65. when we roll die first time,there is 1/6 probability that we get 6 and 5/6 probability of not getting 6.in case of 5/6, we have to start all over again,so now our number of steps are (E+1). if we get 6 in first roll , then there are 3 cases. (a) we roll a 5, (b), we roll a 6, or (c) we start all over again. in (a) we get our 65 so here our additional number of steps are 1 and probability is 1/6. in (b) we got 6 so will roll again to get 5 . so our additional number of steps now are (E-1) in (c) we have to start all over again so our additional number of steps are (E+2)
so my equation is E = (5/6)(E+1)+(1/6)(1+(1/6)(1)+(1/6)(E-1)+(4/6)(E+1))
by solving this , i got 40 but ans given ans is 30. where did i wrong?please help. thanks
To complete the calculation along the lines you started:
Let $E$ be the answer. Let $E_6$ be the expected number of turns it takes, given that the last roll was a $6$.
Then we get $$E=\frac 16\times (E_6+1)+\frac 56\times (E+1)\implies E = E_6+6$$ and $$E_6=\frac 16\times 1+\frac 16 \times (E_6+1)+\frac 46\times (E+1)\implies 5E_6=4E+6$$
This system is easily solved and yields $E=36$. (Note: it also yields $E_6=30$ which, possibly?, is the source of the error in the "official" answer).