Jumpy, Patient Frog Named Sally And A Math Problem About Her Life-IV

$This\quad is\quad very\quad similar\quad to\quad the\quad random\quad walk.\\ We\quad know\quad that\quad D\quad is\quad at\quad a\quad distance\quad of\quad 3\quad sides\\ or\quad units\quad from\quad A\quad and\quad that\quad it\quad doesn't\quad matter\quad how\\ the\quad frog\quad gets\quad there,\quad so\quad the\quad distance\quad squared\quad is\\ going\quad to\quad be\quad 9.\quad In\quad the\quad random\quad walk,\quad the\quad expected\\ value\quad of\quad D_{ n }^{ 2 },\quad i.e,\quad the\quad square\quad of\quad the\quad distance\quad travelled\\ form\quad the\quad origin\quad is\quad nothing\quad but\quad N.\quad So,\quad N\quad =\quad lifetime\\ in\quad seconds\quad =\quad 9.\\ \therefore \quad In\quad minutes\quad it's\quad gonna\quad be\quad 9/60=\boxed { 0.15 } \quad$

From the previous problem , we know that the probability of landing on $D$ after $2k+1$ seconds is $\frac{1}{4} \times (\frac{3}{4})^{k-1}$ .

Sally's expected lifespan will then be:

$N=\sum_{k=1}^\infty (2k+1) \times \frac{3^{k-1}}{4^k}$ .

This can be calculated using a generating function like so:

$f(t)=\frac{1}{3} \times \sum_{k=1}^\infty \frac{3^k}{4^k}t^{2k+1}=\frac{t^3}{4} \times \sum_{k=0}^\infty (\frac{3t^2}{4})^k=\frac{t^3}{4-3t^2}$ .

Indeed, for $N=f'(1)=9$ seconds, or $\boxed {0.15}$ minutes.