Let $f(n)$ be the minimal number of moves necessary for Bob to guarantee winning. We will show that if $n = 5$ Bob cannot win and
$$f(n) = \begin{cases} 2 & \text{if } n = 3, 4, n \ge 12 \\ 3 & \text{if } 6 \le n \le 11 \end{cases}$$

Let the vertices be $v_1, v_2, \dots, v_n$ in clockwise direction. The Bob's choice and Alice's answer at move number $i$ will be denoted by $B(i)$ and $d(i)$ respectively. Below always $B(1) = v_1$ and the obvious cases when $d(i) = 0$ will be omitted.

$n = 3$. If $d(1) = 1$, $B(2) = v_2$ and Bob locates the bead after the second move. Therefore, $f(3) = 2$.

$n = 4$. If $d(1) = 2$ then Bob locates the bead after the first move. If $d(1) = 1$, $B(2) = v_2$ and Bob locates the bead after the second move: for $d(2) = 1, 2$ the bead is located at $v_3$ or $v_4$ respectively. Therefore, $f(4) = 2$.

$n = 5$. If $d(1) = 2$, then w.l.o.g. $B(2) = v_2, v_3$. If $B(2) = v_2$, $d(2) = 2$ is possible. If $B(2) = v_3$ then $d(2) = 1$ is possible. In either case the location is not possible and the situation repeats. Therefore, Bob cannot win for $n = 5$.

$n = 6$. If $d(1) = 1$ then $B(2) = v_2$ and Bob locates the bead after the second move. If $d(1) = 2$, then w.l.o.g. $B(2) = v_1, v_2, v_3, v_4$. If $B(2) = v_1, v_2$ or $v_4$ then $d(2) = 2$ is possible and two moves will not be enough for winning. $B(2) = v_3$. If $d(2) = 2, 3$ then Bob locates the bead. If $d(2) = 1$ then $B(3) = v_4$ and wins after the third move. If $d(1) = 3$ Bob locates the bead after the first move. Therefore, $f(6) = 3$.

$n = 7$. If $d(1) = 1$ then $B(2) = v_2$ and locates the bead after the second move. If $d(1) = 2$, then w.l.o.g. Bob can choose $v_1, v_2, v_3$ or $v_4$. If Bob chooses $v_1, v_2$ or $v_4$ then $d(2) = 2$ is possible and two moves will not be enough for winning. $B(2) = v_3$. If $d(2) = 2$ then Bob locates the bead at $v_5$. If $d(2) = 1$ then $B(3) = v_2$ and Bob wins after the third move. If $d(2) = 3$ then $B(3) = v_1$ and Bob locates the bead after the third move. Therefore, $f(7) = 3$.

$n = 8$. If $d(1) = 1$ then $B(2) = v_2$ and Bob locates the bead after the second move. If $d(1) = 2$ then w.l.o.g. Bob can choose $v_1, v_2, v_3, v_4$. If $B(2) = v_1, v_2, v_4$ then $d(2) = 2$ is possible and two moves will not be enough for winning. $B(2) = v_3$. If $d(2) = 4$ Bob locates the bead at $v_7$. If $d(2) = 1, 3$ Bob chooses $v_1$ and wins after the third move. If $d(1) = 4$ Bob locates the bead after the first move. Therefore, $f(8) = 3$.

$n = 9$. If $d(1) = 1, 4$ then Bob chooses $B(2) = v_3, v_4$ respectively and locates the bead after the second move. If $d(1) = 2$ then $B(2) = v_4$ locates the bead after the second move. If $d(1) = 3$ two moves will not be enough for winning. $B(2) = v_4$. If $d(2) = 3$, Bob locates the bead after the second move. If $d(2) = 1$ then $B(3) = v_2$; If $d(2) = 2$ then $B(3) = v_1$ locates the bead. Therefore, $f(9) = 3$.

$n = 10$. If $d(1) = 1, 4$ then Bob chooses $B(2) = v_3, v_4$ respectively and locates the bead after the second move. If $d(1) = 2, 3$ two moves will not be enough for winning. If $d(1) = 2$ then $B(2) = v_4$ for all possible values of $d(2)$ except $d(2) = 4$ Bob locates the bead, for $B(2) = v_4$ there are two possibilities $v_8$ and $v_{10}$ and Bob finishes by $v_1$. Therefore, $f(10) = 3$.

$n = 11$. As above, if $d(1) = 1, 2, 4, 5$ then Bob locates the bead after the second move. If $d(1) = 3$ then two moves will not be enough for winning. $B(2) = v_4$. If $d(2) = 1, 2, 4, 5$ then $B(3) = v_1$ locates the bead. If $d(2) = 3$, $B(3) = v_2$ locates the bead. Therefore, $f(11) = 3$.

$n \ge 12$. If $d(1) < \lfloor \frac{n}{2} \rfloor$ then $B(2) = v_{d+2}$, if $d(1) \ge \lfloor \frac{n}{2} \rfloor$ then $B(2) = v_d$ locates the bead after the second move. Therefore, $f(n) = 2$.