Probability puzzle on three people rolling a die Given three players A, B, C, taking turns to roll a fair die in the order A B C A B C A B ....  What's the probability of that the occurrence of second 6 is in A's turn? My approach was  Let $F_6$ and $S_6$ be the r.v's for the first and second occurrences of 6 respectively.  $$\begin{align}P(S_6 = A) &amp;= P(S_6 = A \cap F_6 = A)  + P(S_6 = A \cap F_6 = B)   + P(S_6 = A \cap F_6 = C)\\  &amp;= P(F_6=A)P(S_6=A | F_6=A) +P(F_6=B)P(S_6=A | F_6=B)\\ &amp;+P(F_6=C)P(S_6=A | F_6=C)\\ &amp;= 3(\frac{5}{6} P(F_6 = A))^2 \end{align}\\$$  Here I am using the following relations   $$\begin{align}P(F_6=B) &amp;= \frac{5}{6}P(F_6=A),\\ P(F_6=C)&amp;=(\frac{5}{6})^2P(F_6=A),\\\text{and}\quad P(S_6=A | F_6=A)&amp;=(\frac{5}{6})^2P(F_6=A)\\P(S_6=A | F_6=B)&amp;=\frac{5}{6}P(F_6=A)\\P(S_6=A | F_6=C)&amp;=P(F_6=A)\end{align}$$  My reasoning for the first relation is that  $P(\text{first occurrence of B in seq ABCABC...}) = P(\text{A doesnt throw 6}) P(\text{first occurrence of B in seq BCABCABC...})$\  By symmetry $P(\text{first occurrence of B in seq BCABCABC...})=P(\text{first occurrence of A in seq ABCABCAB...})$  I am unsure if my method is correct. Please help me verify this. Thanks in advance.
What is the probability that a person is the second person to roll a 6 for the first time? Suppose $A$, $B$, and $C$ roll a die in turns. What is the probability that $A$ is the second person to roll a $6$ for the first time?
Is there any canon information in Harry Potter that proves one way or another whether a wizard always owns only one wand as a matter of routine?  What I am talking about is a wizard buying and owning one wand and also a second wand (for backup, or simultaneous use, or separate purpose, or ceremonial+work wand), akin to people buying more than one pair of boots, guns or hammers.  A proof would be either canon info showing someone owning 2 wands simultaneously; or a statement showing why that does not happen except in rare exceptions (see below).    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