Question
A game marshal publicly displays two numbered dials, each marked 0,1,2,3,4,5. First the marshal chooses a starting number A uniformly from the six marks. Separately, they draw an independent offset K, also uniform on 0 through 5, and set the second displayed number to \(B=(A+K)\) modulo 6. Modulo 6 means that counting beyond 5 wraps around to 0.
Do the displayed numbers A and B turn out to be independent even though B was constructed using A? Find \(P(A=a,B=b)\) for any fixed \(a,b\) and the probability that \(B=2\) after seeing \(A=4\).
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Hint 1
For a fixed starting value a, how many of the six offsets produce a particular second value b?
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Solution
A and K are independent uniform variables on the six residues 0,...,5. Fix any displayed pair (a,b). The modular equation \(b\equiv a+K\pmod6\) has exactly one solution \(K\in\{0,\ldots,5\}\), namely \(K=(b-a)\) modulo 6. Therefore
\(P(A=a,B=b)=P(A=a,K=(b-a)\bmod6)=(1/6)(1/6)=\frac1{36}\).
Summing this joint probability over six values of the other variable yields \(P(A=a)=P(B=b)=\frac16\). Because \(\frac1{36}=(1/6)(1/6)\) for every \(a,b\), the outputs A and B are independent.
In particular, after seeing \(A=4\), the second dial shows 2 precisely when \(K=(2-4)\bmod6=4\), which occurs with probability \(\frac16\). Dependence cannot be inferred merely from the fact that one random variable appears in another's construction: the independent uniform offset hides all information about the base modulo 6.
Assumptions
Both underlying draws are discrete uniform on all six residues, and are mutually independent.
The offset is chosen afresh for each game, and the modulus is exactly 6.
Final Answer
A and B are independent; every pair \((a,b)\) has probability \(\frac1{36}\), both marginals are uniform at \(\frac16\), and \(P(B=2\mid A=4)=\frac16\).