Problem

A fair coin and fair die are used. A wins on \(H\) with die \(1\text{--}4\) or \(T\) with die \(1\text{--}2\); otherwise B wins. Compute A's probability with weighted branches, then evaluate fairness.

Big Picture

What this problem is really about

A win can be reached through more than one complete branch, and each branch must carry the probability of reaching it. Multiply within a path, then add the disjoint paths that lead to the same participant. Only those final participant totals—not the unweighted rates on individual branches—can be compared for fairness.

Answer choices

Choose your path.

Preview only

Use only the heads branch: P(A)=4/6, so P(B)=2/6 and the game favors A.

Use the weighted result P(A)=P(B)=1/2, but call the game unfair because it has two stages.

Add unweighted branch rates: P(A)=4/6+2/6=1, so P(B)=0 and the game is unfair.

P(A) = (1/2)(4/6) + (1/2)(2/6) = 1/2, so B also has 1/2 and the game is fair.

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A fair coin and fair die are used. A wins on \(H\) with die \(1\text{--}4\) or \(T\) with die \(1\text{--}2\); otherwise B wins. Compute A's probability with weighted branches, then evaluate fairness.

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