2008 AMC 10A Problem 22

The first term needs no flip, so the fourth term is determined by three flips: $2^3 = 8$ equally likely sequences. Write $H$ for "double, then subtract $1$" and $T$ for "halve, then subtract $1$," and trace every path from $6$.

After flip 1: $H \to 11$, $T \to 2$.

After flip 2:
- from $11$: $H \to 21$, $T \to 4.5$;
- from $2$: $H \to 3$, $T \to 0$.

After flip 3:
- from $21$: $H \to 41$, $T \to 9.5$;
- from $4.5$: $H \to 8$, $T \to 1.25$;
- from $3$: $H \to 5$, $T \to 0.5$;
- from $0$: $H \to -1$, $T \to -1$.

The eight fourth terms are $41,\ 9.5,\ 8,\ 1.25,\ 5,\ 0.5,\ -1,\ -1$. Five of them are integers.

The probability is $\frac{5}{8}$, so the answer is $\boxed{\textbf{(D)}\ \frac{5}{8}}$.

With only eight outcomes, an organized tree is faster and safer than any clever shortcut. The structure to notice while tracing: $H$ preserves integers, $T$ preserves integers only from an even input, and $H$ can rescue a half-integer ($4.5 \to 8$). Because that rescue exists, "integer or not" is not enough state to carry; you must track the actual values (or at least the fractional part).

• Assuming a non-integer term can never become an integer again; doubling 4.5 gives 8, and missing that path yields 4/8 = 1/2 (choice C).
• Using four flips (sixteen paths) because the question says "fourth term," forgetting the first term needs no flip.
• Forgetting that halving an even number such as $2$ or $0$ still gives an integer, and discarding the $T$ branches too early.