Game Description

Players: Alice and Bob
They take turns choosing one number at a time from the following set, without replacement:

\[ \left{ \frac{1}{16},\ \frac{1}{8},\ \frac{1}{4},\ 1,\ 2,\ 4,\ 8,\ 16 \right} \]

The first player to obtain three numbers whose product is 1 wins.

Alice moves first.

Step 1: Change the Problem to Additive

Take the base-2 logarithm of all numbers:

Number \( \log_2 \) value
\( \frac{1}{16} \) \( -4 \)
\( \frac{1}{8} \) \( -3 \)
\( \frac{1}{4} \) \( -2 \)
\( 1 \) \( 0 \)
\( 2 \) \( 1 \)
\( 4 \) \( 2 \)
\( 8 \) \( 3 \)
\( 16 \) \( 4 \)

So instead of finding three numbers whose product is 1, we’re now finding three numbers whose sum of logs is:

\[ \log_2(a) + \log_2(b) + \log_2(c) = \log_2(abc) = \log_2(1) = 0 \]

Step 2: Reformulate

The problem becomes:

Alice and Bob take turns choosing one number from the set
\[ {-4, -3, -2, 0, 1, 2, 3, 4} \]
without replacement.
Whoever gets three numbers that sum to 0 wins.

Step 3: Recognize the Hidden Tic-Tac-Toe

There are exactly 8 numbers, and the win condition is:

Find 3 elements from the set that sum to 0.

This is equivalent to 3-in-a-line in a magic square or Tic-Tac-Toe on a suitable representation of these 8 values.

This game has been studied — and it turns out the structure is isomorphic to standard Tic-Tac-Toe.

Step 4: Result

In standard Tic-Tac-Toe, the first player can force a win or draw — but not always win if the second player plays optimally.

Hence, in this game:

  • Alice cannot always guarantee a win.
  • If Bob plays optimally, the game will end in a draw.
  • But if Bob makes a mistake, Alice can win.

Conclusion

\[ \boxed{ \text{Alice cannot always win, but she can force at least a draw with optimal play.} } \]

Reference