Flip out three fair die at a time, find the probability such that  the product of spots is divisible by 10.

2012 Tokyo Institute of Technology entrance exam, problem 1-B
To find the probability that the product of the spots on three fair dice is divisible by 10, we need to consider the prime factorization of 10, which is \(10 = 2 \times 5\). Therefore, the product of the spots must include at least one factor of 2 and one factor of 5.

1. **Identify the numbers on a die that are divisible by 2 and 5:**
   - The numbers on a die that are divisible by 2 are 2, 4, and 6.
   - The number on a die that is divisible by 5 is 5.

2. **Calculate the total number of possible outcomes:**
   - Since each die has 6 faces, the total number of possible outcomes when rolling three dice is:
     \[
     6^3 = 216
     \]

3. **Calculate the number of favorable outcomes:**
   - We need to count the number of outcomes where the product of the spots is divisible by both 2 and 5. We will consider three cases:

   **Case (a): Exactly one die is divisible by 2 and one die is divisible by 5:**
   - Choose 1 die to show a number divisible by 2 (3 choices: 2, 4, 6).
   - Choose 1 die to show the number 5 (1 choice).
   - The remaining die can show any of the 6 numbers.
   - The number of ways to arrange these three dice is \(3!\).
   - Therefore, the number of favorable outcomes in this case is:
     \[
     3 \times 1 \times 6 \times 3! = 3 \times 1 \times 6 \times 6 = 108
     \]

   **Case (b): Two dice are divisible by 2 and one die is divisible by 5:**
   - Choose 2 dice to show numbers divisible by 2 (3 choices each: 2, 4, 6).
   - Choose 1 die to show the number 5 (1 choice).
   - The number of ways to arrange these three dice is \(\frac{3!}{2!}\) (since two dice are identical in terms of being divisible by 2).
   - Therefore, the number of favorable outcomes in this case is:
     \[
     3 \times 3 \times 1 \times \frac{3!}{2!} = 3 \times 3 \times 1 \times 3 = 27
     \]

   **Case (c): One die is divisible by 2 and two dice are divisible by 5:**
   - Choose 1 die to show a number divisible by 2 (3 choices: 2, 4, 6).
   - Choose 2 dice to show the number 5 (1 choice each).
   - The number of ways to arrange these three dice is \(\frac{3!}{2!}\) (since two dice are identical in terms of showing 5).
   - Therefore, the number of favorable outcomes in this case is:
     \[
     3 \times 1 \times 1 \times \frac{3!}{2!} = 3 \times 1 \times 1 \times 3 = 9
     \]

4. **Sum the number of favorable outcomes:**
   - Total number of favorable outcomes is:
     \[
     108 + 27 + 9 = 144
     \]

5. **Calculate the probability:**
   - The probability that the product of the spots is divisible by 10 is:
     \[
     \frac{144}{216} = \frac{2}{3}
     \]

The final answer is \( \boxed{ \frac{2}{3} } \)