Probability of not guessing any number in lotto drawing 7/34

Question

I'm trying to calculate the probability of not guessing any numbers in a 7/34 lotto drawing. My approach was to multiply the probabilities of guessing a single number 7 times in a row while removing 1 from the total number of balls left:

$$\frac{7}{34} \cdot \frac{7}{33} \cdot \frac{7}{32} \cdot \frac{7}{31} \cdot \frac{7}{30} \cdot \frac{7}{29} \cdot \frac{7}{28} $$

However looking through an old question someone posted this answer:

$$ \frac{7}{34} \cdot \frac{6}{33} \cdot \frac{5}{32} \cdot \frac{4}{31} \cdot \frac{3}{30} \cdot \frac{2}{29} \cdot \frac{1}{28}$$

I'm not sure what's the idea behind the second solution. Am I missing something?

Answer 1

Method 1: You select seven of the $34$ numbers. If you do not select any of the winning numbers, you must select seven of the other $24 - 7 = 27$ numbers.

The number of ways of selecting seven of the $34$ numbers is $$\binom{34}{7}$$ The number of ways of selecting none of the winning numbers is $$\binom{27}{7}$$ Thus, the probability of selecting none of the winning numbers is $$\frac{\dbinom{27}{7}}{\dbinom{34}{7}}$$

Method 2: Since there are $27$ non-winning numbers, the probability that the first number you pick is not among the winning numbers is $7/27$. That leaves six more numbers to pick and $26$ non-winning numbers from which to pick. Hence, the probability that the second number you pick is also a non-winning number is $6/26$.

Continuing in this way, the probability that you pick none of the seven winning numbers is $$\frac{7}{27} \cdot \frac{6}{26} \cdot \frac{5}{25} \cdot \frac{4}{24} \cdot \frac{3}{23} \cdot \frac{2}{22} \cdot \frac{1}{21}$$