For each potential card there is a $\frac{12}{13}$ chance that it isn't an ace. Therefore in order for the first ace to occur at position 30, the first 29 cards can't be an ace. The probability of this is $(\frac{12}{13})^{29}$.
The probability of card 30 being an ace among the 23 remaining cards in the deck is $\frac{4}{23}$, and so the probability of card 30 being the first ace is $(\frac{12}{13})^{29} * \frac{4}{23} \approx 0.017$.
Is my reasoning correct?
Best Answer
The positions of the four aces is a random subset of size four taken from the index set $1,2,\dots, 52$. There are ${52\choose 4}$ equally likely outcomes.
The number of outcomes that include the index "30" and three larger indices is $22\choose 3$, since there are 22 indices in the range $31,32,\dots, 52$.
The required probability is $${{22\choose 3}\over{52\choose 4}}={44\over 7735}=0.005689.$$