# Drawing at least 1 ace from 13 drawing with replacement in a deck of 52 cards

I need to calculate the probability of drawing at least 1 ace from 13 drawing in a deck of 52 cards with replacement. Ok So i have tried $13 \cdot \frac{4}{52}$ but this gives me $100%$ of chance and I know it's wrong. (Can someone explain me what it corresponded to)
Next I have tried $(\frac{4}{52})^{13}$ but this was giving me the probability of having at least 1 ace on every 13 drawing.
So I made a quick python program to simulate $100.000$ test and I found in average 64700 drawing with a least 1 ace.
I can't manage to get the exact probability using a formula.. I have seen binomial coefficient and formula with or without replacement but I can't manage to use them. Thanks

• Your calculation is incorrect because it assumes that the $13$ events are mutually exclusive which they are not. Hint: it is easier to compute the probability that none of the $13$ cards are Aces. – lulu Sep 24 '17 at 17:05

Consider the complement: what is the probability of having at least NO ace on any draw? Then the desired probability is $$p=1-\left(\frac{48}{52}\right)^{13}$$
Note that drawing at least one ace is the complement of drawing no aces. The probability of drawing no aces in the 13 draws is $\left (\frac{12}{13} \right )^{13}$, so the probability of drawing at least one ace is $1-\left (\frac{12}{13} \right )^{13}$.