What are the odds of drawing 23 cards and there is at least one ace?

Question

So my friends dad asked him this question and he asked it to me. None of us know the answer, and while I tried to solve it, I'm not to sure if my math behind is right.

I solved it a number of ways each way gave me a different answer, so i'll just show you the last way i solved it was, and then if someone can tell me how its solved.

The odds that one is an ace is 1 - P(not an ace) Which is 1 - ( 48C23 / 52C23) = ~90%

Im not too confident about this because its odd that Id get such a high percentage for only 23 cards.

Answer 1

That's correct. Don't be surprised about the high probability: 1/13 of all the cards are aces. Therefore, om average, you'd already expect 1 ace when drawing 13 cards, and nearly 2 when drawing 23.

Answer 2

Your calculation is correct, though the final approximation could be better:

$$1-\frac{\binom{48}{23}}{\binom{52}{23}}\approx0.91227\;.$$

In fact if we’re a little bit clever, we can even get the fraction exactly without using a very high-precision calculator:

$$\frac{\binom{48}{23}}{\binom{52}{23}}=\frac{\frac{48\cdot47\cdot\ldots\cdot26}{23!}}{\frac{52\cdot51\cdot\ldots\cdot30}{23!}}=\frac{29\cdot28\cdot27\cdot26}{52\cdot51\cdot50\cdot49}=\frac{29\cdot9}{17\cdot25\cdot7}=\frac{261}{2975}\;,$$

so the desired probability is

$$1-\frac{261}{2975}=\frac{2714}{2975}\;.$$