Probability of getting at least 1 ace in 5-card hand.

Question

I know the correct answer is $1 - P(\text{no aces}) = 1 - \frac{\binom{48}{5}}{\binom{52}{5}}$

But I cannot think of why $\frac{{\binom 4 1}{\binom {51} 4}}{\binom {52} 5}$ is wrong. Does this formula even make sense? What am I missing?

Answer 1

Your solution is wrong because of the word atleast. When your dealing with atleast, you cannot use the combination $\binom{4}{1}$, because this says that you want exactly 1 ace. Instead, you have to do it the converse way, like the correct answer.

You answer is wrong because atleast one ace cannot be calculated via combinations. It must be calculated by $1 - P(\text { no ace ) }$.

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Suppose you want exactly 1 ace. Well ,there are 4 aces in a deck, and we you want to choose exactly 1. $\binom{4}{1}$. Now that we have chosen a ace, there are 52-4 = 48 cards left in the deck, since we only want 1 ace. So out of these 48 cards, we want to choose 4 more. That's $\binom{48}{4}$. Then we divide by $\binom{52}{5}$, because that's the total number of cases.

Hope this helps.

The probability of an ace from a 5-card hand?

Read the correct answer for this.

Answer 2

I realized that I am double counting some hands, for example (A1 A2 K1 K2 K3) and (A2 A1 K1 K2 K3) are counted separately.

Answer 3

I don't think this formula makes sense. If you are doing it using at least in view. Then you have to make 4 cases and add them.

Using this formula you are adding some redundant cases also.