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?

One quick check for problems like these is that the UPPER entries in the numerator (so, $4$ and $51$) will sum to the UPPER entry in the denominator. $$4 + 51 \neq 52$$ so you've done something wrong... — The Chaz 2.0, Feb 25 '17 at 17:11

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

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 $1 - P(\text { no ace ) }$.

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.

score 0 · Answer 2 · answered Feb 25 '17 at 16:56

0

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.

answered Feb 25 '17 at 16:56

Jack Black

181

If no one answers and you have found the answer to your question, you can delete this quetsion. – K Split X Feb 25 '17 at 16:58

score 0 · Answer 3 · answered Feb 25 '17 at 17:02

0

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.

answered Feb 25 '17 at 17:02

Amar

847

score 0 · Answer 4 · answered Jul 02 '19 at 03:35

Based on the subsequent answer that you posted, you have recognized that your reasoning was incorrect.

Now, taking a closer look at

$$ \frac{{\binom 4 1}{\binom {51} 4}}{\binom {52} 5}, $$

your numerator suggests that you want to count the number of combinations on $4$ cards taken $1$ at a time. I interpret this to mean that you want to count the number of ways of selecting exactly one ace from the four. Then, if you multiply this by ${\binom {48} 4}$ you will have the number of ways of getting exactly $1$ ace in a five-card hand.

But, since you want the probability of getting at least one ace in a five-card hand, you must add to this the number of ways of getting exactly two aces in five cards, the number of ways of getting exactly three aces in five cards, and the number of ways of getting exactly four aces in five cards; then --- divide by the denominator that you've indicated.

This is the direct, but certainly the long way to do the problem.

The far more expedient approach is to subtract P(no aces) from $1$.

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

4 Answers4