Does knowing that you drew at least $1$ ace increase the probability that you drew at least $2$ aces ($5$ card hand)?

Question

So say I draw $5$ cards from a standard deck at random without replacement and without looking at them. If someone told me that the hand contains at least $1$ ace, would that change the probability that my hand contains at least $2$ aces?

I did the math and I believe the probability of drawing at least $2$ aces is: $1 - \dfrac{\binom{48}{5}}{\binom{52}{5}} - \dfrac{\binom{4}{1}\binom{48}{4}}{\binom{52}{5}}$.

Additionally, what if that person then told me that the hand contains the ace of spades; would knowing that there is at least one ace and that it's the ace of spades change the probability of having at least $2$ aces in my hand?

Answer 1

It is not correct to say that whether you are told that an ace is in the hand or whether you are told that the ace of spade is in the hand, the probability of having at least two aces is the same

Case 1: You are told that you have at least one ace

Note that in the following formula, the sample space shrinks to $\binom{52}5 - \binom{48}5$ to take into account that at least one ace must be present, and that in the numerator we need to ensure that one ace is there

$Pr = 1 - \dfrac{\binom41\binom{48}4}{\binom{52}5 - \binom{48}5},\;\approx 0.1222$

Case 2: You are told that you have the ace of spades

Sinca ace of spades is assured, the sample space is now simply $\binom{51}{4}$, and from the numerator, we just subtract $\binom{48}4$ to indicate that more aces must be present.

Thus $Pr = 1 - \dfrac{\binom{48}4}{\binom{51}4}\; \approx 0.2214$

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Added approach

Since probability problems are often "tricky", here is an approach simpler to understand, though longer computationally.

Basically, the numerator will count hands with $2$ or more aces, and the denominator, hands with $1$ or more aces

Case 1: You are told that you have at least one ace

$Pr = \dfrac{\binom42\binom{48}3 + \binom43\binom{48}2 +\binom44\binom{48}1}{\binom41\binom{48}4+\binom42\binom{48}3 + \binom43\binom{48}2 +\binom44\binom{48}1}\approx 0.1222$

Case 2: You are told that you have the ace of spades

Bear in mind that you have the ace of spades and are left to pick from $51$ cards including $3$ aces

$Pr = \dfrac{\binom31\binom{48}3 + \binom32\binom{48}2 +\binom33\binom{48}1}{\binom30\binom{48}4+\binom31\binom{48}3 + \binom32\binom{48}2 +\binom33\binom{48}1} \approx 0.2214$

Answer 2

Intuitively, you need to draw at least one ace to draw two or more aces, so knowing you have drawn at least one increases the probability you have drawn more than one. The event that your hand contains a spade also implies you drew at least one ace so the probability of two aces must also increase (edit: I should clarify that, as Ross and true blue nail have pointed out, the probability of two aces is not the same in this case as in the previous).

Here is a math argument for the first case you asked about. The conditional probability of drawing at least two aces '$A>1$' given we drew at least one ace '$A>0$' is $$P(A>1|A>0) = \frac{P(A>1,A>0)}{P(A>0)} = \frac{P(A>1)}{P(A>0)} $$ since $A>1$ implies $A>0$ and so $P(A>1,A>0) = P(A>1)$. Whereas the probability of drawing two aces is just $P(A>1)$ which is less that $\frac{P(A>1)}{P(A>0)}$ since $P(A>0)$ is less than $1$ (because not every hand contains an ace).