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

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?

• I disagree with the assertion for the second part that knowing that its is an ace of spades does not change the situation. I have put up an answer that explains this. Apr 21 at 4:04
• Have a look at the added approach. Apr 21 at 7:18

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

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

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

• This is correct for the second part of the problem, which I think is the more nonintuitive part. +1 and I wish I could do more. Apr 21 at 1:51
• I agree with this calculation now. But intuitively it makes no sense. If you show someone you have an Ace but cover up the suit with your finger, then revealing the suit should have no effect on the probability. But it does. Why is this? Apr 21 at 12:53
• @AdamRubinson: covering up and revealing the suit is different from specifying one in advance. Half of the cases with exactly two aces include the ace of spades while only one quarter of the cases with exactly one ace do. The ace of spades reduces the denominator more than it does the numerator. Apr 21 at 13:41

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).

• Ooh that makes a lot of sense. So the calculation I made was really for P(A>1), and I need to divide it by the probability of drawing at least 1 ace? Apr 20 at 12:09
• Yes, you can just divide the calculation you made for $P(A>1)$ by $P(A>0) = 1 - \frac{\binom{48}{5}}{\binom{52}{5}}$. In general, when you want the conditional probability $P(A|B)$ of event $A$ given event $B$ you need to divide the probability $P(A, B)$ of both $A$ and $B$ happening by the probability $P(B)$ of $B$ happening. The situation in your question where you can just divide $P(A)$ by $P(B)$ occurs whenever event $A$ implies event $B$. Apr 20 at 12:27
• You imply that changing the condition from an unknown ace to the ace of spades does not change the probability of two aces. This is incorrect. The ace of spades is present in half the cases of exactly two aces but only one quarter of the cases of one ace. That almost doubles the probability of two aces given one. Apr 21 at 1:55
• @RossMillikan You are correct and my answer could be better if I was more specific in the spades case. I think true blue nail has covered that case well already now though. I only meant to imply that in both cases the probability of two aces was different than from the situation where you have no knowledge of the number of aces in your hand, not that they were exactly the same. Apr 21 at 9:37