# How to find the probability of drawing a certain two cards from a pack

Could I please ask for help on the last part of this question:

Two cards are drawn without replacement form a pack of playing cards. Calculate the probability:
a) That both cards are aces
b) that one (and only one) card is an ace
c) That the two cards are of different suits
d) Given that at least one ace is drawn, find the probability that the two cards are of different suits.

Here's my attempt (for parts a, b, and c I get the answer given in the book):

Let:

$$A =$$ Event that both cards are aces

$$B =$$ Event that one and only one card is an ace

$$C =$$ Event that the two cards are of different suits.

a) $$P(A) = \frac{4}{52} \cdot \frac{3}{51} = \frac{1}{221}$$ (as must pick an ace AND another ace)

b) $$P(B) = \frac{4}{52} \cdot \frac{48}{51} + \frac{48}{52} \cdot \frac{4}{51} = \frac{32}{221}$$ (as can pick ace then not ace, or not ace than ace)

c) $$P(C) = \frac{13}{52} \cdot \frac{39}{51} \cdot 4 = \frac{13}{17}$$ (as can pick any given suit first, followed by not that same suit, and this can be done in four ways, one for each suit).

d) Let $$D =$$ Event that at least one ace is drawn.

$$P(D) = P(A) + P(B)$$ (because at least one ace is drawn only "if both cards are aces" or "one and only one card is an ace")

so $$P(D) = \frac{1}{221} + \frac{32}{221} = \frac{33}{221}$$

Now, I need to calculate $$P(C \mid D) = \frac{P(C \cap D)}{P(D)}$$

So if I can calculate $$P(C \cap D)$$ then I can divide this by $$P(D)$$ to get the answer.

I (wrongly it appears!) reasoned like so:

To end up with two cards where "at least one is an ace and both are of different suits" you can only have this by either having "the first card be an ace and the second a card of a different suit from that ace" OR having "the first card be of a certain suit and the second card an ace of another suit".

Let @ stand for any suit.

so \begin{align*} P(C \cap D) & = P(\text{ace of @}) \cdot P(\text{not @}) + P(\text{@}) \cdot P(\text{ace not of @})\\ & = \frac{4}{52} \cdot \frac{39}{51} \cdot 4 + \frac{13}{52} \cdot \frac{3}{51} \cdot 4\\ & = \frac{5}{17} \end{align*}

Well this leads to $$P(C \mid D) = \frac{5}{17} \cdot \frac{221}{33} = \frac{65}{33}$$.

Answer given in book is $$\frac{25}{33}$$.

Thanks for any help.

Two cards are drawn without replacement from a standard deck of playing cards. Given that at least one ace is drawn, find the probability that the two cards are of different suits.

Let $$f(k)$$ denote the probability that $$k$$ Aces were drawn, where $$k \in \{0,1,2\}.$$

• $$\displaystyle f(2) = \frac{1}{221}$$.
• $$\displaystyle f(1) = \frac{32}{221}.$$
• $$\displaystyle f(0) = 1 - f(1) - f(2) = \frac{188}{221}.$$

Now, use Bayes Theorem. The (relative) probability that exactly one card was an Ace, given that at least one card was an Ace is

$$\frac{32}{32 + 1} = \frac{32}{33}.$$

So, the probability that two Aces were drawn is

$$\frac{1}{32 + 1} = \frac{1}{33}.$$

Let $$p(k)$$ denote the probability that the two cards are of different suits, given that $$k$$ Aces were drawn $$~: ~k \in \{1,2\}.$$

Then, the desired computation will be

$$\left[\frac{32}{33} \times p(1)\right] + \left[\frac{1}{33} \times p(2)\right]. \tag1$$

If exactly $$1$$ Ace was drawn, there are $$52$$ cards remaining, of which $$3$$ must be rejected as impossible : the other $$3$$ Aces. Therefore, when exactly $$1$$ Ace is drawn, there will be $$36$$ cards remaining in different suits, and $$(12 \times 4) = 48$$ cards remaining in the deck that may have been drawn.

Therefore $$p(1) = (3/4).$$

When both cards drawn were an Ace, then they had to be different suits. Therefore, $$p(2) = 1.$$

Feeding these numbers back into the expression in (1) above gives

$$\left[\frac{32}{33} \times (3/4)\right] + \left[\frac{1}{33} \times 1\right] = \frac{25}{33}$$

Favorable events are
[two aces] $$+$$ [one ace and one non-ace of different suit]

Sample space is
[all combos of two] $$-$$ [combos of two without an ace]

Thus $$Pr = \dfrac{\binom42 +\binom41\binom{36}1}{\binom{52}2-\binom{48}2} = \dfrac{25}{33}$$

• Thanks so very much to everyone for their help. Very much appreciate it. Commented May 24, 2022 at 13:06
• You're welcome ! Commented May 24, 2022 at 13:28

Here's how I would think about these: a) Initially there are 52 cards in the deck, 4 of them aces. The probability the first card you draw is and ace is 5/52= 1/13. Now there are 51 cards in the deck, 3 of them aces. The probability the second card is an ace is 3/51= 1/17. The probability the two cards are both aces is (1/13)(1/17)= 1/221.

b) Initially there are 52 cards in the deck, 4 of them aces. The probability the first card you draw is and ace is 5/52= 1/13. Now there are 51 cards in the deck, 51- 3= 48 of them not aces. The probability the second card is NOT an ace is 48/51= 16/17. The probability the two cards are an ace and not an ace, in that order is (1/13)(16/17)= 16/221. The probability the first card you draw is NOT an ace is 48/52= 12/13. Now there are 51 cards in the deck, 4 of them aces. The probability the second card is an ace is 4/51. The probability the two cards are not an ace and an ace, in that order, is (12/13)(4/51)= (4/13)(4/17)= 16/221. The probability of an ace and not an ace, in either order, is 16/221+ 16/221= 32/221.

c) The first card MUST be of some suit. There are then 51 cards left, 52- 13= 39 (or 3*13= 39) of a different suit from the first. The probability the second card is of a different suit from the first is 39/51= 13/17. The probability the two cards are of different suits is 13/17.

d) The probability the two cards are both aces is, as in (a), 1/221, and, in that case, they must be of different suits. If one of the cards is an ace there are 52- 4= 48 "non-aces" and 3*12= 36 of them are not of the same suit as the ace. The probability of that is 36/48= 3/4. The probability, given that at least one card is an ace, that they are of different suits is 1/221+ 3/4= 1/884+ 664/884= 665/884.

• Your answer to the last question is incorrect. Commented May 24, 2022 at 21:33

Two cards are drawn without replacement from a standard deck of playing cards. Given that at least one ace is drawn, find the probability that the two cards are of different suits.

Method 1: We correct your attempt.

As you observed, the probability that at least one ace is drawn is $$\Pr(\text{at least one ace is drawn}) = \frac{4 \cdot 3 + 4 \cdot 48 + 48 \cdot 4}{52 \cdot 51} = \frac{33}{221}$$

If we draw two aces, they must be of different suits. This event can occur in $$4 \cdot 3$$ ways.

If we draw one of the four aces with the first draw, there are $$51 - 3 - 12 = 36$$ cards left in the deck which are not aces and are of a different suit than the ace. Hence, the number of ways of drawing an ace and then drawing a non-ace of a different suit is $$4 \cdot 36$$.

If we draw one of the $$48$$ non-aces with the first draw, there are $$3$$ aces left in the deck with a different suit than the non-ace we drew first. Hence, the number of ways of drawing a non-ace and then drawing an ace of a different suit is $$48 \cdot 3$$.

Hence, the probability that the two cards are of different suits and at least one ace is drawn is $$\Pr(\text{of different suits} \cap \text{at least one ace is drawn}) = \dfrac{4 \cdot 3 + 4 \cdot 36 + 48 \cdot 3}{52 \cdot 51} = \frac{25}{221}$$

Hence, the probability that the two cards are of different suits given that at least one ace is drawn is \begin{align*} \Pr(\text{of different suits} \mid & \text{at least one ace is drawn})\\ \qquad & = \frac{\Pr(\text{of different suits} \cap \text{at least one ace is drawn})}{\Pr(\text{at least one ace is drawn})}\\ \qquad & = \frac{\dfrac{4 \cdot 3 + 4 \cdot 36 + 48 \cdot 3}{52 \cdot 51}}{\dfrac{4 \cdot 3 + 4 \cdot 48 + 48 \cdot 4}{52 \cdot 51}}\\ \qquad & = \frac{4 \cdot 3 + 4 \cdot 36 + 48 \cdot 3}{4 \cdot 3 + 4 \cdot 48 + 48 \cdot 4}\\ \qquad & = \frac{25}{33} \end{align*}

Method 2: We work with combinations to avoid considering the order in which the cards are drawn, which simplifies the calculations.

The probability of drawing at least one ace when two cards are drawn is $$\Pr(\text{at least one ace}) = \frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{48}{1}}{\dbinom{52}{2}}$$ since we must either select two of the four aces or one of the four aces and one of the $$48$$ non-aces while selecting two of the $$52$$ cards in the deck.

The probability that the two cards are of different suits and at least one ace is drawn is $$\Pr(\text{of different suits} \cap \text{at least one ace is drawn}) = \frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{36}{1}}{\dbinom{52}{2}} = \frac{25}{221}$$ since either two aces are drawn or one ace and one of the $$36$$ non-aces that are of a different suit than the ace are drawn when two cards are drawn from the deck of $$52$$ cards.

The probability of drawing two cards of different suits given that at least one ace has been selected is \begin{align*} \Pr(\text{of different suits} & \mid \text{at least one ace is drawn})\\ \qquad & = \frac{\Pr(\text{of different suits} \cap \text{at least one ace is drawn})}{\Pr(\text{at least one ace is drawn})}\\ \qquad & = \frac{\frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{36}{1}}{\dbinom{52}{2}}}{\frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{48}{1}}{\dbinom{52}{2}}}\\ \qquad & = \frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{36}{1}}{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{48}{1}}\\ \qquad & = \frac{25}{33} \end{align*}

Note: Observe that in both methods, after we cancel the common denominators, the numerator reduces to the number of cases in which at least one ace is drawn and cards are of different suits and the denominator reduces to the number of cases in which at least one ace is drawn. This is because the sample space for the conditional probability that the cards are of different suits given at least one ace is drawn is the set of cases in which at least one ace is drawn.