Tricky Cards !

Probability Level 3

Two cards are drawn at random from a pack of 52 cards. The probability of getting at least a spade and an ace , is

8/221 3 / 34 29/442 1/26

2 solutions

Brian Charlesworth
Dec 2, 2014

In order to avoid duplication, the combinations we are looking for are

(i) the ace of spades along with any other card, or

(ii) a non-spade ace along with a non-ace spade.

There are $51$ possible combinations in case (i), and $3*12 = 36$ possible combinations in case (ii). Since in general there are $\binom{52}{2} = 1326$ possible combinations of two cards, the probability of getting at least a spade and an ace is

$\dfrac{51 + 36}{1326} = \boxed{\dfrac{29}{442}}$ .

Sir, what is wrong with this, $\begin{cases}\underline{SPADE}\\1(ace)\\2\\3\\4\\5\\6\\7\\8\\9\\10\\J\\Q\\K\end{cases}\begin{cases}\underline{ACES}\\heart\\diamond\\club\\spade\end{cases}.$ Now,let us take the $1$ of spades which is also the ace of spades.Now,we can pair it with the $2,3,4....10,J,Q,K$ of spades and also with the aces(heart,diamond,club).Thus,we have $12+3=15$ pairs.Now,let us take the $2$ of spades,we can pair it with the ace of hearts,diamonds and club.Similarly for the other cards of spades.Thus,total number of cases $=15+12*3=51.$ Total number of ways of choosing two cards $=\dbinom{52}{2}=51*26.$ Thus,the required probability $=\dfrac{1}{26}.$

Adarsh Kumar - 6 years, 6 months ago

If you have the ace of spades, then you already have an ace and a spade, all in one card. Thus you have satisfied all the conditions, and thus can pair it with any of the remaining $51$ cards. The second part of your reasoning is correct, so the total number of pairs that satisfy the conditions is $51 + 36 = 87$ , giving a probability of $\frac{87}{51*26} = \frac{29}{442}$ .

Brian Charlesworth - 6 years, 6 months ago

wow!that is super amazing! $\text{I}$ $\heartsuit$ $\text{ Brian Charlseworth's brain.}$

Adarsh Kumar - 6 years, 6 months ago

@brian charlesworth

Adarsh Kumar - 6 years, 6 months ago

Thanks for reviewing it ! Big fan ! :D

Keshav Tiwari - 6 years, 5 months ago

Brock Brown
Dec 27, 2014

from fractions import Fraction as f
cards = set()
for suit in "cdhs":
    for val in "234567890jqka":
        cards.add(suit+val)
top = 0
bottom = 0
for a in cards:
    for b in cards-set((a,)):
        bottom += 1
        if 'a' in a+b and 's' in a+b:
            top += 1
print f(top, bottom)

Tricky Cards !

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