# Thread: how many hands should I use for this question?

1. ## how many hands should I use for this question?

Hi folks,

in probability questions that ask the probability of getting say, 3 queens in a hand of 5 cards, the probability P is the ratio of the favourable outcomes to the total number of possible outcomes/hands. In a normal 52 card deck the total number of 5 card hands is the number of possible selections i.e. $^{52}C_{5} = 52!/47! \times 5!$ or 2,598,960. If the hands are of only 2 cards then the number of such hands is $^{52}C_{2}$ or 1326.

I have a question in which two cards are drawn (without replacement) from a deck of 52 and we are asked to find the probability that the first card is a spade and the second card is the king of spades.

As you see, it is slightly puzzling because of the order: What happens if I pull the king of spades with the first card? I decided to ignore this possibility. So the first card must be one of the other 12 spades, so there are 12 ways of selecting the first card and only one way of selecting the king of spades so favourable outcomes = 12 and the total Probability should be 12/1326 = 2/221 and this is wrong.

Of course, one could say that there are 12/52 ways of selecting the first card and 1/51 ways of selecting the second card which is 12/51.52 = 1/221 and get the right answer, but my question is what is wrong with the first method?

2. ## Re: how many hands should I use for this question?

Because $_{52}C_2$ is the number of ways of drawing two cards in any order. The number of ways of drawing two cards where order matters is $_{52}P_2=\dfrac{52!}{50!}$

3. ## Re: how many hands should I use for this question?

BUT... does order matter in this case? Is it because the first card must be a spade and the second card a king of spades?

There is a website that explains how to calculate the odds of getting certain poker hands. For example a royal flush is A, K, Q, J, 10 in any of the 4 suits. The probability of getting one is $4 / ^{52}C_5$ not $4 / ^{52}P_{5}$ In every case (full house, 4 of a kind etc) the combination rather than the permutation is used to calculate the number of possible hands. Why should I start using the permutation just for this problem? Is it because in a normal poker hand the cards can be in any order but in this case the order is specified?

4. ## Re: how many hands should I use for this question?

Originally Posted by s_ingram
Is it because in a normal poker hand the cards can be in any order but in this case the order is specified?
Exactly. The royal flush needs those cards in any order, so you do not care what order they are drawn. Your problem specifies that you care about the order the cards are being drawn. This is different from drawing a hand and seeing if the cards (in any order) have certain properties. You are being asked a specific question about the order that the cards are drawn. When asked about orderings, you are dealing with permutations, not combinations. When asked about a poker hand, the order the cards are drawn does not affect the hand, so you are dealing with combinations, not permutations.

5. ## Re: how many hands should I use for this question?

Great. Thanks as usual for making this Forum a great place.