Card game - Spades

Question

In the card game Spades each player is dealt 13 cards. The player with the 2 of clubs starts the game and plays that card. The person to the left then plays their lowest club, followed by the next person playing their lowest club and then the last person plays their lowest club.

What is the probability that the first 4 cards to be played are the 2, 3, 4 and 5 of clubs?

Answer 1

We can figure this using conditional probability or combinatorics.

Once the 2 of clubs is selected, the next person has 13 cards to select out of 51 not the 2 of clubs, so the probability of selecting the 3 of clubs is 13/51. For the next person, the probability of selecting the 4 of clubs will be 13/50, and for the last, the probability of selecting the 5 of clubs is 13/49.

13/51*13/50*13/49 = 13^3/P(51,3) = 2197/124950 = .0176.

Using combinatorics, there are

C(52,13)C(39,13)C(26,13) = 52!/((13^4)^4)

To have the 2 of clubs, then 3 of clubs, then 4 of clubs, then 5 of clubs means, for each, after selecting a particular club, being able to select all cards still available minus the 4 clubs (the first person will have 48 available, the second 48 - the 12 others selected by the first = 36, the third will have 48 - 2 * 12 = 24). There are 4 different people that can have the 2 of clubs. Thus, we have

4 * C(48,12)*C(36,12)*C(24,12) = 4 * 48!/((12!)^4)

Thus, our probability is 4 * 48!/((12!)^4) /(52!/((13^4)^4) =

4/(52*51*50*49/13^4) =

13^3/(51*50*49) = 2197/124950 = .0176.

This is the answer provided above.

We can complicate this problem by considering whether the 3, 4, and 5 would have been played if available.

We can also note that our answer is a little above (1/4)^3 = .015625, which is what we would get under independence.

Answer 2

Well, someone gets the 2. What is the probability that the  3 went to someone else? This will be  39/51 as there are 51 remaining equally likely positions in players hands that the  3 could occupy,  39 of which are in a hand different than the hand holding the 2.

Given that the two and the three are in different players' hands what is the probability that the 4 went to someone else? This will be 26/50.

Finally, given that the two three and four all went to different players' hands, the probability that the  went to the final remaining hand will be 13/49

The overall probability that each of these cards went to different players is then the product of these:

39 * 26 * 13

--------------

51 * 50 * 49