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.

I previously posted a question but realized I didn’t frame the question properly so I’m hopeful this is more clear.

If after the cards are dealt I have the 5 of clubs as my lowest club, what is the probability that the three other players have the 2, 3 and 4 as their lowest club (in no particular order).

I play online spades with friends and we are trying to figure out the odds of budding zero when having the 5 of clubs as your lowest club and then taking that trick because the other three players have lower clubs. It feels like it’s about 20% but I don’t know how to do the math.

Thank you!

Jon

Answer 1

We use C() for combinations and P() for permutations below.

It's interesting to start with an argument assuming independence, then see how closely you come.

If 3 people have equal chances of a selection and we make 3 selections, the probability of each getting 1 is

3!/(1!1!1!)(1/3)^3 = 2/9. We should expect our answer to be a little greater than this.

The number of ways that we can have 1 player with the 2, 1 with the 3, 1 with the 4, and 1 with the 5 is

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

The number of ways that we can have 1 player with the 5 of clubs and nothing lower is 4 * C(48,12) (any of the 4 can have this hand).

Then, the next player can have C(39,13) hands, and the next player C(26,13) hands. C(39,13)*C(26,13) = 39!/(13!)^3.

Thus, the total number of hands with one player having the 5 and nothing lower is

4 * C(48,12)*39!/(13!)^3

Thus, given this hand with the 5 and nothing lower, the chance of the other 3 having the 2, 3, and 4 in any order is

4! * 48!/(12!)^4/(4 * 48!/(36!12!)*39!/(13!)^3)) =

6*13^3/P(39,3) =

13^2/(19*37) =

169/703 =

.240398

As we hypothesized, this is a little bit greater than 2/9 = .222222, the probability assuming independence, as we expected.