Probability of one player having more of a certain suite, with different hand sizes

[mattthr]

Hi,

I am not a mathematician and this is not homework: I'm trying to write an article on card game strategy. This has involved me learning some combinatorics, but aspects of it are still defeating me.


Here's the question I want to answer. Imagine two players being dealt card from a standard 52 card deck. One player gets 10 cards, the other gets 5. What are the odds that the second player will have more of any single suite than the first player?


It would be helpful to understand if there is an equation for this sort of problem, and how it is derived? As far as I can see the basic "law of probability" that I've learned - P(A or B) = P(A) + P(B) - P(A and B) - is not that helpful here? At least I can't work out how to use that equation to solve the answer to this problem.

Any assistance greatly appreciated, thanks!

 

[Deleted member 4993]

Hi,

I am not a mathematician and this is not homework: I'm trying to write an article on card game strategy. This has involved me learning some combinatorics, but aspects of it are still defeating me.


Here's the question I want to answer. Imagine two players being dealt card from a standard 52 card deck. One player gets 10 cards, the other gets 5. What are the odds that the second player will have more of any single suite than the first player?


It would be helpful to understand if there is an equation for this sort of problem, and how it is derived? As far as I can see the basic "law of probability" that I've learned - P(A or B) = P(A) + P(B) - P(A and B) - is not that helpful here? At least I can't work out how to use that equation to solve the answer to this problem.

Any assistance greatly appreciated, thanks!

First you need to consider the number of ways you can distribute the cards 10 & 5? What did you get?

 

[mattthr]

First you need to consider the number of ways you can distribute the cards 10 & 5? What did you get?

Really, I'm not sure how one would calculate this. I'm reading up about it as I go along.

Let's see, five cards is simplest. The possible distributions are:

1-1-1-2
0-1-2-2
0-1-1-3
0-0-1-4
0-0-0-5

Each has four different possibilities, right because there are 4 suites? So 5x4 = 20 different possible ways the suites could fall out in a five card hand?

I'm going to assume that it's no co-incidence that there are 5 ways of arranging the suites in a 5 card hand, and presume there are 10 ways of arranging the suites in a 10 card hand. So that would be 10x4 = 40 different ways they could fall out in a ten card hand?