Calculating Ways to Distribute Cards with At Most 1 Void in a Suit

[bodensee9]

Hello:

I have a question about the number of ways to distribute a 6 hand card with at most 1 void in a suit (so you can have void in either spades, clubs, diamonds, heart, but you can't have a void in both spades and clubs, etc).

Would the solution just be the ways to distribute a 6 hand card (52 choose 6) - the ways to distriute cards with a void in 2 suits? (4 choose 2)*(26 choose 6)?

The reasoning is say the we have a universe of the total number of ways to distribute 6 cards, with no conditions. Included in that universe is the number of ways to distribute 6 hands with voids in 1 suit, and then included within that subset is the number of ways to distribute a 6 hand card with voids in 2 suits, etc. So to find the number of ways to distribute cards with either no void in any suite or with 1 void in a suite, wouldn't we just subtract out the number of ways to distribute cards with voids in 2 suits since that subset also includes the number of ways to distribute cards with voids in 3 suits?

Thanks.

 

[Billy Bob]

Pretty good reasoning, but it has a very subtle error which I admit took me a long time to figure out.

You have overcounted the number of hands with 2 or more voids, specifically the number of hands with exactly 3 voids.

Here's why. Using your scheme for counting the number of hands with 2 or more voids, suppose the two suits chosen to be used are hearts and clubs. Then suppose when you pick 6 of those 26 cards, you choose 6 hearts.

Those hands get counted again if the two suits chosen to be used are hearts and diamonds, and then you pick 6 hearts.

I don't know if it's easier to try to count those overcounts or to start over with a somewhat different scheme. The basic method of using 52 C 6 - (number of hands with 2 or more voids) is good, either way.

 

[bodensee9]

You're right! Thanks.
I decided to do it like this:
the total would be => number with no voids + number with 1 void. This way, there shouldn't be overlap. And I would use the Inclusion/Exclusion principle for both of these.
Thanks!