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

• bodensee9
In summary, the conversation discusses the number of ways to distribute a 6 hand card with at most 1 void in a suit. The solution is to use the Inclusion/Exclusion principle to calculate the total number of hands with no voids and the total number of hands with 1 void, without any overlap.
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.

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.

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!

1. How many cards are in a standard deck?

A standard deck of playing cards contains 52 cards in total. This includes 4 suits (clubs, diamonds, hearts, and spades) each with 13 cards (Ace, 2-10, Jack, Queen, and King).

2. How many cards should be dealt to each player?

It depends on the card game being played. In most standard card games, each player is dealt 5-7 cards. However, certain games may require more or less cards to be dealt to each player.

3. Is it possible to distribute cards evenly among players?

Yes, it is possible to distribute cards evenly among players as long as the total number of cards is divisible by the number of players. For example, if there are 52 cards and 4 players, each player will receive 13 cards.

4. How do you shuffle and distribute cards?

To shuffle and distribute cards, start by shuffling the deck thoroughly to ensure that the cards are randomized. Then, starting from the player to the left of the dealer, deal one card at a time until each player has the appropriate number of cards.

5. Can the dealer also participate in the game?

Yes, the dealer can also participate in the game. In fact, in some card games, the dealer is required to play against the other players. However, in other games, the dealer may simply facilitate the game and not actively participate in it.

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