1. ## Playing cards

We are playing a card game, where we have a big discusion about statistic. Hope one of you can help solving the problem.

I have the card spades 2. 3 other persons each recieve a single card - 3 in total.

How big is chance that one of the 3 persons recieve a spades? All 52 cards are used - after I have received spades 2 there are 12 spades back of a total of 51 card.

Kind Regard
Kim

2. ## Re: Playing cards

Each of them has a probability of 12/51 of having another spade.

3. ## Re: Playing cards

I agree, but how big is the chance in total, that one of the three have a spade? I think it is 36/51. Right?

4. ## Re: Playing cards

It is a bit hard to follow your setup. Say you have one card, a spade, then you deal one card to each of two other people. I think that you want to know the probability their getting at least one spade.
The probability their getting no spade is $\displaystyle \frac{39\cdot 38}{51\cdot 50}.$

So the probability their getting at least one spade is $\displaystyle 1-\frac{39\cdot 38}{51\cdot 50}.$

5. ## Re: Playing cards

Hi Plato

I play a card game with 4 players involved. I have a spade, and want to know how big (percent) a chance there is, that one of the 3 other players getting a spade.

6. ## Re: Playing cards

Originally Posted by KimN
I play a card game with 4 players involved. I have a spade, and want to know how big (percent) a chance there is, that one of the 3 other players getting a spade.
How many cards are dealt? Are you thinking one card each?
So four cards in all. If so it is done exactly as I did with three people.

7. ## Re: Playing cards

I have one card - eg. spade 2. Then there is 51 cards back. Of these 51 cards i give one card to 3 persons. Then 48 cards back. How many percent chance is it that one of the 3 persons have a spade?

8. ## Re: Playing cards

Originally Posted by KimN
I have one card - eg. spade 2. Then there is 51 cards back. Of these 51 cards i give one card to 3 persons. Then 48 cards back. How many percent chance is it that one of the 3 persons have a spade?
$\displaystyle 1-\frac{39\cdot 38\cdot 37}{51\cdot 50\cdot 49}$
That is the probability that at least one of the other three players gets a spade.

9. ## Re: Playing cards

Originally Posted by Plato
$\displaystyle 1-\frac{39\cdot 38\cdot 37}{51\cdot 50\cdot 49}$
I agree with Plato.
That comes out to .56115..., or approximately 11/20

Your question could be stated this way:
a spade is missing from a regular 52cards deck;
3 cards are pulled at random from the remaining 51 cards:
what is probability that at least one card pulled is a spade?