# B Card-drawing probability problem

1. Aug 25, 2016

### Kavorka

I've been trying how to figure out how to figure out the probability of different situations in card-choosing, and I am having trouble getting my head around how to approach this:

Let's say we have two people drawing 26 of the 52 cards in a deck. What is the probability that either team has the Ace, King and Queen of one particular suit?

If you were picking 3 cards it would be simple to find the numerator of the probability ratio: 12/52*1/51*1/50. If you were trying to just get any ace, king and queen it would be a simple combination in the numerator, but in this case where you have a situation constraint as well as many picks I'm not sure how to approach this.

Last edited: Aug 25, 2016
2. Aug 25, 2016

### Heinera

Do you mean a particular suit, like "what's the probability that either team has the Ace, King and Queen of Spades", or do you mean just the probabilty that they are of any suit (but all of the same suit)?

3. Aug 25, 2016

### Kavorka

The second one, basically the chance of picking the ace, king and queen of any suit with 26 picks.

4. Aug 25, 2016

### Merlin3189

It sounds like you are asking, after each team has picked 26 cards, that for any suit, A,K & Q of that suit are not in the same half pack. Is that right?

If so (and maybe in other ways of analysing it) you can start back to front with the idea that each card has a 0.5 chance of being in either half pack.
So p(A, K & Q of spades are all in my hand) would be (1/2)3
Etc.

5. Aug 25, 2016

### Kavorka

So how would you find the combined chance that in picking half the pack you end up with the A,K,Q of 1 or more suits? It seems that with 26 picks the chance would be decent, I asked on another forum and someone said they found 20.71% but didn't explain how

6. Aug 25, 2016

### Deedlit

You can use the Principle of Inclusion-Exclusion. Define

A = the event that you get A,K,Q of spades
B = the event that you get A,K,Q of hearts
C = the event that you get A,K,Q of diamonds
D = the event that you get A,K,Q of clubs

then

$$P(A \cup B \cup C \cup D) = P(A) + P(B) + P(C) + P(D) - P(A \cap B) - P(A \cap C) - P(A \cap D) - P(B \cap C) - P(B \cap D) - P(C \cap D) + P(A \cap B \cap C) + P(A \cap B \cap D) + P(A \cap C \cap D) + P(B \cap C \cap D) - P(A \cap B \cap C \cap D)$$

$$= 4 * \frac{C^{26}_3}{C^{52}_3} - 6 * \frac{C^{26}_6}{C^{52}_6} + 4 * \frac{C^{26}_9}{C^{52}_9} - \frac{C^{26}_{12}}{C^{52}_{12}}$$

$$= \frac{56058703}{138046426} \approx 0.40608587$$

Last edited: Aug 25, 2016
7. Aug 26, 2016

### Merlin3189

I'm not sure I can work out where all the terms of your expression come from, but it does show I am not correct with my simplistic p=1/2 for the location of each card, because you can't end up with more (or less) than 26 cards in each hand. After putting the ace of spades, say, in one hand, the chances of the king of spades ending up in the same hand is slightly less than 1/2.
So I think the combinations approach must be needed.