# Likelihood of getting flush, straight, etc

On Planet X, cards can take on a numerical value from $$1$$ to $$7$$ (inclusive) and their suit can be either red or blue. In a game of poker, each player gets three cards.

1) What is the probability of a straight (e.g. three consecutive cards)? Note that wrap-arounds count, meaning that $$7, 1, 2$$ would count as a straight

2) What is the probability of getting a flush that is NOT a straight?

3) What is the probability of getting a straight flush?

4) What is the probability of getting a pair? Note that a pair cannot also be a straight or a flush.

5) What is the probability of getting a hand that is not a straight, nor a flush, nor does it contain $$2$$ of a kind?

I am sort of new to probability and combinatorics so I wanted someone to help me verify these problems. Here are my attempts:

$$1$$) There are $$7 \cdot 2 = 14$$ total cards in the deck, so there are $${14\choose 3}$$ possible hands. Then I need to count the number of ways to get a straight. I sort of just tried to list them all out... but then it gets complicated since there are two different suits.

I would appreciate it if someone can please help me with this problem. I know that the denominator will be $${14\choose 3}$$, but I don't have a clever way to enumerate the numerator.

Also, I think the answer to $$5$$ will just be the $${14\choose 3}$$ minus the sum of everything we've done so far all over $${14\choose 3}$$, but I'm not sure about that again.

• Any one of the 7 ranks can be the op card of a straight so there are 7 different hinds of straights. Once you have picked the kind of straight (say 6-high) how many ways can you pick the individual cards that make up the hand? – saulspatz Apr 22 at 22:25
• So there are $2^{3}$ internal orderings for each straight right? Since there are $2$ choices for each suit? So is the answer for $1$ equal to $(2^{3} \cdot 7)/{14\choose 3}$? – effunna9 Apr 23 at 0:29
• Yes, that's correct – saulspatz Apr 23 at 4:24
• How can I do #2? So it's the probability of getting a flush and not a straight. There are $2$ possible suits and we need all three to be identical. I think this is the same as picking $3$ cards from either $7$-suited-decks. So I think the answer is $2{7\choose 3}/{14\choose 3}$ is that right? – effunna9 Apr 23 at 10:17
• You have computed the probability of getting a flush. You need to subtract the the straight flushes in the numerator. – saulspatz Apr 23 at 13:45