# Math Help - Poker game

1. ## Poker game

A man is dealt a poker hand (5 cards) from an ordinary deck. In how many ways can he be dealt

(i) a straight.

The answer in the back of my book says $10\cdot4^5-40$ (I know 40 is deducted from the number of straight flush, but I don't know what the $4^5$ is.

(ii) a pair of aces (I could not find the poker rule for this, so have no idea what this is)

The answer provided was ${4\choose2}{12\choose3}\cdot4^3.$ ( again I do know what the reason is for ${12\choose3}\cdot4^3$)

I have never played a card game, and know nothing about poker. I understand it would be tough to explain these to me, but I would appreciate the time and effort that you give me.

2. There are 13 cards per suit. So it you have a hand of five cards, two of which are aces, there are three spots left, and you have 12 cards to chose from (2 through King). Therefore the number of ways to arrange those last 3 spots is 12 choose 3.

Not sure myself what their four-cubed is referring to.

3. Hello, novice!

A man is dealt a poker hand (5 cards) from an ordinary deck.
In how many ways can he be dealt

(i) a straight.

The answer in the back of my book says $10\cdot4^5-40$
(I know 40 is the number of straight flushes, but I don't know what the ${\color{blue}4^5}$ is.)

A Straight is five cards whose values are consecutive, but not all the same suit.
. . (That would be a Straight Flush.)

There are 10 possible sequences: A-2-3-4-5, 2-3-4-5-6, . . . 10-J-Q-K-A.

Suppose the sequence is 5,6,7,8,9.

. . There are 4 choices for the 5: $\{5\heartsuit,\:5\spadesuit,\:5\diamondsuit,\:5\cl ubsuit\}$
. . There are 4 choices for the 6: $\{6\heartsuit,\:6\spadesuit,\:6\diamondsuit,\:6\cl ubsuit\}$
. . There are 4 choices for the 7: $\{7\heartsuit,\:7\spadesuit,\:7\diamondsuit,\:7\cl ubsuit\}$
. . There are 4 choices for the 8: $\{8\heartsuit,\:8\spadesuit,\:8\diamondsuit,\:8\cl ubsuit\}$
. . There are 4 choices for the 9: $\{9\heartsuit,\:9\spadesuit,\:9\diamondsuit,\:9\cl ubsuit\}$

So there are: . $4^5$ ways to get a "5-6-7-8-9" Straight.

There are 10 possible sequences, so there are: . $10\cdot4^5$ possible Straights.

But this includes the 40 Straight Flushes (which are rarer and more valuable).

Therefore, the number of Straights is: . $10\cdot4^5 - 40$

(ii) a pair of aces (I could not find the poker rule for this, so have no idea what this is)

The answer provided was ${4\choose2}{12\choose3}\cdot4^3.$ ( again I do know what the reason is for ${12\choose3}\cdot4^3$

A "pair of Aces" means the hand contains exactly two Aces
. . and three other unmatched cards.
There are four Aces and we want two of them: . ${4\choose2}$ choices.
The other 3 cards must not be Aces.
. . We pick 3 of the other 12 values: . ${12\choose3}$ choices.

For each of these three cards, there are 4 choices of suits: . $4^3$ choices.

Got it?

4. Soroban,

That was a very clear instruction. I understood it completely for the first time. Thank you for your time and taking the pain to do it.

5. Originally Posted by ANDS!
There are 13 cards per suit. So it you have a hand of five cards, two of which are aces, there are three spots left, and you have 12 cards to chose from (2 through King). Therefore the number of ways to arrange those last 3 spots is 12 choose 3.

Not sure myself what their four-cubed is referring to.
Thank you for trying. Now you get to see Soroban's good instruction. We are lucky.

6. I'm dumb. Haha - I didn't even remember the different suits.