# poker hand question

• October 7th 2008, 08:18 AM
kuntah
poker hand question
I have a small problem

I read that the there are 10x(4^5) possibilities for a straight (royal and straight flush included)

But I thought that there were 10x(20x16x12x8x4) possibilites

My highest card can be 5,6,7,...,K
So i have 10 possibilities for the highest card

Now I want to draw a straight 1,2,3,4,5
I have 20 possibilities for the first one:1 till 5 and there are 4 suits
for example I draw 1(spades), then I have 16 possibilities for the next card (4 suits and 4 cards).
What is wrong with my thinking?

Thanks guys
• October 7th 2008, 08:42 AM
Plato
In a straight $C_1$ is the 'low' card and $C_5$ is the high card.
There are 10 values for $C_1$: Ace to ten.

There are 4 suits possible for $C_1$.
The same can be said of each $C_i ,\;\; 0\le i \le 5$

So $10(4^5)$.
• October 7th 2008, 08:54 AM
kuntah
I understand that, but what is wrong with my reasoning
that there should be 20x16x12x8x4(x10) possibilities?
• October 7th 2008, 09:06 AM
Plato
Quote:

Originally Posted by kuntah
I understand that, but what is wrong with my reasoning
that there should be 20x16x12x8x4(x10) possibilities?

$\frac{20x16x12x8x4(x10)}{5!} = 10(4^5)$