How many $6$ card hands have at least half of them an odd rank?

Question

Say you have the usual $52$ card deck, $13$ ranks and $4$ suits. If I have a hand of $6$ cards, how many hands are there where at least half the cards have an odd number as rank? they can be of any suit.

This is what I'm doing:

So $13$ ranks (assume they're numbered from $1$ to $13$ and the ace is a one) , this means that there are $7$ odd ranks ($1$, $3$, $5$, $7$, $9$, $11$, $13$). We want half meaning at least $3$ of the $6$ cards in my hand need to be odd. So:

I want the cases where there $3$ odd cards $+ 4$ odd cards $+ 5$ odd cards $+ 6$ odd cards. Is this how to represent it?

$$\frac 73 + \frac74 + \frac 75 +\frac76 $$

I feel like I'm missing something

Answer 1

First on your working:

If you are ranking $13$ cards of each suit from $1$ to $13$, there are $7$ odd number cards in each suit. So in total there are $28$ odd number cards and $24$ even number cards.

Hence number of hands of $6$ cards where there are at least $3$ odd number cards is be given by,

$\displaystyle {28 \choose 3} {24 \choose 3} + {28 \choose 4} {24 \choose 2} + {28 \choose 5} {24 \choose 1} + {28 \choose 6}$

But if the question considers odd number cards as $1, 3, 5, 7, 9$ as in a usual deck (Ace considered $1$), then there are $20$ odd number cards and $32$ of rest. You can apply the same logic as earlier to calculat number of hands with at least $3$ odd number cards.