Texas hold 'em odds If I have a $6,7$ spades in my hand, and the flop shows $4$ clubs, $5$ spades and $Q$ spades. What is the probability I will hit my straight or flush with $2$ cards still to come?  I see $15$ cards that help me with $32$ that don't help me. So about $32\%$ with one card. But with $2$ yet to come?

"That's not easy math at the table in your head"
There is the Rule of $2$ and $4$ which is a useful rule of thumb when you are at the table. 
Number of outs $\: \times \: 2 + 1 =$ probability of improvement with one card left, the river.
Number of outs $\: \times \: 4 =$ probability of improvement with two cards left, on the turn or river.
If you have an open-ended straight draw ($8$ outs), then the rule of thumb for the river would give you a percentage of winning of $17$% ($2 \times 8 + 1$). Calculated exactly, the probability is $17.4$%. In an all-in situation on the flop the rule of thumb gives you a chance of winning of $32$% ($4 \times 8$), calculated exactly it is $31.5$%.
You should though be aware of that this rule of thumb works worse the more outs you have. 
But 
$1)$ those scenarios dont come up often and 
$2)$ being aware of that you can deduct a few percent.
So when you have $15$ outs, as in your question, this rule of thumb is a bit off.
