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?
 A: "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.
A: To do this, let's find the chances of missing your draw first, because that's actually easier. Let's break this entire scenario down into three situations: either the first card helps, so the second card doesn't matter (good); the first card doesn't help but the second card does (also good); or neither card helps (bad). The chances of the first card not helping are $32\over47$ (number of cards not helping over total number of cards available at this point); this would rule out the first scenario but leaves open the second. Given this, the chances of the second card also not being helpful is $31\over46$, to rule out the second scenario. Thus, the chances of missing your draw are $32\times31\over47\times46$, making the chances of hitting your draw $1-\frac{32\times31}{47\times46}$, or about 54.12%.
A: The question is asking what the probability is that at least one of those 15 cards shows up in the next 2.  We can do this via complementary probability (not sure if this is the actual name): finding the probability that it does not happen and then subtracting that probability from 1.
There are 32*31 pairs of cards (where order is important) that could show up in the next so so that neither one helps, so the probability that this happens is $\frac{32*31}{52*51}$ so the answer is $1-\frac{32*31}{47*46}=\boxed{\frac{585}{1081}}$
