Poker + probabilities

Question

In a 2-player game of Texas Hold 'em poker, player vs dealer, the dealer shuffles the deck of cards and deals himself $7\clubsuit,7\spadesuit$ and the player $6\heartsuit,6\diamondsuit$. He then draws the flop. What is the probability that these 3 cards will give the player a better chance statistically of winning the game? My solution so far: There are $\binom {48}3 = 17,296$ possible combinations of these three cards.This will be the denominator to find the probability but I am stuck on how to find the $3$ cards that will give the player a $>50\%$ chance of winning after the flop has been drawn.

Answer 1

In order for the player to win, he must hit a 3 of a kind or a straight or a flush.

3 of a kind: What is the chance a 6 falls in the flop?

And other contingencies to consider... yes, the dealer could also flop a 3rd 7.

We need to see a 6 but no 7. Or 6-6-7 on the flop.

Straight or flush: The player can do no better than 4 cards toward a straight or a flush. If he hits this on the flop, what is the chance that he converts with the remaining two cards? If it is less than 50% you can dismiss these cases.